Page 98 - Solutions Manual to accompany Electric Machinery Fundamentals

4-6. The internal generated voltage E A of a 2-pole, -connected, 60 Hz, three phase synchronous generator is
14.4 kV, and the terminal voltage V T is 12.8 kV. The synchronous reactance of this machine is 4 , and
the armature resistance can be ignored.
(a) If the torque angle of the generator δ = 18°, how much power is being supplied by this generator
at the current time?
(b) What is the power factor of the generator at this time?

(c) Sketch the phasor diagram under these circumstances.
(d) Ignoring losses in this generator, what torque must be applied to its shaft by the prime mover at
these conditions?
SOLUTION

(a) If resistance is ignored, the output power from this generator is given by
3VE  3 12.8 kV 14.4 kV 
P   A sin  sin18  42.7 MW
X S 4 

(b) The phase current flowing in this generator can be calculated from

E  V jX I
A  S A
E  V
I A  A 
jX S

 

14.4 18 kV 12.8 0 kV

I A   1135  11.4 A
j 4 
 0.98 lagging .
Therefore the impedance angle   11.4 , and the power factor is cos 11.4
(c) The phasor diagram is


E A  14.4 18 kV

jX I
SA
  V 12.8 0 kV



 I 1035   11.4 A 
A
(d) The induced torque is given by the equation
P conv   
ind
m
With no losses,
P 42.7 MW
 app   ind  conv   113,300 N m


m 2  hz 60
4-7. A 100-MVA, 14.4-kV, 0.8-PF-lagging, 50-Hz, two-pole, Y-connected synchronous generator has a per-
unit synchronous reactance of 1.1 and a per-unit armature resistance of 0.011.
(a) What are its synchronous reactance and armature resistance in ohms?

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