3-Ph Induction Motor

8/10/2019 3-Ph Induction Motor

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3-ph Induction Motor3-ph Induction Motor


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3-ph Induction Motor3-ph Induction Motor


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Theory

1. Three-phase induction motor is the mostpopular type of a. c. motor.

2. It is very commonly used for induction

drives since it is cheap, robust, efficientand reliable.

3. It has good speed regulation and high

starting torque.4. It requires little maintenance.5. It has a reasonable overload capacity.


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The basic principle of operation is inductionand hence thename Induction Motor.

Induction is a phenomenon of an induced voltage in a coil due

to changing flux.

This flux is established by either another coil(as a generalcase) or the same coil.


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Slot

Tooth

Stator core

or Stamping

IM consists of: 1. stator2. rotor.

Rotor

Shaft


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Tooth

Stator core

or Stamping

Rotor

Shaft

The stator is the stationarypart.

and the rotor is a part.rotating

IM consists of: 1. stator2. rotor.

Slot


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Rotor

Shaft

The laminations are slottedon the inner peripheryand are insulated from each other.

The insulated stator conductorsare placed in these slots.


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Rotor

Shaft

The stator conductors are connected to form athree-phase winding.

The phase winding may be either star or deltaconnected.


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Rotor

Shaft

The clearance between the stator and the rotor iscalled an air gap.The air gap is kept as small aspossible to

a) Reduce leakage reactance& no load currentandb) Improve the power factor.


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Rotor

Shaft

The rotor is built up of thin laminationsof thesame material as that of stator.The laminated cylindrical core is mounted directly

on the shaftor a spider carried by the shaft.


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Th t t l i ti t d i t t


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Rotor

Shaft

The stator laminations are supported in a statorframeof caste ironor fabricated steel plate.

Stator frame is connected to coverings.

The coverings are rested on bearingswhich are mounted onshaft.


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Air Gap

STATOR

ROTOR

Stator Core

Rotor Core

Shaf

tRotor wdg

Stator wdg

Bearings

Stator Frame

Base

STATOR

Common constructional features forALL

rotating electrical machines

Slip rings

Brushes


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There are TWOtypes of induction motor dependingon the types of rotor.

End RingsRotor Bars(Slightly skewed)

Rotor Bars(Slightly skewed)

Fig. 1. Cage Rotor

1. Squirrel-cagerotor or simply cage rotor. (SCIM).

The slots nearly parallel to the shaft axis or skewed .

2.Phase wound or wound rotor or slip ring rotor (SRIM).

Each slot contains an uninsulated bar conductor ofaluminium or copper.


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Fig. 1. Cage Rotor


At each end of the rotor, the rotor barconductors are short circuited by heavy end ringof the same material.


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Fig. 1. Cage Rotor


2.Phase wound or wound rotor or slip ring rotor (SRIM).

The conductors and the end rings look like a cage of a birdor form a cage of the type which was once commonly used for

keeping squirrel;

Hence this rotor is known as the squirrel cage rotor.


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The important features of this squirrel cage rotorare:1. The skewed bar reduces harmonics.2. This gives uniform torque and less noise.

3. The locking tendency is reduced.4. It has a compact and ruggedconstruction.5. The end rings can be projected for fanning actionfor

cooling.6. It requires no slip rings.

7. It has less loss and more efficiencyas compared to SRIM

8. It is not possibleto add extra rotor resistance to changethe torque and speed. This is the only disadvantage.

Ha Ha Ha, We are important in Electrical Machines

2 Ph d d li i


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ROTOR

OR

SHAFT

Slip Rings

Brushes

StartMax

RunMin

Fig. 2. Slip Ring Rotor

2.Phase wound or wound rotor or slip ring rotor(SRIM).

Rotor windingIn delta or Star

The wound rotor consists of a slotted armature.Insulated conductors are put in the slots and connected toform a three phase distributed double layer winding

similar to the stator winding.

2 Ph d d t li i t


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ROTOR

OR

SHAFT

Slip Rings

Brushes

StartMax

RunMin




The rotor windings are connected in star or in delta.The three ends of rotor windings are brought outside therotor and connected to three insulated slip rings.

The slip rings are mounted on the shaft withbrushes resting on them.

2 Ph d d t li i t


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ROTOR

OR

SHAFT

Slip Rings

Brushes

StartMax

RunMin




The resistors enable to increase each rotorphase resistance to serve the following purposes:

1. to increase the starting torque.2. to decrease the starting current.3. to improve the starting power factor.

4. to decrease the speedof the motor.

P i i l f O ti


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Principle of Operation

R

Y

B

R Y B


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R

Y

B

R Y B


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R

Y

B

R Y B


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0 60 120 180 240 300 360

r y b

r

y

b


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0 60 120 180 240 300 360

r y b

r

y

b

C n id n n l 0o (3/2) S


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0 60 120 180 240 300 360

r y b

Consider an angle 0o

-y0={(3)/(2)}m

b0={(3)/(2)m

r

y

b -y0b0

0=(3/2)m

R2

R1Y2

Y1

B2

B1

N

S

B1=+ & B2 = Y1= & Y2 =+

R1= 0 & R2 =0

At an angle 60o


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0 60 120 180 240 300 360

r y b

At an angle 60o

-y60={(3)/(2)}m

r60={(3)/(2)m

r

y

b-y0

r60

60=(3/2)m

R1=+ & R2 = Y1= & Y2 =+

B1= 0 & B2 =0

R2

R1Y2

Y1

B2

B1

N

S

At an angle 120o


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0 60 120 180 240 300 360

At an angle 120o

-b120={(3)/(2)}m

r120={(3)/(2)m

r

y

b

-b120

r120

120=(3/2)m

R2

R1Y2

Y1

B2

B1N

S

R1=+ & R2 = Y1= 0 & Y2 = 0

B1= & B2 =+

At an angle 180o

N


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0 60 120 180 240 300 360

At an angle 180o

-b180={(3)/(2)}m

y180={(3)/(2)m

r

y

b

-b180Y180

180=(3/2)m

R2

R1Y2

Y1

B2

B1

N

S

R1=0 & R2 = 0Y1= + & Y2 = B1= & B2 =+

At an angle 240o


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0 60 120 180 240 300 360

At an angle 240o

-r240={(3)/(2)}m

y240={(3)/(2)m

r

y

b

-r240

Y240

R2

R1Y2

Y1

B2

B1N

S

R1= & R2 = +Y1= + & Y2 = B1= 0 & B2 = 0

At an angle 300o


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0 60 120 180 240 300 360

At an angle 300o

-r300={(3)/(2)}m

b300={(3)/(2)m

r

y

b

-r300

b300

300

=(3/2)m

R2

R1Y2

Y1

B2

B1S

N

R1= & R2 = +Y1= 0 & Y2 = 0

B1= + & B2 =

At an angle 360o

=(3/2) S


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0 60 120 180 240 300 360

At an angle 360

-y360

={(3)/(2)}m

b360={(3)/(2) }m

r

y

b-y360b360

360=(3/2)m

R2

R1Y2

Y1

B2

B1

N

S

B1=+ & B2 =

Y1= & Y2 =+

R1= 0 & R2 =0

Phasor is

same asthat ofangle 0o

S


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0 60 120 180 240 300 360

R2

R1Y2

Y1

B2

B1

N

S

R

Y

B

For Clockwise RYB,Flux rotate Clockwise,Poles rotate Clockwise,

R Y B

S


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0 60 120 180 240 300 360

R2

R1

B2

B1

Y2

Y1

N

S

R

B

Y

For anticlockwise phase sequence, Flux rotate anticlockwise,Poles rotate anticlockwise,

Now change thephase sequenceof motor to RBY

R Y B

With samesupply phasesequence

Rotation of Rotor


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R

Y

ROTOR

Rotor ConductorSTATOR

Three Phase Supply is givenFlux is set up in the stator and passes from

stator to rotor

BFIRSTMETHOD

Rotation of Rotor


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Rotation of Rotor


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R

Y

BFIRSTMETHOD

Rotation of Rotor

FluxDirection

Motion ofConductor

w .r. t.StationaryField

By Faradays lawof electromagnetic induction, a voltagewillbe induced in the conductor.

Since the rotor is complete, either through the end rings oran external resistance, the induced voltage causes a currentto flowin the rotor conductor.

