Induction Motor II

Equivalent Circuit Model

No Load Test, Blocked Rotor Test and DC resistance of Stator

Windings

Equivalent Circuit of the Induction Motor

IEEE-Recommended Equivalent Circuit

Thevenin equivalent circuit

12

121

VXXR

XV

m

mth

2121 mXXRIf

11

1

VKVXX

XV th

m

mth

thth

m

mth jXR

XXjR

jXRjXZ

11

11

If , then, 2121 mXXR

12

1

2

1

RKRXX

XR th

m

mth

1XX th

PERFORMANCE CHARACTERISTICS

s

RIT

synmech

222

1

s

R

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

syn

mech **1

2

2

s

R

XXsRR

VT

thth

th

synmech

22

22

2

2

*/

*1

Linear Characteristic near

synchronous speed

At larger values of slip,

22 XX

s

RR thth

s

R

XX

VT

th

th

syn

mech2

2

2

2

**1

s

R

XXsRR

VT

thth

th

synmech

22

22

2

2

*/

*1

Torque varies inversely

with s

Torque-speed profile at different voltages.

Maximum Torque

0/ dsdT

2222

max

XXRS

Rthth

T

s

R

XXsRR

VT

thth

th

synmech

22

22

2

2

*/

*1

222

2

maxXXR

RS

thth

T

Then

222

2

max *2

1

XXRR

VT

ththth

th

syn

Torque speed characteristics for varying2R

.

222

2

max *2

1

XXRR

VT

ththth

th

syn

1RIf is small (hence thR is negligibly small)

222

2

maxXXR

RS

thth

T

222

2

max *2

1

XXRR

VT

ththth

th

syn

2

2

max XX

Rs

th

T

2

2

max *2

1

XX

VT

th

th

syn

Then

Then

maxmax

*/

/2

2

2

2

2

2

2

2max

TthTth

thth

s

s

XXsRR

XXsRR

T

T

1RIf is small (hence thR is negligibly small)

maxmax

*/

/2

2

2

2

2

2

2

2max

TthT

th

s

s

XXsR

XXsR

T

T

maxmax

max */2

//2

2

2

2

2

2max

TT

T

s

s

sR

sRsR

T

T

ss

ss

T

T

T

T

**2max

max

22

max

Find Tst/Tmax

Effect of Rotor Resistace

ST max R2' Rext

Rotor I2R loss dissipated in external

resistance

Rotor heating lower during starting and

acceleration period

222

2

maxXXR

RS

thth

T

3

=163.11 N.m

%5.87100*4.32022

3.28022motor

(c) (i)

(c) (ii)

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 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

o

ostI

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 o

oFL83.1056.53

83.10959.4

6.2651

Then the power factor is: 9822.083.10cos o lag.

.sec/08.1572*60

1500radsys

Vj

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 68.228

2.02.04.5198.0*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

4033.0

2.02.0198.0

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

22

max

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:

0.12 j0.25

j10

j0.25

205.0

1.0

ee XRjZ 25.012.01

o

j

jjjZ 55.231314.2

25.102

25.02*1025.012.01

1.120

3

2081 V V

o

oI 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

Tag

.6682.925.188

9485.17467

5.188

NmP

Tshaft

shaft 9127.855.188

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:VV

XX

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

0.06589 j0.1947 j0.2

s

05.0

257.7V

(b)

NmTst 7.624

2.01947.005.006589.025.94

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)

22

2121

2

maxR

XXR

RsT

or 4.005.0*1249.0

1*

122

max

Rs

sR

T

start

start

Then 243.02.1/05.04.0 2 extR

Starting of Squirrel Cage Motor

Draws a large current at start - Low Torque and Power Factor

∼5 to 8 time the Full Load Current

Large Current leads to overheating of motor

Reduced Voltage Starting

Auto-Transformer

R: Running contact

S: Starting contacts

Start: S are closed & R is open

Running: S are open & R is closed

Starting of Squirrel Cage Motor

Star-Delta 1: Star 2: Delta

Running - Delta

Starting - Star

Solid State Voltage Controller

Stator Windings

Control Speed

Reduce Voltage ⇒ Reduce Torque