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322

ELECTRICAL ENGINEERING

 

Example 8.13. A 3-phase 400 V, 50 Hz induction motor takes a starting current which is

4 times full load current at rated voltage. It's full load slip is 4%. Determine the auto-trans­ former ratio which will have starting current not more than 21h times the full load current at starting. Also determine the starting torque under this condition.

 

Solution. Suppose the ratio

 

transformation or tap ratio for auto-transformeris then

where is the short circuit current. Dividing by full load current

 

we have

 

 

 

 

of

 

 

 

 

 

 

 

 

 

 

 

Ir,

 

x,

 

 

Ifl =

 

 

Ifl

 

 

or

 

 

 

 

2.5

=

 

 

x 4

 

= 0.79

 

 

i.e. the tap setting required is 79%

 

=x2X

-

 

 

 

 

 

 

 

 

Isc

Ist

 

 

Isc

 

 

 

 

 

 

 

 

 

 

Now

TJ.t

 

 

XIJc

 

2

 

sfl

 

 

...(8.24)

 

 

 

st

 

 

2

 

fl

 

 

 

 

 

 

 

 

 

1(1

 

(

 

 

 

:2s

)

 

.:2

 

 

 

 

or

 

-

 

 

 

 

 

 

 

 

 

T81

==

0.x279

 

x 4

x 0.04x

 

 

 

 

 

= 0.399 Tfl

 

40% offull load torque.

 

 

 

:::::0.4

Tft

 

or

 

 

 

 

 

 

 

 

Tfl

 

 

 

 

Example 8.14. Determine the starting torque of a 3-phase induction motor in terms of

 

 

 

 

 

 

 

Isc

 

 

 

 

 

 

 

 

full load torque when started by (i) star-delta starter (ii) an auto-transformer starter with 70. 7% tapping. The starting current of motor is 5 times the full load current at rated voltage and the full load slip is 4%.

 

Solution. Starting torque with

star-delta starter

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 3

(

 

 

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

_!

(5)

2

x

0.04

= 0.33

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

I z

 

 

r

 

z

 

 

 

 

 

 

 

 

 

or

33.3% offull load torque.

 

 

 

 

 

 

 

 

 

r

 

 

 

 

 

 

 

voltageTfl

 

 

 

 

 

 

(ii) Since the voltage applied is 70.7% of the normalTfl

and torque is

 

V2

and also

 

starting current will be

0. 707

_of!

 

 

st

current2

 

 

 

 

 

 

 

 

 

 

 

 

 

theI

when normal voltage is applied and hence

 

again usingequation (8.24) we have

 

 

 

 

 

I

 

 

 

 

 

oc

 

 

 

 

 

 

 

X 1 1

 

2

 

 

 

T.

 

 

 

0.707

 

 

 

 

T.

 

 

 

 

T =

 

(

 

8

)

 

 

 

 

 

 

= (

 

 

st )2

 

O04

 

 

 

 

 

 

 

 

Sfl

f

 

 

 

X

fl

 

 

 

 

= "21 xfl5

 

 

 

 

 

= 0.5Jfl

 

 

·

 

 

 

 

 

x 0.04 l

 

 

 

 

 

 

 

 

The starting torque is 50% offull2load torque.

 

 

 

 

 

 

 

 

 

 

 

Example 8.15.

 

 

 

 

 

 

 

 

 

 

 

 

 

induction motor develops 4000 watts at 1 440

 

 

A 3-phase, 4 pole, 50 HzT

 

 

Tfl

 

 

 

 

 

 

 

 

rpm. Determine the stator inputst

if the stator lossflis 320 watts.

 

 

 

 

 

 

 

 

 

 

 

 

J

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution. The synchronous speed ofthe induction motor is

n

s

=

120f

p

=

120 x

4

50

=

1500

r.p.m.

= 1000 r.p.m.

THREE-PHASE INDUCTION MOTORS

 

323

and hence the slip is

 

 

s =

1500 - 1440

= 0.04

 

1500

 

Sincethe powerdevelopedis 4000 watts and the slip is 0.04, the powerinput tothe rotoris

4000

4000

 

1 - s =

0.96 = 4167 watts

-- --

 

Also the statorpower loss is givento be 320 watts hence, input to the stator is 4167 + 320 = 4487 watts. Ans.

Example 8.16. A 6-pole 3-phase 50 Hz motor develops mechanicalpower of40 hp and has mechanical loss of 1500 watts at a speed of 960 rpm when connected to 500 volts. The power factor is 0. 8 lag. Determine (i) rotor copper loss (ii) total input to stator if stator loss is 1800 watts (iii) the line current (iv) efficiency.

Solution. The synchronous speed of the motor for 6 pole and 50 Hz supply is

120 x 50

6

Hence the s1ip. is. 1000 - 960 = 0.04 1000

Mechanical power developed= 40 x 735.5 = 29420 Watt.

Power input to rotor

=

29420

=

29420

= 30645.8 w

1- s

0.96

(i)Now since powerinput to rotor is 122R218, the rotor copper loss is s . 122R2/s = 122R2

=0.04 x 30645.8 = 1226 Watts.

(ii)Power input to stator = Power input to rotor + stator loss

=30646 + 1800 = 32446 Watts.

(iii)Hence the line current

J3 32446 = 46_8 A x 500 x 0.8

(iv)The output available at the shaft

=Mechanical power developed - Mechanical loss 294201500 = 27920 Watts

Hence

% 11 =

27920

x 100 = 86% Ans.

32446

Example 8.17. A 3-phase 50 Hz 400 V, 25 hp, 4 pole induction motor has the following impedances referred to stator.

R1 = 0. 5 Q/phase, R2 = 0.35 Q, X1 = X2 = 1.2 Q, Xm = 25 Q.

The combined rotational losses (mechanical and core losses) amount to 800 W and are assumed to remain constant. For a rotor slip of 2.5% at rated voltage and rated frequency, determine (i) the motor speed (ii) the stator current (iii) the p.f and (iv) the efficiency.