A04 induction motor Speed control - Home | Department of …work6000/A04 induction motor... · · 2010-10-29A 3 phase, 6-pole, 380 Volt, 50Hz star connected induction motor has

M.Sc.(Eng) in building services MEBS 6000 2010 Utilities services

Department of Electrical & Electronic Engineering University of Hong Kong

K.F. Chan (Mr.) Page 1 of 32 July 2010

Induction Motor speed control

Not long ago, induction machines were used in applications for which adjustable

speed is not required. Before the power electronics era, and the pulse width

modulation in particular, speed control of induction machines was limited to highly

inefficient methods with a narrow range of speed. With the advances in solid state

devices and variable frequency power convertors, different approaches to inductions

motor drive systems have emerged and developed that result in more sophisticated

operations. Induction machines can now be used in high performance applications

where precise movement is required.

The speed control of an induction motor requires more elaborate techniques than the

speed control of dc machines. The speed torque characteristics of an induction motor

is given by

+

+

==2

2

21

22

'

'3

eqs

gd

Xs

RRs

RVPT

ωω

where sω is the synchronous speed in rad/s,

1R is the stator resistance,

2'R is the rotor resistance referred to the stator,

eqX is the combined inductive reactance of the motor, referred to the stator,

s is the slip, and

V is the terminal voltage input to each stator winding. In HK for LV

installation, it is 220V and 380V for star and delta connected motors

respectively.

By examining this equation, it can be seen that the slip, s, or the speed, ω , can be

controlled if at least one of the following variables or parameters is altered:

- armature or rotor resistance,

- armature or rotor inductance,

- magnitude of terminal voltage,

- frequency of terminal voltage.

We will look at speed control by changing voltage and frequency in detail.




Controlling speed using rotor resistance

If we assume steady state operation, and the small slip approximation, the developed

torque equation can be reduced to

2

2

'

3

R

sVT

sd ω

≈

If the voltage, frequency, and torque are kept constant, the increase in 2'R results in

an increase in the slip, in other words, a decrease in speed.

The following figure shows the motor characteristics for the case when a resistance

addR is added to the rotor circuit. The increase in rotor resistance does not change the

synchronous speed or the magnitude of the maximum torque, it only skews the

characteristics so the maximum torque occurs at a lower speed.

Adding a resistance to the rotor circuit does not cause the motor speed to change by

any appreciable value at light loading conditions. The difference in speed between

points 1 and 2 in the above figure is rather small. Although at heavy loading

conditions, 2T , the motor speed may change by a wider range – from point 3 to point

4 – the speed range is still narrow. Therefore, controlling the motor speed by changing

the rotor (or stator) resistance is not considered a realistic option. In addition, this

method increases the motor losses substantially.




Controlling speed using inductance

Adding inductance to the motor windings is an unrealistic option for the following

reasons:

1) The physical size of the inductance required to make a sizable change in speed is

likely to be larger than the motor itself.

2) Unlike variable resistance, variable inductance requires expensive and elaborate

design.

3) The insertion of inductance reduces the starting torque.

4) The insertion of inductance consumes reactive power that further lowers the

already low power factor of induction motors.




Controlling speed by adjusting the stator voltage

Several techniques can be used to change the stator voltage of the motor. Among them

are fixed pulse modulation (FPM) or the phase control by SCRs shown here below:

As seen in the torque equation, the motor torque is proportional to the square of its

stator voltage. For the same slip and frequency, a small change in motor voltage

results in a relatively large change in torque. A 10% reduction in voltage causes a 19%

reduction in developed torque as well as the starting and maximum torque.

The following figure shows characteristics of the motor under voltage control.

It shows 2 curves for 2 different values of the stator voltage. Note that the slip at the

maximum torque remains unchanged since it is not a function of the voltage:

221

2max

'

eqXR

Rs

+=

For normal operation in the linear region, the above figure shows that the motor speed

can be modestly changed when the voltage is altered. However, a wide range of speed




control cannot be accomplished by this technique alone. Nevertheless, it is an

excellent method for reducing starting current and increasing efficiency during light

loading conditions. The losses are reduced, particularly core losses, which are

proportional to the square of the voltage.

