Page 94 - Solutions Manual to accompany Electric Machinery Fundamentals



E  11544 23.1 V
A
The input power to this generator is equal to the output power plus losses. The rated output power
is
P OUT   50 MVA 0.9    45 MW

P CU  3I A 2 A  R   3 2092 A 2   0.2   2.6 MW

P F&W  1 MW

P core  1.5 MW
P stray  (assumed 0)

P  IN P OUT  P  CU P F&W  P core  P stray  50.1 MW

P 45 MW
h = OUT ´ 100% = ´ 100% = 89.8%
P 50.1 MW
IN
(b) If the generator is loaded to rated MVA with lagging loads, the phase voltage is


V = f 7967 0 V and the internal generated voltage is E A  11544 23.1 V . Therefore, the phase




voltage at no-load would be V   11544 0 V . The voltage regulation would be:

11544 7967
VR   100% 44.9%

7967


(c) If the generator is loaded to rated kVA with leading loads, the phase voltage is V = f 7967 0 V and
the internal generated voltage is
E  V R I  jX I
A  A A S A
E A  7967 0     0.20   2092    25.8 A  j 2.5  2092 25.8 A  


 E 7793 38.8 V
A
The voltage regulation would be:

7793 7967
VR   100%   2.2%
7967
(d) If the generator is loaded to rated kVA at unity power factor, the phase voltage is

V = f 7967 0 V and the internal generated voltage is

E  V R I  jX I
A  A A S A

E A  7967 0     0.20   2092  A   0 j 2.5   2092 0  A 


 E 9883 32 V
A
The voltage regulation would be:


VR  9883 7967  100% 24%
7967
(e) For this problem, we will assume that the terminal voltage is adjusted to 13.8 kV at no load
conditions, and see what happens to the voltage as load increases at 0.9 lagging, unity, and 0.9 leading
power factors. Note that the maximum current will be 2092 A in any case. A phasor diagram
representing the situation at lagging power factor is shown below:

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