Page 114 - Solutions Manual to accompany Electric Machinery Fundamentals

The resulting plot is:

4-17. (a) What are the rated current and internal generated voltage of this generator? (b) What field current
does this generator require to operate at the rated voltage, current, and power factor?

SOLUTION
(a) The rated line and armature current for this generator is
S 500 kVA
I  A I  L   90.2 A
3 T 3  V 3200 V


The power factor is 0.9 lagging, so I A  90.2 25.8 A . The rated phase voltage is V = 3.2 kV / 3 =

1850 V. The saturated synchronous reactance at rated conditions was found to be 15.7  in the previous
problem. Therefore, the internal generated voltage is
E  A V   R I  A A jX I
S A
 E 1850    0 0.7   90.2   A   25.8 j 15.7   90.2   25.8  A 
A


 E 2814 26.3 V
A


(b) This internal generated voltage corresponds to a no-load terminal voltage of 32814 = 4874 V.
From the open-circuit characteristic, the required field current would be 3.2 A.
4-18. What is the voltage regulation of this generator at the rated current and power factor?
SOLUTION The voltage regulation is

V  V 4874 3200

VR  T ,nl T ,fl  100%   100% 52.3%

V T ,fl 3200
4-19. If this generator is operating at the rated conditions and the load is suddenly removed, what will the
terminal voltage be?

108