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# Basic_Electrical_Engineering_4th_edition

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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.