Page 51 - Solutions Manual to accompany Electric Machinery Fundamentals

2.00  j 7.98 

13.3 k j 2.66 k

  R 0.015 133     2.00 
EQ, S

X   0.06 13   7.98 3 
EQ, S

R  100 133   13.3 k
C

X   20 133  2.66 k   
M
(d) The per-unit losses in the series resistance are
P  EQ I R EQ   0.8 2   0.015  0.0096 pu
2
and the actual losses in the series resistance are:

P  EQ S base EQ ,pu  0.0096 pu 100 MVA  0.96 MW
P
The per-unit losses in the excitation branch are:
2  V 1.039 2
P EX ,pu    0.0108 pu
R EX 100
and the actual losses in the excitation branch are:

P  EX base EX ,pu   S 0.0108 pu  100 MVA  1.08 MW
P
The per-unit power supplied to the load
P 80 MW
P  load.   0.80 pu
load.pu
S base 100 MVA
Therefore, the transformer’s efficiency is
P 0.80

  OUT  100%   100% 97.5%


P IN 0.80 0.0096 0.0108
2-12. Three 20-kVA 24,000/277-V distribution transformers are connected in -Y. The open-circuit test was
performed on the low-voltage side of this transformer bank, and the following data were recorded:
V line,OC  480 V I line,OC  . 4 10 A P 3,OC  945 W

The short-circuit test was performed on the high-voltage side of this transformer bank, and the following
data were recorded:
V line, SC  1400 V I line, SC  . 1 80 A P 3,SC  912 W

(a) Find the per-unit equivalent circuit of this transformer bank.
(b) Find the voltage regulation of this transformer bank at the rated load and 0.90 PF lagging.
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