Induction Motors.pdf

Three-Phase Induction Motors

1

2

Main Features of 3-phase Induction Motors

• A induction machine can be used as either an induction

generator or a induction motor.

• The summation of all 3-phase powers is a steady value;

so, three phase machines are more advantageous than

their single-phase counter-parts. Single-phase machines

have oscillating power/torque which have negative

excursions too.

• Induction motors are popularly used in the industry

because they are rugged, reliable, maintenance free and

less expensive.

• Main disadvantages: speed control is not easy; starting

current is way too high

• Almost 80% of the motors used in the industry are

induction motors. 70% of the electricity produced is used

by the induction motors.

500 HP Induction Motor

0.5 HP Induction Motor

Range of Induction Motors

Three Phase Induction Motor

Stator of 3-Ø induction motor is similar to the stator of a

Synchronous generator with three phase windings 120° phase

shifted from each other.

7

Rotor Construction

The rotor is the rotating part of the electromagnetic circuit of the

induction motor.

It can be found in two types:

◦ Squirrel cage

◦ Wound rotor

However, the most common type of rotor is the squirrel cage rotor.

Squirrel cage rotor has rotor bars or thick conductors embedded in

slots short-circuited by end rings on both sides.

Wound rotor has 3-phase windings wound and placed in rotor slots

along the outer-periphery of the rotor just similar to the stator

structure.

The terminals are brought out to the external world through slip

rings and brushes; they can be short-circuited external to the

machine.

So, the rotor characteristics can be modified by including R or X in

the rotor circuit in a wound rotor; this is not possible in a cage

rotor.

Rotating Magnetic Field

FIELD GENERATION IN INDUCTION MOTOR

13

Revolving Magnetic Field Theory

Consider an induction motor with three-phase windings (concentrated

winding is shown for simplicity). The three-phase currents applied are

shown side-by-side.

14

The three-phase currents are

At time t=t0 or t4

At time t=t1 At time t= t2 At time t=t3

16

From the Figure shown in the previous slide, it can be seen that the

resultant MMF of the 3-phases will be of the same magnitude at each

and every time instant; but, the orientation is displaced by an angle θ in

space which is equal to ωt where ω is the angular frequency of the

three-phase currents that are impressed on the three-phase stator

windings.

Thus, the revolving magnetic field speed = 2πf radians/sec

Here, the winding is wound only for 2 poles; i.e, one conductor for A

phase carrying +ve current and another conductor carrying –ve current.

If the winding is wound for 4 poles, then there will be 2 conductors for A

phase carrying +ve current (displaced from each other by 180 degrees)

and 2 conductors carrying –ve current.

In that case, a 180 degree displacement mechanically will amount to

360 degree displacement electrically; i.e., from one +ve peak of the

current in A phase to another +ve peak.

Thus mechanical speed = 2πf /(p/2) radians/sec

where p= no. of poles for which the stator winding is wound.

17

Analytical Method

The MMF along a direction θ is

Rotor conductors get an induced emf because of this rotating magnetic field

when the rotor is stationary; the rotor is short circuited and hence, rotor

currents will flow. This will induce another revolving flux. The interaction

between the stator and rotor flux causes torque production. By Lenz’s law, this

torque will oppose its own cause. The cause is the relative velocity between

stator rotating filed and the stationary rotor; so the rotor starts rotating in the

same direction to reduce the relative speed.

20

Stator and Rotor induced Voltages

Revolving magnetic field induces voltage in the stator windings whose

frequency corresponds to ωs (synchronous frequency)

Nph is the no. of turns in stator winding per phase; ϕpole is the flux per pole, f

is the frequency.

As the rotor gains speed, the relative speed between the revolving magnetic

field and the rotor conductors is ωs – ωr where ωr is the rotor speed.

Rotor Induced EMF at standstill condition E2=(ns/nr)*Erms

where ns=No. of stator turns per phase and nr=No. of rotor turns

Rotor induced emf at any other rotor speed ωr will be related to slip’s’.

Slip ‘s’= (ωs – ωr) / ωs

Thus rotor induced emf magnitude = s*E2

Its frequency = sf

Equivalent Circuit of Induction Machines

Conventional equivalent circuit

● Never use three-phase equivalent circuit. Always use per-

phase equivalent circuit.

