Page 100 - Solutions Manual to accompany Electric Machinery Fundamentals

(e) If the field current is held constant, what is the maximum power possible out of this generator?
How much reserve power or torque does this generator have at full load?
(f) At the absolute maximum power possible, how much reactive power will this generator be
supplying or consuming? Sketch the corresponding phasor diagram. (Assume I F is still unchanged.)

SOLUTION
(a) The speed of rotation of this generator’s shaft is

120 120  f 50 Hz
n  se   300 r/min
sync
P 20
(b) The per-unit phase voltage at rated conditions is V   1.0 0 and the per-unit phase current at
rated conditions is I A  1.0   25.8  (since the power factor is 0.9 lagging), so the per-unit internal
generated voltage is

E  V R I  jX I
A  A A S A

E A  1  0  0.1 1  25.8  j 0.9 1 25.8 


 E 1.69 27.4 pu
A
The base phase voltage is
V  ,base  12 kV / 3  6928 V

so the internal generated voltage is



 E   1.69 27.4 pu 6928 V    11,710 27.4 V
A
(c) The torque angle of the generator is   27.4 .
(d) The base impedance of the generator is
3 V 2   3 6928 V 2
Z   ,base   0.72 
base
S 200,000,000 VA
base
Therefore the synchronous reactance is

 X  0.9 0.72  0.648   
S
and the armature resistance is


 R  0.1 0.72  0.072   
A
(e) If the field current is held constant (and the armature resistance is ignored), the power out of this
generator is given by
3VE
P   A sin
X
S
The max power is given by
3VE   3 6928 V 11,710 V 
P max   A sin90   376 MW
X S 0.648 


Since the full load power is P   200 MVA 0.85  170 MW , this generator is supplying 45% of the
maximum possible power at full load conditions.

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