Question

A 0.15-m-diameter pulley turns a belt rotating the driveshaft of a power plant pump. The torque applied by the belt on the pulley is 200 N ∙ m, and the power transmitted is 7 kW. Determine the net force applied by the belt on the pulley, in kN, and the rotational speed of the driveshaft, in RPM.

Answer 1

F= 2666.66 N

N=334.22 RPM

Step-by-step explanation:

Given that

Diameter of pulley ,d= 0.15 m

Radius ,r= 0.075 m

Torque ,T= 200 N.m

Power ,P = 7 kW

Lets take force = F

T = F .r

`F=(T)/(r)`

Now by putting the values

`F=(200)/(0.075)\ N`

F= 2666.66 N

Lets take rotational speed = N RPM

We know that

`P=(2\pi N\ T)/(60)`

`N=(60* P)/(2\pi \ T)`

Now by putting the values in the above equation

`N=(60* 7000)/(2\pi * 200)\ RPM`

N=334.22 RPM