Page 167 - Solutions Manual to accompany Electric Machinery Fundamentals

I A
R 1 jX 1 jX 2 R 2

+ j0.175 ? 0.07 ?
0.10 ? j0.175 ?
 1  s
j8.33 ? jX M R  
V ? 2  s 

1.33 ?
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.

I A
R 1 jX 1 jX F R F

+
0.10 ? j0.175 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX M is:
1 1
Z  F 1 1  1 1  1.308  j 0.386 1.36 16.5    

jX M  Z 2 j 8.33   1.40  j 0.175

The line voltage must be derated by 5/6, so the new line voltage is V  173.3 V . The phase voltage is
T
173.3 / 3 = 100 V, so line current I L is
V 100 0 V

I  L I  A  
R  1 jX  1 R  F jX F 0.10   j 0.175   1.308   j 0.386 

I  L I  A 66.0  21.7 A
(b) The stator copper losses are

P SCL  3I A 2 1  R   3 66 A 2   0.10   1307 W
R
2
(c) The air gap power is P  3I 2 2  3I R
2
AG
s A F
R
2
/
(Note that 3IR is equal to 3I 2 2 , since the only resistance in the original rotor circuit was R s ,
F
A
2
s 2
and the resistance in the Thevenin equivalent circuit is R . The power consumed by the Thevenin
F
equivalent circuit must be the same as the power consumed by the original circuit.)
R 2
P  3I 2 2  3I 2  R   3 66 A  1.308    17.1 kW
AG
2
s A F
161