Page 148 - Solutions Manual to accompany Electric Machinery Fundamentals

So E = 489 V at rated conditions.
A
(c) The maximum power that the motor can produce at rated speed with the value of E A from part (b)
is

3VE  3 277 V 489 V 
P     A  226 kW
max
X S 1.8 
(d) Since E A must be decreased linearly with frequency, the maximum value at 300 r/min would be

 10 Hz 
E A ,300     489 V  97.8 V
 50 Hz 

(e) If the applied voltage V  is derated by the same amount as E , then V = (10/50)(277) = 55.4 V.

A
Also, note that X = (10/50)(1.8 ) = 0.36 . The maximum power that the motor could supply would
S
be
3VE    3 97.8 V  55.4 V
P  A   45.1 kW
max
X 0.36 
S
(f) As we can see by comparing the results of (c) and (e), the power-handling capability of the
synchronous motor varies linearly with the speed of the motor.

5-14. A 2300-V, 400-hp, 60-Hz, eight-pole, Y-connected synchronous motor has a rated power factor of 0.85
leading. At full load, the efficiency is 90 percent. The armature resistance is 0.8 , and the synchronous
reactance is 11 . Find the following quantities for this machine when it is operating at full load:
(a) Output torque
(b) Input power

(c) n m

(d) E A

(e) I A

(f) P conv

(g) P mech  P core  P stray

SOLUTION
(a) Since this machine has 8 poles, it rotates at a speed of

120 120  f 60 Hz
n  se   900 r/min
m
P 8
At full load, the power supplied by this machine is

 P  400 hp 746 W/hp    298 kW
out
If the output power is 298 kW, the output torque is
P  298,000 W

  out   3162 N m
load
 m  900 r/min  2 rad    1 min 
  1 r     60 s  
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