Keep in mind that the terminal voltage cannot exceed the rated value to prevent the

damage of the windings’ insulation. This technique is thus only suitable for speed

reduction below the rated speed.




Example 1

A 3 phase, 6-pole, 380 Volt, 50Hz star connected induction motor has a rotor

resistance referred to the stator of 0.5Ω .

The induction motor is driving a travellator requiring constant load torque of 120 Nm.

Ignore the rotational losses, calculate the motor speed at full load. Also calculate the

motor speed if the voltage is reduced by 20%. Assume that the supply frequency is

kept constant.

Answer

+

+

==2

2'2

1

'2

23

eqs

dd

Xs

RRs

RVPT

ωω

where V is input voltage

Near full speed, s is small, so we may assume that

2'2

1 eqXs

RR >><<

so 2'2

'2

23

≈

s

Rs

RVT

s

d

ω

'2

23

R

sVT

sd ω

=⇒

5.060

10002

3

3803

120

2

=⇒

π

s

0435.0=s

speed at full voltage is




RPMsnn s 956)0435.01(1000)1( =−=−=

sV

sV

T

T newnew

d

newd2

2

1==

so 068.0)0435.0(8.0

12

=

=news

New speed at 80% voltage

= 100(1-0.068)

= 932 RPM




Controlling speed by adjusting the supply frequency

In steady state, the induction motor operates in the small slip region where the speed

of the motor is always close to the synchronous speed of the rotating flux:

=

pfns

260

where f is the supply frequency,

p is the number of poles of the motor.

Since the synchronous speed is directly proportional to the frequency of the stator

voltage, any change in frequency results in an equivalent change in motor speed.

The following figure shows torque-speed characteristics of an induction motor under

different supply frequency.

The effect of frequency on motor current is determined by the following formula:

22

21

2

''

eqXs

RR

VI

+

+

=

Under the same speed, or same s, as frequency is reduced, eqX is also reduced, thus

2'I is increased. The overall effect is depicted in the following diagram:




Frequency manipulation appears to be an effective method for speed control that

requires a simple dc/ac convertor with variable switching intervals similar to the ones

as shown below:

However, there are severe limitations to this method: very low frequencies may cause

motor damage due to excessive currents, and large frequencies may stall the motor.




Effect of excessively high frequency

The decrease in supply frequency results in the following 5 changes:

1) A decrease in the no-load speed (synchronous speed).

2) An increase in the maximum torque. The maximum torque is given by

++

=22

11

2

max

2

3

eqs XRR

VT

ω

It shows that the maximum torque is inversely proportional to both the synchronous

speed sω and the equivalent reactance of the winding eqX . Each of these

quantities increases by increasing the frequency. Hence the maximum torque

increases when the frequency of the supply voltage decreases.

3) An increase in the starting torque. The starting torque of an induction motor,

22

2 '3

eqs

stX

RVT

ω≈

It can be seen that starting torque increases when the synchronous speed and

equivalent reactance decrease, as supply frequency decreases.

4) A decrease in speed at the maximum torque. Due to decrease in frequency, the slip

at maximum torque 22

1

2max

'

eqXR

Rs

+= increases when the equivalent reactance

decreases. The synchronous speed also decreases on frequency reduction, so the

combined effect is reduction in speed at the maximum torque.

5) An increase in the starting current. The starting current can be computed by

setting s=1 in the motor current formula:

22

21

2

''

eqXs

RR

VI

+

+

= , or

( ) 2221

2

''

eq

st

XRR

VI

++=

When the frequency decreases, the equivalent reactance decreases and the

starting current increases. On the other hand, at high frequencies, the resistance

of the motor windings may also increase due to the skin effect thus further




increasing the starting current.

Now let us examine the case when the increase in frequency is excessive.

The above diagram shows 2 characteristics for 2 different values of stator frequency.

Assume that the load torque is constant, the motor operates initially at frequency f1.

