Page 32 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) The rated kVA of the transformer is 100 kVA, and the rated voltage on the secondary side is 277 V,
so the rated current in the secondary side is 100 kVA/277 V = 361 A. Therefore, the base impedance on
the primary side is

V 277 V

Z  base   0.767
base
I base 361 A
Since Z pu  Z actual / Z base , the resulting per-unit equivalent circuit is as shown below:

0.0078 j0.0094 0.0065 j0.0078

78.2 j15.6

(c) To simplify the calculations, use the simplified equivalent circuit referred to the secondary side of the
transformer:

0.0005 j0.0132
0.011

60  j12 

The secondary current in this transformer is
100 kVA


 I   31.8 A 361   31.8 A
S
277 V
Therefore, the primary voltage on this transformer (referred to the secondary side) is
  
S
P V S  V EQ jX EQ  R I
 



P 277 0 V     V j 0.0132 0.011  31.8 A    361 283 0.4 V
The voltage regulation of the transformer under these conditions is
283- 277

VR   100% 2.2%
277
(d) Under the conditions of part (c), the transformer’s output power copper losses and core losses are:
P OUT  S cos   100 kVA 0.85  85 kW

P CU   S I 2 R EQ    361 2   0.11  1430 W

26