Page 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
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Submersible Motor
Table of Content
Section Content Page
1 General 3
2 Motor construction 3
21 Stator
22 Rotor
23 Rotor bearing
24 Motor thrust bearing
25 Pothead
3 Electromagnetism 8
31 Magnetic field
32 Magnetic flux and flux density
33 Magnetic field due to current in a solenoid
34 Changing polarity
35 Induced voltage
36 Electromagnetic attraction
4 Start coil arrangement 15
5 Power supply 15
51 Start
52 Time 1
53 Time 2
54 360 degree rotation
6 Mathematical analysis of rotating magnetic
field due to 3 phase current 19
7 Slip 24
8 Rotor current frequency 24
9 Magneto-motive force and magnetic field
Strength 25
10 Force in current carrying conductor in
magnetic field 26
11 Torque on a current carrying coil in
magnetic field 27
12 Theory of operation 28
13 Motor configurations 30
14 Motor current 32
15 Motor rating 32
16 Motor protection 35
17 Application of ESP motor 37
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18 Fundamentals of electricity 39
19 Equivalent circuit of induction motor 63
191 Effective circuit of induction motor at
Standstill
192 Effective circuit of induction motor under
Operating conditions (rotor is shorted)
193 Power relations
20 Determination of motor parameters 70
21 NEMA standard for squirrel cage IM 75
22 Torque of squirrel cage IM 77
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Submersible Motor
31 General
Motor is an electric machine which converts electric
energy into mechanical energy
Three phase induction motors are the most frequently
encountered in industry They are simple rugged low priced
and easy to maintain They run at essentially constant speed
from zero to full load The speed is frequency dependent
however variable speed electronic drives are being used more
and more to control the speed of the motors
ESP motor Classified as 3 phases squirrel cage 2 pole
induction Alternating current motor
The position of the motor in ESP integrity is just below the
protector (seal)
32 Motor Construction
The induction motor is a three phase squirrel cage two
pole induction design consists of
1 Stator which supports windings which receive energy from
the mains circuit
2 Rotor which carries windings in which working current is induced
3 Shaft which transfers the mechanical energy to the pump 4 Bearings 5 Housing 6 Insulated Magnet Wire 7 Thrust bearing
Fig (31) Most of the motor construction
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321 Stator
The stator is the stationary electrical part of the motor
The stator core of a National Electrical Manufacturers
Association (NEMA) motor is made up of several hundred thin
laminations
Fig (32) stator laminations
Stator laminations are stacked together forming a hollow
cylinder (fig 32) Coils of insulated wire are inserted
into slots of the stator core (fig 33)
Fig (33) stator core
Electromagnetism is the principle behind motor operation
Each grouping of coils together with the steel core it
surrounds form an electromagnet The stator windings are
connected directly to the power source
The stator winding consists of three individual windings
which overlap one another and are offset by an electrical
angle of 120deg (fig 34) When it is connected to the power
supply the incoming current will first magnetize the
stator This magnetizing current generates a rotary field
which turns with synchronous speed Ns
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Chapter 3
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Fig (34) stator windings
When the alternating current passes through a coil group a
magnetic field of fixed shape and sinusoidally varying
amplitude will result A magnetic pole is formed at the
center of this coil group The internal stator winding
connections determine the number of poles the voltage
applied to individual windings and the direction of
rotation In a three phase induction motor rotating
magnetic field is obtained by three separate single phases
with currents that differ in phase by 120 degrees
Three phases reach their maximum and minimum in a rapid
succession sequence As currents change the effect is to
rotate the magnetic fields The magnetic field rotates
continuously at a constant speed determined by the line
frequency and number of poles
The laminations wound with three very big loops of wire one
for each phase When current is flowing through a phase
magnetic flux is induced as shown in fig (35)
Fig (35) induced magnetic flux due to current flow
Because of this configuration the inside of the stator
holds a strong magnetic field
The strength of the field will depend on the amount of
current flowing through the wire loop (ie the phase
winding)
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Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
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7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
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Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
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9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
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Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
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Chapter 3
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332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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Chapter 3
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Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
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Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
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build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
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15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
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Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
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Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
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18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
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Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
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Chapter 3
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(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
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The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
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Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
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Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
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Chapter 3
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For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
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77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 2
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Chapter 3
2
18 Fundamentals of electricity 39
19 Equivalent circuit of induction motor 63
191 Effective circuit of induction motor at
Standstill
192 Effective circuit of induction motor under
Operating conditions (rotor is shorted)
193 Power relations
20 Determination of motor parameters 70
21 NEMA standard for squirrel cage IM 75
22 Torque of squirrel cage IM 77
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Chapter 3
3
Submersible Motor
31 General
Motor is an electric machine which converts electric
energy into mechanical energy
Three phase induction motors are the most frequently
encountered in industry They are simple rugged low priced
and easy to maintain They run at essentially constant speed
from zero to full load The speed is frequency dependent
however variable speed electronic drives are being used more
and more to control the speed of the motors
ESP motor Classified as 3 phases squirrel cage 2 pole
induction Alternating current motor
The position of the motor in ESP integrity is just below the
protector (seal)
32 Motor Construction
The induction motor is a three phase squirrel cage two
pole induction design consists of
1 Stator which supports windings which receive energy from
the mains circuit
2 Rotor which carries windings in which working current is induced
3 Shaft which transfers the mechanical energy to the pump 4 Bearings 5 Housing 6 Insulated Magnet Wire 7 Thrust bearing
Fig (31) Most of the motor construction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
4
321 Stator
The stator is the stationary electrical part of the motor
The stator core of a National Electrical Manufacturers
Association (NEMA) motor is made up of several hundred thin
laminations
Fig (32) stator laminations
Stator laminations are stacked together forming a hollow
cylinder (fig 32) Coils of insulated wire are inserted
into slots of the stator core (fig 33)
Fig (33) stator core
Electromagnetism is the principle behind motor operation
Each grouping of coils together with the steel core it
surrounds form an electromagnet The stator windings are
connected directly to the power source
The stator winding consists of three individual windings
which overlap one another and are offset by an electrical
angle of 120deg (fig 34) When it is connected to the power
supply the incoming current will first magnetize the
stator This magnetizing current generates a rotary field
which turns with synchronous speed Ns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
5
Fig (34) stator windings
When the alternating current passes through a coil group a
magnetic field of fixed shape and sinusoidally varying
amplitude will result A magnetic pole is formed at the
center of this coil group The internal stator winding
connections determine the number of poles the voltage
applied to individual windings and the direction of
rotation In a three phase induction motor rotating
magnetic field is obtained by three separate single phases
with currents that differ in phase by 120 degrees
Three phases reach their maximum and minimum in a rapid
succession sequence As currents change the effect is to
rotate the magnetic fields The magnetic field rotates
continuously at a constant speed determined by the line
frequency and number of poles
The laminations wound with three very big loops of wire one
for each phase When current is flowing through a phase
magnetic flux is induced as shown in fig (35)
Fig (35) induced magnetic flux due to current flow
Because of this configuration the inside of the stator
holds a strong magnetic field
The strength of the field will depend on the amount of
current flowing through the wire loop (ie the phase
winding)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
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Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
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54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
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77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 3
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Chapter 3
3
Submersible Motor
31 General
Motor is an electric machine which converts electric
energy into mechanical energy
Three phase induction motors are the most frequently
encountered in industry They are simple rugged low priced
and easy to maintain They run at essentially constant speed
from zero to full load The speed is frequency dependent
however variable speed electronic drives are being used more
and more to control the speed of the motors
ESP motor Classified as 3 phases squirrel cage 2 pole
induction Alternating current motor
The position of the motor in ESP integrity is just below the
protector (seal)
32 Motor Construction
The induction motor is a three phase squirrel cage two
pole induction design consists of
1 Stator which supports windings which receive energy from
the mains circuit
2 Rotor which carries windings in which working current is induced
3 Shaft which transfers the mechanical energy to the pump 4 Bearings 5 Housing 6 Insulated Magnet Wire 7 Thrust bearing
Fig (31) Most of the motor construction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
4
321 Stator
The stator is the stationary electrical part of the motor
The stator core of a National Electrical Manufacturers
Association (NEMA) motor is made up of several hundred thin
laminations
Fig (32) stator laminations
Stator laminations are stacked together forming a hollow
cylinder (fig 32) Coils of insulated wire are inserted
into slots of the stator core (fig 33)
Fig (33) stator core
Electromagnetism is the principle behind motor operation
Each grouping of coils together with the steel core it
surrounds form an electromagnet The stator windings are
connected directly to the power source
The stator winding consists of three individual windings
which overlap one another and are offset by an electrical
angle of 120deg (fig 34) When it is connected to the power
supply the incoming current will first magnetize the
stator This magnetizing current generates a rotary field
which turns with synchronous speed Ns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
5
Fig (34) stator windings
When the alternating current passes through a coil group a
magnetic field of fixed shape and sinusoidally varying
amplitude will result A magnetic pole is formed at the
center of this coil group The internal stator winding
connections determine the number of poles the voltage
applied to individual windings and the direction of
rotation In a three phase induction motor rotating
magnetic field is obtained by three separate single phases
with currents that differ in phase by 120 degrees
Three phases reach their maximum and minimum in a rapid
succession sequence As currents change the effect is to
rotate the magnetic fields The magnetic field rotates
continuously at a constant speed determined by the line
frequency and number of poles
The laminations wound with three very big loops of wire one
for each phase When current is flowing through a phase
magnetic flux is induced as shown in fig (35)
Fig (35) induced magnetic flux due to current flow
Because of this configuration the inside of the stator
holds a strong magnetic field
The strength of the field will depend on the amount of
current flowing through the wire loop (ie the phase
winding)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
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Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
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54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
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77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 4
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
4
321 Stator
The stator is the stationary electrical part of the motor
The stator core of a National Electrical Manufacturers
Association (NEMA) motor is made up of several hundred thin
laminations
Fig (32) stator laminations