The steady state operation is represented by point 1. Now assume that the frequency

of the stator voltage increases to a higher value, f2, where the new maximum torque of

the motor is less than the load torque. In this case, no steady state operating point can

be achieved, and the motor eventually stalls or even operates under braking. One

solution to this problem is to increase the supply voltage when the frequency is

increased.




Example 2

A 380V, 2 pole, 50Hz, star connected indication motor has an inductive reactance of

4Ω and a stator resistance of 0.2Ω . The rotor resistance referred to the stator is

0.3Ω . The motor is driving an escalator having a constant torque load of 40 Nm at

a speed of 2900 RPM. Assume that this torque includes the motor rotational losses.

a) Calculate the maximum frequency of the supply voltage that would not result

in stalling the motor.

b) Calculator the motor current at 50Hz.

c) Calculate the motor current at the maximum frequency computed in (a) above.

d) Calculate the power delivered to the load at 50Hz.

e) Calculate the power delivered to the load at the frequency computed in (a)

above.

Assume that the supply voltage is kept constant.

Answer

a) )(2

322

11

2

max

eqs XRR

VT

++=

ω

Near full speed, as R1 is 0.2Ω , while Xeq is 4Ω , we may assume that R1 << Xeq ,

so

eqsX

VT

ω2

3 2

max ≈

Let the new frequency be f. The motor inductive reactance at f is .50 eqXf

The maximum torque at f is

=

eqs Xff

VT

50502

3 2

max

ω

Now, in order not to stall the motor, Tmax must be at least 40 Nm, so

=

45060

30002

502

340

2

ffV

π




where3

380=V

Hzf 92.59=⇒

b) 033.03000

29003000 =−=s

2

2'2

'2

eqXs

RR

VI

+

+

=

A9.21

4033.0

3.02.0

3380

22

=

+

+

=

c) Now 22

1

'2

max

eqXR

RS

+=

At the new frequency, 79.4)4(50

92.59 ==eqX

0626.079.42.0

3.022max =

+=S

Current at new frequency is

AI 71.31

79.40626.0

3.02.0

3380

22

'2 =

+

+

=




d) kWTP dd 1.1260

2900240 =×== πω

e) New speed at the new frequency is

(1-0.0626)(59.92)(60) = 3370 RPM

kWTP dd 1.1460

3370240 =×== πω




Effect of excessively low frequency

Reducing the supply frequency reduces the speed of an induction motor. However,

frequency reduction may result in an increase in motor current

22

21

2

''

eqXs

RR

VI

+

+

=

At very low frequencies, the equivalent reactance of the motor eqX is very low.

Since Xeq is the limiting parameter for motor current at starting, its large

reduction could lead to an excessive current beyond the ratings of the machine.




Example 3

A 480V, 2 pole, 60Hz, star connected indication motor has an inductive reactance of


0.3Ω . The motor is driving an escalator having a constant torque load of 40 Nm at a

speed of 3500 RPM. Assume that this torque includes the motor rotational losses.

Calculate the motor speed and starting current if the frequency is decreased to 50Hz.

The supply voltage is kept constant.

Answer

At 60Hz, the starting current is

( ) 22'21

2'

eqXRR

VI

++=

where V is input voltage to the winding, i.e., phase voltage in this case

Amp75.68

4)5.0(

3480

22

=+

=

At 50Hz,

Ω=

= 33.346050

eqX

AmpI st 2.8233.35.0

3480

22

'2 =

+=

This is about 20% increase in starting current but at 16% drop in speed.

Now, at 50Hz, the synchronous speed is 3000 RPM, i.e. 314.16 rad/s.

At small slip,




'2

23R

sVT

sd ω

≈

where V is the phase to neutral voltage for star connection.

Thus )3.0(16.314

48040

2 s=

0164.0=⇒ s

Thus new speed at 50Hz is 3000(1-0.0164) = 2951 RPM.

This is about 16% drop in speed.




Voltage/frequency control

Now increase in supply frequency increases the motor speed and reduces the

maximum torque of the motor. Increase in voltage results in an increase in the

maximum torque of the motor. If we combine these two features, we can achieve a

control design by which the speed increases and the torque is kept the same. This is

known as voltage/frequency control, v/f.

