Page 130 - Solutions Manual to accompany Electric Machinery Fundamentals

(c) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is n = 1200
m
r/min. The induced torque is
P 298.4 kW
  OUT   3166 N m

ind
 m  900 r/min  1 min   2 rad  
  60 s     1 r  
The maximum possible induced torque for the motor at this field setting is the maximum possible power
divided by 
m

3VE  3 480 V 406 V 

 ind,max   A   10,340 N m

m X S  900 r/min   1 min     2 rad     0.6 
 60 s   1 r 
The current operating torque is about 1/3 of the maximum possible torque.

(d) If the magnitude of the internal generated voltage E A is increased by 30%, the new torque angle
can be found from the fact that E A sin  P  constant .

E A 2  1.30 E  A 1  1.30 406 V   487.2 V

 E  1  406 V 
 sin  1  A 1 sin  1   sin   sin    17.8    14.8
2
E A 2    487.2 V 
The new armature current is


V  E 480 0 V 487.2  14.8 V


I   A 2   208  4.1 A
A
2
jX S j 0.6 
The magnitude of the armature current is 208 A, and the power factor is cos (-24.1) = 0.913 lagging.
(e) A MATLAB program to calculate and plot the motor’s V-curve is shown below:

% M-file: prob5_1e.m
% M-file create a plot of armature current versus Ea
% for the synchronous motor of Problem 5-1.

% Initialize values
Ea = (0.90:0.01:1.70)*406; % Magnitude of Ea volts
Ear = 406; % Reference Ea
deltar = -17.8 * pi/180; % Reference torque angle
Xs = 0.6; % Synchronous reactance (ohms)
Vp = 480; % Phase voltage at 0 degrees
Ear = Ear * (cos(deltar) + j * sin(deltar));

% Calculate delta2
delta2 = asin ( abs(Ear) ./ abs(Ea) .* sin(deltar) );

% Calculate the phasor Ea
Ea = Ea .* (cos(delta2) + j .* sin(delta2));

% Calculate Ia
Ia = ( Vp - Ea ) / ( j * Xs);

% Plot the v-curve

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