Page 309 - Solutions Manual to accompany Electric Machinery Fundamentals

The phase voltage of this machine is V   V T /3  7967 V . The internal generated voltage of the
machine is
I
 E A   V R A A  I jX q A


  E 7967 0    0.20  2092  25.8 A   j 1.8  2092  25.8 A 
A


E   A 10485 17.8 V
Therefore, the torque angle  is 17.8. The direct-axis current is
  I
I d A sin        90 
I d   2092 A sin 43.6      72.2 

I d  1443   72.2  A
The quadrature-axis current is

 
I q  I A  cos   

I q   2092 A cos 43.6 17.8    


I q  1515 17.8 A
Therefore, the internal generated voltage of the machine is
E  V  R I  jX I  jX I q

d
A
q
d
A
A

 E 7967 0     0.20   2092    25.8  j 2.5   1443    72.2 j    1.8   1515 17.8  
A


E A  11496 17.8 V
E is approximately the same magnitude here as in Problem 4-2b, but the angle  is quite different.
A
(c) The power supplied by this machine is given by the equation
3VE 3V  2 X  X 
P   A sin     d q  sin 2
X d 2   X X q  
d
  3 7967 11496    3 7967  2 2.5 1.8  

P  sin 17.8     sin 35.6  

2.5 2    2.5 1.8   
303