Page 243 - Solutions Manual to accompany Electric Machinery Fundamentals

load p86_mag.dat;
if_values = p86_mag(:,1);
ea _values = p86_mag(:,2);
...
Ea = interp1(if_values,ea_ values,if,'spline')
(a) If R = 120 , the total field resistance is 320 , and the resulting field current is
adj
V 240 V
I  T   0.75 A
F
R  F R adj 200 120   
This field current would produce a voltage E of 245 V at a speed of n = 3000 r/min. The actual E
Ao o A
is 240 V, so th e actual speed will be

E 240 V
n  A  n   3000 r/min  293 9 r/min
E Ao o 245 V

(b) At full load, I  I  L I  F 100 A 0.75 A  99.25 A , and
A
E  A V  T I R  A A   240 V 99.25 A 0.14    226.1 V

re
The fore, the speed at full load will be
E 226.1 V
n  A n   3000 r/min  2769 r/min
E Ao o 245 V
(c) The speed regulation of this motor is


n  n 2939 r/min 2769 r/min
SR  nl fl  100%   100% 6.16%

n fl 2769 r/min
(d) A MATLAB program to calculate the torque-speed characteristic is shown below:

% M-file: prob8_16.m
% M-file to create a plot of the torque-speed curve of the
% the shunt dc motor with and without armature reaction.

% Get the magnetization curve. Note that this curve is
% defined for a speed of 3000 r/min.
l oad p86_mag.dat;
if_values = p86_mag(:,1);
ea_values = p86_mag(:,2);
n_0 = 3000;

% First, initialize the val ues needed in this program.
v_t = 240; % T erminal voltage (V)
r _f = 200; % Field resistance (ohms)
r_adj = 120; % Adjustable resistance (ohms)
r_a = 0.14; % Armature resistance ( ohms)
i_l = 0:1:100; % Line currents (A)
n_f = 1500; % Number of turns on shunt field

% Calculate the armature current for each lo ad.
i_a = i_l - v_t / (r_f + r_adj);

% Now calculate the internal generated voltage fo r
% each armature current.
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