By right-hand rulewe can determine the directionof induced

current in the conductor.

Rotation of Rotor


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R

Y

BFIRSTMETHOD

Rotation of Rotor

By this rule the direction of the induced current is outwards,represented by dot.The current in the rotor conductor produces its own

magnetic fieldwhich is opposite to stator field on right handside and addition of flux on left hand side.This flux opposes the cause of it and cause is stator rotatingmagnetic field. (Lenzs Law)

FluxDirection

Motion ofConductor

w .r. t.StationaryField


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Rotation of Rotor


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FIRSTMETHOD

Rotation of Rotor

So rotor condror rotor moves in the same dirnof stator field.

R

Y

B

By this rule the direction of the induced current is outwards,represented by dot.The current in the rotor conductor produces its own

magnetic fieldwhich is opposite to stator field on right handside and addition of flux on left hand side.This flux opposes the cause of it and cause is stator rotatingmagnetic field.

Rotation of Rotor


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R

Y

BFIRSTMETHOD

Rotat on of Rotor

If rotor moves in opposite directionthen the speeddifferencebetween rotor and stator rotating magnetic field

increases.This increases the opposition. (Not opposing the cause of it)

Rotation of Rotor


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R

Y

BFIRSTMETHOD

f

If rotor moves in same directionthen the speed differencebetween rotor and stator rotating magnetic field decreases.

This decreases the opposition. (opposing the cause of it)

Rotation of Rotor


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The current in the rotor conductor produces its own magneticfield which is opposite to stator field on right hand side andaddition of flux on left hand side.

Rotor conductor moves towards right due to tension action offlux. (Catapult)It is seen that the force acting on the conductor is in the

same directionas the direction of the rotating magnetic field.

MoreFlux

Less

Flux

SECONDMETHOD

Rotation of Rotor

R

Y

B

Rotation of Rotor


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R

Y

BTHIRDMETHOD

Rotation of Rotor

FluxDirection

Direction ofCurrent

When a conductor carrying currentis put in a magnetic field aforceis produced on it. The direction of force can be found byleft hand Rule.

It is seen that the force acting on the conductor is in thesame directionas the direction of the rotating magnetic field.

Force on Conductor

S

Rotation of Rotor


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R2

R1

Y2

Y1

B2

B1

N

S

3 ph magnetic field rotates in clockwisedirection forclockwise RYBApply RHR to rotor conductor

Motion

Flux

Mark the Poles formed in rotor

N SN S

Now consider N of stator and N of rotor, Repulsion

Now consider N of stator and S of rotor, Attraction

So rotor has to rotate in clockwise direction

FOURTHMETHOD

Rotation of Rotor

Similarly S of stator and S of rotor,

S

Rotation of Rotor


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R2

R1

Y2

Y1

B2

B1

N

S


Motion

Flux

Mark the Poles formed in rotor

N SN S

Now consider N of stator and N of rotor, Repulsion



FOURTHMETHOD

Rotation of Rotor


S

Rotation of Rotor


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R2

R1

Y2

Y1

B2

B1

N

S

N SN S

FOURTHMETHOD

Rotation of Rotor

Flux

Motion


Mark the Poles formed in rotorNow consider N of stator and N of rotor, Repulsion




SRotation of Rotor


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R2

R1

Y2

Y1

B2

B1

N

S

N S

FOURTHMETHOD

Rotation of Rotor


Mark the Poles formed in rotorNow consider N of stator and N of rotor, Repulsion




Torque


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R

Y

BTorque

Since the rotor conductor is in a slot on the circumference ofthe rotor, this force acts in a tangential directionto the rotorand develops a torque on the rotor.

Similar torques are produced on all the rotor conductors.

BTorque


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R

Y

BTorque

Since the rotor conductor is in a slot on the circumference ofthe rotor, this force acts in a tangential directionto the rotorand develops a torque on the rotor.

Similar torques are produced on all the rotor conductors.

Hence rotor rotates

Three-phase induction motor is self starting.

and thus

BRotor Rotation


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R

Y

Bw. r. t. StatorFlux

Rotor Current

Ns

Nr

fr=sf

s=(Ns-Nr)/Ns=0.04

sf=0.04x50=2Hz

Stator Current, f=50Hz, t=20msec

Rotor Current, f=2Hz, t=500msec

t=500msec

BRotor Rotation


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Y

B

R

B

R

Y

w. r. t. StatorFlux

Ns

Ns

In this case Nr =slip = 0;No Speed difference between Rotorand Rotating Magnetic Field.No cutting of flux;No voltage; No current;

Nr

Nr

Ns;


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BRotor Rotation


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R

Y


Ns

Nr

R

Y

B Ns

Nr

As soon as speed decreases, rotor current flowsand


Ns;

No Rotor Flux.No torque. Speed decreases.

Torque is produced

BRotor Rotation


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R

Y


Ns

Nr

R

Y

B Ns

Nr

As soon as speed decreases, rotor current flowsand


Ns;

No Rotor Flux.No torque. Speed decreases.

Torque is produced

BRotor Rotation


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R

Y


Ns

Nr

An induction motor cannotrun at synchronous speed.

Therefore the rotor speed is slightly less thanthesynchronous speed.

Since the operation of this motor depends on the inducedvoltagein its rotor conductors, it is called an induction motor.

An induction motor may also be called as asynchronous


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y ymotor as it does not run at synchronous speed.

The difference between the synchronous speed and the

actual rotor speed is called slip speed Ns-Nr.

The slip is defined as the ratio of slip speed to thesynchronous speed, s =(Ns-Nr)/Ns.

The slip at full load varies from 2 to 5%.

Frequency of rotor emf = f2

120

&rotorfluxrotatingbetnspeedrelativeP

120

rs NNP

120

ssNP 1sf

Speed of rotor

f2120sf1120 sN


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flux w r t itself P

P ssN

Rotor itself rotates at a speedNr = (1-s) Ns, wrt stator

Speed of rotor flux wrt stator=sNs+ Nr

=NsThus stator androtorfluxrotate at synchronous speed wrt

stator

Speed of rotor

f2120sf1120 sN


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flux w r t itself P

P ssN



=NsThus stator androtorfluxrotate at synchronous speed wrt

statorN

swrt stator

Nswrt stator

N

S

Nrwrt stator

N S

Speed of rotor

f2120sf1120 sN


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flux w r t itself P

P ssN



=Ns

Thus stator androtorfluxrotate at synchronous speed wrt

stator

Relative speed is zero. MMF are stationarywrt each other

This producesuniform torque

The rotor freqquantity acts at stator freqwhen referred tostator

Induction Motor & Transformer


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Induction Motor is also called as

Induction Motor TransformerStator Primary

Rotor Secondary

Generalized TransformerTherefore, equivalent circuitdiagram of transformer is

applicable to IMVoltage equation of stator, V1= -E1+I1(r1+jx1) =V1

+I1(r1+jx1)

jXmRc

I1

I0

Ic I

V1

r1x1

f1

Voltage equation of rotor,


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V2=E2I2(r2+jx2), for transformer

= sE2I2(r2+jsx2), for IM

= 0, for shorted rotor (IM)

= sE2I2(r2+jsx2), for IM

Rotor currentI2= ,22

2

jsxr

sE

at slip frequencyf2I2

sE2f

2

=sf1

jsx2

r2

Rotor currentI2= ,

22

2

jx

s

r

E

at frequencyf1I2

E2f1

jx2

sr /2

E2 = I2(r2+jsx2),


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)(' 22222

221 xjas

ra

a

IaEEE

2 2( 2 j 2),

)''

(' 22

11 jxs

rIE

I1

E1f1

Jx2

sr /'2

Complete Equivalent Circuit Diagram


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p q g

f1

sr /'2jx1 Jx2

jXmRc

r1 I1 I1

I0

Ic I

V1

Complete Equivalent Circuit Diagram


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p q g

f1

jx1Jx2

jXmRc

r1 I1 I1

I0

Ic I

V1 sr /'2

s

s1rr-

s

r22

2


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Power Balance


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jx1jx2

jXmRc

r1 I1 I1

I0

Ic I

V1

r2

s

s1r2

StatorInput

powerV1I1cos1

Stator

I12r1loss

Stator

coreloss

Rotor Input powerE2 I2cos2=I22r2/s

(Air Gap Power Pg)

Rotor

I22r2loss

Mechanical

PowerDeveloped inRotor Pm

Rotor Input power= V1I1 cos 1- I12r1- Pi


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p p 1 1 1 1 1 iAir gap power,Pg=I22r2+ Pm

=E2I2 cos 2

and

2

2

2

2

22

xs

r

EI

2

2

2

2

2

2

xs

r

sr

cos

2

2

2

2

2

22

xs

r

sr

IE

Pg

srI 222Pg

s

s1rIrI 2

2

22

2

2Pg

gg s)P(1sP Pg

P

s1

rI 22

2


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Pm

TeMechanical

s

rI 22

gPs)(1

Internal (or Gross) torquedeveloped per phase is

SecondsperRadianinSpeedRotor

RotorinDevelopedPowerInternal

Mechanical

r(m)

m

P

s(m)

g

s)(1Ps)(1

secWatts/rad/

P

s(m)

g

Tecan also be expressed as Synchronous Watts

due toPgand s.