Stator laminations are stacked together forming a hollow
cylinder (fig 32) Coils of insulated wire are inserted
into slots of the stator core (fig 33)
Fig (33) stator core
Electromagnetism is the principle behind motor operation
Each grouping of coils together with the steel core it
surrounds form an electromagnet The stator windings are
connected directly to the power source
The stator winding consists of three individual windings
which overlap one another and are offset by an electrical
angle of 120deg (fig 34) When it is connected to the power
supply the incoming current will first magnetize the
stator This magnetizing current generates a rotary field
which turns with synchronous speed Ns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
5
Fig (34) stator windings
When the alternating current passes through a coil group a
magnetic field of fixed shape and sinusoidally varying
amplitude will result A magnetic pole is formed at the
center of this coil group The internal stator winding
connections determine the number of poles the voltage
applied to individual windings and the direction of
rotation In a three phase induction motor rotating
magnetic field is obtained by three separate single phases
with currents that differ in phase by 120 degrees
Three phases reach their maximum and minimum in a rapid
succession sequence As currents change the effect is to
rotate the magnetic fields The magnetic field rotates
continuously at a constant speed determined by the line
frequency and number of poles
The laminations wound with three very big loops of wire one
for each phase When current is flowing through a phase
magnetic flux is induced as shown in fig (35)
Fig (35) induced magnetic flux due to current flow
Because of this configuration the inside of the stator
holds a strong magnetic field
The strength of the field will depend on the amount of
current flowing through the wire loop (ie the phase
winding)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
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54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 5
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
5
Fig (34) stator windings
When the alternating current passes through a coil group a
magnetic field of fixed shape and sinusoidally varying
amplitude will result A magnetic pole is formed at the
center of this coil group The internal stator winding
connections determine the number of poles the voltage
applied to individual windings and the direction of
rotation In a three phase induction motor rotating
magnetic field is obtained by three separate single phases
with currents that differ in phase by 120 degrees
Three phases reach their maximum and minimum in a rapid
succession sequence As currents change the effect is to
rotate the magnetic fields The magnetic field rotates
continuously at a constant speed determined by the line
frequency and number of poles
The laminations wound with three very big loops of wire one
for each phase When current is flowing through a phase
magnetic flux is induced as shown in fig (35)
Fig (35) induced magnetic flux due to current flow
Because of this configuration the inside of the stator
holds a strong magnetic field
The strength of the field will depend on the amount of
current flowing through the wire loop (ie the phase
winding)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
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Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 6
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
6
The more copper that is in the stator the more the winding
losses are reduced making the motor more efficient
The winding is two pole because two magnetic poles are
created (one North and one South) Motors can be wound
differently to create more than two poles such as a four
pole motor
Remember that the direction of the magnetic field in the
stator depends on the direction of current flowing in the
wire
With AC or Alternating Current the direction of current
flow is changing 60 times every second for 60 Hz power (or
50 times per second for 50 Hz power)
322 Rotor
The rotor is the rotating part of the electromagnetic
circuit
The most common type of rotor is the ldquosquirrel cagerdquo rotor
Fig (36) squirrel cage rotor
The rotor consists of a stack of steel laminations
The squirrel cage rotor consists of copper or aluminum bars
accommodated in slots of rotor core (fig 37)
Fig (37) rotor core
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
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Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 7
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
7
At each end the bard are connected to heavy conducting end
rings which serve the purpose of short circuit bars In the
absence of the end rings emfs would be induced in the rotor
bars but no current would flow through them and no torque
would be produced
The wound rotor has a 3 phase winding placed in the slots of
the rotor core The rotor is wound for the same number of
poles as the stator
The terminals of the rotor winding are brought out of three
slip rings mounted on the machine shaft
During running condition the slip ring are short circuited
so as to close the rotor circuit
The air gap or more accurately the clearance between the
stator and rotor should be as small as possible in order to
the primary and secondary leakage fluxes to minimum
323 Rotor bearing
Rotor Bearings are one of the most vital parts of the
motor The Bearing Material is Babbitt-lined steel and
machined after processing There are fluid holes to insure oil
circulation and wide angle oil grooves on the OD to distribute
lubrication evenly over the entire length of the bearing
surface
Fig (38) rotor bearing
The bearing sleeve is a bronze material for the sleeve
construction of the bearing This part is keyed to the shaft
and the hole on the sleeve is aligned with the hole on the
shaft to insure proper cooling and lubrication
324 Motor thrust bearing
The motor thrust bearing is installed at the top of the
rotor string It is designed to hold the weight of the entire
rotor string
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
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Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
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Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 8
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
8
Fig (39) motor thrust bearing
Currently three types of motor thrust bearings are used
Babbitt
Glacier
KMC Bronze pads The thrust bearing limits on the system will indicate the type
of load required for the selected bearing material
325 Pothead
Pothead is the place where the motor lead extension
cable is connected to the motor three phase windings (fig
310) There are two types of pothead they are
1 Tape in type where tape wrapped around individual connector leads inside motor
2 Plug in type where mating block mounted in motor
Fig (310) motor pothead
33 Electromagnetism
331 Magnetic field
When an electric current is passed through a conductor
a magnetic field is set up around the conductor The direction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
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The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
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leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
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353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
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Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
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(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
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The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
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At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
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Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
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Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
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For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
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mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
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If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
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T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
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The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
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313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
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Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
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32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
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33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
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Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
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35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
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36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
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37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
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38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
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39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 9
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
9
of the magnetic field can be found by using right hand rule or
the right hand screw The right hand rule states ldquoGrasp the
wire in the right hand with the thumb pointing in the
direction of the current The fingers will curl around the
wire in the direction of the magnetic fieldrdquo
The right hand screw is explained in this way as a wood screw
is turned clockwise it progresses into the wood The
horizontal direction of screw is analogous to the direction of
current in a conductor The circular motion of the screw shows
the direction of the magnetic flux around the conductor (fig
311)
Fig (311) Magnetic field around the conductor carrying current
In fig 312 (a) the dot inside the circle is the standard
symbol used to show that the direction of current flow is out
of the page Then by right hand rule the magnetic field is
counterclockwise In fig 312 (b) the cross is the standard
symbol used to show that the current is entering the page The
magnetic field is clockwise The strength of the magnetic
field is proportional to the current ie if the current is
doubled the magnetic field will be doubled
(a) Current coming out of the page (b) Current entering the page
Fig (312) Magnetic field surrounding the conductor
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Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
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51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 10
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
10
Since a current carrying conductor has a magnetic field around
it when two current carrying conductors are brought close
together there will be interactive between the fields When
the currents in the two conductors are in opposite direction
the fields are as shown in fig 313 (a) and the force of
repulsion is experienced
When the currents are in the same direction the field are as
shown in fig 313 (b) and a fore of attractive is experienced
Fig (313) force between parallel current carrying conductors
Consider a single turn coil carrying current As shown in
figure 35 the hole of the magnetic flux generated by electric
current passed through the centre of the coil Therefore the
coil acts like a little magnet and has a magnetic field with
identifiable N and S poles The coil may also have more than
one turn The flux generated by each of the individual turns
of the coil tends to link up and pass out of one end of coil
and back into the other end Such an arrangement is known as
solenoid (figure 36) and has a magnetic field pattern very
similar to that of bar magnet The right hand rule for
determining the direction of flux from solenoid states ldquoGrasp
the solenoid in the right hand such that the fingers point in
the direction of current flow in the coil The thumb will
point towards N pole of fieldrdquo
As discussed above a current flow in the conductors produces
a magnetic field The converse is also possible that is a
magnetic field can produce a current in a conductor This is
known as the phenomenon of electromagnetic induction (this
will be discussed later)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
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Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 11
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11
332 Magnetic flux and flux density
The total lines of force in a magnetic field are called
magnetic flux Flux density is the flux per unit area of cross
section
Weber (Wb) is the SI unit of magnetic flux Tesla (T) is the
SI unit of flux density and represents Wbm2 Flux density is
also known as magnetic induction From its definition
AB
Where B is the flux density in teslas φ is the total flux in
webers and A is the cross sectional area in m2
333 Magnetic field due to current in a solenoid
When an electric current passed through a solenoid the
resultant magnetic flux is very similar to that of a bar
magnet The magnetic flux lines make complete circuit inside
and outside the coil each line is a closed path The side at
the flux emerges is the North Pole the other side where the
magnetic flux reenters is the South Pole
The strength of the magnetic field in the DC electromagnet can
be increased by increasing the number of turns andor current
in the coil The greater the number of turns the stronger the
magnetic field will be See fig (314) and (315)
Fig (314) Magnetic field in coils of different number of
turns
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12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
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Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
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16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
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Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
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Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
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Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
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Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
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Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
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Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
