Page 141 - Solutions Manual to accompany Electric Machinery Fundamentals

S 2798 kVA
I  base   392 A
L ,base
3V L ,base   3 4120 V
I 392 A
I  ,base  L ,base   226 A
3 3

Therefore, the line current of 300 A in per-unit is
I 300 A
I  L   0.765 pu
,pu
L
I 392 A
L ,base
and the final per-unit current is
I  pu 0.765 31.8 pu


The internal generated voltage in per-unit is
E  V R I  jX I
A  A A S A

1
E A   0  0.1 0.765 31.8    j  0.765 31.8  
1.1
 E 1572  28.7 pu
A
In volts, the internal generated voltage is

 E    1.572  28.7 pu 4120 V  6477   28.7 V
A
And the torque angle δ is -28.7.
5-9. Figure P5-2 shows a synchronous motor phasor diagram for a motor operating at a leading power factor
with no R A . For this motor, the torque angle is given by

XI cos 
tan   SA
V  X I sin 

SA
 XI cos  
  tan  -1 SA 
 V  sin  
  X I 
SA
Derive an equation for the torque angle of the synchronous motor if the armature resistance is included.
SOLUTION The phasor diagram with the armature resistance considered is shown below.

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