● The equivalent circuit always based on the Y connection

regardless of the actual connection of the motor.

● Induction machine equivalent circuit is very similar to the

single-phase equivalent circuit of transformer. It is

composed of stator and rotor equivalent circuits

21

Step1 Rotor winding is open (The rotor will not rotate)

Note: ◦ the frequency of E2 is the same as that of E1 (provided the

no. of turns in stator and rotor are the same) since the rotor is at standstill. At standstill s=1.


E2 V2

i1 i

2

r1 X1

rc Xm

r2 X

2

im

ic

I2

E1

N2N1

Air gapStator CircuitRotor Circuit

f1 f1

22


23

Step2 Rotor winding is shorted

(Under normal operating conditions, the rotor winding is

shorted. The slip is s)

Note:

◦ the frequency of E2 is fr=sf because rotor is rotating.


N2N1


V1

R1 X1

RcXm

I0

Ic Im

R2scI2sc

X2sc

f1 sf1

E1

sE2

24

Step3 Eliminate f2

Keep the rotor current same:

2 2 2

2 222 2 2 2

2

scsc

sc sc

E sE EI I

RR jX R jsXjX

s


N2N1


V1

R1 X1

RcXm

I0

Ic Im

R2/sI2

X2

f1 f1

E1

E2

25

Although I2 value is the same (mathematically) whether we write sE2 in the numerator OR

divide by ‘s’ both quantities in the numerator and the denominator; the former gives slip

frequency (fr) current but the latter indicates a frequency of ‘f’.

Equivalent Circuit of Induction Machines Step 4 Referred to the stator side Now that both stator and rotor electrical quantities are at

synchronous frequency they can be combined

Note:

◦ X’2 and R’2 will be given or measured. In practice, we do not have to calculate them from above equations.

◦ Always refer the rotor side parameters to stator side.

◦ Rc represents core loss, which is the core loss of stator side. Sometimes, it may be clubbed together with rotational losses.


V1

R1 X1

RcXm

I0

Ic Im

R’2I2X’2

f1 f1

E1

E2

R’2(1-s)/s

' 2

2 2' 2

2 2

' 22

X a X

R a RI

Ia

26

IEEE recommended equivalent circuit At the air gap terminals Thevenin’s equivalent is taken with

Vth=V1(jXm)/{R1+j(X1+Xm)}

And Zth= (R1+jX1) in parallel with (jXm)

Rc is omitted. The core loss is lumped with the rotational loss.

(R2/s)can be separated into 2 PARTS

◦ The former refers to rotor Cu loss and the latter is the mechanical output. So, Rotor i/p : Rotoro/p : Rotor Cu loss =1 :(1-s) :s


V1

R1 X1

Xm

I0

Im

R’2I2X’2

f1 f1

E1R’2(1-s)/s


2 22

(1 )R R sR

s s

27

28

Determination of motor parameters

Due to the similarity between the induction

motor equivalent circuit and the transformer

equivalent circuit, same tests are used to

determine the values of the motor parameters. ◦ DC test: determine the stator resistance R1

◦ No-load test: determine the rotational losses and

magnetization current (similar to no-load test in

Transformers).

◦ Locked-rotor test: determine the rotor and stator

impedances (similar to short-circuit test in

Transformers).

29

DC test

◦ The purpose of the DC test is to determine R1. A

variable DC voltage source is connected between two

stator terminals.

◦ The DC source is adjusted to provide approximately

rated stator current, and the resistance between the

two stator leads is determined from the voltmeter and

ammeter readings.

30

◦ then

◦ If the stator is Y-connected, the per phase stator

resistance is

◦ If the stator is delta-connected, the per phase stator

resistance is

DC

DC

DC

VR

I

12

DCRR

1

3

2DCR R

DC test

31

No-load test

1. The motor is allowed to rotate freely

2. The only load on the motor is the friction and windage

losses, so all Pconv is consumed by mechanical losses

3. The slip is very small

32

No-load test

4. At this small slip, R2/s becomes infinity and hence

rotor carries hardly any current.

The equivalent circuit reduces to…

33

No-load test

5. Combining Rc & RF+W we get……

34

6. At the no-load conditions, the input power

measured by meters must equal the losses in the

motor.

