Page 99 - Solutions Manual to accompany Electric Machinery Fundamentals

(b) What is the magnitude of the internal generated voltage E A at the rated conditions? What is its
torque angle  at these conditions?
(c) Ignoring losses in this generator, what torque must be applied to its shaft by the prime mover at
full load?

SOLUTION

SOLUTION The base phase voltage of this generator is V  ,base  14,400 / 3 8314 V . Therefore, the base
impedance of the generator is
3 V 2   38314 V 2
Z   ,base   2.074 
base
S 100,000,000 VA
base
(a) The generator impedance in ohms are:

R  0.011 2.074     0.0228 
A

 X  1.1 2.074  2.281   
S
(b) The rated armature current is
S 100 MVA
I  A I  L   4009 A
3 T 3  V 14.4 kV


The power factor is 0.8 lagging, so I A  4009 36.87 A . Therefore, the internal generated voltage is
E  V R I  jX I
A  A A S A
 E    8314 0 0.0228   4009   A   36.87 j 2.281   4009   36.87  A 
A


E A  15,660 27.6 V
Therefore, the magnitude of the internal generated voltage E = 15,660 V, and the torque angle  = 27.6.
A
(c) Ignoring losses, the input power would equal the output power. Since
P OUT   0.8 100 MVA    80 MW

and

120 f 120 50 Hz 
n  se   3000 r/min
sync
P 2
the applied torque would be
80,000,000 W

     254,700 N m




app ind  3000 r/min 2 rad/r 1 min/60 s 
4-8. A 200-MVA, 12-kV, 0.85-PF-lagging, 50-Hz, 20-pole, Y-connected water turbine generator has a per-
unit synchronous reactance of 0.9 and a per-unit armature resistance of 0.1. This generator is operating in
parallel with a large power system (infinite bus).
(a) What is the speed of rotation of this generator’s shaft?
(b) What is the magnitude of the internal generated voltage E A at rated conditions?
(c) What is the torque angle of the generator at rated conditions?

(d) What are the values of the generator’s synchronous reactance and armature resistance in ohms?

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