Shaft PowerPsh=Pm -F & W Loss


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Tsh

=Te - Lost Torque due to constant loss

r(m)

sh

P

- Iron LossPi (If not considered

earlierin Equi Ckt Diagram)

=Pm -Constant Loss

Efficiency= 100%PPP

P

cuconstantm

m

0.5%is deducted from calculated efficiency

in order to consider STRAY load losses.

Torque- SlipCharacteristics


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jx1jx2

jXmRc

r1 I1 I1

I0

Ic I

V1 /sr2

Apply Thevenins Theorem at points Aand B.

Obtain VTH, ITH, RTH andXTH

r1andx1are in parallel withXm

B

A



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jXTHjx2RTH ITH

VTH /sr2

q p

Thevenins Equivalent Circuit Diagram

B

A

I1 =I2



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q p

s

rI

m

PT 22

2ss

g

e

Nms

r

xXsrR

V

m 2

22TH

22

TH

2TH

s

Slip1 0

Ns0 Speed

Te

Motoring Mode

Test

TemTop

TLUnstable

Stable (Shunt ch)

222TH2

2TH

THest r

xXrRKT

22

THTH

THem

XRR2

KT

xXX TH

2

THop r

KsT

(T-s linear ch)

KMotor torque Tein terms of Tem


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em

e

TT

sr

xX

s

rR

K 22

2TH

2

2TH

TH

Assume r1 very small, so neglected.

Xsr mt2

2

2THTH

TH

XRR2

K

XxXxXRs

r2TH2THTH

mt

2 22

)(

RTHr1, RTHis also neglected.

em

e

T

T 2X

22

mt Xs

Xs

s

Xsmt

22THTH XRR2

22TH2

2TH xXs

rR

sr2

2

mt

mt

s

s

s

s

e 2T


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mt

mtem

e

ss

s

s2

T

T

mt

mtem

s

s

s

sT

2

s

1

1

s

eTTest

Analysis of equivalent circuit


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y q

r2/s

jx1 jx2

jXmRc

r1 I1 I2

I0Ic I

V1

ss1rr

sr 222


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jx1 jx2

jXmRc

r1 I1 I2

I0Ic I

V1

r2

s

s1r2

With this, circuit is equivalent to transformer.

At standstill, s=1,

At synchronous speed, s=0,

ckt becomes equivalent to shortcircuited transformer.

ckt becomes equivalent toopen circuited transformer.

Equivalent Circuit parameters


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q pThe equivalent circuit parameters can be

determined from

1.DC resistance measurement method2.No Load Test or Running Light test or Open Circuit Test3.Blocked Rotor test or Short Circuit Test

1.DC resistance measurement methodThe DC stator resistance r1, at room temperature ismeasured by circulating a suitable DC currentmeasuring voltage dropbetween stator terminals

The hot resistance at 750is

rdc=Vdc/Idc

r1=(1.1 to 1.3) rdc

A

V

2.No Load Test or Running Light Test orOpen Circuit Test


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Open Circuit TestThis test gives

1. Core loss2. F & W loss3. No load current I0

5. Ic, Rc, I, Xm6. Mechanical faults, noise

Rated per voltage V0,with

rated freq is given to stator.

Motor is run at NO LOAD

STATOR

A

I0

VV0

R

YB

ROTOR

N

W0

P0, I0and V0are recorded

P0 = I02r1+Pc+Pfw

4. No load power factor

No load power factor is small,0 05 t 0 15

00 IV

PC os


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0.05 to 0.15

1. Ic=I0cos0 2. I=I0sin0

3.

On No load, Motor runs near to syn speedSo, s zero 1/s=or open circuit

4.

00IV

)(, 11000c

0c jxrIVEI

ER

I

EX 0m

r1

r2/s

jx1 jx2I2

jXmRc

I0I0

Ic I

V0

open

cir

cuit

provided x1is known

The F & W loss Pfw, can be obtained from thistest


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test.Vary input voltage and note input power

Input Power

Input Voltage

Pfw

Thus Pc=P0 - I02r1 - Pfw

R t i bl k d S d 0 li 1

3.Blocked Rotor test or Short Circuit Test


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Rotor is blocked, Speed = 0, slip = 1

A

Isc

V Vsc

R

YB

N

Wsc

3-ph Variac

I M

Rotor is blocked or held stationary by

belt pulleyor by hand

Low voltageis applied upto rated statorcurrent

Voltage Vsc, Current Iscand Power Pscare measured.

Since slip is 1, secondary is shortcircuited

j


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Mechanical loss =0

Rc and Xm >> r2+jx2

Therefore, Zsc

= Vsc

/ Isc

=Rsc+jXsc

This test gives copper loss

jx1 jx2

jXmRc

r1 Isc

I0

Ic I

Vsc

r2

s

s1r2

Core loss negligible

Hence omittedscscIV

Pcos scsc =0.8 to 0.9

= r1+r2Rsc= Psc/Isc2

R


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Class of motor x1 x21. Class A(normal Tstand Ist) 0.5 0.5

For wound rotormotor, x1 = x2 = Xsc /2

22scscsc RZX

r2= Rscr1

21 xx

For squirrel cagemotor,

2. Class B(normal Tstand low Ist) 0.4 0.6

3. Class C(high Tstand low Ist) 0.3 0.7

4. Class D(high Tstand high slip) 0.5 0.5

Circle Diagram of Ind Motor


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But the advantage of circle diagram is that

torque and slipcan be known from circle diagram

The circle diagram is constructed with the help of

Graphical representation

The equivalent ckt., operating ch. can be obtained

by computer quickly and accurately

1. No load test (I0& 0)

2. Blocked rotor test (Isc& sc)

g

extremities or Limits of stator current, Power,

y


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xI00

Isc

sc

1. Draw x and y axes(V1on y axis)2. Draw I0and Isc(=V1/Zsc)3. Draw parallel line to x axis from I0.

This line indicates constant lossvertically

V1

Line I0Isc isoutput line

4. Join I0 and Isc

Output line

O

y


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xI00

sc

C

Output line

L1

T

V1

5. Draw perpendicular bisector to output line

6. Draw circle with Cas a centre

7. Draw perpendicular from Isc on x axis..

8. Divide IscL1in such a way that. LossCuStator

LossCuRotor

r

'r

LT

TI

1

2

1

sc

Isc

L2O


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y

R

P


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xI00

sc

C

Output line

11. FromR, draw line parallel to output linecrossing at P & P.

P is operating point

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

12. Join O and P. Cos1is operating pf.

1

Lebel O, T , L1 and L2

13. From P draw perpendicular on x axis

O

y

R

P


91/246

xI00

sc

C

Output line

14. Determine the following1.Constant Losses and copper losses

=Core loss + F & W loss

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

L1L2=L1L2=constant losses

no load current I0

1

O

y

R

P


92/246

xI00

sc

C

Output line

At standstill, input power = IscL2 L1L2=Constant Loss

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

Constant loss= Stator core loss +rotor core loss (f)

F & W loss=0

1

O

y R

P


93/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

At operating point P, input power = PL2,

1

L1L2=Constant Loss

Constant loss = Stator core loss + F & W loss

Rotor core loss 0 (sf)

Thus L1L

2=L

1L

2= Constant loss

O

y

R

P


94/246

xI00

sc

C

Output line

At standstill,Stator Cu loss=TL1

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

rotor Cu loss = IscTAt P, stator Cu loss =TL1 and

1

rotor Cu loss = OT

O

y R

P

P


95/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L22. Output Power and Torque

1

Output Power = OP

The gap betnoutput line and circle is OUTPUTPower.

