Question

The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotational speed of 620 rpm. Determine the torque of the input shaft shaft, in N-m.

Answer 1

Torque at input shaft will be 176.8695 N-m

Step-by-step explanation:

We have given input power
`P_(IN)=42.6KW=42.6* 10^3W`

Angular speed = 2300 rpm

For converting rpm to rad/sec we have multiply with
`(2\pi )/(60)`

So
`2300rpm=(2300* 2\pi )/(60)=240.855rad/sec`

We have to find torque

We know that power is given by
`P=\tau \omega`, here
`\tau` is torque and
`\omega` is angular speed

So
`42.6* 10^3=\tau * 240.855`

`\tau =176.8695N-m`

So torque at input shaft will be 176.8695 N-m