Page 126 - Solutions Manual to accompany Electric Machinery Fundamentals

(a) What is the synchronous reactance of the generator in ohms?
(b) What is the internal generated voltage E of this generator under rated conditions?
A
(c) What is the armature current I A in this machine at rated conditions?

(d) Suppose that the generator is initially operating at rated conditions. If the internal generated
voltage E A is decreased by 5 percent, what will the new armature current I be?
A
(e) Repeat part (d) for 10, 15, 20, and 25 percent reductions in E A .

(f) Plot the magnitude of the armature current I A as a function of E A . (You may wish to use
MATLAB to create this plot.)

SOLUTION
(a) The rated phase voltage of this generator is 14.4 kV / 3 = 8313 V. The base impedance of this
generator is

3V 2   3 8313 V 2
Z   ,base   2.07 
base
S base 100,000,000 VA
Therefore,
R 
0 (negligible)
A

 X  1.0 2.07  2.07   
S
(b) The rated armature current is
S 100 MVA
I  A I  L   4009 A
3 T 3  V 14.4 kV


The power factor is 0.8 lagging, so I A  4009 36.87 A . Therefore, the internal generated voltage is
E  V R I  jX I
A  A A S A
 E 8313   j  0 2.07   4009   A  36.87
A


E A  14,858 26.54 V

(c) From the above calculations, I A  4009 36.87 A .
(d) If E A is decreased by 5%, the armature current will change as shown below. Note that the infinite
bus will keep V  and  constant. Also, since the prime mover hasn’t changed, the power supplied by
m
the generator will be constant.
E E
A2 A1

I
A2 V jX I

I S A
A1
Q  I sin 
A

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