Page 46 - Solutions Manual to accompany Electric Machinery Fundamentals

The equivalent circuit is

R EQ,P  0.018  X EQ,P  j 0.075 
R  not specified X M  120 
C
(b) If the load on the secondary side of the transformer is 150 MVA at 0.8 PF lagging, and the referred
secondary voltage is 15 kV, then the referred secondary current is

I   P LOAD  150 MVA  12,500 A

S
V S  PF 15 kV 0.8 


I  12,500  36.87 A
S
The voltage on the primary side of the transformer is
 
 V  V I Z
P S S EQ, P

 15,000 0 V     V  36.87 A 12,500 0.018  j 0.075   15755 2.23 V  
P
Therefore the voltage regulation of the transformer is
15,755-15,000
VR   100%  5.03%
15,000
(c) This problem is repetitive in nature, and is ideally suited for MATLAB. A program to calculate the
secondary voltage of the transformer as a function of load is shown below:

% M-file: prob2_8.m
% M-file to calculate and plot the secondary voltage
% of a transformer as a function of load for power
% factors of 0.8 lagging, 1.0, and 0.8 leading.
% These calculations are done using an equivalent
% circuit referred to the primary side.

% Define values for this transformer
VP = 15000; % Primary voltage (V)
amps = 0:125:12500; % Current values (A)
Req = 0.018; % Equivalent R (ohms)
Xeq = 0.075; % Equivalent X (ohms)

% Calculate the current values for the three
% power factors. The first row of I contains
% the lagging currents, the second row contains
% the unity currents, and the third row contains
% the leading currents.
I = zeros(3,length(amps));

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