Induction Motor

•Why induction motor (IM)? –Robust; No brushes. No contacts on rotor shaft –High Power/Weight ratio compared to Dc motor –Lower Cost/Power –Easy to manufacture –Almost maintenance-free, except for bearing and other mechanical parts

•Disadvantages –Essentially a “fixed-speed” machine –Speed is determined by the supply frequency –To vary its speed need a variable frequency supply

ConstructionStator

Construction

RotorSquirrel Cage

Construction

p

ff

Pn

12060*

2

s

s

n

nns

ss snnnrpmSlip 12120120

sfsnp

nnp

f ss

ssnp

sf

p

fn 12

2120120

Performance of Three-Phase Induction Motor

Example 5.1 A 3-phase, 460 V, 100 hp, 60 Hz, four‑pole induction machine delivers rated output power at a slip of 0.05. Determine the:

(a) Synchronous speed and motor speed.(b) Speed of the rotating air gap field.(c) Frequency of the rotor circuit.(d) Slip rpm.(e) Speed of the rotor field relative to the (i) rotor structure. (ii) Stator structure. (iii) Stator rotating field.(f) Rotor induced voltage at the operating speed, if the stator‑to‑rotor turns ratio is

1 : 0.5.

Solution:

rpmp

fns 1800

4

60*120120

rpmnsn s 17101800*05.011 (b) 1800 (same as synchronous speed)

Equivalent Circuit of the Induction Motor

Equivalent Circuit of the Induction Motor

22

22

jsXR

sEI

2222 RIP

22

22 / jXsR

EI

s

RIs

s

RRIPP ag

222

22

22 1

ss

RIPmech 122

2 agmech PsP *1

2

1P

s

sPmech

agsPRIP 2222

ssPPP mechag 1::1:: 2

IEEE‑Recommended Equivalent Circuit

IEEE recommended equivalent circuit.

Theveninequivalent circuit

12

121

VXXR

XV

m

mth

2

121 mXXRIf

111

VKVXX

XV th

m

mth

ththm

mth jXR

XXjR

jXRjXZ

11

11

If , then, 21

21 mXXR

12

1

2

1

RKRXX

XR th

m

mth

1XX th

Tests To Determine The Equivalent Circuit

No‑load test

2/1 LLRR

PhaseVV

V LL /3

1 1

1

I

VZNL 2

13I

PR NL

NL

22NLNLNL RZX

NLm XXX 1

Locked‑rotor test

BL

BLBL

I

PR

213

BL

BLfBLBL I

VZ

1

1

22BLfBLBLfBLBL RZX

testrotorblockedatFrequency

FrequencyRatedXX

fBLBLBL *

21 XXX BL

NLm XXX 1 1XXX NLm

Blocked‑rotor equivalent circuit for improved value for 2R

222

22

2

RXXR

XR

m

m

RX

XXR

m

m

2

22

2

2

2

RXX

XR

m

m

1RRR BL

Example 4.2 A no‑load test conducted on a 30 hp, 835 r/min, 440 V, 3‑phase, 60 Hz squirrel‑cage induction motor yielded the following results:No‑load voltage (line‑to‑line): 440 VNo‑load current: 14 ANo‑load power: 1470 WResistance measured between two terminals: 0.5 The locked‑rotor test, conducted at reduced volt age, gave the following results:Locked‑rotor voltage (line‑to‑line): 163 VLocked‑rotor power: 7200 WLocked‑rotor current: 60 ADetermine the equivalent circuit of the motor.

Solution:Assuming the stator windings are connected in way, the resistance per phase is:

25.02/5.01R

From the no‑load test:

PhaseVV

V LL /2543

440

31

143.1814

254

1

1

I

VZNL

5.214*3

1470

3 221I

PR NL

NL

97.175.2143.18 2222 NLNLNL RZX

97.171 NLm XXX

6667.060*3

7200

3 221

BL

BLBL

I

PR

From the blocked‑rotor test

The blocked‑rotor reactance is:

42.16667.05685.1 2222BLBLBL RZX

42.121 XXX BL

71.021 XX

26.1771.097.171XXX NLm

4167.025.06667.01RRR BL

4517.04167.0*

26.17

26.1771.022

22 R

X

XXR

m

m

• Example 5.3 The following test results are obtained from a three-phase 60 hp, 2200 V, six‑pole, 60 Hz squirrel‑cage induction motor.