The above diagram shows 3 characteristics; one is used as our reference at voltage V1

and frequency f1. For the arbitrary fan type load in the figure, the reference operating

point is 1. If we increase the frequency of the supply to f2 while keeping the voltage

V1 unchanged, the speed of the motor increases and the maximum torque decreases.

The load torque in this case is higher than the maximum torque provided by the motor.

Thus, no steady state operating point can be achieved and the motor eventually stalls.

Now let us keep the supply frequency to the new value of f2, but also increase the

magnitude of the voltage to V2. The motor characteristics in this case stretch and the

maximum torque increases. The motor operates at point 2 and a new steady state

pointy is achieved.

Change in voltage and frequency is a powerful method for speed control. Note that

both frequency and voltage can change simultaneously by the pulse width modulation

technique.




(Adopted from CARROW R.S. Electrician's technical reference. Variable-frequency drives)

PWM technique is the most common type of induction motor speed control used in

building services application nowadays. There are several variations where the v/f

ratio is also adjusted to provide a special operating performance. The most common

method, though, is the fixed v/f ratio.




The above diagrams show the characteristics of an induction motor operating under

constant v/f control.

Now

++

=22

11

2

max

2

3

eqs XRR

VT

ω

If we assume that the equivalent inductive reactance Xeq, at frequencies near the

rated value, is much larger than the armature resistance, the equation can be

approximated as

( )eqeqs fLf

p

V

X

VT

ππω2

42

3

2

3 22

max

=≈




It can be thus be seen that

2

max

∝

f

VT

It is now clear that when the v/f ratio is kept constant, the maximum torque is

unchanged. However, it must be kept in mind that this approximation may not be

valid at very low frequencies when Xeq is not much larger than the R1.

Another feature of the constant v/f control is that the magnitude of the starting current

is almost constant. At s=1

( ) 2221

2

''

eq

st

XRR

VI

++=

Assume that ( )221

2 'RRXeq +>> (this assumption is valid for frequencies close to

the rated frequency), the starting current can thus be approximated as

f

V

LX

VI

eqeqst π2

1'2 ==

This equation again shows that when v/f is kept constant, the starting current remains

unchanged; this is another advantage of v/f control.

Almost all PWM frequency inverters used for induction motor speed control in

building services applications employs this fixed v/f control.

When the change in voltage is used to control the induction machine, whether it is a

voltage control or v/f control, one must be careful not to increase the voltage

magnitude beyond the ratings of the motor. Excessive voltage can cause instant

damage to the insulation of the motor’s windings, leading to shorts and internal faults.

Usually, the voltage should be kept below 110% of the rated value.

Most motors used for building services applications in HK are suitable for 380V/50Hz

or 480V/60Hz power supply. Therefore, it is possible to increase the motor output by

increasing the frequency up to 60Hz. However, it is dangerous to increase the supply

voltage beyond 110% of the rated voltage of 480V, unless the motor is specially

designed.




PWM frequency inverter drives

PWM frequency inverter can be constant voltage source (VSI) or current source (CSI).

Almost all modern day frequency inverter drives are VSI.

Voltage source inverter allows a variable frequency supply to be obtained from a dc

supply. One example of solid state VSI frequency inverter control schematics is

shown herebelow:

[Adopted from HERMAN, Stephen L. Industrial Motor Control]

The stepped wave inverter illustrated above uses SCRs in the ac/dc conversion section,

i.e. the power supply section, and junction transistors in the dc/ac inversion. SCRs

control the output voltage by chopping the incoming waveform. This can cause

harmonics on the line. When bipolar junction transistors, which are current controlled

devices, are employed as switches, they are generally driven into saturation by

supplying them with an excessive amount of base-emitter current. Saturating the

transistor casues the collector-emitter voltage to drop to between 0.04 and 0.03V. This

small voltage drop allows the transistor to control large amounts of current without

being destroyed. When a junction transistor is driven into saturation, however, it

cannot recover or turn off as quickly as normal. This limits the frequency response of

the transistor. Therefore, new design of VSD employs MOSFETs or IGBTs (for larger

ratings).




