Page 149 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) The input power is
P 298 kW
P  OUT   331 kW
IN
 0.90
(c) The mechanical speed is
n  900 r/min
m
(d) The armature current is
P 331 kW
I  A I  L IN   97.8 A

3 V T  PF  3 2300 V 0.85 


 I 97.8 31.8 A
A


The phase voltage is V    2300 V / 3 0   1328 0 V . Therefore, E A is
E  V R I  jX I
A  A A S A
 E 1328 0 V   0.8   97.8 31.8 A    j 11  97.8 31.8 A  
A

 E 2063   27.6 V
A
(e) The magnitude of the armature current is 97.8 A.
(f) The power converted from electrical to mechanical form is given by the equation P  P  P
conv IN CU
P  3I 2  R   3 97.8 A 2   0.8   23 kW
CU A A


P  P  P  331 kW 23 kW 308 kW
conv IN CU
(g) The mechanical, core, and stray losses are given by the equation


P mech  P core  P stray  P conv  P OUT  308 kW 298 kW 10 kW
5-15. The Y-connected synchronous motor whose nameplate is shown in Figure 5-21 has a per-unit
synchronous reactance of 0.70 and a per-unit resistance of 0.02.
(a) What is the rated input power of this motor?
(b) What is the magnitude of E at rated conditions?
A
(c) If the input power of this motor is 12 MW, what is the maximum reactive power the motor can
simultaneously supply? Is it the armature current or the field current that limits the reactive power
output?
(d) How much power does the field circuit consume at the rated conditions?

(e) What is the efficiency of this motor at full load?
(f) What is the output torque of the motor at the rated conditions? Express the answer both in
newton-meters and in pound-feet.

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