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Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
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Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
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Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 12
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
12
Fig (315)
Magnetic field in coils of different currents
The magnetic flux density in the interior of a solenoid
carrying an electric current depends on the current intensity
passing through the coil (I) and number of turns per unit
length (n) ie B is proportionally change with I an n
InB
Where μ is the permeability of the core material The equation
can be written as follows
Where N is the number of turns of a solenoid and l is its
length
334 Changing polarity
The magnetic field of an electromagnet has the same
characteristics as a natural magnet including a north and
South Pole However when the direction of current flow
through the electromagnet changes the polarity of the
electromagnet changes The polarity of an electromagnet
connected to an AC source will change at the same frequency as
the frequency of the AC source This can be demonstrated in
the following illustration (fig 37) At Time 1 current flow
is at zero There is no magnetic field produced around the
electromagnet At Time 2 current is flowing in a positive
direction A magnetic field builds up around the
electromagnet The electromagnet assumes a polarity with the
South Pole on the top and the North Pole on the bottom At
Time 3 current flow is at its peak positive value The
strength of the electromagnetic field is at its greatest
value At Time 4 current flow decreases and the magnetic field
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
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18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 13
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
13
begins to collapse until Time 5 when current flow and
magnetic field are at zero Current immediately begins to
increase in the opposite direction At Time 6 current is
increasing in a negative direction The polarity of the
electromagnetic field has changed The north pole is now on
top and the south pole is on the bottom The negative half of
the cycle continues through Times 7 and 8 returning to zero
at Time 9 This process will repeat 50 times a second with a
50 Hz AC power supply (fig 316)
Fig (316)
335 Induced voltage
A conductor moving through a magnetic field will have a
voltage induced into it This electrical principle is used in
the operation of AC induction motors In the following
illustration an electromagnet is connected to an AC power
source Another electromagnet is placed above it The second
electromagnet is in a separate circuit There is no physical
connection between the two circuits Voltage and current are
zero in both circuits at Time 1 At Time 2 voltage and current
are increasing in the bottom circuit A magnetic field builds
up in the bottom electromagnet Lines of flux from the
magnetic field building up in the bottom electromagnet cut
across the top electromagnet A voltage is induced in the top
electromagnet and current flows through it At Time 3 current
flow has reached its peak Maximum current is flowing in both
circuits The magnetic field around the coil continues to
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 14
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
14
build up and collapse as the alternating current continues to
increase and decrease As the magnetic field moves through
space moving out from the coil as it builds up and back
towards the coil as it collapses lines of flux cut across the
top coil As current flows in the top electromagnet it creates
its own magnetic field (fig 317)
Fig (317) magnetic field increases as the current increases
336 Electromagnetic attraction
The polarity of the magnetic field induced in the top electromagnet is opposite the polarity of the magnetic field
in the bottom electromagnet Since opposite poles attract the
top electromagnet will follow the bottom electromagnet when it
is moved (fig 318)
Fig (318)
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Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
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Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
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Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 15
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
15
34 Start coil arrangement
The following schematic (fig 319) illustrates the
relationship of the coils In this example six coils are used
two coils for each of the three phases The coils operate in
pairs The coils are wrapped around the soft iron core
material of the stator These coils are referred to as motor
windings Each motor winding becomes a separate electromagnet
The coils are wound in such a way that when current flows in
them one coil is a north pole and its pair is a south pole
For example if A1 were a north pole then A2 would be a south
pole When current reverses direction the polarity of the
poles would also reverse
Fig (319)
35 Power supply
The stator is connected to a 3-phase AC power supply In
the following illustration phase A is connected to phase A of
the power supply Phase B and C would also be connected to
phases B and C of the power supply respectively
Fig (320)
Phase windings (A B and C) are placed 120deg apart In this
example a second set of three-phase windings is installed
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
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44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
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45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
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46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
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47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
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51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
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71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
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Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
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Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
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75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
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76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
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Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
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78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 16
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Chapter 3
16
The number of poles is determined by how many times a phase
winding appears In this example each phase winding appears
two times This is a two-pole stator If each phase winding
appeared four times it would be a four-pole stator
Fig (321) 2 poles stator winding
When AC voltage is applied to the stator current flows
through the windings The magnetic field developed in a phase
winding depends on the direction of current flow through that
winding The following chart is used here for explanation
only It will be used in the next few illustrations to
demonstrate how a rotating magnetic field is developed It
assumes that a positive current flow in the A1 B1 and C1
windings result in a north pole
Winding
Current Flow
Direction
Positive Negative
A1 North South
A2 South North
B1 North South
B2 South North
C1 North South
C2 South North
351 Start
It is easier to visualize a magnetic field if a start
time is picked when no current is flowing through one phase
In the following illustration for example a start time has
been selected during which phase A has no current flow phase
B has current flow in a negative direction and phase C has
current flow in a positive direction Based on the above
chart B1 and C2 are south poles and B2 and C1 are north
poles Magnetic lines of flux leave the B2 North Pole and
enter the nearest South Pole C2 Magnetic lines of flux also
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Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
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Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
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Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
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Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
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Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
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Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
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Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
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Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
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Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
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Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
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Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
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Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
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Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 17
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
17
leave the C1 North Pole and enter the nearest South Pole B1
A magnetic field results as indicated by the arrow fig
(322)
Fig (322) start
352 Time 1
If the field is evaluated at 60deg intervals from the
starting point at Time 1 it can be seen that the field will
rotate 60deg At Time 1 phase C has no current flow phase A has
current flow in a positive direction and phase B has current
flow in a negative direction Following the same logic as used
for the starting point windings A1 and B2 are north poles and
windings A2 and B1 are south poles fig (323)
Fig (323) time 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
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43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
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44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
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45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
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46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
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47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
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49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
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51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
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52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 18
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
18
353 Time 2
At Time 2 the magnetic field has rotated 60deg Phase B
has no current flow Although current is decreasing in phase A
it is still flowing in a positive direction Phase C is now
flowing in a negative direction At start it was flowing in a
positive direction Current flow has changed directions in the
phase C windings and the magnetic poles have reversed
polarity
Fig (324) time 2
354 360 degree rotation
At the end of six such time intervals the magnetic
field will have rotated one full revolution or 360deg This
process will repeat 60 times a second on a 60 Hz power supply
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 19
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
19
Fig (325) time 1
36 Mathematical analysis of rotating magnetic field due to 3 phase current
When a 3-phase winding is energized from a 3-phase supply
a rotating magnetic field is produced This field is such that
its poles do no remain in a fixed position on the stator but
go on shifting their positions around the stator
For this reason it is called a rotating magnetic field It
will be shown that magnitude of this rotating field is
constant and is equal to 15 fm where fm is the maximum flux
due to any phase
To see how rotating field is produced consider a 2-pole 3
phase winding as shown in fig 326(i) The three phases A B
and C are energized from a 3-phase source and currents in
these phases are indicated as IA IB and IC Referring to Fig
326 (ii) the fluxes produced by these currents are given by
Here φm is the maximum flux due to any phase We shall now
prove that this 3-phase supply produces a rotating field of
constant magnitude equal to 15 φm
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 20
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
20
(i) (ii)
Fig (326)
At start fig 326 (ii) and fig 327 (i) the current in phase
A is zero and currents in phases B and C are equal and
opposite The currents are flowing outward in the top
conductors and inward in the bottom conductors This
establishes a resultant flux towards right The magnitude of
the resultant flux is constant and is equal to 15 φm as
proved under
At start ωt = 0deg Therefore the three fluxes are given by
(i) (ii)
Fig (327)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
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Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
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Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 21
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
21
The phasor sum of - φB and φC is the resultant flux φR (fig 328) It is clear that
Fig (328)
At time 1 fig (15 (ii)) ωt = 60deg Therefore the three
fluxes are given by
The phasor sum of - φB and φA is the resultant flux φR [See Fig (318) It is clear that
Fig (329)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 22
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
22
At time 2 fig (26 (ii) ωt = 120deg Therefore the three fluxes are given by
The phasor sum of ndash φC and φA is the resultant flux φR (Fig 330) It is clear that
It follows from the above discussion that a 3-phase supply
produces a rotating field of constant value (= 15 φm where φm
is the maximum flux due to any phase)
Fig (330)
We shall now use another useful method to find the magnitude
and speed of the resultant flux due to three-phase currents
The three-phase sinusoidal currents produce fluxes φ1 φ2 and
φ3 which vary sinusoidally The resultant flux at any instant
will be the vector sum of all the three at that instant
The fluxes are represented by three variable magnitude
vectors fig (331)
In fig (331) the individual flux directions are fixed but
their magnitudes vary sinusoidally as does the current that
produces them To find the magnitude of the resultant flux
