A fourpole 10 kW DC machine operating at 1000 rotationsminut

A four-pole, 10 kW DC machine operating at 1000 rotations/minute (RPM) has a generated electro-motive-force (emf) of 100 V.

If the rotational speed is reduced to 50% of its original value, and the pole magnetic flux is doubled, determine:

The induced electro-motive-force (emf)

The frequency of the square-wave voltage in the DC machine’s armature

Ans) P=4; no of poles

N1=1000 RPM

Electro-motive force Eb1=100 V

Eb is proportional to flux*speed so

new Eb with N2=50% of 1000 i.e 50*1000/100=500 RPM

N2=500 RPM flux phi2=2*phi(old)

So new Eb2=Eb1*(N2*phi2/N1*phi)=100*(500*2/1000)=100 V

Eb2=100 V same as with speed 1000 RPM

Mechanical speed N2=500 RPM in second case

Angular speed in rad/sec W=2*pi*N/60=2*pi*500/60=52.36 rad/sec

Angular frequency (electrical)=(P/2)*W=(4/2)*52.36=104.72 rad/sec

Frequency in Hertz=F=Angural frequency/(2*pi)=104.72/(2*pi)=16.67 Hz

F=16.67 Hz Frequency of square wave