Page 146 - Solutions Manual to accompany Electric Machinery Fundamentals

P 74.6 kW
I  motor   112 A

A
3V  cos   3 277 V 0.8 


Since the power factor is 0.8 leading, the current Ι A  112 36.87 A , and the internal generated voltage
is
E  V R I  jX I
A  A A S A
 E   277 0 V    0.023   112 36.87 A    j 2.51  112 36.87 A  
A
 E 498  27.0 pu
A
The field current must corresponding to 498 V is 1.86 A.

(c) The torque angle of the motor is    27
(d) The real, reactive, and apparent power supplied to the plant is


P TOT  P  1 P  2 P motor  100 kW 64 kW+74.6 kW  238.6 kW


Q TOT  Q  1 Q  2 Q motor  48.4 kvar 48 kvar 56 kvar  40.4 kvar
S TOT  P TOT 2  TOT 2   Q 238.6 kW 2   40.4 kVA 2  242 kVA

The line current is given by
S 242 kVA
I  L TOT   291 A
3 L  V  3 480 V
(e) If the field current is increased to 2 A, the magnitude of the internal generated voltage will rise from
498 V to 517 V (see the OCC curve). The power supplied to the load will remain constant (because the
load torque and rotational speed are unchanged), so the distance E A sin  P will remain constant.
Therefore the torque angle becomes
 E  1 498 V 

 sin  1  A 1 sin  1   sin    sin   27    25.9
2
E A 2    517 V 
The new armature current is




V  E 277 0 V 517  25.9 V


I   A 2   117 39.8 A
A
2
jX S j 2.51 

The current angle is 39.8, so the impedance angle    39.8 . The real power supplied by the motor is

P motor  3V I  cos   3 277 V 117 A cos  39.8   74.6 kW


Q motor  3V I  cos   3 277 V 117 A sin  39.8     62.2 kvar


(f) The torque angle of the motor is    39.8
(g) The power factor of the motor is cos  39.8   0.768 leading .
(h) The real, reactive, and apparent power supplied to the plant is

P TOT  P  1 P  2 P motor  100 kW 64 kW+74.6 kW  238.6 kW

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