Question

8. A diesel engine that develops 30 kW at 3600 rpm delivers its power through a 4:1 reduction gearbox whose efficiency is 95 percent. What is the output torque?  

Answer 1

The torque developed by the motor:

$$T=P/\omega=P/(2\pi n).$$

The output torque due to the reducer is

$$T_r=rT=rP/(2\pi n).$$

The total output torque:

$$T_o=rT=\eta rP/(2\pi n),\\ T_o=0.95·4(30·10^3)/(2\pi(3600/60))=302\text{ N}·\text{m}.$$