Page 259 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) At an armature current of 20 A, the internal voltage drop in the armature resistance is
 20   A 0.18    V 6 . 3 . As shown in the figure below, there is a difference of 3.6 V between E A and

V T at a terminal voltage of about 116 V.

A MATLAB program to locate the position where there is exactly 3.6 V between the E A and V T lines is
shown below. This program created the plot shown above. Note that there are actually two places where
the difference between the E A and V T lines is 3.6 volts, but the low-voltage one of them is unstable. The
code shown in bold face below prevents the program from reporting that first (unstable) point.

% M-file: prob8_25b1.m
% M-file to create a plot of the magnetization curve and the
% field current curve of a shunt dc generator, determining
% the point where the difference between them is 3.6 V.

% Get the magnetization curve. This file contains the
% three variables if_values, ea_values, and n_0.
clear all
load p87_mag.dat;
if_values = p87_mag(:,1);
ea_values = p87_mag(:,2);
n_0 = 1800;

% First, initialize the values needed in this program.
r_f = 20; % Field resistance (ohms)
r_adj = 10; % Adjustable resistance (ohms)
r_a = 0.18; % Armature + series resistance (ohms)
i_f = 0:0.05:6; % Field current (A)
n = 1800; % Generator speed (r/min)

% Calculate Ea versus If
Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If
Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two
% lines is 3.6 V. This will be the point where the line
% line "Ea - Vt - 3.6" goes negative. That will be a
% close enough estimate of Vt.
diff = Ea - Vt - 3.6;

% This code prevents us from reporting the first (unstable)
% location satisfying the criterion.
was_pos = 0;
for ii = 1:length(i_f);
if diff(ii) > 0
was_pos = 1;
end
if ( diff(ii) < 0 & was_pos == 1 )
break;
end;
end;

% We have the intersection. Tell user.

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