A DC shunt motor rotating at 1560 RPM is supplied from a 240

A DC shunt motor rotating at 1560 RPM is supplied from a 240-V source. The line current supplied to the motor is equal to 27 A. The shunt field resistance of the motor is equal to 120 Ohm and the armature resistance is equal to 0.6 Ohm. Calculate the following: The armature current of the machine. The back-emf. The output mechanical power and the net torque developed by the motor if its mechanical losses are equal to 280 W.

Solution

N=1560 RPM

VS=240 VOLTS

LINE CURRENT I=27 AMPS

RF=120 OHMS

Ra=0.6 OHMS

IF=240/120=2 AMPS

I=IF+Ia

ARMATURE CURRENT Ia=I-IF=27-2=25 AMPS

WE KNOW THE MOTOR EQUATION

VS=Eb+Ia*Ra

Eb=240-25*0.6=225 volts.

output power =(Eb*Ia-mechanical losses)=225*25-280=5625-280=5345 watts.

we know

output power= w*T

T=(5345*60)/(2*3.14*1560)=32.7 N-m.