INDUCTION MOTORSASSIGNMENTS

1. A 3­ slip ring induction motor gives a reading of 55V across slip ringsΦ

on open circuit when at rest with normal stator voltage applied. The rotor is star connected and has impedance of (0.7+j5) /ph. Find the rotor current and p.f. whenΩ the machine is a) At standstill with the slip rings joined to a star­connected starter with a phase

impedance of (4+j3) .Ωb) Running normally with 5% slip.

2. A 4­pole 50Hz 3 induction motor has a rotor resistance of 0.024 /ph and standstillΦ Ω reactance of 0.6 /ph. Determine the speed at which the maximum torque isΩ developed.

3. A 6­pole, 3 , 50Hz induction motor runs on full load with a slip of 4%. Given theΦ rotor standstill impedance per phase as (0.01+j0.05) . Calculate the availableΩ maximum. Torque in terms of full load torque. Also determine the speed at which the maximum torque occurs.

4. The rotor resistance and standstill reactance per phase of a 3­ slip ring Φ

induction motor are 0.02 and 0.1 respectively. What should be the value of theΩ Ω external resistance per phase to be inserted in the rotor circuit to give maximum torque at starting?

5. A 4­pole, 50Hz, 3­ induction motor has rotor resistance and reactanceΦ

of 0.03 and 0.12 per phase respectively. What is the value of the speed atΩ Ω which maximum torque occurs? Find the amount of external rotor resistance per phase to be inserted to obtain 75% of maximum torque at starting.

6. The maximum torque of a 3­ induction motor is twice the full loadΦ

torque and starting torque is equal to full load torque. Calculate the full load speed and the slip at which maximum torque occurs.

7. An 8­pole alternator runs at 750 rpm and supplies power to a 6­poleinduction motor which has full load slip of 3%. Find the full load speed of the induction motor and the frequency of its rotor emf.

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8. A 500 hp (373 KW), 3­ , 440V , 50Hz induction motor has a speed of Φ 950 rpm on full load. The machine has 6­poles. Calculate the slip. How many complete alternations will the rotor voltage make per minute?

9. A 6­pole, 3­ , 50Hz induction motor runs on full load with a slip of 4%. Φ

Given that the rotor standstill impedance per phase as (0.01+j0.05) , CalculateΩ the available maximum torque in terms of full load torque. Also determine the speed at which the maximum torque occurs

10. The resistance and standstill reactance of each phase of a 3­ inductionΦ

motor with Y­connected rotor are 0.06 and 0.4 respectively. The full load slip isΩ Ω 4%. Calculate the resistance per phase of a Y­connected rheostat which when connected to the rotor, will give a pull out max. Torque at ½ of the full load speed. What is then the power factor?

11. A 3­ , 50Hz induction motor draws 50Kw from the mains. If the stator Φ

losses are 2KW and the rotor emf is observed to make 100 complete oscillations (alternations) per minute, determine

(i) rotor copper loss(ii) gross mechanical output

12. The power input to the rotor of a 3­ , 50Hz, 6­pole, slip ring inductionΦ

motor is 40KW and the motor runs at 960 rpm. The rotor resistance per phase is 0.25 . Determine the value of rotor current per phase.Ω

13. The power input to a 3­phase, 50Hz induction motor is 50KW. The total stator loss is 800W. Find the total mechanical power developed and the rotor copper losses per phase, if it is observed that the rotor e.m.f. makes 90 complete cycles per minute.

14. The rotor resistance and reactance per phase of a 4­pole, 50Hz, 3­phase induction motor are 0.025 and 0.12 respectively. Find the speed at Ω Ω

maximum torque. Also find the value of additional rotor resistance per phase required to give ¾ of maximum torque at starting. Neglect stator impedance.

15.

2

. Fig. (given above) shows the torque­speed characteristics of a 3­phase induction motor. Why does the motor operate at point ‘b’ rather than point ‘a’ even

though point ‘a’ is reached earlier when the machine is accelerating? TL is a constant load torque applied to the machine and Tst is the starting torque.