• (1) No‑load test:• Supply frequency = 60 Hz, Line voltage = 2200 V• Line current = 4.5 A, Input power = 1600 W• (2) Blocked‑rotor test:• Frequency = 15 Hz, Line voltage = 270 V• Line current = 25 A, Input power = 9000 W• (3) Average DC resistance per stator phase: 2.8 • (a) Determine the no‑load rotational loss.• (b) Determine the parameters of the IEEE‑recommended equivalent

circuit• (c) Determine the parameters (Vth, Rth, Xth) for the Thevenin

equivalent circuit of Fig.5.16.

PhaseVV /2.12703

22001 27.282

5.4

2.1270

1

1

I

VZNL

34.265.4*3

1600

3 221I

PR NL

NL

(a) No-Load equivalent Circuit (b) Locked rotor equivalent circuit

28134.2627.282 2222NLNLNL RZX

2811 NLm XXX

= 281.0 .

8.425*3

9000

3 221I

PR BL

BL

28.28.412 RRR BL

impedance at 15 Hz is:

24.625*3

270

1

1

I

VZBL

The blocked‑rotor reactance at 15 Hz is 98.38.424.6 22BLX

Its value at 60 Hz is 92.1515

60*98.3BLX

21 XXX BL

96.72

92.1521 XX at 60 Hz

04.27396.7281mX 28.28.41RRR BL

12.2204.273

04.27396.72

2R

)c (

11 97.004.27396.7

04.273VVVth

63.28.2*97.097.0 21

2 RRth

96.71XX th

PERFORMANCE CHARACTERISTICS

sR

ITsyn

mech22

21

sR

XXsRR

VT

thth

th

synmech

22

22

2

2*

/*

1

ss

RITP mechmechmech 122

260

2 nmech

synmech s 1

snsyn

mech 1260

P

f

P

fsyn

14

60

2*

120

agsynmech Ps

RIT 222

agsyn

mech PT

1

Where

synmech nsn 1or

At low values of slip,

ththth Rs

RandXX

s

RR

2

22

sR

VT th

synmech **

1

2

2

sR

XXsRR

VT

thth

th

synmech

22

22

2

2*

/*

1

At larger values of slip,

22 XX

s

RR thth

s

R

XX

VT

th

th

synmech

22

2

2

**1

sR

XXsRR

VT

thth

th

synmech

22

22

2

2*

/*

1

Torque‑speed profile at different voltages.

Maximum Torque

0/ dsdT

22

22

max

XXRS

Rthth

T

sR

XXsRR

VT

thth

th

synmech

22

22

2

2*

/*

1

22

2

2max

XXR

RS

thth

T

Then

22

2

2

max *2

1

XXRR

VT

ththth

th

syn

Torque speed characteristics for varying 2R

.

22

2

2

max *2

1

XXRR

VT

ththth

th

syn

1RIf is small (hence thR is negligibly small)

22

2

2max

XXR

RS

thth

T

22

2

2

max *2

1

XXRR

VT

ththth

th

syn

2

2max XX

Rs

thT

2

2

max *2

1

XX

VT

th

th

syn

Then

Then

maxmax

*/

/2

22

2

22

22max

TthTth

thth

s

s

XXsRR

XXsRR

T

T

1RIf is small (hence thR is negligibly small)

maxmax

*/

/2

22

2

22

22max

TthT

th

s

s

XXsR

XXsR

T

T

maxmax

max */2

//2

2

22

22max

TT

T

s

s

sR

sRsR

T

T

ss

ss

T

T

T

T

**2max

max

22max

Efficiency

Power flow in an induction motor.

111 cos3 IVPin The power loss in the stator winding is:

1211 3 RIP

2222 3 RIP

in

out

P

P

inag PP agsPP 2

sPPP agmecjout 1

sP

P

in

outideal 1

Ideal Efficiency

Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole

wound-rotor induction motor has the following parameters per

phase:

1R = 0.25 , 2.02 R , 5.021 XX , 30mX

The rotational losses are 1700 watts. With the rotor terminals

short-circuited, find

(a) (i) Starting current when started direct on full voltage.

(ii) Starting torque.

(b) (i) Full-load slip.

(ii) Full-load current.

(iii) Ratio of starting current to full-load current.

(iv) Full-load power factor.

(v) Full-load torque.

(iv) Internal efficiency and motor efficiency at full load.

(c) (i) Slip at which maximum torque is developed.

(ii) Maximum torque developed.

(d) How much external resistance per phase should be

connected in the rotor circuit so that maximum torque occurs at

start?

