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

Answer:

F= 2666.66 N

N=334.22 RPM          

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

$F=\dfrac{T}{r}$

Now by putting the values

$F=\dfrac{200}{0.075}\ N$

F= 2666.66 N

Lets take rotational speed = N RPM

We know that  

$P=\dfrac{2\pi N\ T}{60}$

$N=\dfrac{60\times P}{2\pi \ T}$

Now by putting the values in the above equation

$N=\dfrac{60\times 7000}{2\pi \times 200}\ RPM$

N=334.22 RPM