[Adopted from HERMAN, Stephen L. Industrial Motor Control]

The above diagram shows another type of VSI frequency inverter using IGBTs in the

dc/ac inverter section. IGBTs have an insulated gate very similar to some types of

FETs. However, the gate is ‘insulated’ and has very high impedance, so the IGBT is a

voltage not current controlled device. This type of frequency drives does not use SCR

in the power supply, i.e. the ac/dc converter section. This has the advantage of greatly

reducing the line harmonics associated with SCRs. The great disadvantage is that the

fast switching rate of the transistors can cause voltage spikes in the range of up to

2000V to be applied to the motor.

At high switching frequency, capacitance between the cables connecting the

frequency inverter and the motor becomes a concern. Line length from the inverter to

the motor is thus of great concern with drives using IGBTs. Short line lengths are

preferred.

Generally, MOSFET is used in low voltage and low power inverters, IGBT used up to

medium power levels and GTO’s and insulated gate commutated thyristors are used

for high power levels.




Example 4

A 480V, 2 pole, 60Hz, star connected induction motor has an inductive reactance of


0.3Ω . The motor is driving a constant torque load of 40Nm at a speed of 3500 RPM.

Assume that this torque includes the motor rotational losses.

The motor is now supplied by constant voltage to frequency ratio control and the

frequency is reduced to 50Hz.

Calculate the new motor speed and the starting current.

Answer

For fixed v/f ratio, when the frequency is reduced to 50Hz, the voltage should be

Volt400)60

50(480 =

At 400V, 50Hz,

srads /16.314)50(2 == πω

( )[ ]222'21

'2

23

sXRsR

sRVT

eqs

d++

=ω

( ) ( )[ ]222

2

43.02.016.314

3.040040

ss

s

×++×××=⇒

On simplification, this becomes

009.07.304.16 2 =+− ss

Note that for a quadratic equation:

02 =++ cbxax

the solution can be written as:

a

acbbx

2

42 −±−=

So s = 0.203 or 0.0276

Motor speed is thus

(1-0.0276)3000 = 2917RPM




Alternatively we can assume '2

23R

sVT

sd ω

≈ where V is the input voltage.

thus )3.0(16.314

)400(40

2 s=

s = 0.0236

The slip now is higher than by reducing the frequency alone (0.0164, Example 3).

The starting current at 480V, 60Hz is

22'21

'2

)( eq

stXRR

VI

++=

( )A75.68

43.02.0

3480

22=

++=

At 400V, 50Hz,

33.3)4(60

50 ==eqX

So starting current becomes

AmpI st 5.6833.3)3.02.0(

3400

22

'2 =

++=

Note that the starting current is almost unchanged in this case while the starting

current will increase if frequency is reduced alone (82.2Amp, Example 3).




CSI (current source inverter)

The PWM frequency converter we have seen is called voltage source inverter (VSI)

because the DC link is a constant voltage supply. There is another type of inverter

called current source inverter.

A thyristor CSI is shown herebelow. Diodes D1 to D6 and capacitors C1 to C6 provide

commutation of thyristors T1 to T6, which are fired with a phase difference of 60o in

sequence of their numbers. This inverter behaves as a current source due to the

presence of large inductance Ld in the dc link.

(Adopted from DUBEY, G.K. Fundamentals of Electrical Drives)

The major advantage of CSI is its reliability. In the case of VSI, a commutation failure

will cause two devices in the same leg to conduct. This connects conducting devices

directly across the source causing current through devices suddenly rises to

dangerous values. This problem would not hurt a CSI chiefly because that the

inductor will limit the current.




When a CSI is used, the induction motor exhibits different characteristics as compared

to the VSI. Most noticeable is the low starting torque. This is primarily due to the high

rotor current I’2 at starting, which reduces the magnetizing current Im, because the

current source I1 is constant:

ms IIII +== 21 '

The low magnetizing current at starting reduces the flux of the motor; hence, it

reduces the starting torque.