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SOLUTIONS:1.

a) At standstill Induced emf per phase in rotor winding

VE 8.313

55 ==

Resistance per phase in rotor circuit Ω=+= 7.447.02RReactance per phase in rotor circuit Ω=+= 8352XImpedance per phase in rotor circuit

( ) Ω=+=+= 28.987.4 2222

222 XRZ

Rotor current per phase AZEI 425.3

28.98.31

2

02 ===

Power factor )(506.028.97.4

2

22 lagging

ZRCos ===φ

b) Normal running at 5% slip

Induced emf in rotor winding per phase VSEE 59.18.3105.002 =×==

Rotor impedance per phase, ( ) ( ) Ω=×+=+= 7433.0505.07.0 222

022

22 XSRZ

Rotor current per phase AZEI 14.2

7433.059.1

2

22 ===

Power factor, )(942.07433.0

7.02

22 lagging

ZRCos ===φ

2.Rotor resistance per phase Ω= 024.02RRotor standstill reactance per phase Ω= 6.00XSince the torque under running condition is maximum at that value of the slip which makes rotor reactance per phase equal to the rotor resistance per phase

Therefore, slip corresponding to max. Torque 04.06.0

024.02

2max ===

XRS

Speed corresponding to maximum Torque

( ) ( ) ( ) rpmSP

fSNN S 144004.012

50601601 maxmax =−×=−=−=

3. Rotor resistance per phase R2=0.01ΩRotor standstill reactance per phase X0=0.05ΩFull load slip S=4%=0.04

4

(i) Ratio of max. Torque to full load torque

( )6.2

016.00416.0

04.005.001.02

04.005.001.0

2

22

0

2

22

0

2

==××

+

=

+

=

f

f

SXR

SXR

Therefore, maximum torque Tmax=2.6Tf

(ii) Slip corresponding to max. Torque, 2.005.001.0

0

2max ===

XRS

Speed corresponding to maximum torque,

rpmSNN S 800)2.01(3

5060)1( max =−×=−=

4. Let external resistance per phase added to the rotor circuit be of r ΩRotor circuit resistance per phase 2 (0.02 )R r= + ΩRotor circuit standstill reactance per phase Ω= 1.00x

Torque at starting will be max. if 22 0 0

0

1 (0.02 ) 0.1R S or R x or r xx

= = = + = =

0.1 0.02 0.08r = − = Ω

5. Slip corresponding to maximum torque, 25.012.003.0

0

2 ===XRSm

Synchronous speed rpmP

fNS 15002

506060 =×==

Speed corresponding to maximum torque( ) ( ) rpmSNN mS 112525.0115001 =−=−=

Let r be the external resistance introduced per phase in rotor circuit

Ratio of starting torque to maximum torque 122

max +=

aa

TTst

Where 12.003.0

0

2 rxRa +==

2 2max

2 3 2 1 4 1

stT a aorT a a

= =+ + Therefore, starting torque is 4

3 of Tmax

or 3a2­8a+3=0

or 4515.06

33488 2

=××−±=a reject higher value

5

Substituting a=0.4515 we get

4515.012.0

03.0 =+= ra

or r=0.05418­0.03=0.02418Ω

6. Tmax=2Tf

Tst=Tf

Ratio of starting torque to maximum torque

5.0max

=TTst , but 1

22

max +=

aa

TTst where

0

2

XRa =

=> 125.0 2 +

=a

a or a2­4a+1=0

322

444 2

−=−±=a (reject higher value)

Let full load slip be Sf

Since the ratio of full load torque to maximum torque

( )( ) 22 2 2max

2 2 32 12 2 3

ff f

f f

ST aSor

T a S S

−= =

+ − +

or ( ) ( ) ff SS 32432 22−=+−

or ( ) ( ) 03232422 =−+−− ff SS

or ( ) ( ) ( ) ( )222

322

3243216324−=

−−−±−=fS

6

=0.072 (rejecting higher value)

Full load speed Nf =NS (1­Sf) =N (1­0.072) =0.928NS

i.e. 0.928 times synchronous speed

Slip corresponding to maximum torque aXRSm ==

0

2

268.032max =−=S

7. HzNPf aa 50607504

60=×==

03.0%3 ==fS

rpmP

fNm

S 10003

506060 =×==

Full load speed N=NS(1­S)=1000(1­0.03)=970rpm

HzSffr 5.15003.0 =×==

8. Supply frequency, f=50Hz

rpmP

fNS 10003

506060 =×==

Actual speed N=950 rpm

%5........05.01000

9501000 orN

NNSS

S =−=−=

9. Rotor resistance per phase, R2=0.01ΩRotor standstill reactance per phase X2=0.05ΩFull load slip, Sf=4%=0.04Ratio of maximum torque to full load torque

