Page 110 - Solutions Manual to accompany Electric Machinery Fundamentals

S 120 MVA
I  I    5020 A
A L
3 T 3  V 13.8 kV

The power factor is 0.8 lagging, so I  5020 36.87 A . The phase voltage is 13.8 kV / 3 = 7967
A
V. Therefore, the internal generated voltage is
E  V R I  jX I
A  A A S A
 E 7967 0    0.1   5020   36.87  A   j 1.2   5020   36.87  A 
A


E A  12,800 20.7 V
The resulting voltage regulation is

12,800 7967
VR   100% 60.7%

7967
(b) If the generator is to be operated at 50 Hz with the same armature and field losses as at 60 Hz (so
that the windings do not overheat), then its armature and field currents must not change. Since the
voltage of the generator is directly proportional to the speed of the generator, the voltage rating (and
hence the apparent power rating) of the generator will be reduced by a factor of 5/6.

5
  V 13.8 kV  11.5 kV
T ,rated 6
5
  S 120 MVA  100 MVA
rated 6
Also, the synchronous reactance will be reduced by a factor of 5/6.

5
S  X  1.2   1.00 
6
(c) At 50 Hz rated conditions, the armature current would be
S 100 MVA
I  A I  L   5020 A
3 T 3  V 11.5 kV


The power factor is 0.8 lagging, so I A  5020 36.87 A . The phase voltage is 11.5 kV / 3 = 6640
V. Therefore, the internal generated voltage is
E  V R I  jX I
A  A A S A
 E 6640 0    0.1   5020   36.87  A   j 1.0   5020   36.87  A 
A


E A  10,300 18.8 V
The resulting voltage regulation is


10,300 6640

VR   100% 55.1%
6640
Problems 4-16 to 4-26 refer to a six-pole Y-connected synchronous generator rated at 500 kVA, 3.2 kV, 0.9 PF
lagging, and 60 Hz. Its armature resistance R A is 0.7 . The core losses of this generator at rated conditions are
8 kW, and the friction and windage losses are 10 kW. The open-circuit and short-circuit characteristics are shown
in Figure P4-2.

104