Page 73 - Solutions Manual to accompany Electric Machinery Fundamentals

P    25 W 
  cos  1 SC   cos    1     76.4
 V SC I SC    10.0 V  10.6 A 

Z EQ  R EQ  jX EQ  0.943 76.4   0.222  j  0.917
The resulting per-unit impedances are

0.222  0.917 
R   0.00963 pu X   0.0398 pu
23.04  23.04 
EQ EQ
The per-unit equivalent circuit is
I P I S
jX EQ
R EQ

+ 0.00963 j0.0398 +

V P R C jX M V S

263 j53.1

- -

At rated conditions and unity power factor, the output power to this transformer would be P IN = 1.0 pu.
The input voltage would be
V  V I Z
P S S EQ
1 0 V 
V P   1 0  0.00963 j 0.0398   1.01 2.23 pu  

The core losses (in resistor R C ) would be

2  V 1.01 2
P core    0.00388 pu
R C 263

The copper losses (in resistor R EQ ) would be

P  CU I R EQ   1.0 2  0.00963   0.00963 pu
2
The input power of the transformer would be


P  IN P OUT  P  CU P core  1.0 0.00963 0.00388 1.0135

and the transformer efficiency would be
P 1.0

  OUT  100%   100% 98.7%
P IN 1.0135
The voltage regulation of the transformer is

1.01 1.00

VR   100% 1.0%
1.00
67