Page 44 - Solutions Manual to accompany Electric Machinery Fundamentals

116.2-115

VR   100% 1.04%
115
(3) 0.8 PF Leading:
  8.7 36 87 A



P V S  V Z EQ S  115 0 V I  j 0.126 0.476    .



P  V 113.5 2.0 V
113.5-115
VR   100%   1.3%
115
(c) At rated conditions and 0.8 PF lagging, the output power of this transformer is

P OUT  V I  cos   115 V 8.7 A 0.8     800 W
SS
The copper and core losses of this transformer are
P  CU I S 2 EQ,S   R 8.7 A 2   0.126   9.5 W
  2 2

V
P core    P 118.4 V  4.1 W
R C 3383 

Therefore the efficiency of this transformer at these conditions is
P 800 W
  OUT  100%   98.3%


P OUT  P  CU P core 800 W 9.5 W 4.1 W
2-7. A 30-kVA 8000/230-V distribution transformer has an impedance referred to the primary of 20 + j100 .
The components of the excitation branch referred to the primary side are R  C 100 k and

X M  20 k .
(a) If the primary voltage is 7967 V and the load impedance is Z = 2.0 + j0.7 , what is the secondary
L
voltage of the transformer? What is the voltage regulation of the transformer?

(b) If the load is disconnected and a capacitor of –j3.0  is connected in its place, what is the secondary
voltage of the transformer? What is its voltage regulation under these conditions?

SOLUTION

(a) The easiest way to solve this problem is to refer all components to the primary side of the
transformer. The turns ratio is a = 8000/230 = 34.78. Thus the load impedance referred to the primary
side is
 2 
L Z  34.78 2.0  j 0.7   2419  j 847 
The referred secondary current is



 7967 0 V 7967 0 V

I    3.045   21.2 A
S  100  20    j 2419  j 847   2616 21.2  
and the referred secondary voltage is
  

 A 2419 
V S  I S Z   21.2 3.045  j 847   7804 1.9 V
L
The actual secondary voltage is thus

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