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

Answer:

Torque at input shaft will be 176.8695 N-m

Explanation:

We have given input power $P_{IN}=42.6KW=42.6\times 10^3W$

Angular speed = 2300 rpm

For converting rpm to rad/sec we have multiply with $\frac{2\pi }{60}$

So $2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec$

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 $42.6\times 10^3=\tau \times 240.855$

$\tau =176.8695N-m$

So torque at input shaft will be 176.8695 N-m