Page 311 - Solutions Manual to accompany Electric Machinery Fundamentals

2
3 VE 3 V  X  X 
P   A sin     d q  sin 2




X d 2  XX q 
d

2
3  7621 9890  3 7621  . 0 62  . 0 40 
P  sin      sin  2

. 0 62   2  62 . 0 . 0  40 
P  364 7 . sin   77 3 . sin 2  MW
A plot of power supplied as a function of torque angle is shown below:

The peak power occurs at an angle of 70.6, and the maximum power that the generator can supply is
392.4 MW.

(d) If this generator were non-salient, P MAX would occur when  = 90, and P MAX would be 364.7
MW. Therefore, the salient-pole generator has a higher maximum power than an equivalent non-salint
pole generator.

C-3. Suppose that a salient-pole m achine is to be used as a motor.
(a) Sketch the phasor diagram of a salient- pole synchronous machine used as a motor.
(b) Write the equations describing the voltages and currents in this motor.

(c) Prove that the torque angle  between E and V on this motor is given by

A


IX cos I R sin 
 tan -1 A q A A
V   I X q sin   I R A cos 
A
A
SOLUTION

305