Question

A 200 kW, 400 V, 14-pole dc machine has lap wound armature with 1100 conductors. The pole arc to pole pitch ratio is 0.6. The number of pole face conductors of the compensating winding in each pole so as to obtain uniform air gap density under pole faces is

• A. 3

Answer 1

The correct option is A 3
$\text{Full load armature current}=\frac{200\times 1000}{400}=500A$

$\text{Armature ampere conductors per pole}=\frac{500\times 1100}{14\times 14}$

$\text{Ampere conductors per pole to be compensated by pole face winding}$

$=\frac{\text{Pole arc}}{\text{Pole Pitch}}\times \frac{500\times 1100}{14\times 14}$

$\text{Compensating winding conductors per pole}$

$=\frac{0.6\times 500\times 1100}{14\times 14}\times \frac{1}{500}$

$=3.36\simeq 3$