Page 45 - Solutions Manual to accompany Electric Machinery Fundamentals



V 7804 1.9 V

 V S   224.4  1.9 V
S
a 34.78
The voltage regulation is
7967-7804
VR   100% 2.09%

7804
(b) The easiest way to solve this problem is to refer all components to the primary side of the
transformer. The turns ratio is again a = 34.78. Thus the load impedance referred to the primary side is
 Z  2   3.0    3629 
L 34.78 j j
The referred secondary current is



 7967 0 V 7967 0 V


I S   100  20   j  j 3629     3529   89.7    2.258 89.7 A
and the referred secondary voltage is
   

S  S Z  I L  V  2.258 89.7 A  j 3629     8194  0.3 V
The actual secondary voltage is thus


V 8914 0.3 V

 V S   256.3  0.3 V
S a 34.78
The voltage regulation is

7967 8914
VR   100%   10.6%
8194
2-8. A 150-MVA 15/200-kV single-phase power transformer has a per-unit resistance of 1.2 percent and a per-
unit reactance of 5 percent (data taken from the transformer’s nameplate). The magnetizing impedance is
j80 per unit.
(a) Find the equivalent circuit referred to the low-voltage side of this transformer.
(b) Calculate the voltage regulation of this transformer for a full-load current at power factor of 0.8
lagging.
(c) Calculate the copper and core losses in transformer at the conditions in (b).

(d) Assume that the primary voltage of this transformer is a constant 15 kV, and plot the secondary
voltage as a function of load current for currents from no-load to full-load. Repeat this process for
power factors of 0.8 lagging, 1.0, and 0.8 leading.
(a) The base impedance of this transformer referred to the primary (low-voltage) side is

2  V 15 kV 2
Z base  base   1.5 
S base 150 MVA
so R EQ  0.012 1.5   0.018 

X EQ   0.05   0.075   1.5 
X M   1.5  80   120 

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