DEPARTMENT OF ELECTRICAL ENGINEERING, INDIAN INSTITUTE OF TECHNOLOGY DELHI

EEL203: ELECTROMECHANICS

SOLUTIONS TUTORIAL TEST-III

Q.1. (a) A 415V, 25 hp, 50 Hz, 4-pole induction motor has its speed of 1460 rpm at rated load. (i) What is the slip

of the motor? (ii) What is the frequency of the rotor current? (iii) What is the angular velocity of the stator

produced air gap flux wave with respect to the stator? (iv) What is the angular velocity of the rotor produced

air gap flux wave with respect to the rotor? [20]

(b)A 15 hp, 6-pole, 415V, 50 Hz, delta connected, 3-phase induction motor has a full load slip of 4%. Its

rotational loss (friction and windage losses) is 2% of its output power. When this motor operates at full load,

calculate (i) the rotor copper loss, (ii) the air-gap power, and (iii) the shaft torque. [30]

Solution:

(A)

(i) Synchronous speed is 1500 r/min. Therefore,

a. s =1500 − 1460

1500

b. = 0.026 = 2.6%

(ii) Rotor currents are at slip frequency, fr = s*50 = 1.33 Hz.

(iii)The stator flux wave rotates at synchronous speed with respect to the stator (1500 r/min). It rotates

at slip speed ahead of the rotor (s*1500 = 40r/min).

(iv) It rotates at slip speed ahead of the rotor (s*1500 = 40 r/min).

(B)

Output power = 15hp = 15*750 = 11.25kW.

Friction and windage losses = 0.02*output power = 0.02*11.25 = 0.225kW.

Power developed = output power+friction and windage losses = 11.475kW.

(i) Air gap power = Pd/(1-s) = 11.95kW.

(ii) Rotor copper losses = Pg-Pd=0.478kW.

(iii) Rotor speed = Ns(1-s) = 1500*(1-0.04) = 1440rpm.

output power*60 11.25k*60shaft torque = 74.6

2* * 2* *1440r

NmN

Q.2. The following tests data are achieved on a 230V, 60 Hz, 4-pole, star connected, three-phase squirrel cage

induction motor. No load test data are: power input=115W, line current=0.45A at the rated voltage. Blocked

rotor test data are: power input=65W, line current=1.2A at reduced voltage of 47V. The stator winding

resistance between any two lines is 4.1 ohms. Calculate all six equivalent circuit parameters and draw

equivalent circuit of it. [50]

Solution:

115No load power factor,cos = 0.6415

3 * * 3 *230*0.45

230Rm = 460

3 * *cos 3 *0.45*0.6415

230Xm = 384.73

3 * *sin 3 *0.45*0.767

NoNo

No No

No

No No

No

No No

P

V I

Vohms

I

Vohms

I

SC 2 2

1

2

2 2 2 2

SC SC

1 2

From short circuit test:

65R = 15.05

3* 3*1.2

4.1R 2.05

2

R 15.05 2.05 13

47Z 22.612

3 * 3 *1.2

X Z -R 22.612 -15.05 16.88

X8.44

2

SC

SC

SCSC

SC

SC

SC

Pohms

I

ohms

ohms

Vohms

I

ohms

X X ohms

Q.3. A delta-connected, 5hp, 400V, three-phase, 4-pole, 50-Hz squirrel-cage induction motor has the following

equivalent-circuit parameters per phase referred to the stator: R1 = 5.39Ω, R2 = 5.72 Ω, X1 = 8.22Ω, X2 =

8.22Ω. Neglect the shunt branch in the equivalent circuit. (a) Calculate the starting current and starting torque

for this motor when connected to a 400V, 50 Hz, 3-phase ac source. (b) Also calculate the starting current,

starting torque and power factor for this motor when connected to a 200V, 25 Hz, 3-phase ac source. [50]