7. Rotor copper loss is negligible because I2 is

extremely small since R2/s is very large.

8. The input power equals

Where

&

2

1 13

in SCL core F W

rot

P P P P

I R P

&rot core F WP P P

No-load test

35

9. The equivalent input impedance is thus

approximately

If X1 can be found, in some other fashion, the

magnetizing impedance XM will be known

1

1,

eq M

nl

VZ X X

I

No-load test

36

Blocked-rotor test

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.

37

The AC voltage applied to the stator is adjusted

so that the current flow is approximately full-

load value.

The locked-rotor power factor can be found as

The magnitude of the total impedance

cos3

in

l l

PPF

V I

LR

VZ

I

Blocked-rotor test

38

Where X’1 and X’2 are the stator and rotor reactance's

at the test frequency respectively

'

cos sin

LR LR LR

LR LR

Z R jX

Z j Z

1 2

' ' '

1 2

LR

LR

R R R

X X X

2 1LRR R R

'

1 2

rated

LR LR

test

fX X X X

f

Blocked-rotor test

Analysis of Induction Machines For simplicity, let

assume

Is=I1 , IR=I2

(s=stator, R=rotor)

ZR Zm

Zs

Vs1

Is1 Im1 IR1

22

'' ;

// ;;

;//

R

m c m c

m m c

s s s

Total s m R

RZ jX

sZ R jX R neglectedZ jX R neglectedZ R jXZ Z Z Z

1

1

s

s

T

VI

Z

39

ZR Zm

Zs

Vs1

Is1 Im1 IR1

Analysis of Induction Machines

1 1

1 1

,

Rm s

m R

mR s

m R

Current Dividing Rules

ZI I

Z Z

ZI I

Z Z

1 1

1

1

1

1

,

//

,

R mRM s

T

RM

R

R

RM

m

m

Voltage Dividing Rules

Z ZV V

Z

VHence I

Z

VI

Z

OR

40

Power Flow Diagram

Pin (Motor)

Pin (Stator)

Pcore loss

(Pc)

Pair Gap

(Pag)

Pdeveloped

Pmechanical

Pconverted

(Pm)

Pout, Po

Pstator copper

loss, (Pscu)

Protor copper

loss (Prcu) Pwindage, friction,

etc

(P - Given)

3 coss sV I

2 '3 ' R

R

RI

s

2

3 RM

c

V

R

23 ' 'R RI R

2 13 ' 'R R

sI R

s

1 746hp W

23 s sI R

Pin (Rotor)

41

Power Flow Diagram Ratio:

Pag Prcu Pm

2 '3 ' R

R

RI

s

23 ' 'R RI R2 1

3 ' 'R R

sI R

s

1

s

11

s1

1 s 1 s

Ratio makes the analysis simpler to find the value of a particular

power if we have another value of power. For example:

1rcu

m

P s

P s

42

Efficiency

100%

,

,

3 cos746 746

out

in

losses

o in losses

o m

in s s

out

P

Pif P are givenP P PP P P

otherwise

P V IP x hp W x Watt

43

Example

A 480-V, 60 Hz, 50-hp, three phase induction motor is

drawing 60A at 0.85 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 Pconv.

3. The output power Pout.

4. The efficiency of the motor.

44

Solution

3 cos

3 480 60 0.85 42.4 kW

in L LP V I

42.4 2 1.8 38.6 kW

AG in SCL coreP P P P

70038.6 37.9 kW

1000

conv AG RCLP P P

&

60037.9 37.3 kW

1000

out conv F WP P P

1.

2.

3.

45

37.350 hp

0.746outP

100%

37.3100 88%

42.4

out

in

P

P

4.

46

Torque-Equation

Torque, can be derived from power equation in term of

mechanical power or electrical power.