=163.11 N.m

%5.87100*4.32022

3.28022motor

%7.96100*0333.01100*1int sernal

(c) (i)

(c) (ii)

Note that for parts (a) and (b) it is not necessary to use Thevenin equivalent circuit. Calculation can be based on the equivalent circuit of Fig.5.15 as follows:

A three-phase, 460 V, 60 Hz, six-pole wound-rotor induction motor

drives a constant load of 100 N - m at a speed of 1140 rpm when

the rotor terminals are short-circuited. It is required to reduce the

speed of the motor to 1000 rpm by inserting resistances in the

rotor circuit. Determine the value of the resistance if the rotor

winding resistance per phase is 0.2 ohms. Neglect rotational

losses. The stator-to-rotor turns ratio is unity.

Example The following test results are obtained from three

phase 100hp,460 V, eight pole star connected induction machine

No-load test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is

100V, 60Hz, 140A 8kW. Average DC resistor between two stator

terminals is 0.152

(a) Determine the parameters of the equivalent circuit.

(b) The motor is connected to 3 , 460 V, 60 Hz supply and runs

at 873 rpm. Determine the input current, input power, air

gap power, rotor cupper loss, mechanical power developed,

output power and efficiency of the motor.

(c) Determine the speed of the rotor field relative to stator

structure and stator rotating field

Solution: From no load test:

64.640

3/460NLZa

875.040*3

4200

*3 221I

PR NL

NL

58.6875.064.6 22NLX

58.61 mXX

From blocked rotor test:

136.0140*3

80002BLR 076.0

2152.0

1R

412.0140

3/100BLZ

389.0136.0412.0 22BLX

1945.02389.0

21 XX

3855.61945.058.6 mX

06.0076.0136.01RRR BL

0637.006.0*3855.6

3855.61945.02

2

R

0.076 j0.195

j6.386

j0.195

s

0637.0

389.021 XX

rpmP

fnb s 900

8

60*120120

03.0900

873900

s

s

n

nns

123.203.0

0637.02

s

R

Input impedance

o

j

jjjZ 16.27121.2

195.0386.6123.2

195.0123.2386.6195.0076.01

o

Z

VI 16.2722.125

16.2712.2

3/460

1

11

Input power:

kWP oin 767.8816.27cos22.125*

3

460*3

Stator CU losses:

kWPst 575.3076.0*22.125*3 2 Air gap power kWPag 192.85575.3767.88

Rotor CU losses kWsPP ag 556.2192.85*03.02

Mechanical power developed :

kWPsP agmech 636.82192.85*03.011

rotmechout PPP

From no load test: WRIPP NLrot 2.3835076.0*40*34200*3 21

21

kWPout 8.782.383510*636.82 3

%77.88100*767.88

8.78100*

in

out

P

P

Example A three phase, 460 V 1450 rpm, 50 Hz, four pole

wound rotor induction motor has the following parameters per

phase ( 1R =0.2, 2R=0.18 , 21 XX =0.2, mX =40). The

rotational losses are 1500 W. Find,

(a) Starting current when started direct on full load voltage.

Also find starting torque.

(b) (b) Slip, current, power factor, load torque and efficiency

at full load conditions.

(c) Maximum torque and slip at which maximum torque will

be developed.

(d) How much external resistance per phase should be

connected in the rotor circuit so that maximum torque

occurs at start?

phaseVV /6.2653

4601

o

j

jjZ 59.4655.0

2.4018.0

2.018.0*402.02.01

oost I

VI 3.4691.482

59.4655.0

6.265

1

1

0333.01500

14501500 s

4.50333.0

18.02

s

R

o

j

jjjZ 83.10959.4

4.454.5

2.04.5*402.02.01

AI ooFL

83.1056.5383.10959.4

6.2651

Then the power factor is: 9822.083.10cos o lag.

.sec/08.1572*60

1500radsys

V

j

jV o

th 285.0275.2642.402.0

40*6.265

Then,

2.0198.0285.45281432.0

2.402.0

2.02.0*40j

j

jjZ o

th

NmT 68.2282.02.04.5198.0*08.157

4.5*275.264*322

2

Then, WTP sysag 1.3592108.157*68.228*

Then, WsPP ag 11971.35921*0333.02

And, WPsP agm 7.347231

Then, WPPP rotmout 7.3322315007.34723

WPin 419179822.0*56.53*6.265*3

Then, %26.7941914

7.33223

in

out

P

P

NmTm 56.862

2.02.0198.0198.05.188*2

275.264*32/122

2

4033.02.02.0198.0

18.02/122max

Ts

(d) 2/122

2

2.02.0198.01

max

ext

T

RRs

Then, 446323.02 extRR

Then, 26632.018.0446323.0 extR

Example 5.6 The rotor current at start of a three-phase, 460 volt,

1710 rpm, 60 Hz, four pole, squirrel-cage induction motor is six

times the rotor current at full load.