Solution:

st2 2 2 2

1 2 1 2

2 2

st 2

(A) Delta connected at rated V and f

3 * 3 *400Starting line current, I = 34.92

( ) ( ) (5.39 5.72) (8.22 8.22)

I * 34.92 *5.72Starting Torque = 44.04

157.08

(B) Delta connected

Line

s

VA

R R X X

RNm

st2 2 2 2

1 2 1 2

2 2

st 2

1

at half the rated V and f

3 * 3 *200Starting line current, I = 25.065

( ) (0.5( )) (5.39 5.72) (0.5(8.22 8.22))

I * 25.065 *5.72Starting Torque = 45.76

78.54

Power factor =

Line

s

VA

R R X X

RNm

R 2

2 2 2 2

1 2 1 2

5.39 5.720.804

( ) (0.5( )) (5.39 5.72) (0.5(8.22 8.22))

R

R R X X

Q.4. A three-phase squirrel-cage induction motor develops a maximum torque of 210 percent and a starting torque

of 165 percent of the rated torque when operated at the rated voltage and frequency. If the effect of stator

resistance is neglected, determine (a) the slip at full load, (b) the slip at the maximum torque, and (c) the rotor

current at the starting in per unit of the full load rotor current. [50]

Solution:

Solution I:

maxmax

max

2fl

fl T

T fl

T

s sT

s s

max max

max2 2

max max max

maxmax

max

2 21.650.485

1 2.1 1

2Also,

1 20.122

0.4852.1

0.485

st T T

T

T T

fl

fl T

T fl

fl

fl

fl

T s ss

T s s

T

s sT

s s

ss

s

3.975 4.099

3.693 369.3%0.485 1.11

j

j

Q.5. The starting current of a 415V, three phase, delta connected, induction motor is 60 A when rated voltage is

applied to stator winding and developed starting torque is 120% of the rated torque. Determine the starting

current and the developed starting torque in terms of the rated torque when the applied voltage is reduced to

240 V. If the starting is not to exceed 25 A, what should be the applied voltage and developed starting torque

in terms of the rated torque? [50]

Solution:

At starting since s=1

so, and

i.e. and .

For applied voltage 240 V,

= 34.69 A and , i.e., starting torque is

40.1% the rated torque.

For new starting current 25 A, applied voltage is

= 172.91 V and , i.e, starting torque

is 20.8% of the rated torque.

Q.6. The inner and outer cages of a 3-phase double cage induction motor have standstill impedances of (0.5+j2)

ohm/phase and (2+j0.5) ohm/phase. Find the ratio of the torques developed and currents drawn by the two

cages (a) at starting and (b) at a slip of 0.05. [50]

Solution:

for constant slip,

so, ratio of torque developed by the outer & inner cages is

a. at standstill, s=1

so,

b. at s=0.05

so,

Q.7. A 120V, 60Hz, 4-pole, single-phase, capacitor start induction motor is tested to yield the following data: With

auxiliary winding open: No-load test: 120V, 2.7A, 56W. Blocked rotor test with auxiliary winding open: 120V,

15A, 1175W. Blocked rotor test with main winding open: 120V, 5.2A, 503.4W.The main winding resistance is

2.5 Ω and the auxiliary winding resistance is 12.5 Ω. (a) Obtain the parameters of the equivalent circuit of the

motor referred to the main winding. (b) Obtain the parameters of equivalent circuit of the motor referred to the

auxiliary winding. (c) Determine the no-load rotational losses and turns ratio between the auxiliary to the main

windings. [50]