resolve each flux into horizontal and vertical components and
then find their vector sum
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 23
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
23
Fig (331)
The resultant flux is given by
Thus the resultant flux has constant magnitude (= 15 φm) and
does not change with time The angular displacement of φR relative to the OX axis is
So
Thus the resultant magnetic field rotates at constant angular
velocity ω(= 2πf) radsec For a P-pole machine the rotation
speed (ωs) is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 24
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
24
Thus the resultant flux due to three-phase currents is of
constant value (= 15 φm where φm is the maximum flux in any phase) and this flux rotates around the stator winding at a
synchronous speed of 120fP rpm
For example for a 2-pole 50 Hz 3-phase induction motor N
= 120x502 = 3000 rpm It means that flux rotates around the
stator at a speed of 3000 rpm
37 Slip
We have seen above that rotor rapidly accelerates in the
direction of rotating field In practice the rotor can never
reach the speed of stator flux If it did there would be no
relative speed between the stator field and rotor conductors
no induced rotor currents and therefore no torque to drive
the rotor The friction and windage would immediately cause
the rotor to slow down Hence the rotor speed (N) is always
less than the suitor field speed (Ns) This difference in speed
depends upon load on the motor
The difference between the synchronous speed Ns of the rotating
stator field and the actual rotor speed N is called slip It
is usually expressed as a percentage of synchronous speed
ie
The quantity Ns-N is sometimes called slip speed
When the rotor is stationary (ie N = 0) slip s = 1 or 100
In an induction motor the change in slip from no-load to full-load is hardly 01 to 3 so that it is essentially a
constant-speed motor
38 Rotor current frequency
The frequency of a voltage or current induced due to the
relative speed between a vending and a magnetic field is given
by the general formula
Where
n = Relative speed between magnetic field and the winding
P = Number of poles
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 25
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
25
For a rotor speed N the relative speed between the rotating
flux and the rotor is Ns-N Consequently the rotor current
frequency fr is given by
ie Rotor current frequency = slip x Supply frequency
When the rotor is at standstill or stationary (ie s = 1) the frequency of rotor current is the same as that of supply
frequency (fr = sf = 1 fr = f)
As the rotor picks up speed the relative speed between the rotating flux and the rotor decreases Consequently the slip
S and hence rotor current frequency decreases
39 Magneto-motive force and magnetic field strength
An emf causes a current to flow in an electric circuit
Similarly a magneto motive force (mmf) symbol F produces a
magnetic flux in a magnetic circuit The mmf of a coil is the
product of current in the coil and the number of turns of the
coil and has the unit of ampere turns (AT)
The magnetic flux which can be set up in a magnetic circuit
depends on the mmf and the length of the magnetic circuit If
the length is large the mmf has to act over a long distance
and the resulting magnetic flux is small The magnetic field
strength H is defined as the mmf per unit length of magnetic
circuit ie
Where I is the current in amperes and N is the number of
turns and l is the length of the magnetic circuit in meter
Example
The total flux of an electro magnet is 4x10-4 Wb
a If the cross sectional area of the core is 1 cm2 find the flux density in the core
b The electromagnet has 50 turns and a current of 1 A flow through the coil If the length of the magnetic circuit is
20 cm find the mmf and the magnetic field strength
Flux density 4101
10424
4
m
Wb
AB
T
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 26
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
26
mmf 50150 NIF AT
Field strength 2501020
502
m
AT
l
NIH ATm
310 Force in current carrying conductor in magnetic field
Figure 332 (a) shows a current carrying conductor (the
current entering the page) laying in magnetic field flux
density B The current in the conductor sets up a flux in a
clockwise direction around the conductor When the external
field is in the vertically downward direction the field of
the conductor assists the external field on the right hand
side of the conductor The effect of this is to produce a
force that pushes the conductor to the lift If the direction
of the current is reversed as shown in figure 36 (b) the flux
around the conductor is in counterclockwise direction and the
resulting force pushes the conductor to the right In both
cases maximum force is generated if the conductor is at right
angle to the direction of the magnetic flux The force is
always in a direction perpendicular to both the conductor and
the field
The magnitude of the force F is given by
Where B is the flux density in telsas I is the current in
amperes and l is the length of the conductor in meters
A force of one Newton is exerted on a 1 meter long conductor
carrying a current of 1 ampere and situated at right angle to
a magnetic field having a flux density of 1 tesla
(a) Conductor current entering the page (b) Conductor current coming out of the page
Fig (332) Force on a current carrying conductor in a magnetic
field
+ Force
Flux set up
by current
in on
conductor Flux set up
by current
in on
conductor
Force
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
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Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
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Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 27
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
27
If the conductor is not perpendicular to magnetic field but
inclined at angle Ө to the magnetic field the force is given
by
311 Torque on a current carrying coil in magnetic field
Figure 333 shows a current carrying coil placed in
magnetic field From the discussion in section 35 it follows
that a downward force is exerted on the left hand conductor
and an upward force is exerted on the right hand conductor
The force on each conductor is given by the equation 33 The
total force is given by
Fig (333) Force on a coil carrying current in a magnetic
field
If the coil has N turns the force is
Since the force is acting at a radius r meters the torque on
the coil is
The above simple arrangement is the basic part of electric
measuring instrument The operation of an electric motor is
also based on this principle
Example
A 30 cm long conductor is carrying a current of 10 A and
situated at a right angle to a magnetic field having flux
density of 08 T Calculate the force on the conductor
F = 08 x 10 x 30 x 10-2 = 24 N
Example
A 200 turn coil having an axial length of 3 cm and radius of 1
cm is pivoted in magnetic field having a flux density of 08
S
N
I
2r
Flux
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 28
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
28
T The coil carries a current of 05 A Calculate the torque
acting on the coil
Torque = 2 x 08 x 05 x 3 x10-2 x 200 x 1 x10
-2 = 48 x 10
-2 Nm
Figure 334 is the same as figure 333 but with other
direction of the magnetic flux
Fig (334) Force on a coil in a magnetic field
If the loop is in line with the magnetic field the secondary
magnetic field will be perpendicular to the main field This
will cause two equal and opposite forces (a torque) on the
loop causing it to rotate until the forces balance (fig 335)
The forces will reach a steady state and hold the magnet in
place as long as current is applied
To cause rotation the field must rotate This is accomplished
with the alternating current where the field is rotated
This is accomplished with the alternating current going
through the windings in the stator of the induction motor
Fig (335) Force on a coil in a magnetic field
312 Theory of operation
An induction motor consists of a stator and rotor The
stator carries a 3 phase winding in its slots and is connected
to a 3 phase supply The rotor carries a cage winding and it
is free to rotate within the stator
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 29
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
29
The 3 phase currents flowing in the stator winding produce a
rotating field The rotor winding cuts the rotating field and
an emf is induced in the rotor winding When the rotor is at
rest the frequency of this emf is the same as the supply
frequency If the rotor circuit is closed a current flows in
the rotor winding The rotor current produces an mmf which
rotates and is directed in opposition to stator mmf The
interaction of the stator and the rotor fields produces a
torque which causes the rotor to rotate in the direction of
the rotor field
If the rotor shaft is not loaded the machine has only to
rotate itself against the mechanical losses and the rotor
speed is very close to the synchronous speed However the
rotor speed cannot become equal to the synchronous speed
because if it does so the emf induced in the rotor winding
would become zero and there will be no torque Hence the rotor
speed remains slightly less than the synchronous speed
If the motor shaft is loaded the rotor will slow down and the
relative speed of rotor with respect to the stator rotating
field will increase The emf induced in the rotor winding will
increase and this will produce more rotor current which will
increase the electromagnetic torque produced by the motor
Conditions of equilibrium are attained when the rotor speed
has adjusted to a new value so that the electromagnetic torque
is sufficient to balance the mechanical or load torque applied
to the shaft The speed of the motor when running under full
load conditions is somewhat less than the no load speed
The speed of rotation of the field mmf is called synchronous
speed is related to the frequency and number of poles by the
expression
np
f s
2
Where
f is the frequency in Hz
p is the number of poles and
ns is the synchronous speed in revolution per second
An alternative form of above expression is
p
fN s
120
Where
Ns is the synchronous speed in revolution per minute (rpm)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 30
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
30
313 Motor configurations
Motors come in single sections (head and base) as well as
tandem configurations
The tandems can include the UT (upper tandem - head but no
base) the CT (center tandem - no head or base) or the LT
(lower tandem - base but no head)
An upper tandem motor can be used as a single section if it is
completed on the bottom with either a Universal Motor Base
(UMB) or Downhole Monitoring System (Sensor)
If additional horsepower is required over what can be achieved
in one piece a CT or LT motor can be added
Submersible electric motors can be designed in tandem
configuration to create the desired Horsepower required for
each application
So based on the above there are four different motor
sections they are
1 Single section
Where the motor has head and base a certain horsepower we
cannot increase and could not attach any equipment below
2 Upper Tandem motor (UT)
UT motor has head no base (open circuit) and the
horsepower can be increased by adding another central tandem
(CT) or lower tandem (LT) The circuit must be closed either
by Universal Motor Base (UMB) or sensor
3 Central Tandem (CT)
CT motor has no head no base and cannot use alone UT must
be attached on top of it Another central tandems or LT can
be attached on the bottom of it
4 Lower Tandem (LT)
Where the motor has a base no head and cannot use alone
it must be attached with upper tandem (as there is no head
attached)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 31
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
31
Single UT CT
Fig (336) motor configurations
Notes
1 For any particular horsepower the product of the volts and amps will be essentially constant For example in a
particular Hp if we have a 1000 V 50 A motor a 2000 V
motor would be 25 amps and a 500V motor would be 100 amps
In other words KVA is constant
2 When putting more than one motor together in tandem
combinations always keep the sections the same Hp and
voltage For example a 300 Hp 540 motor should be made of
two 150 Hp motors
3 With two motors we double the Hp (add the two Hps
together) We also double the voltage but the amperage
remains the same With three motors we triple the Hp and
voltage but the amperage still does not change
For example a 140 Hp 1299 V 695 A UT motor coupled to a
140 Hp 1299V 695 CT motor would give us a 280 Hp 2598 V
695 A motor
4 Always take care when adding motors together so that the total voltage does not exceed the system limits ie do not
try to put 3500 volts on a 3 kV cable Surface controllers
transformers wellhead feedthru mandrels etc will all have
voltage limits we need to be concerned with
5 For any given Hp there will be several voltages and
amperages available why have more than one voltage
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 32
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
32
The answer is not in the motor but in the power cable Lower
voltage means higher current and this results in higher
voltage lost in the power able
So even though the motor efficiency does not change the
overall system efficiency will decrease with higher
amperage
If the amperage is too high the motor may not even be able
to start as we will see when we discuss power cable chapter
6 Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
314 Motor current
Induction motor current consists of reactive (magnetizing)
and real (torque) components
The current component that produces torque (does useful work) is almost in phase with voltage and has a high power factor
close to 100
The magnetizing current would be purely inductive except
that the winding has some small resistance and it lags the
voltage by nearly 90deg
The magnetizing current has a very low power factor close to zero
The magnetic field is nearly constant from no load to full load and beyond so the magnetizing portion of the total
current is approximately the same for all loads
The torque current increases as the load increases
At full load the torque current is higher than the
magnetizing current
For a typical motor the power factor of the resulting
current is between 85 and 90
As the load is reduced the torque current decreases but the magnetizing current remains about the same so the resulting
current has a lower power factor
The smaller the load the lower the load current and the
lower the power factor Low power factor at low loading
occurs because the magnetizing remains approximately the same
at no load as at full load
315 Motor rating
If we look in the manufacturer catalog we will find several pages of motors which give the horsepower rating and break
down the various sizes into several voltages and amperages as
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 33
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
33
the table below
KMH 562 SERIES MOTORS
HP Volt Amp
60 HZ 50 HZ 60 HZ 50 HZ