2, , ( / )

6060

,2

nPower P T where rad s

PHence T

n

,60

,2

60,

2

mm

r

oo

r

ThusP

Mechanical Torque Tn

POutput Torque T

n

47

48

Torque, power and Thevenin’s Theorem

Thevenin’s theorem can be used to transform the

network to the left of points ‘a’ and ‘b’ into an

equivalent voltage source VTH in series with equivalent

impedance RTH+jXTH

49


1 1( )

M

TH

M

jXV V

R j X X

1 1( ) //TH TH MR jX R jX jX

2 2

1 1

| | | |( )

M

TH

M

XV V

R X X

50

Since XM>>X1 and XM>>R1

Because XM>>X1 and XM+X1>>R1

1

M

TH

M

XV V

X X

2

1

1

1

M

TH

M

TH

XR R

X X

X X


51

Then the power converted to mechanical (Pconv)

22

22

2( )

TH TH

T

TH TH

V VI

Z RR X X

s

2 2

2

(1 )3conv

R sP I

s

And the internal mechanical torque (Tint)

intconv

m

PT

(1 )

conv

s

P

s

2 2

23AG

s s

RI

Ps


52

2

2int

2

222

3

( )

TH

s

TH TH

V RT

sRR X X

s

2 2

int 2

222

31

( )

TH

s

TH TH

RV

sT

RR X X

s


53

Complete Speed-torque characteristics

- Smax T

Smax T 1

1

54

Maximum torque

Maximum torque occurs when the power

transferred to R2/s is maximum.

This condition occurs when R2/s equals the

magnitude of the impedance RTH + j (XTH + X2)

max

2 22

2( )TH TH

T

RR X X

s

max

2

2 2

2( )T

TH TH

Rs

R X X

55

The corresponding maximum torque of an

induction motor equals

The slip at maximum torque is directly proportional

to the rotor resistance R2

The maximum torque is independent of R2

2

max2 2

2

31

2 ( )

TH

s TH TH TH

VT

R R X X

Maximum torque

56

Rotor resistance can be increased by inserting

external resistance in the rotor of a wound-rotor

induction motor.

The value of the maximum torque remains

unaffected

but

the speed at which it occurs can be altered.

Maximum torque

57

Maximum torque

Effect of rotor resistance on torque-speed characteristic

58

Example

A two-pole, 50-Hz induction motor supplies 15kW to a

load at a speed of 2950 rpm.

1. What is the motor’s slip?

2. What is the induced torque in the motor in N.m

under these conditions?

3. What will be the operating speed of the motor if its

torque is doubled?

4. How much power will be supplied by the motor

when the torque is doubled?

59

Solution

1.

2.

120 120 503000 rpm

2

3000 29500.0167 or 1.67%

3000

e

sync

sync m

sync

fn

P

n ns

n

3

no given

assume and

15 1048.6 N.m

22950

60

f W

conv load ind load

conv

ind

m

P

P P

P

60

3. In the low-slip region, the torque-speed curve is

linear and the induced torque is direct proportional

to slip. So, if the torque is doubled the new slip will

be 3.33% and the motor speed will be

4.

(1 ) (1 0.0333) 3000 2900 rpmm syncn s n

int

2(2 48.6) (2900 ) 29.5 kW

60

conv mP T

Solution

Torque–Speed Characteristic

61

Torque–Speed Characteristic

Depending on various design

features , the torque-speed

characteristics of the

induction machine can be

modified to suit particular

application.

Motor designers can modify the shape of the

torque–speed curve by

changing various aspects of the machine design

such as the cross section and depth of the rotor

conductors. 62

Example

A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction

motor has the following impedances in ohms per phase

referred to the stator circuit:

R1= 0.641 R2= 0.332

X1= 1.106 X2= 0.464 XM= 26.3

The total rotational losses are 1100 W and are assumed to

be constant. The core loss is lumped in with the

rotational losses. For a rotor slip of 2.2 percent at the

rated voltage and rated frequency, find the motor’s

1. Speed

2. Stator current

3. Power factor

63

Solution

1.

2.

120 120 601800 rpm

4

e

sync

fn

P

(1 ) (1 0.022) 1800 1760 rpmm syncn s n

2

2 2

0.3320.464

0.022

15.09 0.464 15.1 1.76

RZ jX j

s

j

2

1 1

1/ 1/ 0.038 0.0662 1.76

112.94 31.1

0.0773 31.1

f

M

ZjX Z j

64

3.

4.

0.641 1.106 12.94 31.1

11.72 7.79 14.07 33.6

tot stat fZ Z Z

j

j

1

460 0

318.88 33.6 A

14.07 33.6tot

VI

Z

cos33.6 0.833 laggingPF

3 cos 3 460 18.88 0.833 12530 Win L LP V I

2 2

1 13 3(18.88) 0.641 685 WSCLP I R

12530 685 11845 WAG in SCLP P P

Solution

65

5.

