Page 223 - Solutions Manual to accompany Electric Machinery Fundamentals
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E n
A
E Ao n o
E 240 V
n A n o 1200 r/min 1061 r/min
E Ao 271.5 V
The field current when R = 400 is:
adj
V 240 V 240 V
I T 0.505 A
F
R adj R F 400 75 500
From Figure P8-1, this field current would produce an internal generated voltage E of 167 V at a speed
Ao
n of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
o
E n
A
E Ao n o
E 240 V
n A n o 1200 r/min 1725 r/min
E Ao 167 V
8-6. What is the starting current of this machine if it is started by connecting it directly to the power supply
V T ? How does this starting current compare to the full-load current of the motor?
SOLUTION The starting current of this machine (ignoring the small field current) is
V 240 V
I T 1260 A
,start
L
R A 0.19
The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current. This
much current is extremely likely to damage the motor.
8-7. Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
full-load armature reaction of 1200 Aturns. (Assume that the armature reaction increases linearly with
increases in armature current.)
SOLUTION This problem is best solved with MATLAB, since it involves calculating the torque-speed
values at many points. A MATLAB program to calculate and display both torque-speed characteristics is
shown below.
% M-file: prob8_7.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 1200 r/min.
load p81_mag.dat
if_values = p81_mag(:,1);
ea_values = p81_mag(:,2);
n_0 = 1200;
% First, initialize the values needed in this program.
v_t = 240; % Terminal voltage (V)
r_f = 75; % Field resistance (ohms)
r_adj = 175; % Adjustable resistance (ohms)
r_a = 0.19; % Armature resistance (ohms)
i_l = 0:1:110; % Line currents (A)
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