Page 230 - Solutions Manual to accompany Electric Machinery Fundamentals

At full load conditions, the armature current is

I  A I  L I  F 110 A 0.739 A  109.2 A
The internal generated voltage E is
A
E  A T I A  V  A R S R   240 V   109.2 A 0.21   217.1 V

The equivalent field current is
N 14 turns
I *  I  SE I  0.739 A   109.2 A  1.305 A
F
F
N A 2700 turns
F
From Figure P9-1, this field current would produce an internal generated voltage E of 268 V at a speed
Ao
n o of 1200 r/min. Therefore,
 E   217.1 V 
n   A  n o     1200 r/min  972 r/min
 E Ao   268 V 

The new full-load speed is higher than the full-load speed in Problem 8-10.

For Problem 8-12, the motor is now connected differentially compounded as shown in Figure P8-4.
8-12. The motor is now connected differentially compounded.
(a) If R = 175 , what is the no-load speed of the motor?
adj
(b) What is the motor’s speed when the armature current reaches 20 A? 40 A? 60 A?

(c) Calculate and plot the torque-speed characteristic curve of this motor.
SOLUTION

(a) At no-load conditions, E  A V  T 240 V . The field current is given by
V 240 V 240 V
I  F    0.96 A
R  F adj R F 175   75  275 

From Figure P8-1, this field current would produce an internal generated voltage E of 241 V at a speed
Ao
n o of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be

E A  n
E n
Ao o
 E   240 V 
n   A  n o    1200 r/min  1195 r/min
 E Ao   241 V 

(b) At I A = 20A, the internal generated voltage E is
A
E  A T I A  V  A R S R   240 V   20 A 0.21   235.8 V

The equivalent field current is
N 14 turns
I F *  I  F SE I  A 0.96 A   20 A  0.856 A
N F 2700 turns

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