Page 129 - Solutions Manual to accompany Electric Machinery Fundamentals

Chapter 5: Synchronous Motors

5-1. A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole -connected synchronous motor has a synchronous
reactance of 0.6  and negligible armature resistance. Ignore its friction, windage, and core losses for the
purposes of this problem. Assume that E is directly proportional to the field current I (in other
A
F
words, assume that the motor operates in the linear part of the magnetization curve), and that E = 480
A
V when I = 4 A.
F
(a) What is the speed of this motor?
(b) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the magnitudes and angles
of E A and I A ?

(c) How much torque is this motor producing? What is the torque angle  ? How near is this value
to the maximum possible induced torque of the motor for this field current setting?

(d) If E is increased by 30 percent, what is the new magnitude of the armature current? What is
A
the motor’s new power factor?
(e) Calculate and plot the motor’s V-curve for this load condition.

SOLUTION

(a) The speed of this motor is given by
120 120  f 60 Hz
n  se   900 r/min
m
P 8
(b) If losses are being ignored, the output power is equal to the input power, so the input power will be
 P  400 hp 746 W/hp   298.4 kW
IN
This situation is shown in the phasor diagram below:

V

I
A jX I
S A
E
A
The line current flow under these circumstances 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  406   17.8 V
A
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