6.

(1 ) (1 0.022)(11845) 11585 Wconv AGP s P

& 11585 1100 10485 W

10485= 14.1 hp

746

out conv F WP P P

1184562.8 N.m

1800260

AG

ind

sync

P

1048556.9 N.m

1760260

out

load

m

P

10485100% 100 83.7%

12530

out

in

P

P

Solution

66

67

Starting of Induction Motors

Direct On-Line Starting

Squirrel cage induction motors

are frequently started by

connecting them directly to the

supply-lines.

Suitable for low rating motors

Starting current of the order of

450-800 percent of full load

current

68

The current drawn by a motor when starting is excessive because of a lack of counter EMF at the instant of starting.

Tst/Trated =(I2st/Irated)2 *sFL

Normally FL slip < 0.05

Even when DOL current is 6 times the rated current,

Starting torque = 62*0.04=

1.44*FL torque

Direct On-Line Starting

69

Autotransformer starting uses a tapped 3phase autotransformer to

provide reduced-voltage starting. If the ratio of applied voltage to

output voltage is n:1, then the torque as well as line currents are

1/n^2 times the DOL values.

70

In an autotransformer starting circuit, reduced voltage is applied to the motor circuit by having a low voltage tapping on the transformer winding to provide reduced current when starting.

71

Windings on a wye-delta motor may be joined to form a wye or

delta configuration.

72

A delta-connected motor has each coil winding

directly connected across two power lines so each

winding receives the entire source voltage of 208

V.

73

A wye-connected motor has two power lines

connected across two sets of windings. Line

current and torque in Y configuration will be

1/3rd of the values in Delta configuration.

74

The control circuit of a typical wye-delta starting circuit consists of two motor starters, a contactor, and an ON-delay timer.

75

Impedance Starting

An Impedance (R+jX) is connected in series with the source in each

phase to make sure that the total current drawn by the motor is

limited.

Correspondingly the torque also becomes limited.

All these starting methods are not good for applications that require

high starting torque.

They are OK for fan or pump type loads where power is proportional

to the cube of speed and the torque is proportional to the square of

speed.

Wound rotor induction motor with rotor resistance

starting can offer very good starting torque with

limited current and a good power factor during

starting.

76

A solid-state starter ramps up voltage, reduces inrush current, minimizes starting torque, and smoothes acceleration.

77

Wound Rotor Motor Starting

1. Startup of high inertia loads

1. Heat in external resistor – not rotor – so, easy to

dissipate

2. Vary external resistor – alter torque/speed curve

2. Variable speed drive – converter on rotor circuit

1. Inefficient – heat lost in external resistor

78

Rotor Resistance Control of IM

Three Ph. AC

Supply

WRIM

Rex

RexRex

• Starting by increasing resistance R by changing the modulation index of

chopper switch.

• The highest value of rotor resistance is chosen to limit current at zero speed

within the safe value

• As the motor accelerates, the external resistance are cut out gradually to limit

the rotor current between specified limits

Speed Control

79

Induction Motor

Squirrel Cage Rotor Wound Rotor

Voltage

Injection in

rotor circuit

Rotor

resistance

control

Pole

changing

control

Voltage/

frequency

control

80

Pole Changing Control of IM The speed control of an induction motor can be achieved

by changing the synchronous speed.

The synchronous speed of IM can be changed by

changing number of stator poles.

The stator windings are designed such that simple

changes in coil connection can change the number of

poles.

This method requires same number of poles on the rotor,

therefore it is suitable for squirrel cage rotors.

With two independent sets of stator windings, each

arranged for pole changing, four synchronous speeds can

be obtained in a squirrel cage induction motor.

81

Stator Voltage Control of IM

The speed of an induction motor can be controlled by

varying the amplitude of stator voltage.

The internal torque developed by an induction motor is

proportional to the square of the stator voltage.

The variation of speed can be achieved by varying the

stator voltage as shown in the torque speed

characteristic.

This method of speed control is commonly employed with

small sq. cage motors driving fans, blowers etc where

cost is an issue and the inefficiency of high slip operation

can be tolerated.