At I0, o/p=0, at Isc, o/p=0

O

Pmax

Max output power=PmaxO

Slip1 0

Ns0 Speed

Pmax

TL1L2O

y R

P

P


96/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

Output Torque = TPThe gap betntorque line and circle is OUTPUTtorque.At I0, torque=0, but at

Isc, torque=T Isc

Pmax

=Starting torque

Tmax

Max output torque=TmaxT

2. Output Power and Torque

Slip1 0

Ns0 Speed

Tmax

OTL1L2

O

TL1L2O

y R

P

P


97/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

Max Power and Max Torque are not occurring at same time

Contradictionto max power transfer theorem

Pmax

Tmax

2. Output Power and Torque

OTL1L2

O

TL1L2O

y R

P

P


98/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

Pmax

Tmax

OTL1L2

O

TL1L2

Air gap power Pg= Input power Stator Cu loss- core loss

=PL2-TL1-L1L2

3. Slip, Power factor and Efficiency

=PT

s = rotor Cu loss/Pg =OT/PT"

""

TP

TOsmp

max

'"

'"'"

TT

TOsmt

max

O

y R

P

P


99/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

Pmax

Tmax

OTL1L2

O

TL1L2

3. Slip, Power factor and Efficiency

O

Power factor cos1= PL2/OP

PO/PL2Efficiency=


100/246

y R

P

P


101/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

s=0

O

Pmax

Tmax

T

5. Induction Generator

O

s=1

s=

braking torque

y R

P

P


102/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

s=0

(Generator)

O

Pmax

Tmax

T


O

s=1

s=

braking torque

s= -ve

G

G

OG=Gen CurrentOG=Mech I/p

L2G=Active power

OL2=reactive powerPGmax

y R

P

P


103/246

xI00

sc

C

Output line

Torque line

V1 Isc

T

P

P

L2

L1TO

L1

L2

1

s=0

(Generator)

O

Pmax

Tmax

T


O

s=1

s=

braking torque

s= -ve

G

G

OG=Gen CurrentOG=Mech I/p

L2G=Active power

OL2=reactive powerPGmax

Slip 0 -1Speed 2Ns

Ns0Slip

Speed

Te

1

(Double Cage Motors)High Torque Cage Motors


104/246

Lower starting current

Lower full load speed

(Double Cage Motors)

Highrotor resistance at the time of STARTgives

Higher starting torque

Better power factor

Poor speed regulation

More rotor ohmic loss

Reduced efficiency

Advantages

Disadvantages

Lowrotor resistance at the time of STARTOn the other hand


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Higher starting current

Higher full load speed

results in

Lower starting torque

Poor power factor

Better speed regulation

Less rotor ohmic loss

More efficiency

Disadvantages

Advantages

Therefore, it may be concludedthatHighrotor resistance at starting

Lowrotor resistance at running

are the desirable features.

conditions are obtained by usingIn SRIM or Wound Rotor IM these two


106/246

which are called as High Torque Rotors.

1. Deep Bar Rotor

yexternal resistance

But in SCIM it is achieved by special rotor arrangement

There are two types of rotors

2. Double cage Rotor or Boucherot Rotor

1. Deep Bar Rotor


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A

B

Slot Leakage Flux

More Flux lines

Less Flux lines

Air Gap

MagneticMaterial

Types of deep bar


108/246

yp p

Parallelsided bar

Conductor

Trapezoidalbar

L-bar T-bar

Portion B has more leakage fluxlines

More leakage reactance x


109/246

A

B

Slot Leakage Flux

More Flux lines

Less Flux lines

Air Gap

MagneticMaterial

More leakage reactance xb

Less leakage reactance xa

At the time of startRotor freq. =f

More current at top (A)

Less current at B (skin effect)

This non-uniformdistribution of current

increases resistanceof lower part (B)

So highstarting torqueis developed

Under running condition,

Freq = sf


110/246

A

B

Slot Leakage Flux

More Flux lines

Less Flux lines

Air Gap

MagneticMaterial

Freq = sfLeakage reactances of both parts decrease

Current is distributed uniformly

Resistance decreases.

Thus better performance

at start and at normalspeed

Equivalent Circuit Diagram


111/246

r2/s

jx1 jx2

jXmRc

r1 I1 I2

I0Ic I

V1

Same phasor diagram of induction motorBut for calculations, r2 and x2must be consideredaccordingly

At start freq f and under running freq sf

2. Double cage Rotor or Boucherot RotorTwo cages topcage less cross section -

h h b l b


112/246

Ax

Bx

a

b

ab

-A

-B

Dumb-bell

slotting

Staggered

slotting

Types

highresistance- brass, aluminium, bronzebottom cage more cross section -lowresistance- copper

SlitHeat

conduction

a= leakage flux of A - rotorb= leakage flux of B-rotor


113/246

Ax

Bx

a

b

ab

SlitHeat

conduction

ab= leakage flux of rotor= mutual flux of A & B

b >> a

At the time of start,

rb+jxb>> ra+jxa

More Current flows through A

Zb>> Za

ra>> rb (ra= 5 to 6 times rb)

More resistance

More starting torque

Under running condition, freq = sfxa& xbare negligible and ra>> rb


114/246

Ax

Bx

a

b

ab

SlitHeat

conduction

Therefore, Za >> Zb

More Current flows through B

Less resistanceGood operating characteristics

jx1 jx2r1 I1 I2

I0


115/246

Equivalent

Circuit Diagram r2/sjXmRc

I0Ic I

V1

With top cage only

With bottom cage only

r3/s

jx1 jx3

jXmRc

r1 I1 I3

I0Ic I

V1

Equivalent Circuit Diagram

For double cage


116/246

r2/s

jx1

jx2

jXm

Rc

r1 I1

I2I0Ic I

V1

r3/s

jx3

I3

All parameters are referred to statorAir gap power, Pg =

s

rI 2

2

2s

rI 3

2

3

Te =s

gP

s

rI

s

rI

ns

32

322

2

2

1


117/246

Torque slip characteristics

First consider top cage(more resistance)


118/246

Slip1 0

Ns0 Speed

Te

Now consider bottom cage(less resistance)

Double cage

Bottom cage

Comparison between single cageanddouble cage IM


119/246

1. T-s characteristics

Slip1 0

Ns0 Speed

Te

Double cage

Single cage

Wide range of torque slip chcan be obtained by choice of

top and bottom cage


120/246

2. Starting performance

Slip1 0

Ns0 Speed

Te

Double cage

Single cage

In case of double cage,higher starting torque canbe obtained

Due to more resistance,

suitable for Direct-On-LinestartingBut there is more heating


121/246

3. Full load performance

Slip1 0

Ns0 Speed

Te

Double cage

Single cage

Double cage has higher r2and x2,

So lower breakdown torquelow power factor

low efficiency

low full load slip

4. Circle diagramAt the time of start, rb+jxb>> ra+jxa, Zb>> Za


122/246

More Current flows through A

x

y

V1

O

A

B

Double Cage Single Cage

ExampleA 4 pole, 50Hz, double cage induction motor has


123/246

the following per phase parameters referred tostator.

Stator: r1= 0.5, x1= 1.5

Rotor: Top cage: r2= 2, x2= 0.6

Bottom cage: r3= 0.4, x3= 3.4

Magnetizing current is zero. The primary in delta is

energized from 400V. Calculate the starting torque and full

load torqueat 4% slip using the approximate equivalent

Circuit. Find also the pfat starting and at full load

SolutionThe equivalent circuit diagram with I0=0 is

jx


124/246

r2/s

jx1

jx2

r1 I1

I2

V1

r3/s

jx3I30.5

1.5

0.6

2/s

3.4

0.4/s

Total impedance at the input terminal is

j42.4

j3.40.4j0.62j1.50.5Z

j2.50431.657

056.513.003

Synchronous speed=2ns =225 =50rad/sec

St t t ti t I 400 133 2A


125/246

Stator starting current, I13.003

400

Starting torque, Test resistancerotorequivalentI

s

21

1.157133.2

50

1 2

133.2A

Nm130.68

Pf at starting Lagging0.5518cos56.51

Total impedance at full load is

j40.042.4

j3.40.040.4j0.6

0.042

j1.50.5Z j3.7698.985

022.767439 .

