Page 279 - Solutions Manual to accompany Electric Machinery Fundamentals

and the reactive power supplied by the ac machine to the ac power system is

Q AC  S  sin   50 kVA sin cos  1   0.8   30 kvar

The power out of the dc motor is thus 40 kW. This is also the power converted from electrical to
mechanical form in the dc machine, since all other losses are neglected. Therefore,

P conv  E I  AA V  T I R A  I  A 40 kW
A
VI  T A I R  A 2 A 40 kW  0

The base resistance of the dc machine is

V 2 230 V 2
R  T   ,base  1.058 
base,dc
P base 50 kW
Therefore, the actual armature resistance is
 R  0.03 1.058    0.0317  
A
Continuing to solve the equation for P conv , we get

VI  T A I R  A 2 A 40 kW  0

Multiplying by -1 and rearranging terms produces
IR  A 2 A V I  T A 40 kW  0

0.0317 I A 2  240I  A 40,000  0

I A 2  7571I  A  1,261,800 0
I  170.5 A
A

and E  A V  T I R  A A 240 V  170.5 A 0.0317    234.6 V .

Therefore, the power into the dc machine is VI  40.92 kW , while the power converted from electrical
A
T
to mechanical form (which is equal to the output power) is EI  AA 236.4 V 170.5 A  40 kW . The
internal generated voltage E of the dc machine is 234.6 V.
A
The armature current in the ac machine is

S 50 kVA
I    60.1 A
A
3   V V  3 480

 I 60.1  36.87 A
A
Therefore, the internal generated voltage E of the ac machine is
A
E  V jX I
A  S A



 E 277 0  j  V  3.0   60.1   36.87 A    411 20.5 V
A
(b) When the field current of the ac machine is increased by 5%, it has no effect on the real power
supplied by the motor-generator set. This fact is true because P  , and the speed is constant (since
the MG set is tied to an infinite bus). With the speed unchanged, the dc machine’s torque is unchanged,
so the total power supplied to the ac machine’s shaft is unchanged.
If the field current is increased by 5% and the OCC of the ac machine is linear, E increases to
A

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