82

T-speed Characteristic T

orq

ue

0 Speed nr2 nr1 100

LoadV1

0.5V1

83

Voltage/Frequency Control of IM

The synchronous speed of an induction motor can be

changed by varying the frequency of stator voltage.

The frequency of stator voltage can be changed using

various power electronic frequency changers.

As the flux density has to be maintained approximately

constant, the stator voltage should also be varied

proportional to the frequency.

This is known as constant voltage per hertz (V/f) control.

The frequency changers mostly used are Voltage Source

Inverters (VSI), Current Source Inverters (CSI),

Cyclo-converters and Matrix Converters.

From Faraday’s law, The air gap component of the

armature voltage in an AC machine is proportional to

the peak flux density in the machine and electrical

frequency.

Thus, neglecting the voltage drop across the armature

resistance and leakage reactance, the stator voltage can

be written as

( )

is the amplitude of the armature voltage;

is the operating frequency; is the peak flux density;

, , are the corresponding

peake

a rated

rated rated

a

e peak

rated rated rated

BfV V A

f B

where V

f B

f B V

rated values.

Analysis

The speed of induction motors can be precisely controlled by

frequency control and can be made independent of variation in

supply voltage, field current and load. Therefore, keeping

Va=Vrated, Eq.(A) can be rewritten as

This demonstrates the constant voltage, variable frequency

operation.

In this mode, a machine operating in saturation at rated voltage

and frequency, any reduction in frequency will lead to further

increase in flux density;

rated

peak rated

e

fB B

f

Higher flux density will result in increased core loss and higher

machine currents;

Therefore, for frequencies less than or equal to rated frequency,

the machine is operated at constant flux density, i.e. Bpeak=Brated.

This makes the Eq. (A) as

This is constant voltage per hertz (V/f) operation.

It is typically maintained from rated frequency to the low

frequency at which armature resistance drop becomes

significant component of the applied voltage.

e a rated

a rated

rated e rated

f V VV V

f f f

For frequencies higher than the rated with the voltage at its

rated value, the airgap flux density will drop below its rated

value {referring Eq (A)};

Thus, to maintain the rated flux density the voltage has to be

increased, which may result in insulation failure.

Therefore, for frequencies above the rated frequency the

terminal voltage is kept at rated value.

Assuming that machine cooling is not affected by rotor speed,

the maximum permissible terminal current will remain constant

at its rated value Irated.

Therefore, for the frequencies below rated frequency the

machine power will be proportional to feVratedIrated.

88

T

Vs= Constant

- b

-T

nnb

Torque-Speed Curve

Constant flux up to base speed and constant voltage above base speed.

Operating Regions

Power

Constant Torque Region Constant Power Region

Torque

Rated Speed Speed

Varying rotor resistance

For wound rotor only

Speed is decreasing

Constant maximum

torque

The speed at which max

torque occurs changes

Disadvantages: ◦ large speed regulation

◦ Power loss in Rext – reduce

the efficiency

T

ns~nNL

T

nr1 nr2 nr3 n

nr1< nr2< nr3 R1 R2 R3

R1< R2< R3

90

91

Rotor Resistance Control of IM

The speed of an induction motor can be controlled by

varying the rotor resistance.

The slip at which maximum torque developed by an

induction motor is proportional to the rotor resistance.

The value of maximum torque remains unchanged for any

value of rotor resistance.

This method of speed control has characteristics similar

to those of DC shunt motor speed control using

resistance in series with the armature.

This method suffers low efficiency at reduced speed and

poor speed regulation with respect to change in load

Varying supply voltage and supply frequency

The best method since

supply voltage and supply

frequency is varied to keep V/f constant

Maintain speed regulation

uses power electronics

circuit for frequency and

voltage controller

Constant maximum torque

T

nNL1

T

nr1 nr2 nr3

n

f decreasing

nNL2 nNL3

92

Selection of Induction Motors

Some of the most important considerations in

selecting an induction motor are:

1. Efficiency

2. Starting torque

3. Pull-out torque

4. Power factor

5. Starting current

93

94

Example

A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-

rotor induction motor has the following impedances

in ohms per phase referred to the stator circuit

R1= 0.641 R2= 0.332

X1= 1.106 X2= 0.464 XM= 26.3

1. What is the maximum torque of this motor? At what

speed and slip does it occur?

