Page 31 - Solutions Manual to accompany Electric Machinery Fundamentals

Chapter 2: Transformers

2-1. A 100-kVA 8000/277-V distribution transformer has the following resistances and reactances:
R  R  S 0.005 
5
P
X  P 6  X  S 0.006 

R  C  50 k X M  10 k

The excitation branch impedances are given referred to the high-voltage side of the transformer.
(a) Find the equivalent circuit of this transformer referred to the low-voltage side.

(b) Find the per-unit equivalent circuit of this transformer.
(c) Assume that this transformer is supplying rated load at 277 V and 0.85 PF lagging. What is this
transformer’s input voltage? What is its voltage regulation?

(d) What are the copper losses and core losses in this transformer under the conditions of part (c)?
(e) What is the transformer’s efficiency under the conditions of part (c)?

SOLUTION
(a) The turns ratio of this transformer is a = 8000/277 = 28.88. Therefore, the primary impedances
referred to the low voltage (secondary) side are

R   R P  5   0.006 
P
a 2  28.88 2
 R 6 
X  P   0.0072 
P
a 2  28.88 2
and the excitation branch elements referred to the secondary side are

R   R C  50 k  60 
C
a 2  28.88 2

 X 10 k
X  M   12 
M
a 2  28.88 2
The resulting equivalent circuit is

0.006  j0.0072  0.005  j0.006 

60  j12 

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