Page 305 - Solutions Manual to accompany Electric Machinery Fundamentals


nP 1200 r/min 6 poles 
f  m   60 Hz
se
120 120
The phase voltage is

V   2N P k p k d  f e  2  turns96 0.981 0.956 0.02 Wb  Hz60   480 V

Therefore, the line voltage is
V L  3V   831 V

B-6. A three-phase Y-connected 50-Hz two-pole synchronous machine has a stator with 18 slots. Its coils
form a double-layer chorded winding (two coils per slot), and each coil has 60 turns. The pitch of the
stator coils is 8/9.
(a) What rotor flux would be required to produce a terminal (line-to-line) voltage of 6 kV?
(b) How effective are coils of this pitch at reducing the fifth-harmonic component of voltage? The
seventh- harmonic component of voltage?

SOLUTION
(a) The pitch of this winding is 8/9 = 160, so the pitch factor is

160
k  sin  . 0 985
p
2
The phase groups in this machine cover three slots each, and the slot pitch is 20 mechanical or 20
electrical degrees. Thus the distribution factor is

 n 20 3  
sin sin
2
k d  2   20  . 0 960
n sin 3 sin
2 2

The phase voltage of this machine will be
V   2N P k p k d   2  coils 6  turns/coi60 0.985 l 0.960  Hz50  
f
e
V   75621 

The desired phase voltage is 6 kV / 3 = 3464 V, so

3464 V
   . 0 046 Wb
75621

160 5  
(b) The fifth harmonic: k  sin  . 0 643
p
2
 160 7  
The seventh harmonic: k  sin   . 0 342
p
2
Since the fundamental voltage is reduced by 0.985, the fifth and seventh harmonics are suppressed
relative to the fundamental by the fractions:

. 0 643
th
5 :  . 0 653
. 0 985

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