F L stator current, I1 A9.743400

F L t T 8 4854001

2


126/246

F L torque, Tefl 8.4859.743

400

50

1

Nm91.05

F L Pf Lagging0.922cos22.760

The starting torque is higher than F L torqueF L pf is better than starting pf

Example

A 6hp, 220V, 50Hz, 6 pole, 3-ph, star connected


127/246

IM gave following test data:

No load test: 220V, 6A, 475W (Line values)Blocked rotor test: 110V, 27A, 1930W (Line values)

Calculate from the circle diagram, for full load condition,

line current, power factor, torque, slip and efficiencyAlso determine the maximum output, maximum torque, slipAlso determine the maximum output, maximum torque, slip

for maximum torque and starting torque.

The stator copper loss at standstill is twice the rotor

copper loss.

17.44A, 0.83 lag, 45.41Nm, 0.046, 79%

8.89hp, 73.14Nm, 0.134, 32.75Nm

Induction motors find wide application in

Starting of 3-ph IM


128/246

Induction motors find wide application in

electrical drives because of

1.constructional advantages2.robustness3. cheapness

In industry, in order to attainhigh productivity

high quality products

it is required to PROPERLY

STARTCONTROL SPEED

STOP the motor

Depends on

Starting of SCIM


129/246

Depends on

1.Size of motor

2.Type of load and3.Capacity of supply lines

Thereare principally TWO methods

1. Fullvoltage starting or Direct-on-Linestarting

2. Reducedvoltage starting

1. Fullvoltage starting or Direct-on-Linestarting

Simplest

inexpensiveAs the name indicates IM is directly connected to

supply mains

pply


130/246

Fig: Direct- On-Line starting

At starting m/c has low pf

This large current may not harmrugged SCIM

but may cause objectionable voltage dropwhich may affect otherequipments

3-ph

Sup

Stator

Rotor

current 5 to 10times F L current

For example dimmingof lamps and tube lights

when refrigerator motor starts in the home.

Therefore, this method is applicable up to 3 HP.If starting timeis more, motor may burn.

Th i f i


131/246

Te s

r

Is

22

2

1

The equation of torque is

Test

Tefl flfl

st

sr

r

I

I

/

1/

2

2

2

2

2

2

fl

fl

st sI

I 2

2

2

If no load current is neglected, then similar to transformer

I1st

I2st=

Rotor TurnsStator turns

I1stx Stator turns= I2stx Rotor turns

I1st= I2stx Ratio of Rotor turns to Stator turns

I1st= I2stx Ratio of Rotor turns to Stator turns


132/246

I1st

I1fl=

fl

2

1fl

1st sII

At the time of start, I1st=Isc

1st 2st f

I1fl= I2flx Ratio of Rotor turns to Stator turns

I2st

I2fl

TestTefl

fl

2

1fl

sc sII

2. Reducedvoltage startingIn SCIM, it is not possible to change the rotor

i t


133/246

Therefore, the starting current can only be reduced by

reducing the statorterminal voltage

There are three methods of reducing voltage

1.Stator reactor(or resistor) starting

resistance.

2.Auto-transformer starting

3.Star-delta starting

1.Stator reactor(or resistor) starting

Three reactors(or resistors) are connected in series withstator winding.The voltage applied to stator wdg is less than the voltage

drop across reactor.

Rotor1V 1xV


134/246

Fig: Stator Reactor starting

The value of x is less than one.

stator and motor is started.

As speed increase, reactor is cut outin steps.

and finally no reactoris in circuit.

Stator

Rotor

Initially more reactoris in circuit, less voltageapplied to

Speed is equal to operating speed

Stator Reactor1V3

1

In place of reactors resistors can be used

Rotor1V 1xV


135/246

Fig: Stator Reactor startingReactors are more costlythan resistors.

1. Low power lossin reactors

2. Effective reductionin voltage applied to stator

Stator

Rotor

But reactors are preferred due to

Stator Reactor1V3

1

V1

I1

VLV1

I1

VR

Vr

Rotor1V 1xV


136/246

Fig: Stator Reactor starting

The starting current is

Stator

Rotor

The short circuit current is

Stator Reactor1V3

1

111st ZxVI /

11sc ZVI /

sc1st xII

fl

2

1fl

1st sII

Test

Teflfl

2

fl

sc2 sIIx

Thus,Starting torque with reactor starting

Starting torque with DOL starting

=x2

2.Auto-transformerstarting

1V


137/246

Fig: Auto-transformer starting

Stator

Rotor

The fraction of xV1is applied to the stator wdg atstarting.

1V

1

xV

1. Voltage is changed by transformer action

2. So power loss and input current are less.

xV1

LIscst xII

As speed increases, gradually voltage is increased

Finally full voltage is applied to the motor.

Advantagesand not by dropping voltage as that of reactor


1V


138/246


Stator

Rotor

The stator starting current is

1V

1

xV

sc11st xIzxVI /

xV1

LIscst xII

For auto-transformer, input VA= output VA

ILV1=Ist (xV1)

IL=xIstIL=x2Isc

Therefore, line current at

inputis x2times the DOLcurrent.

fl

2

1fl

1st sI

I

Test

Tefl

fl

2

fl

sc2 sI

Ix

Thus,


1V V


139/246


Stator

Rotor

Line current at input due to auto-transformerstarting

1V

1

xVxV1

LIscst xII

Line current at input due to stator reactorstarting =x

Stator

Rotor

1V

1xVxV

1

LIscst xII


1V V


140/246


Stator

Rotor

Line current at input due to auto-transformerstarting

1V

1xVxV1

LIscst xII

Line current at input due to stator reactorstarting =x

Starting torque with auto transformerstartingStarting torque with DOLstarting =x

2

Starting torque with auto transformerstartingStarting torque with stator reactorstarting =1

3.Star-Deltastarting

For delta, 6 terminals are required.For star, 3terminals of stator wdg are required.


141/246

Stator

Rotor

q

TPDT

R Y B

2- Run

1- Start - Star

- Delta

Fig.: Star-Deltastarting

Now make delta

Connection.

At starting TPDT to 1, wdg in star

Motor rotates.Reduced voltage is applied to wdg = VL/3

Th st ti t is


142/246

Stator

Rotor

TPDT

R Y B

2- Run

1- Start - Star

- Delta



1Lst.y zVI 3/

L.yI StartingLine current

Now TPDT to 2- Delta

Line voltage appliedto wdg. Motor runs at rated speed



Th st rtin curr nt is


143/246

Rotor

Stator

TPDT

R Y B

2- Run

1- Start - Star

- Delta



1Lst.y zVI 3/



Line voltage appliedto wdg Motor runs at rated speed





144/246

At starting, if, wdg in deltaThe starting current is

1Lst.d zVI / sc.dI

st.dL.d II 3

st.dst.y II3

1

Starting line current with Y-starter

Thus Ist.yin star is one third of that current in delta.

= Ist.y3 Ist.d

= 13Starting line current with stator in


1Lst.y zVI 3/



Line voltage appliedto wdg Motor runs at rated speed

This shows that T in star is one third of starting torque

Starting torque with Y-startingStarting torque with stator in

= (V1/3)2

V12= 1

3


145/246

In case of auto-transformer, if turn ratio x = 1/3

Then starting line current and is starting torque are

This shows that

This shows that Tst.yin star is one third of starting torquein delta.

Star delta starting is equivalent to auto transformer

reduced to one third of their values with delta.

if auto transformer turn ratio x=1/3=0.58 or 58% tapping

This method is cheap, effective and used extensivelyUsed for tool drives, pumps, motor-generator set.

