Page 48 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging,
find the voltage regulation of the transformer. Find its efficiency.
SOLUTION

(a) The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to
directly find the components of the excitation branch relative to the low-voltage side.

21.1 A
Y  G  jB   0.001529
C
EX
M
13.8 kV
P 90.8 kW
  cos  1 OC  cos  1  71.83

VI  13.8 kV 21.1 A 
OC OC
Y EX  G  C jB  M 0.001529   71.83 S   0.0004456  j 0.0013577 S
1
R   2244 
C
G C
1
X   737 
M
B M
The base impedance of this transformer referred to the secondary side is
2 V 13.8 kV 2
Z base  S base  5000 kVA  38.09 
base

so R EQ   0.01 38.09     0.38  and X EQ  0.05 38.09  1.9  . The resulting equivalent circuit
is shown below:

R EQ, s  38  X EQ, s   9 . 1 j
.
0
R Cs  2244  X Ms  737 
,
,
(b) If the load on the secondary side of the transformer is 4000 kW at 0.8 PF lagging and the secondary
voltage is 13.8 kV, the secondary current is

P 4000 kW
I  LOAD   362.3 A

S
V S  PF 13.8 kV 0.8 

I  362.3   36.87 A
S
The voltage on the primary side of the transformer (referred to the secondary side) is

P V S  V I S Z EQ

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