Page 55 - Solutions Manual to accompany Electric Machinery Fundamentals

V  3 V    3 8856 V  15.34 kV
LL,P  ,P
(b) The voltage regulation of the transformer is

8856-8314
VR   100% 6.52%

8314

Note: It is much easier to solve problems of this sort in the per-unit
system. For example, compare this solution to the simpler solution of
Problem 2-9.

(c) The base values of this transformer bank on the primary side are

S base  300 KVA
V LL ,base  V  ,base  14.4 kVA
S 300 KVA
I L ,base  base   12.03 A
3V LL ,base  kV  3 14.4
I 12.37 A
I  ,base  L ,base   7.14 A
3 3

This sort of repetitive operation is best performed with MATLAB. Note that in this case, the problem is
specifying a fixed primary phase voltage of 8314 V, and asking what the secondary voltage will be as a
function of load. Therefore, we must subtract the voltage drop inside the transformer at each load, and
convert the resulting voltage from the primary side to the secondary (low voltage) side.

A suitable MATLAB program is shown below:

% M-file: prob2_13c.m
% M-file to calculate and plot the secondary voltage
% of a three-phase Y-delta transformer bank as a
% function of load for power factors of 0.85 lagging,
% 1.0, and 0.85 leading. These calculations are done
% using an equivalent circuit referred to the primary side.

% Define values for this transformer
VL = 14400; % Primary line voltage (V)
VPP = VL / sqrt(3); % Primary phase voltage (V)
amps = 0:0.01203:12.03; % Phase current values (A)
Req = 18.94; % Equivalent R (ohms)
Xeq = 35.77; % 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.
re = 0.85;
im = sin(acos(re));
I = zeros(3,length(amps));
I(1,:) = amps .* ( re - j*im); % Lagging
I(2,:) = amps .* ( 1.0 ); % Unity
I(3,:) = amps .* ( re + j*im); % Leading

% Calculate secondary phase voltage referred
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