Page 93 - Solutions Manual to accompany Electric Machinery Fundamentals

 E A

E A

jX I
SA


V  V 
I A
 I
A
425 W
(e) The new impedance per phase will be half of the old value, so Z =  . The magnitude of the
phase current flowing in this generator is
E 9526 V 9526 V
I  A    1680 A
A

R  A jX  S Z 0.2  j 2.5 4 25  5.67 
Therefore, the magnitude of the phase voltage is
V   I Z  A 1680 A 4    6720 V

and the terminal voltage is

V  T 3 V   3 6720 V  11,640 V

(f) To restore the terminal voltage to its original value, increase the field current I F .

4-4. Assume that the field current of the generator in Problem 4-2 is adjusted to achieve rated voltage (13.8
kV) at full load conditions in each of the questions below.
(a) What is the efficiency of the generator at rated load?

(b) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0.9-
PF-lagging loads?

(c) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0.9-
PF-leading loads?

(d) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with unity-
power-factor loads?

(e) Use MATLAB to plot the terminal voltage of the generator as a function of load for all three
power factors.

SOLUTION
(a) This generator is Y-connected, so I = I A . At rated conditions, the line and phase current in this
L
generator is

I  I  P  50 MVA  2092 A at an angle of –25.8
A L 3 V
L  3 V 13800
The phase voltage of this machine is V   V T / 3  7967 V . The internal generated voltage of the
machine is

E  V R I  jX I
A  A A S A


 E 7967 0    0.20  251  36.87 A   j 2.5  2092  25.8 A 
A
87