Page 69 - Solutions Manual to accompany Electric Machinery Fundamentals

The resulting equivalent circuit referred to the primary at 50 Hz is shown below:
I P I S
R EQ jX EQ

+ 328 ? j673 ? +

V P R C jX M V S

642 k? j178 k?

- -

(e) The base impedance of this transformer at 50 Hz referred to the primary side is

 V 2  16,667 V 2

Z  P   6.66 k
base,P
S 41.7 kVA
The base impedance of this transformer at 50 Hz referred to the secondary side is
 V 2 400 V 2

Z base,S    S  3.837
S 41.7 kVA
The excitation branch elements can be expressed in per-unit as
642 k 178 k
R   96.4 pu X   26.7 pu
C
6.66 k M 6.66 k
The series impedances can be expressed in per-unit as

328  673 
R   0.0492 pu X   0.101 pu
EQ
6.66 k EQ 6.66 k
The per-unit primary voltage at rated conditions and unity power factor is

V  V I Z
P S S EQ
1 0 V 
V P   1 0  0.0492  j 0.101   1.054 5.49 pu  
The per-unit power consumed by R is
EQ
P  EQ 2  I R  1 pu 2  0.0492 pu   0.0492 pu

The per-unit power consumed by R is
C
2  V 1.054 2
P  C P   0.0115 pu
R C 96.4

Therefore the efficiency of this transformer at rated load and unity power factor is
P P 1.00

  out  100%  out  100%   100% 94.3%


P in P  out P  EQ P C 1.00 0.0492 0.0115
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