PPT Slide
A 1hp, 240 V. DC motor is designed to run at 500 RPM when supplied from a DC source. The motor’s armature resistance is 7.56 W. Its torque and back-emf constants are 4.23 Nm/A and 4.23 V/(rad.sec.) respectively. The motor is driven by a half-wave converter at 200 V., 50 Hz. AC and draws a 2 A. average current at a certain load. While the SCR conducts, the average motor voltage Vm is 120 V. Determine the torque developed by the motor, the motor’s speed and the supply power factor.
Section 13.10: Speed Control of DC Motors cont.
The torque is t = kIa = 4.23 Nm/A´2A. = 8.46 Nm and the back emf is
E = Vm - RaIa = 120V. - 7.56W´2A. = 104.88 V. = 4.23 V/(rad.sec.)´Wm = kWm
Which yields Wm = 24.79 rad./sec. (which is about 237 RPM)
The voltageamperage of the motor is |S|= 200V.´2A. = 400 VA and the average power drawn by the motor is Vm´Ia = 120V.´2A. = 240 W so the power factor is P/|S| = 0.6 lagging.