38 32
43553 36353
87526 72926
131518 109618
57 48
43081 35881
87040 72540
131526 109626
76 63 86553 72153
136034 113334
95 79 84069 70069
133044 110844
114 95
86081 71781
130053 108353
233030 194230
133 111
83098 69298
134560 112160
220537 183837
152 127 134069 111769
232540 193840
171 143 129081 107581
239044 199244
190 158
118598 98898
143081 119281
241548 201348
Most conventional rating systems give a nameplate horsepower to the motor based on some assumptions which typically are
bottom-hole temperature and fluid flow rate past the motor in
this industry
The motor will put out exactly as much horsepower as the pump wants no more and no less
Most motors are designed to be most efficient and have an acceptable speed and power factor at the design pointldquo
In a standard application the surface voltage is fixed and the amperage changes as the load on the motor changes
We in fact use this information in the form of an amp chart
to see how the motor performs at downhole
We can very easily anticipate this relationship by simply
looking at the equation for motor horsepower
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 34
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
34
Note that if everything else is fixed the amps will have to increase if the horsepower does and this should be a linear
relationship
In reality this relationship is not linear since power factor and efficiency are not truly constant and the more they vary
the greater will be the change in amperage
One problem with increasing the amperage too much is that higher amperage will give us greater copper losses in the
motor winding which is not good for efficiency
There will be a practical limit to how far this can continue
We can look at the laminations to understand the basics of
this concept As we increase current on a motor we increase
the flux density induced in the laminations
Fig 337 shows an example of what the flux lines might look like for one phase winding based on a moderate loading of the
motor
Fig (337)
If we place more load on the same motor we get many more flux lines required to generate the necessary horsepower as
shows in fig 338
Fig (338)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 35
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
35
If we keep raising the amperage we will eventually reach a point where we have all the flux lines we can handle this is
called SATURATION
Any more horsepower beyond this point will severely overheat
our motor
Another practical consideration on rating a motor is the
speed We know that the motor will slow down with load If
the motor speed is too low we will lose pump performance
so we must set the Hp at a point where the speed is
acceptable
One of the most important considerations in rating the motor is temperature Heat is generated in the windings which must
be dissipated by the fluid flowing past the outside of the
motor
Another limiting factor will be temperature differential As the motor heats up the components expand and they expand at
different rates since not all the materials are the same
Even if the motor were all one material expansion would vary
since the internal temperature changes within the motor
itself
The motor is designed with certain tolerances to allow this thermal expansion If too much expansion occurs (such as with
(overheating) tolerances might be exceeded and we could have
bearing failures or other damage
316 Motor protection
In this discussion we will address proper protection for ESP motors operating down-hole
Motor controllers can provide very simple protection to very sophisticated protection
Simple controllers will look at overload and under-load
conditions only
More advanced controllers look at all operating parameters including 3 phase current and voltage leg ground power
factor kw back-spin and many others
In either type of controller overload and under-load
protection is of primary importance and it is critical that
these both be set correctly in order to properly protect the
motor from damage
Overloads and under-loads are usually set around +15 and 20 of running current as a rule-of-thumb
While it is not the intent here to give a complete guide to setting overloads and under-loads we will look at some
particular examples where the standard rule-of-thumb settings
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 36
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
36
may not give adequate protection
Lets take a particular case Say we have a 145 stage DN1300 producing a fluid of 10 gravity on a 456 series 50 Hp 885V
355 Amp motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
Say this pump will only require about 45 HP at BEP flow This will place about a 90 load on the motor which means the
motor should draw only 32 amps rather than the 355 of the
nameplate This should be the basis for our calculations and
settings
If we set the under-load at 80 and the overload at 115 these settings would be 256 amps and 368 amps respectively
Would this protect this unit
At shut-in the pump will draw about 28 Hp which is a 56 load on the motor The motor should draw about 62 of NP amps
or 22 amps Since our UL is set at 256 the motor should
trip on UL if the well is shut-in
Note that a shut-in pump is NOT a no-load condition for the
motor What amperage would you expect for this motor if the
pump shaft were broken at the intake (a true no-load
condition)
For the motor to operate unloaded it should draw about 32 of NP amps or in this case 114 amps This is a good
indication of a broken shaft However even if the current
reads higher than this it still could be a broken shaft so
do not rule that out on this basis alone
Lets take another case Say we have a 150 stage DN1750
producing a fluid of 086 gravity on a 456 series 50 Hp
885V 355A motor Should we set the OL at 115 of 355A and
the UL at 80 of 355A
This pump will only require about 47 HP at BEP flow This will place about a 94 load on the motor which means the
motor should draw only 34 amps rather than the 355 amp
nameplate rating This should be the basis for our
calculations
If we set the under-load at 80 and the overload at 115 these settings would be 272 amps and 391 amps respectively
Would this protect this unit
Note that at shut-in the pump will draw about 37 Hp which is a 74 load on the motor The motor should draw about 79 of
NP amps or 28 amps Since our UL is set at 272 the motor
may not trip on UL if the well is shut-in In this case we
should set the UL higher
The point is that these standard rules-of-thumb are not
always perfect Every application should be considered
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 37
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
37
independently to ensure that the settings selected are
adequate to properly protect the down-hole equipment
317 Application of ESP Motors
Motors are in five different series 375 450 540 562 and 738 for different casing sizes see the following examples of
Centrilift and Reda motors
With all these choices which motor should we use for a given application
The process to select the best motor for the application will depend on the economic compromises of the user but in
general after defining the customers objectives and the pump
horsepower load for the application we can resume the
process of selection of the motor as an iterative process
which includes
Motor Series
Motor Type
Motor configuration Voltage and Amperage
Actual motor performance amp Operating Temperature and
compare against max temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
38
All motor Hp ratings are based on 1 ft per second minimum flow past the motor of water
Lower flow rates or higher oil cut can lower the effective Hp rating
After choosing the Motor Series and Type we will know if it will be necessary for a single or tandem motor to match the
HP requirement of the pump
Now we should look at Volts and Amps
For any given Hp there will be several voltages and amperages available
For any particular horsepower the product of the volts and amps will be essentially constant
For example in a particular Hp if we have a 1000V 50A motor a 2000V motor would be 25 amps and a 500V motor would be 100
amps In other words KVA is constant High voltage motors
(single motors) are no more or less efficient than low
voltage motors so why have more than one voltage
The answer is not in the motor but in the power cable Lower voltage means higher current and this results in higher
voltage lost in the power able So even though the motor
efficiency does not change the overall system efficiency
will decrease with higher amperage
If the amperage is too high the motor may not even be able to start as we will see when we discuss power cable
This explains why the various voltages but why such odd
voltages Surface motors for example are rated at 460V
4160V 2300V etc
These motors are made to standard voltages So why do the
motor voltages turn out to be such strange numbers
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 39
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
39
In a way we have already answered this The problem is the voltage drop in the cable ESP motors have to contend with a
very long length of power cable which surface motors do not
If we have a 460 V surface supply we would probably only
want about 430V down-hole (for a low Hp motor) to give us the
necessary 460 V at the surface including the cable loss
So in determining motor voltages we are really limited by surface equipment
Motor control panels come in certain voltage ranges such as
600V 1000V 1500V 2400V etc
Motor voltages are selected assuming a length of cable such
that the total voltage (motor plus cable loss) will fall just
below one of the panel ratings
Remember the higher voltage the better but it cannot be so high that we exceed the control panel rating
Higher voltage motors require smaller gauge wire and very low
Hp motors simply cannot be wound at very high voltages
because the wire would be too small to work with
As mentioned eerily motors come in single sections (head and base) as well as tandem configurations The tandems can
include the UT (upper tandem head but no base) the CT
(center tandem no head or base) or the LT (lower tandem
base but no head) An upper tandem motor can be used as a
single section if it is completed on the bottom with either a
Universal Motor Base (UMB) or Downhole Monitoring System
(Sensor)
If additional horsepower is required over what can be
achieved in one piece a CT or LT motor can be added
318 Fundamentals of electricity
This section is not an attempt to present a course in
electricity but is intended as a review of the terms and
basic formulas associated with ESP applications
Electricity
Since the electrons are normally distributed evenly
throughout a substance a force called electromotive force
(emf) is required to detach them from the atoms and make them
flow in a definite direction This force is also often called
ldquopotentialrdquo or ldquovoltagerdquo The unit of measuring this
electromotive force is the ldquovoltrdquo
The higher the voltage the greater the number of electrons
which will be caused to flow
It has been found experimentally that the charge on a single
electron is 1602e-19 coulomb Hence when a current of 1
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 40
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
40
ampere (or 1 coulombsecond) flows in a conductor the number
of electrons passing any given point must be such that
1602x10-19
x no of electronssecond = 1 coulombsecond
So no of electronssecond = 624e18
Ie when the current in a circuit is 1 ampere electrons are
passing any given point of the circuit at the rate of 624e18
per second
When a potential or voltage of sufficient strength is applied
to a substance it causes the flow of electrons This flow of
electrons is called an electric current The rate of this flow
of current is measured in amperes An ampere is the rate of
flow of electric current represented by movement of a unit
quantity of electricity (coulomb) per second
Every substance is a conductor of electricity but it flow
very easily through some materials such as copper aluminum
iron and called electric conductors Wire and cables are the
common forms of conductors
Materials such as rubber glass certain plastics fibers dry
paper and air allow almost no electricity to pass through
them Such materials are called non-conductors insulators or
dielectrics When an insulator is continuous as for instance
around a wire it is commonly called insulation
The property of any material to oppose the flow of electricity
through it is called impedance The unit of the measurement of
this impedance or opposition to the flow of current is the
ldquoohm(Ω)rdquo Even the best conductors have some impedance poor
conductors have much impedance insulators (dielectrics) have
very high impedance The unit for the measurement of very low
impedance is the ldquomicrohm (microΩ)rdquo and is equal to one-millionth
of an ohm The unit of very high impedance is ldquomegaohm MΩrdquo and
is equal to one million ohm
An element of impedance called resistance in a conductor
varies directly as its length and inversely as its are (see
item 569)
Resistance may be compared to the friction encountered by a
flow of water through a pipe A straight pipe smooth inside
conducts water with little loss of pressure If the pipe is
rough inside and has many bends the loss of pressure and the
rate of flow will be greatly reduced Similarly a good
conductor allows electricity to flow with small loss of
voltage a poor conductor offers a large resistance and so
causes a corresponding large drop in voltage The energy used
in overcoming resistance is converted into heat
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 41
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
41
The voltage required to make a current flow depends upon the
impedance of the circuit A voltage of one volt will make one
ampere flow through an impedance of one ohm
Z
EI or
I
EZ or ZIE
Where
I = Current in amperes
E = Voltage in volts
Z = Impedance in ohms
This formula is known as ldquoohmrsquos lowrdquo
Many alternating current circuits contain coils that produced
magnetic effects These magnetic effects in turn react upon
the current They retard the current and cause it to lag
behind the voltage It means that the voltage has reaches its
maximum and started to fall some time before the current
reaches a maximum Some current will be flowing in the circuit
at the instant when the voltage is zero This magnetic reaction
is called ldquoinductancerdquo or ldquoself inductancerdquo
Another kind of influence on an alternating current is caused
by the presence in the circuit of alternate plates of
conducting material separated by insulation This commonly
referred to as ldquocapacitorrdquo and its effect on the current is
to cause it to lead ahead of the voltage This reaction is
called ldquocapacitancerdquo It tends to counteract the inductance in
a circuit and is useful in overcoming the inductive lag in the
current inherent in most alternating current motors
Therefore in an alternating current circuit there is
resistance inductance and reactance affecting the current
