Page 257 - Solutions Manual to accompany Electric Machinery Fundamentals

The minim um possible field current occurs and minimum speed and field current. The maximum
adjustabl e resistance is R = 30 . The current is
adj
V 120 V
I  F   2.4 A
R  F ,max F R adj 20   30 

From the mag netization curve, the voltage E at 1800 r/min is 93.1 V. Since the actual speed is 1500
Ao
r/min, the m aximum no-load voltage is
E A  n
E Ao n o

n 1500 r/min
E  E   93.1 V  77.6 V
A
n o Ao 1800 r/min
8-23. If the arm ature current of the generator in Problem 8-22 is 50 A, the speed of the generator is 1700 r/min,
and the te rminal voltage is 106 V, how much field current must be flowing in the generator?
SOLUTION The internal generated voltage of this generator is

E  A V  T I R  A A 106 V   50 A 0.18   115 V

at a speed of 1700 r/min. This corresponds to an E at 1800 r/min of
Ao
E A  n
E Ao n o
n 1800 r/min
E  o E    115 V  121.8 V
Ao
n A 1700 r/min
From the magnetization curve, this value of E requires a field current of 4.2 A.
Ao
8-24. Assuming that the generator in Problem 8-22 has an armature reaction at full load equivalent to 400
Aturns of magnetomotive force, what will the terminal voltage of the generator be when I F = 5 A, n m =
1700 r/min, and I A = 50 A?

SOLUTION When I is 5 A and the armature current is 50 A, the magnetomotive force in the generator is
F

 F  F  NI  1  000 turns 5 A  400 A turns   4600 A turns
net F AR

or I F *  net F  F / N 4600 A turns / 1000 turns  4.6 A
The equivalent internal generated voltage E of the generator at 1800 r/min would be 126 V. The actual
Ao
voltage a t 1700 r/min would be
n 1700 r/min
E  E   126 V  1 19 V
A
n Ao 1800 r/min
o
Therefore, the terminal voltage would be
V  T E  A I R  A A 119 V   50 A 0.18   110 V

8-25. The machine in Problem 8-22 is reconnected as a shunt generator and is shown in Figure P8-9. The shunt
field resi stor R adj is adjusted to 10 , and the generator’s speed is 1800 r/min.

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