Solution:

bm bm 2

2 2

rom the blocked rotor test on the main winding with the auxiliary

winding open, we obtain

120 1175Z 8 ; R 5.22

15 15

8 5.22 6.06bm

F

X

1 2 2

2

, 3.03 and R =5.22-2.5=2.72 (all referred to main winding)

the blocked rotor test on the auxiliary winding with the main winding

open, we get

120 503.4Z = 23.07 ; R 18.61

5.2 5.2

2

ba ba

ba

Thus X X

From

X 2 2

1 2 2a

3.07 18.61 13.63

, 6.817 and R =18.61-12.5=6.11 (all referred to auxiliary winding)

rom no-load test data on the main winding with auxiliary winding open,

we have

120 56Z 44.44 ; R

2.7nl nl

Thus X X

F

2

2 2

2

r

2

2

7.682.7

44.44 7.68 43.77

, 2 43.77 1.5 6.06 78.45

P 56 2.7 2.5 0.25 2.72 32.81

the a-ratio, a = 1.49

nl

a

m

X

Hence Xm

and W

Rand

R

Q.8. A single-phase, 230V, 50 Hz, 4-pole, split-phase induction motor has the following standstill impedances.

Main winding: Zm=(10+j12) Ω. Auxiliary winding Za=(16+j10) Ω. (a) Determine the value of the capacitance

to be added in series with the auxiliary winding to obtain maximum starting torque. (b) Compare the starting

torques and starting currents with and without added capacitance in the auxiliary winding circuit when

operated from a 230V, 50 Hz supply. [50]

Solution:

a.

1 1 0

2

1012tan tan 90

10 16

23.33 136.41

main aux

c

c

Z Z

X

X C F

b.

aux without C aux with C

14.72 50

12.18 32 ; 11.04 39.79

sin

sin

14.72 12.18 sin180.3409

14.72 11.04 sin 90

2.933

o

m

main

o o

aux aux c

m auxwithoutc withoutc

withc m aux withc

withc

VI A

Z

V VI A I A

Z Z jX

I IT

T I I

T timesofT

aux without C

aux with C

1.44

0.694

withoutc

mwithoutc

withc m

withc withoutc

I II

I I I

I timesofI

Q.9. A 220V, 50Hz, 0.50 h.p., 4-pole, universal motor runs at 3000 rpm and takes 2.5A when connected to a 220V

dc source. Determine the back emf, the speed, the torque and the power factor of the motor when it is loaded to

take 2.5 A (rms) of the supply current at ac mains of 220 V at 50 Hz. The resistance and its inductance

measured at the terminals of the machine are 10 Ω and 0.275H respectively. [50]

Solution:

22

22

:

| 220 2.5*10 195

:

2 2 *50*0.275 86.393

|

| 2.5*10 220 2.5*86.393

16.84

a dc

a ac a a

a ac

DC operation

E V

AC operation

X fL

E I R V I X

E

V

min (2.5 )

|

|

16.843000* 259.25

195

a dc dc dc

a ac ac ac

ac

Assu g the same flux for the samecurrent Adc and rms

E rpm n

E rpm n

n rpm

, cos

16.84 2.5*100.19

220

16.84*2.5 42.1

42.11.55

259.25*2 / 60

a a a

mech a a

mech

m

E I Rpower factor

V

lag

Mechanical power developed is P E I W

Torquedeveloped is

PT N m

Q.10. An 110V, 60Hz, 2-pole, universal motor operates at a speed of 12500 rpm on full load and draws a current of

5A. The motor parameters are: Ra=5.2Ω, Xa=12.563Ω, Rs=0.5Ω and Xs=2.235 Ω. Determine (a) the induced

emf in the armature, (b) the power output, (c) the shaft torque, (d) the power factor, and (e) efficiency if the

rotational loss is 20W. [50]

Solution:

22

b a rot

( ) *( )

52.89

( ),cos

52.89 5*(5.2 0.5)0.739

110

developed =E *I 264.45 P =244.45W

Shaft Torque = 0.186

cos 406.45

b a a s a a s

b

b a a s

o d

o

m

in a

E I R R V I X X

E V

E I R Rpower factor

V

lag

Power W P P

PNm

P VI W efficien244.45

= *100 60.14%406.45

cy

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