Page 135 - Solutions Manual to accompany Electric Machinery Fundamentals

This internal generated voltage corresponds to a terminal voltage of 3 1370  2371 V . This voltage
would require a field current of 4.54 A.

(b) The motor’s efficiency at full load and unity power factor is
P 746 kW
  OUT  100%   100% 89.7%

P 832 kW
IN
(c) To solve this problem, we will temporarily ignore the effects of the armature resistance R A . If R A
is ignored, then E A sin  is directly proportional to the power supplied by the motor. Since the power
supplied by the motor does not change when I F is changed, this quantity will be a constant.

If the field current is increased by 5%, then the new field current will be 4.77 A, and the new value
of the open-circuit terminal voltage will be 2450 V. The new value of E A will be 2435 V / 3 = 1406
V. Therefore, the new torque angle  will be
 E  1 1370 V 

 sin  1  A 1 sin  1   sin   sin    22.44    23.9
2
E   1406 V 
A 2 
Therefore, the new armature current will be



V  E 1328 0 V 1406  23.9 V


I A   A   227 2.6 A
R  A jX S 0.3 j 2.5 
The new current is about the same as before, but the phase angle has become positive. The new power
factor is cos 2.6 = 0.999 leading, and the reactive power supplied by the motor is
Q  3 V T sin   I L  3 2300 V 227 A sin 2.6     41.0 kVAR
(d) The maximum torque possible at unity power factor (ignoring the effects of R A ) is:

3VE  3 1328 V 1370 V 
 ind,max    X A   1 min   2 rad    5790 N m


m
S
 3600 r/min  60 s     1 r    2.5 
If we are ignoring the resistance of the motor, then the input power would be 7968 kW (note that copper
losses are ignored!). At a power factor of 0.8 leading, the current flow will be

P 796 kW
I  A I  L   250 A

3 V T  PF  3 2300 V 0.8 


so I A  250 36.87 A . The internal generated voltage at 0.8 PF leading (ignoring copper losses) is
E  V R I  jX I
A  A A S A



 E 1328 0  j  V  2.5   250 36.87 A 
A

 E 1775  16.4 V
A
Therefore, the maximum torque at a power factor of 0.8 leading is
3VE  3 1328 V 1775 V 

 ind,max   A   7503 N m

m X S  3600 r/min   1 min     2 rad     2.5 
 60 s   1 r 
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