Used up to rating of 3.3kV,

After this voltage, m/c becomes expensive for delta winding

that the supply current during starting of IM does not

ExampleDetermine the % tappingof the auto-transformer so


146/246

exceed 1.5 times full load current. The shortcircuit current

on normal voltage is 4.5 times the full load currentand the

full load torque.

that the supply currentduring starting of IM does not

Solution

full load slip is 3%.Calculate the ratio of starting torque

Stator

Rotor1V

1xVxV1

LIscst xII

IL=1.5IFLIL=1.5IFL

Isc=4.5IFL

IL/Isc=0.333

In auto-transformer IL/Isc=x2 x=0.577

Hence % tapping is 57%

fl

2

1fl

1st sII

Test

Teflfl

2

fl

sc2 sIIx

Now

0 04 50 2


147/246

Stator

Rotor1V

1xVxV1

LIscst xII

0.034.50.333 2

0.202

its full load current, the stator of which is arranged for star

ExampleThe short circuit line current of a 6hpIM is 3.5times


148/246

delta starting. The supply voltage is 400V, full load effnis

82%and full load power factor is 0.85% (lag).

Neglect magnetizing current.

ts fu a curr nt, th stat r f wh ch s arrang f r star

Solution

Calculate the line currentat the instant of starting.

6hp IM,

Star-delta starting

Isc=3.5IFL

Isc (line) =3.5 IFL

Voltage =400V

FL=82%, pf=0.85 (lag)

P=3 VLILcos

0.854003

1

0.82

7466I

LIFL=

=9.26A (line current fordelta)

=5.34A (phase currentfor delta)

=18.73AIsc=3.5IFL=3.5x5.34

A h i f i d i i


149/246

At the instant of starting, motor wdg is in star

For star, line current is equal to phase current.ILat the instant of start =18.73A for delta (400V)

ILat the instant of start =18.73/3 A for star (400/3)=10.81A

All the methods used for SCIM can also be usedf SRIM

Starting of SRIM


150/246

for SRIM

But the justification of name SR is done if this motor isstarted by external rotor resistance.

This is the simplest and cheapest method

It increases starting torque

It decreases starting currentIt is possible to produce starting torque = maximum torqueIt improves starting power factor.

Rotorin Star

Stator

Rotor

in Delta

Rotor wdg is connected to slip ring, which isStarting of SRIM


151/246

connected to external resistancethrough brushes.

At starting, entire external resistance is connected to rotorwinding. As the motor accelerates, resistance is cut outin

steps so that torque remains maximumduring acceleration.

At operating point, this resistance is fully cut offandand slip rings are shorted.

Rotorin Star

Stator

Rotor

in Delta

Calculation of starter resistance stepsn+1nn-1

R

3

R

2

R

1


152/246

Consider one phase with rotor resistance r2

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1R3

R2

R2R1

R1

Let R1, R2, R3, Rnbe step resistances

Let 1, 2, 3, n, n+1 be studsof starter

Let R1, R2, R3, Rnbe totalresistances of respective studs

R1=R1+ R2+ R3 +

+ Rn+ r2 R2=R2+ R3+ R4 +

+ Rn+ r2

R3=R3+ R4+

+ Rn+ r2 Rn=Rn+ r2

Rn+1

= r2

n-section starter orn-step starterorn+1stud starter

Calculation of starter resistance stepsn+1n

R

n-1

R

3

R

2

R

1


153/246

2. During start up, load torqueremains constant

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1R3

R2

R2R1

R1

1. No load current is neglected

3. Input current fluctuates between I1maxand I1min.

Some assumptionsare made

I1max

t

I1min

Calculation of starter resistance stepsn+1n

R

n-1

R

3

R

2

R

1OFF ON


154/246

The input current shoots up to I1max

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1R3

R2

R2R1

R1

When supply is given, handle is at stud 1

I1max

t

I1min

s1

-Off-total rotor resistance is R1

2TH2

2

1

1TH

11max

Xxs

'RR

VI

statortoreferredareparametersSpeed increases

n+1n

R

n-1

R

3

R

2

R

1

R

OFF

ON


155/246

The input current shoots up to I1max

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1R3

R2

R2R1

R1

When supply is given, handle is at stud 1

I1max

t

I1min

s1

-Off-total rotor resistance is R1

2TH2

2

1

1TH

11max

Xxs

'RR

VI

statortoreferredareparametersSpeed increases

Current decreases to I1min

Slip decreases to s2

s2

N

t

2TH2

2

2

1TH

11min

Xxs'R

R

VI

n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


156/246

Resistance decreases, current increases to I1max

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

I1max

t

I1min

s1

2TH2

2

2

2TH

11max

Xxs

'RR

VI

Now speed increasess2

R1

cut

Handle moves from stud 1 to stud 2

with same speed and slip=s2

n+1n

R

n-1

R

3

R

2

R

1

R

OFF

ON


157/246


Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

I1max

t

I1min

2TH2

2

2

2TH

11max

Xxs

'RR

VI

Now speed increases

R1

cut


with same speed and slip=s2N

t

2

TH2

2

3

2

TH

11min

Xxs

'R

R

VI



s3s1 s2

n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


158/246

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

I1max

t

I1min

2TH2

2

3

3TH

11max

Xxs

'RR

VI

Now speed increases

R1

cut

s3s1 s2




R2

cut

n+1n

R

n-1

R

3

R

2

R

1

R

OFF

ON


159/246

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

I1max

t

I1min

2TH2

2

3

3TH

11max

Xxs

'RR

VI

Now speed increases

R1

cut

s3s1 s2




R2

cut

N

t

2TH2

2

4

3TH

11min

Xxs

'RR

VI



s4

n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


160/246

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

I1max

t

I1min

2TH2

2

4

4TH

11max

Xxs

'RR

VI

Thus I1max& I1minare obtained

R1

cut

s3s1 s2




R2

cut

s4

R3

cut

n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


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Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

1

1

sR '

sm=sflif I1max=Ifl

From all I1maxequations, we can write

2

2

sR '

n

n

3

3

sR

sR '.......'

msr

sR

1n

1n 2'

flsr2

Entire resistance is cut off.

From all I1minequations, we can write

2

1

s

R '

3

2

s

R '

1n

n

4

3

s

R

s

R '.......

'

(1)

m

s

Rn'(3)

mm s

rr

s

sR1,equFrom 1

n 22

1'

(2)

n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


162/246

Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

1

2

ss

Equn 1/3 gives

2

3

ss

n

1n

3

4

ss

ss .......

nssm

'

'

1

2

R

R

2'

3

R

R '

(4)

(5)'

'

n

1n

R

R '

2

nR

r )(, assume

'' 12 RR '' 23 RR '

2

1R

'' 11n RR n

'21

Rnr (6)

From equn 2,m

n

sr 2r2

m

n s

or n

ms /1

(7)


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n+1n

R

n-1

R

3

R

2

R

1

R

OFF ON


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Rotor

Rnr

2

Rn-1

Rn

R3

Rn-1

R3

R2

R2R1

R1

Steps for the design of starter1. Calculate s

m2. Calculate 3. Calculate R1

m

2

s

r

4. Calculate R1= R1(1-)5. Calculate R2= R1

6. Calculate Rn= n-1R1

R1 is referred to stator

In order to calculate actualvalue of rotor resistance

transfer R1to rotor

ExampleDesign a 5-sections of a 6-studsstarter for a 3-ph slip ringIM. The full load slip is 2%and the max starting current


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is limited to twicethe full load current. Rotor resistance per

phase is 0.03. (assume no of turns equal)

SolutionFull load current

2TH22

2TH

11FL

XxsrR

VI

For small slip, r2/s >> RTHand X srV

I2

11FL

Now starting current I1maxis TWICE the full load current

m2

11max sr

VI m2

11FL sr

VI 2m

2

1

2

1 srV(0.02)

rV 2 0.04sm

Here no of sections are 5 = 0.525R1=r2/sm

1/nms )(

= 0.75

65

R5r2

4

R4R

3

R3R R

2

R2R

1

R1

OFF ON0.356 0.187 0.098 0.052 0.027


166/246

The resistance of various elements are

R1 =

R2=

R1(1-) = 0.075(1-0.525) = 0.356

R1= 0.525x0.356= 0.187

R3=2R1= R2 = 0.525x0.187= 0.098

3

R1=R3 = 0.525x0.098= 0.052

R4=

4R1= R4 = 0.525x0.052= 0.027R5=

0.03r2 =

Rotor

r2Rn Rn-1R3 R2 R1

0.03

Total resistance R1=0.075

In industrial application, for wide speed range

and for smooth speed control DC motors are preferred

Speed control of IM


167/246

and for smooth speed control, DC motors are preferred

But it is required to convert AC to DC.The capital cost of DC motor is higher than AC motor.