The combination of any two or all three of these effects is
referred to as ldquoimpedancerdquo of the circuit
Power
Power is defined as the rate of doing work Electric power
is measured in ldquohorsepowerrdquo One horsepower equals 746 watts
One watt is rather small unit of power consequently when
speaking of power required by motors the term ldquokilowattrdquo is
used one kilowatt being thousand watts To obtain the power
delivered to an alternating current apparatus you can not
merely multiply effective amperes by effective volts If the
circuit contains inductance the apparatus circuits always
contain it the product of the effective current and effective
voltage will be greater than the real power This ldquoapparent
powerrdquo is measured in ldquovolt-amperesrdquo or more often in a unit
1000 times as large ldquothe kilovolt-ampererdquo usually abbreviated
to ldquoKVArdquo
In alternating current power system the voltage and current
follow an approximate sine wave They build up from zero to a
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 42
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
42
maximum in one direction then diminish to zero build up again
to a maximum but in the opposite direction and again diminish
to zero thus completing one cycle or two alternations and 360
electrical degree
The power factor is said to be 10 or unity if the voltage and
current reach their respective maximum values simultaneously
However as discussed previously in most alternating current
systems the voltage reaches its maximum value in a given
direction before the current attains its maximum value then
the current is said to lag behind the voltage This lag may
measure in degree The actual current drawn by apparatus of
this class may be considered as having two components one
known as the magnetizing current or that current which must
overcome the choking effect produced by the characteristics of
the apparatus and which lags 90 electric degree behind the
voltage The value of this lagging current is zero when the
voltage has reached its maximum value This lagging or
magnetizing current is called ldquoreactive currentrdquo
The other component is known as ldquoreal currentrdquo and it is in
phase with the voltage This real current and voltage reach
maximum values simultaneously
The actual line current is therefore the vector sum of the
reactive and real currents furthermore it is the current
that would be registered if an ammeter is connected in the
circuit Since there one component lagging 90 electric degree
or at right angles to the voltage the resultant or actual
line current of which this component is a part must
consequently be out of phase with the voltage and lag behind
it The degree or amount that it lags depends upon the
magnitude of this reactive current component and is a measure
of power factor
Resistance
Consider a circuit having resistance R ohms connected
across the terminals of an alternator A as in the following
figure and suppose the alternating voltage to be represented
by the sine wave as in the next figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 43
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
43
If the value of the voltage at any instant is v volts the
value of the current at that instant is given by Rvi
amperes When the voltage is zero the current is also zero
and since the current is proportional to the voltage the wave
from the current is exactly the same as that of voltage Also
the two quantities are in phase with each other that they
pass through their zero value at the same instant and attain
their maximum values in a given direction at the same instant
If Vm and Im are the maximum values of the voltage and the
current respectively it follows that RVI mm
If the instantaneous value of the applied voltage is
represented by
sinVv m
Then the instantaneous value of current in resistive circuit
is
sinR
Vi m
Vector representing the voltage and the current in resistive
circuit is as follows
Inductive Reactance
Let us consider the effect of alternating current flowing
through a coil having an inductance of L henrys and negligible
resistance as the following figure
Current
Voltage
0
Vm
Im i v
-
+
Time
I V
V and I curve for a resistive Circuit
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
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79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 44
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
44
Suppose the instantaneous value of the current through
inductance L henrys to be represented by
Where
t is the time in second
f is number of cycles per second
Suppose the current to increase by di ampere in dt second
then
Instantaneous value of induced emf is
Since f represents the number of cyclessecond the duration
of 1 cycle = 1f second consequently
Hence the wave of the induced emf is represented by the curve
in the figure below lagging the current by a quarter of cycle
(90O)
Since the resistance of the circuit is assumed negligible the
whole of the applied voltage is absorbed in neutralizing the
induce emf
So instantaneous value of the applied voltage is
v L
i
Circuit with inductance only
A Coil
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 45
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
45
Comparison of expression (1) and (3) shows that the applied
voltage leads the current by a quarter of cycle (90O) Also
from (3) it follows that that maximum value Vm of the applied
voltage is
So that
The inductive reactance is expressed in ohms and is
represented by XL
Hence
Capacitance reactive
The property of a capacitor to store an electric charge
when its plates are at different potentials is referred to
capacitance as the following figure
The unit of capacitance is termed the farad (F) Farad is
defined as the capacitance of capacitor which required a
potential difference (pd) of 1 volt to maintain a charge of
one coulomb on that capacitor
Let us consider the effect of alternating current flowing
through a capacitor having a capacitance of C farad and
negligible resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 46
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
46
Circuit with capacitance only
Suppose the instantaneous value of the voltage applied to the
capacitance to be represented by
If the applied voltage increases by dv volt in dt second as in
the following figure
Instantaneous value of current flow through capacitor is
Comparison of (4)and (5) shows that current leads applied
voltage by a quarter of cycle (90O)and the current and voltage
can be represented vectorially as the following figure
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
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54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
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Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
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62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
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64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
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65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
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77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 47
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
47
From expression (5) it follows that the maximum value Im of the
current is
So
The capacitive reactance is expressed in ohms and is
represented by the symbol Xc
Resistance Inductance and Capacitance in series
An actual circuit may have resistance and inductance or
resistance and capacitance or resistance conductance and
capacitance in series Hence if we consider the general case
of R L and C in series we can adapt the results to the
other two cases by merely omitting the capacitive or the
inductance from the expression for general case
In following fig OB vector represents L (2πfLI) leading the
current by 900 and OC vector represents C (I2πfC) lagging the
current by 900
Since OD = OB-OC OB being assumed greater than OC and supply
voltage is the vectorial sum of OA and OD namely OE
I
v
90 0
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 48
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
48
Vector diagram for above figure
From this expression it is seen that
Resultant reactance = inductive reactance -capacitive
reactance
Φ = phase difference between the current and the supply
voltage
B
D
O
E
A I
C
[L]
[C]
φ
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 49
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
49
If the inductive reactance is greater than the capacitive
reactance tan Φ is positive and the current lags the supply voltage by an angle Φ if less tan Φ is negative signifying
that the current leads the supply voltage by angle Φ
If a circuit consists of a coil having a resistance R ohms and
inductance L henrys such a circuit can be considered as a
resistance and inductive in series and
And the phase angle in which the current lags the supply
voltage is given by
Example 1
A coil having a resistance of 12 Ω and inductance of 01 H is
connected across a 100 V 50 cs supply Calculate
a) The reactance and impedance of the coil b) The current c) The phase difference between the current and the apply
voltage
Solution
(a) Reactance = XL = 2ПfL = 2x314159x50x01 = 314 Ω
Impedance = Z = radicR2+XL
2 = radic12
2+314
2 = 336 Ω
(b) Current = VZ = 100 336 = 2975 A
(c) Tan Φ = XLR = 314 12 = 2617
Φ = 690
Example 2
A metal filament lamp rated at 750 watt 100 v is to be
connected in series with a capacitance across a 230 v 60 cs
supply Calculate
a) The capacitance required
b) The phase angle between the current and the supply voltage
Solution
From vector diagram below
(a) V2 = VR
2+VC
2
(230)2 = (100)
2 + VC
2
VC = 207 Volts
Rated current of lamp = 750 w 100 v = 75 A
75 = 2 x 314 x 60 x C x 207
C = 96 x 10-6 farad
= 96 microfarad (μF)
i
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 50
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
50
(b) If φ = phase angle between current I and supply V
Cos φ = VR V = 100 230 = 0435
Φ = 640 12rsquo
Example 3
A resistance of 12 Ω an inductance of 015 H and capacitance
of 100 μF are connected in series across 100 V 50 cs supply
Calculate
a) The impedance
b) The current
c) The voltage across R L and C
d) The phase difference between current and supply voltage
Solution
(a)
fCfLRZ
2
12
2
2
4198631147144
100501415932
10615050141593212
2
2
2z
(b) Current = VZ = 100 194 = 515 A
(c) Voltage across R = VR = 12x515 = 618 V
Voltage across L = VL = 471x515 = 2425 V
Voltage across C = VC = 3185x515 = 164 V
(d) Phase difference between current and supply voltage =
φ = cos-1(VRV)= cos
-1(618100) = 51
0 48rsquo
Or φ = tan-1 (VL-VCVR ) = tan
-1(2425-164)618) = 51
0 48rsquo
Or alternatively
φ = tan-1 [2πfL-(12πfC)]R = [471-3185]12 = 51
0 48rsquo
See vector diagram below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 51
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
51
Power in series RL circuit
From the above analysis it seen that the voltage applied to
a series RL circuit the voltage leads the current by angle φ
It is equally valid to say that the current lags the voltage
by an angle φ If a voltage Emsinωt is applied to a series RL
circuit the current is Imsin(ωt-φ) The instantaneous power
is
p = e x i
= Emsin ωt x Imsin(ωt-φ)
= tIEIE mmmm 2cos
2cos
2
The second term of the right side has an average value of
zero
p = coscos2
EIIE mm -----------(11)
Where E and I are the rms (Root Mean Square) values of
voltage and current and φ is the phase angle between the
voltage and current
The following figure is the plot of instantaneous voltage
current and power The instantaneous power is positive during
the time interval when both e and i are simultaneously
positive or negative During the interval when one of the two
quantities out of e and i is positive and the other is
negative instantaneous power is negative The positive area
is more than the negative area and therefore the average power
is positive
L 471
3185
1525
12
R
C
φ
VL 2425
164
785
618
VR
VC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 52
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
52
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
i=ImSin(ωt-φ)e=EmSinωt
p=EmSinωtImSin(ωt - φ) c o s φ
0 φ
+ ve area
-ve
+ ve area
-ve
0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360
p=EmSinωtImSin(ωt - φ) c o s φ
0
φ
+ ve area
-ve
+ ve area
-ve
p=EIcosφ
Power in RL circuit
Active power reactive power and power factor (single phase)
The average power in the circuit ie EIcosφ is the actual
power supplied by the source to the circuit This is known as
active power of the circuit The active power is measured in
watts The bigger units of active power are KW (kilowatt=103
watts) and MW (megawatt=106 watts) The product of voltage and
current ie EI called apparent power and is measured in
volt-ampere (VA)
The ratio of active power to apparent power equals cosφ This
term cosφ is called power factor of the circuit It is the
factor by which the apparent power (EI) must be multiplied to
give the active power The power factor for purely resistive
circuit is 1 Therefore the apparent power and active power
are equal for purely resistive circuit A circuit may be
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 53
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
53
characterized as having leading or lagging power factor A
leading power factor means that the current leads the voltage
This occurs in the capacitive circuits An inductive circuit
is described as having lagging power factor since the current
lags the voltage The active power in ac circuit can also be
written as I2R
Example
A 10 ohm resistor and 20 mH (mille Henry) inductor are
connected in series across a 230 volts 50 cs supply Find
the circuit impedance current voltage across resistor
apparent power active power reactive power and power
factor
XL = ωL = 2πfL = 2x324x50x20x10-3
= 628 Ω
Z = radicR2+XL
2 = radic10
2+(628)
2 = 118 Ω
Φ = tan-1(XLR) = 321
0
I = EZ = 230118 = 1949 amperes
Voltage across resistor = RI = 10x1949 = 1499 volts
Voltage across inductor = IXL = 1224 volts
Apparent power = EI = 230x 1949 = 44827 VA
Power factor = cos(321) = 0847 lagging
Active power = EIcosφ = 230x1949x0847 = 37968 watts
Reactive power = EIsinφ = 230x1949xsin(321) = 23821 w
Practical importance of power factor (PF)
If an alternator is rated at a given say 2000 A at a
voltage of 400 v it means that these are the highest current
and voltage values of the machine can give without the
temperature exceeding a safe value Consequently the rating of
the alternator is given as 400x20001000=800 KVA The phase
difference between the voltage and the current depends upon
the nature of the load and not upon the generator Thus if the
power factor of the load is unity the 800 KVA are also 800
kW and the engine driving the generator has to be capable of
developing this power together with the losses in the
generator But if the pf of the load is say 05 the power
is only 400 kW so that the engine is only developing about
