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.
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.