Page 280 - Solutions Manual to accompany Electric Machinery Fundamentals

 E  1.05 411 V    432 V
A
The new torque angle  can be found from the fact that since the terminal voltage and power of the ac
machine are constant, the quantity E A sin must be constant.

E A sin  E A sin
 E  1  411 V 
 sin  1  A sin     sin   sin 20.5   18.4
E A    432 V 
Therefore, the armature current will be

E  V 432 18.4 V 277 0 V






I A  A    63.5  44.2 A
jX S j 3.0 
The resulting reactive power is
Q  3 V T sin     3 480 V 63.5 A sin 44.2    36.8 kvar
I
L
The reactive power supplied to the ac power system will be 36.8 kvar, compared to 30 kvar before the ac
machine field current was increased. The phasor diagram illustrating this change is shown below.
E E
A1 A2

I A1 V jX I

I S A
A2
(c) If the dc field current is decreased by 1%, the dc machine’s flux will decrease by 1%. The internal
generated voltage in the dc machine is given by the equation E  K , and  is held constant by the
A
infinite bus attached to the ac machine. Therefore, E on the dc machine will decrease to (0.99)(234.6
A
V) = 232.25 V. The resulting armature current is

V  E 240 V 232.25 V
I  T A   244.5 A
A
,dc
R A 0.0317 
The power into the dc motor is now (240 V)(244.5 A) = 58.7 kW, and the power converted from electrical
to mechanical form in the dc machine is (232.5 V)(244.5 A) = 56.8 kW. This is also the output power of
the dc machine, the input power of the ac machine, and the output power of the ac machine, since losses
are being neglected.
The torque angle of the ac machine now can be found from the equation

3VE
P  X  S A sin
ac


  sin  1 PX S  sin  1 56.8 kW 3.0    29.9
ac

3VE A   3 277 V 411 V 

The new E A of this machine is thus 411 29.9 V  , and the resulting armature current is


E  V 411 29.9 V 277 0 V




I A  A jX S   j 3.0   73.2  21.2 A
The real and reactive powers are now
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