one half of the power which it is capable through the
alternator is supplying its rated output 800 KVA
Similarly the conductors connecting the alternator to the load
have to be capable of carrying 2000 A without excessive
temperature rise Consequently they can transmit 800 kW if the
power factor is unity but only 400 kW at 05 pf for the
same rise of temperature
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 54
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
54
It is therefore evident that the higher the pf of the load
the greater is the power that can be generated by a given
alternator and transmitted by a given conductor
The matter may be put another way by saying that for a given
power the lower the pf the larger must be the size of the
alternator to generate that power and the greater must be the
cross sectional area of the conductor to transmit it in other
words the greater is the cost of generation and transmission
of the electric energy This is the reason why supply
authorities do all they can to improve the pf for their
loads (by ex installing capacitors)
Three phase alternating current
Three phase alternating current is the best suites for long
distance transmission because it may be easily generated at
low to moderately high voltages and can then have the voltage
raised to very high values suitable for efficient
transmission and then the voltage can be reduced to a value
suitable for general use by means of stationary device known
as a transformer The higher the voltage the smaller the wire
required to carry a given amount of power The following
figures are the curve and diagram representing three phase
alternating current
0 30 60 90 120 150 180 210 240 270 300 330 360
phase 1
phase 2 phase 3
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 55
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
55
Since the angle between ENR and EYN is 600
EYNR = 2ENRcos300 = radic3ENR
Line voltage = 173 x phase voltage
In general
If VL = pd between any two line conductors
= line voltage
And VP = pd between a line conductor and neutral point
= phase voltage
And if IL and IP = line and phase current respectively
then for a star connection system
VL = 173 VP
IL = IP
For delta connection
IL = 173 IP
Power in three phase system with balanced load
If Ip = value of the current in each phase
Vp = value of pd across each phase
Power per phase = IpVp x power factor
And total power = 3IpVp x power factor
If IL and VL be the value of the line current and voltage
respectively then for star connection system
VP = VL173 and IP = IL so
Total power in watts = 173 x ILVL x power factor
For delta connection system
120o
120o
30o
ENR
EBNY
ENY
ENB
EYN
EYNR
ERNB
EBN
Vector diagram
Line voltage (VL)
Between any line
conductors
phase voltage (VP)
(voltage to neutral)
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 56
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
56
IP = IL173 and VP = VL so
Total power in watts = 173 x ILVL x power factor
Example
A three phase motor operating off 400 v system is developing
25 bhp at an efficiency of 087 and a power factor of 082
Calculate (a) the line current and (b) the phase current if
the windings are delta-connected
(a) wattsinpowerinput
wattsinpoweroutputEfficiency
FxPVI
bhpx
LL 731
746
820400731
74625870
xxxI
x
L
IL = 378 A
(b) For delta-connected windings
Aline
currentphase 821731
837
731
current
Complex Numbers
The mathematics used in Electrical Engineering to add
together resistances currents or DC voltages uses what are
called real numbers But real numbers are not the only kind
of numbers we need to use especially when dealing with
frequency dependent sinusoidal sources and vectors As well as
using normal or real numbers Complex Numbers were introduced
to allow complex equations to be solved with numbers that are
the square roots of negative numbers radic-1
In electrical engineering this type of number is called an
imaginary number and to distinguish an imaginary number from
a real number the letter j known commonly in electrical
engineering as the j-operator The letter j is used in front
of a number to signify its imaginary number operation
Examples of imaginary numbers are j3 j12 j100 etc Then
a complex number consists of two distinct but very much
related parts a Real Number plus an Imaginary Number
Complex Numbers represent points in a two dimensional complex
or s-plane that are referenced to two distinct axes The
horizontal axis is called the real axis while the vertical
axis is called the imaginary axis
The rules and laws used in mathematics for the addition or
subtraction of imaginary numbers are the same as for real
numbers j2 + j4 = j6 etc The only difference is in
multiplication because two imaginary numbers multiplied
together becomes a positive real number as two negatives make
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 57
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
57
a positive Real numbers can also be thought of as a complex
number but with a zero imaginary part labeled j0
The j-operator has a value exactly equal to radic-1 so successive
multiplication of j (j x j ) will result in j having the
following values of -1 -j and +1 As the j-operator is
commonly used to indicate the anticlockwise rotation of a
vector each successive multiplication or power of j j2
j3 etc will force the vector to rotate through an angle of
90o anticlockwise as shown below Likewise if the
multiplication of the vector results in a -j operator then the
phase shift will be -90o ie a clockwise rotation
Vector Rotation of the j-operator
So by multiplying an imaginary number by j2 will rotate the
vector by 180o anticlockwise multiplying by j
3 rotates
it 270o and by j
4 rotates it 360
o or back to its original
position Multiplication by j10 or by j
30 will cause the vector
to rotate anticlockwise by the appropriate amount In each
successive rotation the magnitude of the vector always
remains the same There are different ways in Electrical
Engineering to represent complex numbers either graphically or
mathematically One such way that uses the cosine and sine
rule is called the Cartesian or Rectangular Form
Complex Numbers using the Rectangular Form
Complex number is represented by a real part and an imaginary
part that takes the general form of
Where
Z is the Complex Number representing the Vector
x is the Real part or the Active component
y is the Imaginary part or the Reactive component
j is defined by radic-1
In the rectangular form a complex number can be represented
as a point on a two-dimensional plane called the complex or s-
plane So for example Z = 6 + j4 represents a single point
whose coordinates represent 6 on the horizontal real axis and
4 on the vertical imaginary axis as shown
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
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Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 58
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
58
Complex Numbers using the Complex or s-plane
But as both the real and imaginary parts of a complex number
in the rectangular form can be either a positive number or a
negative number then both the real and imaginary axis must
also extend in both the positive and negative directions This
then produces a complex plane with four quadrants called
an Argand Diagram as shown below
Four Quadrant Argand Diagram
On the Argand diagram the horizontal axis represents all
positive real numbers to the right of the vertical imaginary
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 59
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
59
axis and all negative real numbers to the left of the vertical
imaginary axis All positive imaginary numbers are represented
above the horizontal axis while all the negative imaginary
numbers are below the horizontal real axis This then produces
a two dimensional complex plane with four distinct quadrants
labeled QI QII QIII and QIV The Argand diagram can also
be used to represent a rotating phasor as a point in the
complex plane whose radius is given by the magnitude of the
phasor will draw a full circle around it for
every 2πω seconds
Complex Numbers can also have zero real or imaginary parts
such as Z = 6 + j0 or Z = 0 + j4 In this case the points
are plotted directly onto the real or imaginary axis Also
the angle of a complex number can be calculated using simple
trigonometry to calculate the angles of right-angled
triangles or measured anti-clockwise around the Argand
diagram starting from the positive real axis
Then angles between 0 and 90o will be in the first quadrant
( I ) angles ( θ ) between 90 and 180o in the second quadrant
( II ) The third quadrant ( III ) includes angles between 180
and 270o while the fourth and final quadrant ( IV ) which
completes the full circle includes the angles between 270 and
360o and so on In all the four quadrants the relevant angles
can be found from tan-1(imaginary componentreal component)
Addition and Subtraction of Complex Numbers
The addition or subtraction of complex numbers can be done
either mathematically or graphically in rectangular form For
addition the real parts are firstly added together to form
the real part of the sum and then the imaginary parts to form
the imaginary part of the sum and this process is as follows
using two complex numbers A and B as examples
Example 1
Two vectors A = 4 + j1 and B = 2 + j3 respectively Determine
the sum and difference of the two vectors in both rectangular
(a + jb) form and graphically as an Argand Diagram
Addition
Subtraction
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 60
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
60
Graphical Addition and Subtraction
Multiplication and Division of Complex Numbers
The multiplication of complex numbers in the rectangular form
follows more or less the same rules as for normal algebra
along with some additional rules for the successive
multiplication of the j-operator where j2 = -1 So for
example multiplying together our two vectors from above
of A = 4 + j1 and B = 2 + j3 will give us the following
result
Mathematically the division of complex numbers in rectangular
form is a little more difficult to perform as it requires the
use of the denominators conjugate function to convert the
denominator of the equation into a real number This is called
rationalizing Then the division of complex numbers is best
carried out using Polar Form which we will look at later
However as an example in rectangular form lets find the value
of vector A divided by vector B
Multiply top and bottom by conjugate (2-j3)
The Complex Conjugate
The Complex Conjugate or simply Conjugate of a complex number
is found by reversing the algebraic sign of the complex
numbers imaginary number only while keeping the algebraic sign
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 61
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
61
of the real number the same and to identify the complex
conjugate of z the symbol z is used For example the
conjugate of z = 6 + j4 is z = 6 - j4 likewise the conjugate
of z = 6 - j4 is z = 6 + j4 The points on the Argand diagram
for a complex conjugate have the same horizontal position on
the real axis as the original complex number but opposite
vertical positions Thus complex conjugates can be thought of
as a reflection of a complex number The following example
shows a complex number 6 + j4 and its conjugate in the
complex plane
Conjugate Complex Numbers
The sum of a complex number and its complex conjugate will
always be a real number as we have seen above Then the
addition of a complex number and its conjugate gives the
result as a real number or active component only while their
subtraction gives an imaginary number or reactive component
only The conjugate of a complex number is an important
element used in Electrical Engineering to determine the
apparent power of an AC circuit using rectangular form
Complex Numbers using Polar Form
Unlike rectangular form which plots points in the complex
plane the Polar Form of a complex number is written in terms
of its magnitude and angle Thus a polar form vector is
presented as Z = A angplusmnθ where Z is the complex number in polar form A is the magnitude or modulo of the vector
and θ is its angle or argument of A which can be either
positive or negative The magnitude and angle of the point
still remains the same as for the rectangular form above this
time in polar form the location of the point is represented in
a triangular form as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 62
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
62
Polar Form Representation of a Complex Number
As the polar representation of a point is based around the
triangular form we can use simple geometry of the triangle
and especially trigonometry and Pythagorass Theorem on
triangles to find both the magnitude and the angle of the
complex number As we remember from school trigonometry deals
with the relationship between the sides and the angles of
triangles so we can describe the relationships between the
sides as
Using trigonometry again the angle θ of A is given as
follows
Then in Polar form the length of A and its angle represents
the complex number instead of a point Also in polar form the
conjugate of the complex number has the same magnitude or
modulus it is the sign of the angle that changes so for
example the conjugate of 6 ang30o would be 6 angndash30o
Converting between Rectangular Form and Polar Form
In the rectangular form we can express a vector in terms of
its rectangular coordinates with the horizontal axis being
its real axis and the vertical axis being its imaginary axis
or j-component In polar form these real and imaginary axes
are simply represented by A angθ Then using our example
above the relationship between rectangular form and polar
form can be defined as
Converting Polar Form into Rectangular Form ( PrarrR )
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 63
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
63
We can also convert back from rectangular form to polar form
as follows
Converting Rectangular Form into Polar Form ( RrarrP )
Polar Form Multiplication and Division
Rectangular form is best for adding and subtracting complex
numbers as we saw above but polar form is often better for
multiplying and dividing To multiply together two vectors in
polar form we must first multiply together the two modulus or
magnitudes and then add together their angles
Multiplication in Polar Form
Multiplying together 6 ang30o and 8 angndash 45o in polar form gives
us
Division in Polar Form
Likewise to divide together two vectors in polar form we
must divide the two modulus and then subtract their angles as
shown
319 Equivalent circuit of induction motor
To analyze the operating and performance characteristics
of an induction motor an Equivalent Circuit can be drawn We
will consider a 3ndashphase Y connected machine the Equivalent
Circuit for the stator is as shown below
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 64
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
64
V1 = Stator Terminal Voltage
I1 = Stator Current
R1 = Stator Effective Resistance
X1 = Stator Leakage Reactance
