Question

A 6-pole generator has a lap-wound armature with 40 slots with 20 conductors per slot.The flux per pole is 25mWb.The terminal voltage is 288v and the voltage regulation of the generator is 4%. Determine the speed of the generator.

Answer 1

$E_g=\frac{P\phi NZ}{60A}$

$E_g=V+0.04V=288+0.04\cdot288=299.52\ (V)$

$N=\frac{60AE_g}{P\phi Z}=\frac{60\cdot6\cdot 299.52}{6\cdot0.025\cdot (40\cdot20)}=682\ (rpm)$ . Answer