Question

A belt conveyor must transmit 135 kW of power. The driving pulley has a diameter of 500 mm, and rotates at 75 r/min. Assume that the contact angle of the driving belt on the pulley is 275° and that the coefficient of friction between the belt and the pulley is 0,25. Calculate the total tensile load on the bearings.

Answer 1

First, we need to calculate the tension in the belt using the power transmitted and the speed of the driving pulley.

Power transmitted, P = 135 kW
Speed of driving pulley, N = 75 r/min
Diameter of driving pulley, D = 500 mm = 0.5 m

Using the formula for power transmitted by a belt:
P = T.v
where T = tension in the belt, v = velocity of the belt

The velocity of the belt can be calculated as:
v = πDN/60
v = (3.1415)(0.5)(75)/60
v = 2.95 m/s

Substitute the values of P and v into the power equation:
135,000 = T(2.95)
T = 45,763 N

To calculate the total tensile load on the bearings, we need to consider the force required to overcome friction. The contact angle is given as 275°, which means the arc of contact is 275°. The normal force is calculated as: W = Tcos(θ), where θ = 275° = 4.799 radians

W = 45,763cos(4.799)
W = 45,763cos(275)
W = 45,763(0.0551)
W = 2,525 N

The frictional force is then calculated as: F = μW
F = (0.25)(2,525)
F = 631.25 N

Therefore, the total tensile load on the bearings is the sum of the tension in the belt and the frictional force:
Total tensile load = T + F
Total tensile load = 45,763 + 631.25
Total tensile load = 46,394 N

Therefore, the total tensile load on the bearings is 46,394 N.