(a) Determine the starting torque as percent of full load torque.

(b) Determine the slip and speed at which the motor develops

maximum torque.

(c) Determine the maximum torque developed by the motor as

percent of full load torque.

Note that the equivalent circuit parameters are not given. Therefore equivalent circuit parameters cannot be used directly for computation.)a) The synchronous speed is

s

RI

s

RIT

syn

2222

22

Example 4.9 A 4 pole 50 Hz 20 hp motor has, at rated voltage

and frequency a starting torque of 150% and a maximum torque of

200 % of full load torque. Determine (i) full load speed (ii) speed

at maximum torque. Solution:

5.1FL

st

T

T and 2max

FLT

T then, 75.0

2

5.1

max

T

Tst

75.01

22

max max

max

T

Tst

s

s

T

T

Then, 075.0275.0maxmax

2 TT ss

Then 21525.2max

Ts (unacceptable) Or 451416.0max

Ts

2*2

max

max

22max

FLT

FLT

FL ss

ss

T

T

But 451416.0max

Ts

Then 2*451416.0*2

451416.0 22max

FL

FL

FL s

s

T

T

0451416.0451416.0*4 22 FLFL ss

0203777.080566.12 FLFL ss

6847.1FLs (unacceptable) or 120957.0FLs

rpmns 15004

50*120

then (a) sFLFL nsn *1

rpmnFL 13191500*120957.01

(b) rpmnsn sTT 8231500*451416.01*1maxmax

Example 4.10 A 3, 280 V, 60 Hz, 20 hp, four-pole induction

motor has the following equivalent circuit parameters.

12.01 R , 1.02 R , 25.021 XX , and 10mX

The rotational loss is 400 W. For 5% slip, determine (a) The

motor speed in rpm and radians per sec. (b) The motor current. (c)

The stator cu-loss. (d) The air gap power. (e) The rotor cu-loss. (f)

The shaft power. (g) The developed torque and the shaft torque.

(h) The efficiency.

rpmns 18004

60*120 , sec/5.1882*

60

1800rads

Solution:

0 .1 2 j0 .2 5

j1 0

j0 .2 5

205.0

1.0

ee XRjZ 25.012.01

o

j

jjjZ 55.231314.2

25.102

25.02*1025.012.01

1.1203

2081 V V

ooI 55.231314.2

55.231314.2

1.1201

A

(c) WP 031.114312.0*3479.56*3 21

(d) WP os 9794.1861055.23cos*3479.56*1.120*3

WPPP sag 9485.174671

(e) WsPP ag 3974.8739785.17467*05.02

(f) WPsP agm 5511.165941

(g) mNP

T ag .6682.925.188

9485.17467

5.188

NmP

T shaftshaft 9127.85

5.188

5511.16194

5.188

(h) %02.87100* s

shaft

P

P

Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction

machine has the following

equivalent circuit parameters:

07.01 R , 05.02 R , 2.021 XX , and 5.6mX

(a) Derive the Thevenin equivalent circuit for the

induction machine.

(b) If the machine is connected to a 30, 460 V, 60 Hz supply,

determine the starting torque, the maximum torque the machine

can develop, and the speed at which the maximum torque is

developed.

(c) If the maximum torque is to occur at start, determine the

external resistance required in each rotor phase. Assume a

turns ratio (stator to rotor) of 1.2.

Solution:

VVXX

XV

m

mth 7.2576.265*

5.62.0

5.6* 1

1

1947.006589.0

5.62.007.0

07.02.0*5.6j

jj

jjjXR thth

0 .0 6 5 8 9 j0 .1 9 4 7 j0 .2

s

05.0

2 5 7 .7 V

(b)

NmTst 7.6242.01947.005.006589.025.94

05.0*7.257*3

22

2

Nm

T

8.2267

2.01947.006589.006589.025.94*2

7.257*322

2

max

1249.0

2.01947.006589.0

05.0

22max

Ts

Speed in rpm for which max torque occurs

= rpmns sT 5.787900*1249.01*1max

(c)

2221

21

2max

RXXR

RsT

or 4.005.0*1249.0

1*

122

max

Rs

sR

T

startstart

Then 243.02.1/05.04.0 2 extR