Page 147 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 147
Q TOT Q 1 Q 2 Q motor 48.4 kvar 48 kvar 62.2 kvar 26.2 kvar
S TOT P TOT 2 TOT 2 Q 238.6 kW 2 26.2 kVA 2 240 kVA
The line current is given by
S 240 kVA
I TOT 288.7 A
L
3 L V 3 480 V
(i) The line current in part (d) is greater than the line current in part (h), because the synchronous motor
is supplying more of the reactive power being consumed by the other two loads, required less reactive
power from the infinite bus.
5-13. A 480-V 100-kW 0.8-PF leading 50-Hz four-pole Y-connected synchronous motor has a synchronous
reactance of 1.8 and a negligible armature resistance. The rotational losses are also to be ignored. This
motor is to be operated over a continuous range of speeds from 300 to 1500 r/min, where the speed
changes are to be accomplished by controlling the system frequency with a solid-state drive.
(a) Over what range must the input frequency be varied to provide this speed control range?
(b) How large is E A at the motor’s rated conditions?
(c) What is the maximum power the motor can produce at rated speed with the E A calculated in part
(b)?
(d) What is the largest value that E A could be at 300 r/min?
(e) Assuming that the applied voltage V is derated by the same amount as E A , what is the
maximum power the motor could supply at 300 r/min?
(f) How does the power capability of a synchronous motor relate to its speed?
SOLUTION
(a) A speed of 300 r/min corresponds to a frequency of
nP 300 r/min 4
f se m 10 Hz
120 120
A speed of 1500 r/min corresponds to a frequency of
nP 1500 r/min 4
f m 50 Hz
se
120 120
The frequency must be controlled in the range 10 to 50 Hz.
(b) The armature current at rated conditions is
P 100 kW
I I 150.3 A
L
A
3 V T PF 3 480 V 0.8
so I A 150.3 36.87 A . This machine is Y-connected, so the phase voltage is V = 480 / 3 = 277 V.
The internal generated voltage is
E V R I jX I
A A A S A
E 277 0 j V 1.8 150.3 36.87 A
A
E 489 26.2 V
A
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