Page 52 - Solutions Manual to accompany Electric Machinery Fundamentals

(c) What is the transformer bank’s efficiency under these conditions?

SOLUTION (a) The equivalent of this three-phase transformer bank can be found just like the equivalent
circuit of a single-phase transformer if we work on a per-phase bases. The open-circuit test data on the
low-voltage side can be used to find the excitation branch impedances referred to the secondary side of
the transformer bank. Since the low-voltage side of the transformer is Y-connected, the per-phase open-
circuit measurements are:

V  ,OC  277 V I  ,OC  4.10 A P  ,OC  315 W
The excitation admittance is given by

I 4.10 A
Y EX   ,OC   0.01480 S
V  ,OC 277 V
The admittance angle is

 P   315 W 
   cos  1   ,OC   cos  1    73.9
V   ,OC I  ,OC   277 V     4.10 A 

Therefore,

Y EX  G  C jB  M 0.01483   73.9   0.00410  j 0.01422
R  1/G  244 
C C
X M  1/ B  M  70.3
The base impedance for a single transformer referred to the low-voltage side is
 V 2  277 V 2
Z   .base,S   3.836 
base,S
S  ,base 20 kVA
so the excitation branch elements can be expressed in per-unit as
244  70.3 
R   63.6 pu X   18.3 pu
3.836  3.836 
C M
The short-circuit test data taken in the high-voltage side can be used to find the series impedances referred
to the high-voltage side. Note that the high-voltage is -connected, so



V  ,SC  V L ,SC  1400 V , I  ,SC  I L ,SC / 3 1.039 A , and P  ,SC  P SC /3 304 W .
V 1400 V
Z   ,SC   1347 
EQ
I  ,SC 1.039 A
 P   304 W 
  cos   1  ,SC   cos   1   77.9
V  ,SC I  ,SC      1  400 V 1.039 A   


Z EQ ,P  R EQ ,P  jX EQ ,P  1347 77.9   282 j  1317
The base impedance referred to the high-voltage side is
 V 2 24,000 V 2
Z     .base,S  23,040 
base,P
S  25 kVA
The resulting per-unit impedances are

46