2. What is the starting torque of this motor?

3. If the rotor resistance is doubled, what is the speed

at which the maximum torque now occur? What is

the new starting torque of the motor?

4. Calculate and plot the T-s c/c for both cases.

95

2 2

1 1

2 2

( )

46026.3

3255.2 V

(0.641) (1.106 26.3)

M

TH

M

XV V

R X X

2

1

1

226.3

(0.641) 0.5901.106 26.3

M

TH

M

XR R

X X

1 1.106THX X

Solution

96

1.

The corresponding speed is

max

2

2 2

2

2 2

( )

0.3320.198

(0.590) (1.106 0.464)

T

TH TH

Rs

R X X

(1 ) (1 0.198) 1800 1444 rpmm syncn s n

Solution

97

The torque at this speed is

2

max2 2

2

2

2 2

31

2 ( )

3 (255.2)

22 (1800 )[0.590 (0.590) (1.106 0.464) ]

60

229 N.m

TH

s TH TH TH

V

R R X X

Solution

98

2. The starting torque can be found from the torque

eqn. by substituting s = 1

2 2

21

22

2

1

2

2

2 2

2 2

2

2 2

31

( )

3

[ ( ) ]

3 (255.2) (0.332)

21800 [(0.590 0.332) (1.106 0.464) ]

60

104 N.m

TH

start ind ss

TH TH

s

TH

s TH TH

RV

s

RR X X

s

V R

R R X X

Solution

99

3. If the rotor resistance is doubled, then the slip at

maximum torque doubles too

The corresponding speed is

The maximum torque is still

max = 229 N.m

max

2

2 2

2

0.396( )

T

TH TH

Rs

R X X

(1 ) (1 0.396) 1800 1087 rpmm syncn s n

Solution

100

The starting torque is now

2

2 2

3 (255.2) (0.664)

21800 [(0.590 0.664) (1.106 0.464) ]

60

170 N.m

start

Solution

101

Induction Motor Braking

Regenerative Braking

Plugging or Reverse Voltage Braking

Dynamic Braking

AC Dynamic Braking

Self Excited Braking using

Capacitors

DC Dynamic Braking

Zero Sequence Braking

102

Induction Braking, Motoring, Generating

Synchronous Speed

(Rotating Field Speed)

Voltage

Current

Braking Region Motoring Region Regenerating Region

+V

e T

orq

ue

-Ve T

orq

ue

Motoring

Regenerating

103

Regenerative Braking

The Power Input to the motor is given by Pin=3VtIcosФ

Ф is the phase angle between stator phase voltage Vt and stator

phase current I

104

For motoring Ф<900

If the rotor speed becomes greater than synchronous speed ,

relative speed between rotor conductors and air-gap rotating

field reverses

This reverses the rotor induced emf, rotor current and

component of stator current which balances the rotor ampere

turns

Consequently, angle Ф becomes greater than 900 and power flow

reverses, giving regenerative braking

Slip s is negative

When fed from a source of fixed frequency, regenerative braking

is possible only for speeds greater than synchronous speed

With a variable frequency source, it can also be obtained for

speeds below synchronous speed

Main advantage of regenerative braking is that generated power

is usually employed

105

Plugging or Reverse Voltage Braking

1- Natural Characteristic Plugging in IV quadrant with

2- With External resistance in rotor large external resistance in rotor

106

When phase sequence of supply of the motor running at a speed

is reversed, operation shifts from motoring to plugging as shown

in Fig in pervious slide

Plugging characteristics are actually extension of motoring

characteristics for negative phase sequence from quadrant III to II

Reversal of phase sequence reverses the direction of rotating field

If slip for plugging is denoted by sn, then

sn={(-ωms- ωm)/(- ωms)}=2-s

It is necessary to disconnect the motor for stopping at or near

zero speed

The braking is suitable for reversing the motor

A special case of plugging occurs when an induction motor

connected to positive sequence voltages is driven by an active

load in the reverse direction

In this method braking energy is wasted in rotor circuit resistance

107

Dynamic Braking

Not a very viable option for induction motors.

Once the stator supply is disconnected, flux also

collapses; so, machine unable to function as a

generator

So, rheostatic or dynamic braking not possible by

connecting resistances to the stator circuit.

108

Thank you!

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