Page 150 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION The base quantities for this motor are:

V T ,base  6600 V
6600 V
V   3811 V
,base

3
I A ,base  I L ,base  1404 A
  16.05 MW
S base  P rated  3 V T PF  I L  3 6600 V 1404 A 1.0
(a) The rated input power of this motor is

  16.05 MW
P  IN 3 V T I L  PF   3 6600 V 1404 A 1.0



(b) At rated conditions, V   1.0 0 pu and I   1.0 0 pu , so E A is given in per-unit quantities as
E  V R I  jX I
A  A A S A


1 0
E A   0.02 1.0 0   j 0.70 1 0 

 E 1.20   35.5 pu
A
The base phase voltage of this motor is V  ,base  3811 V , so E A is

 E   1.20  3811 V   35.5 4573   35.5 V
A
(c) From the capability diagram, we know that there are two possible constraints on the maximum
reactive power—the maximum stator current and the maximum rotor current. We will have to check each
one separately, and limit the reactive power to the lesser of the two limits.
The stator apparent power limit defines a maximum safe stator current. This limit is the same as the rated
input power for this motor, since the motor is rated at unity power factor. Therefore, the stator apparent
power limit is 16.05 MVA. If the input power is 12 MW, then the maximum reactive power load that still
protects the stator current is
Q  2 P  2  S  16.05 MVA 2  12 MW   2  10.7 MVAR

Now we must determine the rotor current limit. The per-unit power supplied to the motor is 12 MW /
16.05 MW = 0.748. The maximum E A is 4573 V or 1.20 pu, so with E A set to maximum and the motor
consuming 12 MW, the torque angle (ignoring armature resistance) is

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