Due to this reasons, preference is given to IM.

IM is 1. Cheaper 2. Robustin construction

3. More economical to operate and maintain.The disadvantages are 1. Restricted small range of speed.

2. Lowoperating pf.

The operating speed of IM is given bynr=ns(1-s)Thus speed can be controlled by

varying syn speed ns=2f/pwhich can be controlled by1. Pole changing 2. freq changing

changing SlipThus speed can also be controlled by

which can be controlled by


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3. Varying line voltage

4. Varying rotor circuit resistance

5. Slip PowerControl

Methods 1, 2 and 3 are carried out from stator side.

Methods 4 and 5 are carried out from rotor side.

Methods 1, 2 and 3are applicable to SCIM and SRIM

Methods 4 and 5 are applicable only to SRIM

a) Cascading of Induction Motors

b) Rotor Voltage Injection

Since no of poles can be changed only by evenb th d t l i t ti b t t d

1. Pole changing methodSpeed is inverselyproportional to no of poles.


169/246

numbers, the speed control is not continuousbut a stepped

one.However, this is very economicalmethod and motor can

operate on quite hardcharacteristics.

The no of poles can be changed by three methods.

1. Use of Multiple Stator Winding

Stator has separatewindings with differentno of poles.

This is not suitable for wide rangeof speed control

Winding having less no of poles is used for high speedandhaving moreno of poles is used for low speed.

TPDTis used for changeover. The unused winding is kept

open to avoid circulating current and heating.

to produce different no of poles for differentd

2. Use of Consequent Pole TechniqueNo multiple stator wdg but wdg is re-connected


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speed.

This method is suitable for SCIM sincerotor wdg adjustautomatically.

For WRIM, rotor wdgmust be designed for same no of poles.

Consider 4 coils of one phase


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12 3

4

Up=oUt of the plane-coming toward U=DOT

dowN=iNthe plane-going away from you=CROSS

Upwardflux

Downwardflux


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1 2 34

If bis connected to c

ab

cd

a b

c d

N

flux

S N S N S N SS

Series connection, 8 poles, low speed

a b

c d Parallel,8 poles,low speed

8 poles

OR


173/246

1 2 34

Now connect bto d

ab

cd


174/246

1 2 34

ab

cd

N S N S

4 poles

Now connect bto d


175/246

1 2 34

Conclusions

ab

cd

N S N S

4 poles

In wdg if current flows from a to b and from c to d,8 poles ( Low speed)are produced

In wdg if current flows from a to b and from d to c,

4 poles ( High speed) are produced

Connection of three phases can be in delta or star

bd

B bc


176/246

a

b

c

Series Delta, 8 poles, Low speed

Y

= ad

R

R

Y

B

a

bd

c

Series Delta, 4 poles, High speed

Y

B

= a

bd

c

R

RY

B

For parallel delta,voltage of wdg

increases

a

bc

S i St 8 l L d

R


177/246

d

Series Star, 8 poles, Low speed

YB a

bd

c

Series Star, 4 poles, High speed

YB

R

ac

Parallel Star, 8 poles, Low speed

R


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bd

, p , p

YB

a

b

d

cParallel Star, 4 poles, High speed

YB

R

This is an elegant and flexiblemethod ofpole changing

3. Pole Amplitude Modulation TechniquePoles are changed by modulationtechnique.


179/246

pole changing

Consider sinusoidally distributed MMF wavesproduced bybalanced 3-ph wdg.

FR ,

2

PsinF FY ,

3

2-

2

PsinF FB

3

2

2

PsinF

where P is no of poles and is the mech angleLet these MMF waves be modulated by another sinusoidal

modulating MMF waves.

FMR ,

2

PsinF

MM FMY ,

3

2-2

PsinF

MM FMB ,

3

2

2

P

sinFM

M

Modulated meansproduct of these two waves

Therefore, modulated MMF waves are given by

FR

2

PsinF2PsinF MM= FR x RMR

2

PPcos

2

P-PcosFF

2

1 MMM


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FY

3

2

-2PsinF32

-2PsinFMM

FB

Consider first termThese are co-phasal in space, hence net MMF is zero.

222

= FY x RMY

3

4-

2

PPcos

2

P-PcosFF

2

1 MMM

= FB x RMB

3

2-

2

PPcos

2

P-PcosFF

2

1 MMM

The second term will produce rotating magnetic field in

opposite direction depending on no of poles P+PM

.

Again if the modulating MMFs areFMR ,

2

PsinF MM FMY ,

3

2

2

PsinF MM

F

2

P

sinF M


181/246

FMB ,

3

2

sinFM

Then, modulated MMFwaves are given by

FR

FY

FB

2

PPcos

2

P-PcosFF

2

1 MMM

2

PPcos3

2-2

P-PcosFF2

1MMM

2

PPcos

3

2

2

P-PcosFF

2

1 MMM

The second terms are co-phasal in space, hence net MMF iszero.

The first term will produce rotating magnetic field in

direction of rotation, depending on no of poles P-PM.

Thus four speeds are possible1. Speed depending on no of poles P

2. Speed depending on no of poles PM


182/246

p p g p M

3. Speed depending on no of poles P+PM4. Speed depending on no of poles P-PM

Therefore, smooth steplessspeed control can be

2. Variation of supply frequencySpeed is directlyproportional to frequency (2f/p)


183/246

possible by smooth variation of frequency.The voltage equation is V=4.44fmTphkw

If f is changed, for constant voltage, flux will change.

So, characteristicwill change.In order to retain high hardness ch, flux should be

maintained constant

The resultant air gap flux is m fVK4.44T1 wph

Thus in order to avoid saturation, V/f is maintained constant.

The variable frequency can be obtained by1. Rotating frequency changer

2. Adjustable frequency generator

3 S lid i f h


184/246

3. Solid state static frequency changer

a). Cycloconverter(output freq lessthan input freq)

b). Rectifier-Inverter(V/f control)(used)

Torque-slip characteristics

V f, Reactance f,speed f

1. Slip at max torque,smt2THTH

2

jxjXR

r

Neglecting stator impedance,smt2

2

x

r

Thus slip at max torque is inversely proportional to freq.

2

2

Lf2

r

mNr

xXrR

V

322

2TH

2

2TH

2

TH

s

2. The starting torque

Test


185/246

22

2TH

2TH

s

rxX

V

n2

3

3f

1K

Thus Testis inversely proportional to f3, if V is constant.

Neglect RTH+r2Test

2

22TH2

2TH r

lLf2

V

2f

P

2

3

Test

The Testis inversely proportional to f, if V/fis constant.

22

THTH

2TH

s XRR2

V

33. The maximum torque

Tem


186/246

2TH

2

TH

s xX2

V

n2

3

2f

1K

Thus Temis inversely proportional to f2, if V is constant.

Neglect RTHTem

2TH

2

TH

lLf22

V

2f

P

2

3

Tem

The Temis unchanged, if V/fis constant.


187/246

The torque-slip characteristicsIf V/f is constant

1. smtisinversely proportional to freq.

2 T i i l t f


188/246

3. Temis independent of f.

2. Testisinversely to f.

Slip1 0Ns0 Speed

Te

f1 f1

smt1

f2f2r2 >TL

N1N2

N3

N4

Thus rotor speeddecreases withincrease in r2

Advantages:1. Simple 2. It is possible to obtain max Test.

3. Low starting currentDisadvantages


200/246

g

1. Speed below syn speedcan only be obtained2. Reduced effnat low speed due to more rotor ohmic loss3. Poorspeed regulation

With negligible r1, the slip at max torque is Xr

s 2mt

This can be written as ,XRs 2mt R2=r2+ additional resistance

The value of smt can be obtained from ,

s

s

ss

2

T

T

mt

mt

emt

e

mt2 sXR

Under normaloperating condition, for load TL1,2

1THL1 r

sKT

For load TL2,2

2THL2 R

sKT

2

2

2

1

L2

L1

r

R

s

s

T

T

L2

L1

1

222 T

T

s

srR


201/246

there is power loss.

