A battery powered direct-current motor operating at steady-state produces output

Question

A battery powered direct-current motor operating at steady-state produces output shaft torque of 12 N-m at a speed of 3600 RPM (377 rad/s). The battery supplies a constant electrical current of 100 amps at 48 volts to the motor. After 30 minutes (1800 s) of operation of steady operation the motor is shut down. Find the motor shaft power output, W& SHAFT (kW) and total work done in 30 minutes (kJ). Find the electrical work output from the battery in the 30 minute period, WELEC (kJ). Qualitatively sketch the battery electrical work done over time ( W(t)ELEC vs t ).

Explanation / Answer

power = Work done/time = Torque *angular displacement/time = 12*(1800*377)/1800 = 4524 Watta

work done = power*time = 8143200 = 8143.2 kJ

electrical work output = v*i*t = 48*100*1800 = 8640000 J = 8640 kJ

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