Z1 = Stator Impedance (R1 + jX1)
Io = Exciting Current (this is comprised of the core loss
component = Ic and a magnetizing current = Im)
The rotor winding can be represented as
Rotor Circuit
I2 = Rotor Current
R2 = Rotor winding Resistance
X2 = Rotor Leakage Reactance
Z2 = Rotor Impedance (R1 + jX1)
E2 = Induced EMF in the rotor (generated by the air gap flux)
The EMF (E2) is equal to the stator terminal voltage less the
voltage drop caused by the stator leakage impedance
Note
Never use three-phase equivalent circuit Always use per- phase equivalent circuit
The equivalent circuit always bases on the Y connection regardless of the actual connection of the motor
Induction machine equivalent circuit is composed of stator circuit and rotor circuit
3191 Effective circuit of induction motor at standstill
Standstill means rotor circuit is open At open
circuits S = 1 accordingly E1 = E2 fs = fr X2 = 2πfrL2
Refer to section 37 and 38
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 65
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
65
Effective circuit at standstill
E2 = V1 - I1 (Z1)
Where E2 is rotor induced emf at standstill
E2 = V1 - I1 (R1 + jX1) = I1 Z1
3192 Effective circuit of induction motor under
operating conditions (rotor winding is shorted)
Now suppose induction-motor is loaded down As motorrsquos load
increases its slip rises because of which the rotor speed
falls
Greater relative motion produces a stronger rotor voltage E2
which in turn produces a larger rotor current I2
Then the rotor magnetic field φ also increases Since rotor
slip is larger rotor frequency rises (fr = s fs ) and rotor
reactance increases (ωLr) Therefore rotor current lags
further behind the rotor voltage
Accordingly the parameters will be changed to be as follows
The stator and rotor sides are in the figure below separated
by an air gap
I2 = Rotor current in running condition
It is important to note that as load on the motor changes the
motor speed changes Thus slip changes As slip changes the
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 66
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
66
reactance of standstill X2 changes to be sX2 which is shown
variable
The rotor impedance can be represented by
Effective circuit of induction motor under operating
conditions
In the running condition the variable resistance R2s can be
rearranged as follows
So the variable rotor resistance R2s has two parts
1 Rotor resistance R2 itself which represents copper loss
2 R2(1 - s)s which represents load resistance RL So it is
electrical equivalent of mechanical load on the motor
Key Point Thus the mechanical load on the motor is
represented by the pure resistance of value R2(1 -s)s
The effective circuit of the induction motor under operating
condition shall be as following
It means that R2s = rotor copper resistance + load resistance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 67
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
67
3193 Power Relations
1 Input Power
2 Stator copper losses
3 Rotor copper losses
4 Air gap power
5 Mechanical power
6 Output power
7 Output torque
Power flow diagram
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 68
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
68
Example 1
A 480-V 60 Hz 50-hp three phase induction motor is drawing
60A at 085 PF lagging The stator copper losses are 2 kW and
the rotor copper losses are 700 W The friction and windage
losses are 600 W the core losses are 1800 W and the stray
losses are negligible Find the following quantities
1 The air-gap power PAG
2 The power converted Pm
3 The output power Pout
4 The efficiency of the motor
Solution
Example 2
A 480V 60 Hz 6-pole three-phase delta-connected induction
motor has the following parameters
R1=0461 Ω R2=0258 Ω X1=0507 Ω X2=0309 Ω Xm=3074 Ω
Rotational losses are 2450W The motor drives a mechanical
load at a speed of 1170 rpm Calculate the following
information
Synchronous speed in rpm
Slip
Line Current
Input Power
Air gap Power
Torque Developed
Output Power in Hp
Efficiency
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 69
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
69
Note the core losses is so small as the IoltltltltltI1 so it is
neglected
Synchronous speed in rpm
Slip
Line Current
Phase current is given by
Note that the machine is delta connected so V1 = VLL = 480 V
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 70
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
70
Therefore IL = radic3 times 466 = 806 A
Input Power
Air gap Power
Air gap power is the input power minus stator losses
Torque Developed
Output Power in Hp
Neglecting friction and windage losses so Pout = Pm
Efficiency
320 Determination of motor parameters
The motor parameters are determined from three tests
Stator DC resistance measurement Determines the stator
resistance value (R1)
o The motor is supplied by DC voltage between two terminals A and B at the figure below
o The dc voltage and current are measured
o If the stator is Y-connected the per phase stator
resistance is
o If the stator is Delta-connected the per phase stator
resistance is
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 71
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
71
No load test
No-load test Provides the magnetizing reactance and core
resistance (Rc and Xm) In this course we will only find Xm
and ignore Rc
o The motor is allowed to spin freely
o The motor is supplied by rated line-to-line voltage V the no-load current I1 and the no load input power P are
measured
o The only load on the motor is the friction and windage
losses so all Pm is consumed by mechanical losses
o The slip of the induction motor at no-load is very low Thus the value of the equivalent resistance R2(1-s)s in the
rotor branch of the equivalent circuit is very high
The equivalent circuit reduces tohellip
Combining Rc amp RF+W we gethelliphellip
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 72
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
72
o At the no-load conditions the input power measured by
meters must equal the losses in the motor
o The PRCL is negligible because I2 is extremely small because R2(1-s)s is very large
o The input power equals
o The equivalent input impedance is thus approximately
o The value of the stator leakage reactance X1 can be
determined from the blocked rotor test The value of the
magnetizing reactance can then be determined
Blocked Rotor Test
o In this test the rotor is locked or blocked so that it cannot move a voltage is applied to the motor and the
resulting voltage current and power are measured
o The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value
o The locked-rotor power factor can be found as
o The magnitude of the total impedance
Where Xrsquo1 and Xrsquo2 are the stator and rotor reactances at the
test frequency respectively
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 73
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
73
X1 and X2 as
function of
XLR
Rotor Design X1 X2
Design A 05
XLR
05
XLR
Design B 04
XLR
06
XLR
Design C 03
XLR
07
XLR
Design D 05
XLR
05
XLR
Example
The following test results are obtained from three phase 100
hp 460 v eight pole star connected induction motor design A
rotor
No load test
460 v 60 Hz 40 A 42 Kw
Blocked rotor test
100 v 60 Hz 140 A 8 Kw
Average DC resistor between two stator terminals is 0152
Determine
1 The parameters of the equivalent cicuit
2 The motor is connected to 3φ 460 v 60 Hz supply and runs at 873 rpm Determine
a The input current
b The input power
c Air gap power
d Rotor copper loss
e Mechanical power developed
f Output power
g Efficiency of the motor
3 The speed of the rotor field relative to stator structure and stator rotating field
Solution
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 74
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
74
From No load test
From block rotor test
As rotor is design A then
Equivalent circuit
Input impedance
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 75
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
75
Input Power
Stator copper loss
Air gap power
Rotor copper loss
Mechanical power developed
From no load test
321 NEMA standard for squirrel cage induction motor
Design A Motor
Hp range 05 ndash 500 hp
Starting current 6 to 10 times full-load current
Good running efficiency 87 - 89
Good power factor 87 - 89
Low rated slip 3 ndash5
Starting torque is about 150 of full load torque
Maximum torque is over 200 but less than 225 of full-load
torque
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 76
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
76
Design B motor
Hp range 05 to 500 hp
Higher reactance than the Design A motor obtained by means of
deep narrow rotor bars
The starting current is held to about 5 times the full-load
current
This motor allows full-voltage starting
The starting torque slip and efficiency are nearly the same
as for the Design A motor
Power factor and maximum torque are little lower than class A
Typical applications constant speed applications where high
starting torque is not needed and high starting torque is
tolerated
Unsuitable for applications where there is a high load peak
Design C motor
Hp range 3 to 200 hp
This type of motor has a double-layer or double squirrel-
cage winding
It combines high starting torque with low starting current
Two windings are applied to the rotor an outer winding having
high resistance and low reactance and an inner winding having
low resistance and high reactance
Operation is such that the reactance of both windings decrease
as rotor frequency decreases and speed increases
On starting a much larger induced currents flow in the outer
winding than in the inner winding because at low rotor speeds
the inner-winding reactance is quite high
As the rotor speed increases the reactance of the inner
winding drops and combined with the low inner-winding
resistance permits the major portion of the rotor current to
appear in the inner winding
Starting current about 5 times full load current
The starting torque is rather high (200 - 250)
Full-load torque is the same as that for both A and B designs
The maximum torque is lower than the starting torque maximum
torque (180-225)
Typical applications constant speed loads requiring fairly
high starting torque and lower starting currents
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 77
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
77
Design D motor
Produces a very high starting torque-approximately 275 of
full-load torque
It has low starting current
High slip 7-16
Low efficiency
Torque changes with load
Typical applications used for high inertia loads
The above classification is for squirrel cage induction motor
322 Torque of squirrel cage induction motor
In order to perform useful work the induction motor must
be started from rest and both the motor and load accelerated
up to full speed Typically this is done by relying on the
high slip characteristics of the motor and enabling it to
provide the acceleration torque
Induction motors at rest appear just like a short circuited
transformer and if connected to the full supply voltage draw
a very high current known as the Locked Rotor Current They
also produce torque which is known as the Locked Rotor
Torque The Locked Rotor Torque (LRT) and the Locked Rotor
Current (LRC) are a function of the terminal voltage to the
motor and the motor design As the motor accelerates both
the torque and the current will tend to alter with rotor speed
if the voltage is maintained constant
The starting current of a motor with a fixed voltage will
drop very slowly as the motor accelerates and will only begin
to fall significantly when the motor has reached at least 80
full speed The actual curves for induction motors can vary
considerably between designs but the general trend is for a
high current until the motor has almost reached full speed
The LRC of a motor can range from 500 Full Load Current (FLC)
to as high as 1400 FLC Typically good motors fall in the
range of 550 to 750 FLC
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 78
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
78
The starting torque of an induction motor starting with a
fixed voltage will drop a little to the minimum torque known
as the pull up torque as the motor accelerates and then rise
to a maximum torque known as the breakdown or pull out torque
at almost full speed and then drop to zero at synchronous
speed The curve of start torque against rotor speed is
dependent on the terminal voltage and the motorrotor design
The LRT of an induction motor can vary from as low as 60 Full
Load Torque (FLT) to as high as 350 FLT The pull-up torque
can be as low as 40 FLT and the breakdown torque can be as
high as 350 FLT Typical LRTs for medium to large motors are
in the order of 120 FLT to 280 FLT
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 79
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
79
Figure above graph shows that starting torque known as locked
rotor torque (LRT) is higher than 100 of the full load torque
(FLT) the safe continuous torque rating
The locked rotor torque is about 175 of FLT for the example
motor graphed above Starting current known as locked rotor
current (LRC) is 500 of full load current (FLC) the safe
running current The current is high because this is analogous
to a shorted secondary on a transformer
As the rotor starts to rotate the torque may decrease a bit
for certain classes of motors to a value known as the pull up
torque This is the lowest value of torque ever encountered by
the starting motor As the rotor gains 80 of synchronous
speed torque increases from 175 up to 300 of the full load
torque This breakdown torque is due to the larger than normal
20 slip
The current has decreased only slightly at this point but
will decrease rapidly beyond this point As the rotor
accelerates to within a few percent of synchronous speed both
torque and current will decrease substantially Slip will be
only a few percent during normal operation For a running
motor any portion of the torque curve below 100 rated torque
is normal
The motor load determines the operating point on the torque
curve While the motor torque and current may exceed 100 for
a few seconds during starting continuous operation above 100
can damage the motor Any motor torque load above the
breakdown torque will stall the motor The torque slip and
current will approach zero for a ldquono mechanical torquerdquo load
condition This condition is analogous to an open secondary
transformer
There are several basic induction motor designs (Figure below)
showing considerable variation from the torque curve above
The different designs are optimized for starting and running
different types of loads The locked rotor torque (LRT) for
various motor designs and sizes ranges from 60 to 350 of
full load torque (FLT) Starting current or locked rotor
current (LRC) can range from 500 to 1400 of full load
current (FLC) This current draw can present a starting
problem for large induction motors
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs
Page 80
Electric Submersible Pumps Mohamed Dewidar 2013
Chapter 3
80
Torque of NEMA designs