210

ELECTRICAL APPARATUS

For the constants:

eQ = 2000 volts.
Zl = 0.1 + 0.3 j.
£2= 1 + 0.5J.

ZQ = 0.5 + 0.5 j.
a = 0.067.

t = 1, that is, the
same number of
turns in stator
and rotor

Then:

Zz = 1.6 + 1.3 j andZ4 = 1.4 + 0.7.7*.
Hence, substituting in equation (9):

ei = \/4 X 106 - (0.7 i+1.6 *i)2 + 1.4 i - 1.3'

2GOO 2400 2200 2000 1800 1COO 1400 1200
5 -9

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0-80-70-60-50-40-30-20-10 0 10 20 30 40 50 60 70 80 9

FIG. 66.—Synchronous induction generator, voltage regulation with power-
factor of load.

thus, for non-inductive load, ?"i = 0:

0i = \/4 X 106 - 0.49 t2 + 1.4 i]
for inductive load of 80 per cent, power-factor ii = 0.6 7; i = 0.8 7:
61 = \/4 X 106 - 2.31 J2 + 0.34 7;
and for anti-inductive load of 80 per cent, power-factor ii =
-0.6 7; i = 0.87:
ei = \/4 X 106 - 0.1672 + 1.9 I.
Comparing the curves of this example with those of the same
machine driven as frequency converter with exciter generator,