Page 89 - Solutions Manual to accompany Electric Machinery Fundamentals

P 50 MVA
I  I    2092 A at an angle of –25.8
A L 3 V
L  3 V 13800
The phase voltage of this machine is V   V T / 3  7967 V . The internal generated voltage of the

machine is
E  V R I  jX I
A  A A S A
 E 7967 0     0.20   2092   A   25.8 j 2.5   2092   25.8  A 
A


E A  11544 23.1 V
(c) The phase voltage of the machine at rated conditions is V  7967 V

From the OCC, the required field current is 10 A.
(d) The equivalent open-circuit terminal voltage corresponding to an E of 11544 volts is
A

V T ,oc   3 11544 V  20 kV
From the OCC, the required field current is 10 A.

(e) If the load is removed without changing the field current, V  E  11544 V . The corresponding

A
terminal voltage would be 20 kV.
(f) 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

Therefore the prime mover must be capable of supplying 50.1 MW. Since the generator is a four-pole 60
Hz machine, to must be turning at 1800 r/min. The required torque is
P 50.1 MW

  IN   265,800 N m
APP
 m  1800 r/min  1 min   2 rad  
  60 s     1 r  
(e) The rotor current limit of the capability curve would be drawn from an origin of
2  3V  3 7967 V 2
Q     76.17 MVAR
X S 2.5 
The radius of the rotor current limit is


 3VE A  3 7967 V 11,544 V 

D    110 MVA
E
X S 2.5 
The stator current limit is a circle at the origin of radius
83