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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 :

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

[tex]F=\dfrac{T}{r}[/tex]

Now by putting the values

[tex]F=\dfrac{200}{0.075}\ N[/tex]

F= 2666.66 N

Lets take rotational speed = N RPM

We know that  

[tex]P=\dfrac{2\pi N\ T}{60}[/tex]

[tex]N=\dfrac{60\times P}{2\pi \ T}[/tex]

Now by putting the values in the above equation

[tex]N=\dfrac{60\times 7000}{2\pi \times 200}\ RPM[/tex]

N=334.22 RPM

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