Page 97 - Solutions Manual to accompany Electric Machinery Fundamentals

4-5. Assume that the field current of the generator in Problem 4-2 has been adjusted so that it supplies rated
voltage when loaded with rated current at unity power factor.
(a) What is the torque angle  of the generator when supplying rated current at unity power factor?
(b) What is the maximum power that this generator can deliver to a unity power factor load when the
field current is adjusted to the current value?
(c) When this generator is running at full load with unity power factor, how close is it to the static
stability limit of the machine?
SOLUTION

(a) The torque  angle can be found by calculating E A :

E  V R I  jX I
A  A A S A

E A  7967 0     0.20   2092  A   0 j 2.5   2092 0  A 


 E 9883 32 V
A
Thus the torque angle  = 32.
(b) The static stability limit occurs at   90 . This generator is a approaching that limit. If we ignore
the internal resistance of the generator, the output power will be given by

3VE
P   A sin
X S

and the output power is proportional to sin . The maximum possible power will occur when sin =
90. That power is

3VE   3 7967 V 9883 V 
P   A sin  sin90  94.5 MW
X S 2.5 

(c) The static stability limit is about twice the rated output power of the generator.
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