Page 131 - Solutions Manual to accompany Electric Machinery Fundamentals

figure(1);
plot(abs(Ea),abs(Ia),'b','Linewidth',2.0);
xlabel('\bf\itE_{A}\rm\bf (V)');
ylabel('\bf\itI_{A}\rm\bf (A)');
title ('\bfSynchronous Motor V-Curve');
grid on;

The resulting plot is shown below

5-2. Assume that the motor of Problem 5-1 is operating at rated conditions.
(a) What are the magnitudes and angles of E and I A , and I ?
F
A
(b) Suppose the load is removed from the motor. What are the magnitudes and angles of E A and I
A
now?
SOLUTION

(a) The line current flow at rated conditions is
P 298.4 kW
I    449 A

L
3 V T PF   3 480 V 0.8 
Because the motor is -connected, the corresponding phase current is I  A 449/ 3  259 A . The angle


of the current is cos  1 0.80  36.87  , so I A  259 36.87 A . The internal generated voltage E A is
E  V jX I
A  S A

 E   480 0 V  j   0.6   259 36.87 A   587   12.2 V
A
The field current is directly proportional to E , with = 480 V when I = 4 A. Since the real E is
A
F
A
587 V, the required field current is
E A 2  I F 2
E I
A 1 F 1

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