SYNCHRONOUS INDUCTION GENERATOR 205

For the constants:

e0 = 2000 volts, £2 = 1 -f 0.5 j,

£1 = 0.1 -f 0.3 j>, £0 = 0.5 + 0.5 j;

hence:

Z = 1.6 + 1.3;/.

Then:

ei = A/4 X 10° - (1.3 i - 1.6 iiY2 - (!.(> i •+ 1.3 fi);
hence, for non-inductive load, 2*1 = 0:

Cl = \/4 X 10s - l.GD i2 - 1.0 f;

400

200

0 100 200 300 100 500 600 700 800 1X30 1000

AMPERES

FIG. 65.—Hynelironous induction generator regulation curves.
for inductive load of 80 per cent, power-factor ii = 0.6 7, i = 0.8 I:

ei = V4 X 106 - 0.006412 - 2.06 I;

and for anti-inductive load of 80 per cent, power-factor ii
-0.6/,i = 0.87:

- 0.5 I.
As seen, due to the internal impedance, and especially the
resistance of this machine, the regulation is Tery poor, and even
at the chosen anti-inductiye load no rise of voltage occurs.
122. Of more theoretical interest is the case (6), where the