Question

A 4-pole, 3-ph, 50 Hz, star-connected alternator has 60 slots with 4 conductors per slot.

Coils are short-pitched by 3 slots. If the phase spread is 60 electrical degrees, find the line

voltage induced for a flux per pole of 0.943 Wb, distributed sinusoidally. All the turns per
phase are in series.​

Answer 1

given,

p=4 pole

f=50 Hz

3 phase Star - connected

S=60 slots

No of conductor per slots =4

flux per pole =0.943 Wb

Find:

The line.

$no of conductor (Z)=60*4\\=240\\No . Of turns (N)=Z/2\\=240/2\\=120\\m= no of slots /pole/phase\\=60/4*3\\=5\\\beta =180/slot/pole\\=180/5\\=36\\Distribution factor\\Kd=sin(m\beta/2 )/m*sin(\beta /2)\\kd=o.6472\\The coil span 2 slots the pole\\\alpha =2\beta \\=2*36\\=72\\kp=cos(\alpha /2)\\=cos(72/2)\\=0.8090\\Eph=4.44*kp*kd*Nf fai$

$Eph = 13153.246 per phase\\line =Eph *\sqrt{3$

line Voltage =22782.079