7

( )6.2

016.00416.0

04.005.001.02

04.005.001.0

22

22

2

2

22

2

222

max ==××

+

=

+

=+

=

f

f

f

f

f SXR

SXR

aSSa

TT

Therefore, maximum torque Tmax=2.6Tf

(ii) Slip corresponding to max. Torque 2.005.001.0

2

2 ===XRSm

Speed corresponding to max. Torque

( ) ( ) rpmSNN mS 8002.013

50601 =−×=−=

10. Let the synchronous speed be NS

Full load slip Sf=0.04Full load speed Nf=NS(1­S)=NS(1­0.04)=0.96NS

½ of the full load speed SSf NNNN 48.096.021

21 =×==

Slip at ½ full load speed, 52.048.0 =−=S

SS

NNNS

Rotor resistance per phase R2=0.06ΩRotor reactance per phase at standstill X2=0.4ΩRotor reactance per phase at ½ full load speed =

SX2=0.52× 0.4=0.208ΩLet r be the resistance per phase of star­connected rheostat connected to the motor. Pull out torque or max. torque will be at ½ full load speed, if the resistance per phase in the rotor circuit is equal to reactance per phase of the rotor at ½ full load speed i.e. 0.06+r=0.208

r=0.148Ω

Power factor =

( ) ( ) ( ) ( )707.0

208.0148.006.0148.006.0

22222

2 =++

+=++

+=SXrR

rRCosφ

)(707.0 lagCos =φ

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11. Supply frequency f=50Hz

Rotor emf frequency, Hzfr 35

60100 ==

Slip 033.05035

===ffS r

Power input to rotor = input to stator­ stator losses = 50­2=48KW

(i) Rotor Cu. Loss = S× input to rotor=0.033× 48=1.6KW(ii) Gross Mechanical power output = rotor input (1­

S)=48(1­0.033)=46.4KW

12. Power input to rotor= 40KW or 40000W

rpmP

fNS 10003

506060 =×==

N=960rpm 04.01000

9601000 =−=−=S

S

NNNS

Total rotor Copper losses = S× power input to rotor = 0.04× 40000=16000W

Rotor Copper loss per phase = 16000

3W

Rotor Current per phase

2 16000 / 3 46.2 tan 0.25

Rotor Copper loss per phaseI ARotor resis ce per phase

= = =

13. frequency Hzf 5.16090

2 ==

Slip 03.050

5.11

2 ===ffS

Air gap power Pg=50000­800=49200W

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Therefore, total mechanical power developed= (1­S)Pg=(1­0.03)×49200=47724WTotal rotor copper loss= SPg=0.03× 49200=1476W

14. rpmN S 15004

50120 =×=

Slip at maximum torque 208.012.0025.0

2

2 ===xrSmt

Therefore, speed at maximum torque Tmax= (1­Smt) NS= (1­0.208)× 1500=1187.5rpm

Again when Tst is equal to ¾ of maximum Torque, then

43

11

2'

'max

=+

=mt

mt

st

SS

TT

OR 43

122

max

=+

=a

aTTst

0383 '2' =+− mtmtS

Where 0

2

xRa =

' 2.21 0.451mtS or=

And R2=0.025+rBut '

mtS cannot be >1, so 'mtS =0.451

then 3a2­8a+3=0 Let the additional rotor resistance a=2.21 or 0.451Per phase be r, then

12.0025.0451.0 r+=

10

12.0025.0025.0

0

' rx

rSmt+=+=

r=0.029Ω

15. point ‘a’ the position is unstable because if there is a tendency of speed rise, the developed torque becomes higher than the load torque causing further rise in speed which brings the motor to the position ‘b’. At ‘b’, the operation is stable because a tendency to rise in speed will be opposed by the fall in developed torque. Similarly, when there is a tendency to fall in speed there will be more developed torque to bring the motor to position ‘b’. Hence, the segment xy of the torque­speed characteristics is unstable region of operation while the segment yz is stable region of operation.

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