The other noticeable difference is that the speed of the motor in the normal operating

region with CSI is stiffer (has a flatter slope) than that of a VSI motor. This is because

the core of the motor is saturated with flux in this region. If the core saturates, small

changes in the magnetizing current tend to have little or no effect on the flux. When

load torque increases, the rotor current tends to increase, which reduces the

magnetizing current (as I1 is constant). When reduction of the magnetizing current

does not reduce the flux (in the saturation region), the speed of the motor remains

almost unchanged.





In summary, the relative merits of VSI and CSI are:

1) CSI is more reliable because

i. conduction of two devices in the same leg due to commutation failure

does not lead to sharp rise of current through them and

ii. it has inherent protection against a short circuit across motor terminals.

2) Because of large inductance in the dc link and large inverter capacitors

required for CSI circuit, CSI drive has higher cost, weight and volume, lower

speed range and slower dynamic response.

3) The CSI drive is not suitable for multi motor drives. Hence, each motor is fed

from its own inverter and rectifier. A single ac/dc converter power supply can

be used to feed a number of VSI dc/ac inverter motor systems connected in

parallel. Similarly a single VSI dc/ac inverter output section can feed a number

of motors connected in parallel.





With advances in IGBT technology, and improvement in reliability of power

electronics, VSI is gaining favour especially in variable speed drives in building

services applications which usually involve maximum rating of a few hundred

horsepower only.

Features of frequency inverter drives

Although the primary purpose of a variable frequency drive is to provide speed

control for an ac motor, most drives provide functions that other types of controls do

not.

Many variable frequency drives can provide the low speed torque characteristic that is

so desirable in dc motors. It is this feature that permits ac squirrel cage motors to be

able to replace dc motors for many applications.

Another feature is ramping. It is used to accelerate and decelerate a motor over some

period of time and permits the motor to bring the load up to speed slowly as opposed




to simply connecting the motor directly to the line.

Other features are current limit, minimum and maximum hertz control.

Inverter rated motors

Due to the problem of excessive voltage spikes caused by high switching frequency of

solid state drives, most noticeably IGBT drives, some manufacturers produce

“inverter rated” motors. These motors are specifically designed to be operated by

variable frequency drives. They differ from standard motors in several ways:

1. Typical motors use a fan connected to the shaft to cool the motor. Some inverter

rated motors contain a separate blower for continuous cooling.

2. Inverter rated motors generally have insulating paper between the windings and

the stator core. The high voltage spikes produce high currents that produce a

strong magnetic field. The increased magnetic field causes the motor windings to

move, because magnetic fields repel each other. This movement can eventually

cause the insulation to wear off the wire and produce a grounded motor winding.

[Adopted from Herman S.L. Industrial Motor Control]

3. Inverter rated motors generally have phase paper added to the terminal leads.

Phase paper is insulating paper added to the terminal leads that exit the motor. The

high voltage spikes affect the beginning lead of a coil much more than the wire

inside the coil. The coil is an inductor that naturally opposes a change of current.




Most of the insulation stress caused by high voltage spikes occurs at the beginning

of a winding.

4. The magnet wire used in the construction of the motor windings has a higher rated

insulation than other motors.

5. The motor casing is larger in size, due to the additional insulating paper. Also

larger casing helps cool the motor.

6. For a given harmonic content in the voltage, the current harmonics are reduced

when the motor has higher leakage inductance – this also reduces derating and

torque pulsations. Therefore, motors specially designed for inverter drive will

have higher leakage inductance than their counterparts used for sinusoidal power

supply.

[Text and figures mostly adopted from EL-SHARKAWI, Mohamed A., Fundamentals

of Electric Drives, Herman S.L. Industrial Motor Control and DUBEY, G.K.

Fundamentals of Electrical Drives,]

A04 induction motor Speed control - Home | Department of …work6000/A04 induction motor... · · 2010-10-29A 3 phase, 6-pole, 380 Volt, 50Hz star connected induction motor has

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