5. Slip Power ControlDue to addition of rotor circuit resistance

Without power loss, slip poweris utilized by suitable slip


202/246

power converter to control the speed.a) Cascade connection of IM

Sometimes called as tandem control

p , p p y p

Not used now-a-days

Only historical importance

MIM1

f1

AIM2

f2=s1f1

SRIM SRIM


203/246

The actual speed is )PPP

(1P

120f

N 21

2

1

1

r

21

1

PP

120f

Speed is dependent on P1+P2


204/246

Speed is dependent on P1+P2.If two torques are in same direction, then the scheme is

called as cumulatively cascadedIf AIM terminals are interchanged, then it is called as

differentially cascadedarrangement and its speed is

)s(1P

120fN 2

2

2r2 )s(1P

f120s2

2

11 )s(1P

120f1

1

1

12

21 PP

Ps

The actual speed is )s(1P

120fN 1

1

1r

21

1

PP

120f

This is possible only when P1 P2

b) Rotor Voltage Injection or

The air gap power is Pg=sPg+(1-s)Pg

sPgis rotor ohmic loss, wasted power, can be used for

Injection of slip frequencyemf in rotor circuit


205/246

g , p ,

speed control. This power can be added to the shaft or

returnedto the supply.

Due to this injected voltage, motor becomes DOUBLY

excited IM. This can be used for speed and pfcontrol.

If this injected voltage, is opposite to rotor emf, speed

decreasesand motor runs at sub-synchronousspeed

If this injected voltage, is additive to rotor emf, speed

increasesand motor runs at super-synchronousspeed

The various methods for injecting slip freq emf are

1) Lablank Exciter

Armwdg

stator

Arm wdg

Stator Ring


206/246

RotoryFrequency Converter (FC)

w g

stator

Coupled to the rotor of IM

DC armwith commutator, brusheson one end andslip ringson the other end

MIM

f

FC

Voltageregulating

device

1) Lablank Exciter

Armwdg

stator

Arm wdg

Stator Ring


207/246

Slip rings of MIMare connected to brushes of FC

g

stator

Slip rings of FC are connected to voltage regulating device

or transformer and then to supply

R YB

MIM

f

FC

Voltageregulating

device

R YB

1) Lablank Exciter

Armwdg

stator

Arm wdg

Stator Ring


208/246

MIM and FC are having equal no of poles

g

statorSame speed

If rotor of FC rotates in opposite direction, thenfield speed

R YB

MIM

f

FC

Voltageregulating

device

R YB

Ns-Nr= Ns-Ns(1-s)=sNsBrush frequency of FC, fb=sNsP/2= sf

1) Lablank ExciterThus at any speed, fbis equal to slip freqand suitable for injectioninto rotor circuit if IM

If brush voltage is less thanIM slip voltage, current flows


209/246

R YB

MIM

f

FC

Voltageregulating

device

R YB

f brush voltage s less than M sl p voltage, current flows

from rotor of IM to brushes of FC to slip rings and supplyMotor runs at Sub-Synchronous speed

If brush voltage is more thanIM slip voltage, current flowsSupply- VRD- slip rings of FC -brushes of FC -rotor of IM

Motor runs at Super-Synchronous speed

Power ischanging

ConstantTorqueDrive

1) Lablank ExciterConstant Power DriveR YB

f

R YB


210/246

MIMSynM/c

FCSynMotor

MIM Syn m/c

=equal no of polesOn one shaft=same speed, Nr

Speed of FC =Ns, due to Syn motor

Nr=Ns(1-s)freq=f(1-s)

Speed of rotating magnetic field of FC=2f(1-s)/P=Ns(1-s)If Nr & Ns are in oppositedirection,Ns-Ns(1-s)=sNs

Freq of emf across brushes of FC=sfSuitable for injectioninto rotor ckt of MIM


f

R YB


211/246

MIMSynM/c

FCSynMotor

This emf can be varied by

excitation and brush angleIf this emf is more thanslip voltagePower flows from FC-Rotor of MIM-Shaft-Syn m/c (g)- FC


MIM operates at Super-Synchronous speedIf brush emf is less than slip voltagePower flows from Rotor of MIM- FC-Syn m/c (m)-Shaft- rotor of MIM

MIM operates at Sub-Synchronous speed


f

R YB


212/246

MIMSynM/c

FCSynMotor

Power is taken and fed to same shaftTherefore, constant power drive


2) Kramer System

R YB

fVoltageregulating


213/246

MIM

f

ACM

device

If brush emf is more than slip voltagePower flows from ACM-Rotor of MIM.

MIM operates at Super-Synchronous speedIf brush emf is less than slip voltage

Power flows from Rotor of MIM- ACM.MIM operates at Sub-Synchronous speedSince power is flowing from one machine to another with one

shaft, it is constant power drive.

3) Scherbius SystemR YB R YB

f


214/246

Power changes

At Super-Synchronous speed, power flows from supply-AIM

(Motor) - ACM -rotor of MIM.

MIM

ACM AIM

Voltageregulatingdevice

Constant torque drive

At Sub-Synchronous speed, power flows from rotor of MIM

- ACM AIM (Gen) - supply.

f

4) Static Slip Power Recovery Scheme(Rectifier Inverter Scheme)Constant torque driveR YB


215/246

from Rotor Bridge A Bridge B Transformer - Supply

Lablank, Kramer,and Scherbius methods require Auxi M/c.Here static, thyristerized ckts, two bridges A & Bare used

If rotor emf is more than bridge A voltage, then power flows

MIM

f Transformer

Bridge A Bridge B

PhaseShifter

Inductor



216/246

from SupplyTransformer Bridge B Bridge A - Rotor

IM runs at sub-synchronousspeed.Bridges A acts as a rectifier, Bridge Bacts as a Inverter

If rotor emf is less than bridge A voltage, then power flows

MIM

f Transformer

Bridge A Bridge B

PhaseShifter

Inductor



217/246

injected emf.

IM runs at super-synchronousspeed.Bridges B acts as a rectifier, Bridge A acts as a Inverter

The phase shifter is used to control the phase of the

MIM

f Transformer

Bridge A Bridge B

PhaseShifter

Inductor

In modern electrical drives, it is frequentlynecessary

Electrical Braking of IM

to stopthe motor quicklyin an exact position


218/246

to decelarate the motorto reverse the direction of rotationof motor

The quality of the productand the productivityof an unit

are often dependent on braking

Types of braking: 1. Electrical, 2. Mechanical

In electrical braking, electrical torqueis opposite to the

direction of rotationof rotor.

Types of electrical braking:1. Regenerativebraking

2. Counter-Current braking or Plugging

3. DC Dynamicor Rheostatic or AC Dynamic Braking

Ns N N N N N

1. RegenerativebrakingRegenerative = returning

useful energy back to

linesThis braking takes place the load forcesto run the motor


219/246

g p f m

above syn speed. For example, cranes or hoists

Slip becomes negative and IM operates as IG.

1 0

Ns0

Te 4 poles8 poles

TLSpeed

Slip

A

If regenerative braking isapplied at A

Poles are changed from 4 to 8A shifts to B

B

B shifts to CC shifts to D

C

D

From B to D, regenerative

brakingand IGoperation

Slip = (OC-OB)/OC =-ve

1. RegenerativebrakingSuitable for SCIM

It is not required to change no of poles of rotor


220/246

1 0

Ns0

Te 4 poles8 poles

TLSpeed

Slip

A

B

C

D

+Te-Te

TL

Two quadrant operation

A

B

D

Ns N N N N N

Ns N N N N N2. Plugging:

Any two leadsare interchanged. The direction of rotating

or Counter-Currentbraking


221/246

magnetic field getsreversed.This is a braking torquewhich stopsmotor quickly.

1 0

Ns0

Te

TLSpeedSlip

A

CD

2

-Ns

The Tegets reversed.

B

Tb=TL+T1

TL

At Csupply should be disconnectedto stop the motor.

EF

Te-Slip Ch for SCIMduring Plugging

T1

otherwise motor starts inreverse direction

Braking

Braking

Motoring

Motoring

Againg ifplugging isdone

Dto E

E to F

and F to A

Pt A to B

B to CC to D


222/246

During plugging slip increasesSlip emfincreases.

To limit this current, extra R2can be inserted in rotor ckt.Currentincreases

Te-Slip Ch for SRIMduring Plugging


223/246

1 0

Ns0

Te

TLSpeedSlip

A

C

D2

-Ns

B

Tb=TL+T1

TL

This R2increases Backward Torque T1.

E

H

T1

Braking

Braking

Moto

3-Ph Induction Motor

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