how to solve 1

Question

3. A conveyor belt drive placed at an inclination angle of 15 degrees has a length of 120m between the loading point and discharge point. The belt speed is 2,5m/s and is subjected to a friction force of 8,5kN. Assume the contact angle on the driving pulley to be 220 degrees. The coefficient of friction is 0,25 and the efficiency of the drive is 85%. The maximum tension in the belt is 27,49 kN. Calculate the quantity of rock transported in per tonne.

Answer 1

Contact Angle 'a' is 220° = 220 * (π/180) = 3.8397 radians

The relation between tension and contact angle is 

Where T1 is Maximum tension in the belt = 27.49 kN .

         a = contact angle =3.8397 radians

           µ is Coefficient of Friction = 0.25

Therefore the Tension in slight side is 10.526 kN

 

Effective Driving Force Fe = T1 - T2

Fe = 27.49 - 10.526

Fe = 16.964 kN

 

Therefore Total Force Acting in lifting the load is

F  = Effective Driving force - Frictional Force

F = 16.964 - 8.5 = 8.464 kN

 

For inclined planes the resultant force is 

F = mg sin ϴ + μ mg cos ϴ

Where ϴ is angle of inclination is = 15°

8.464 = m(g sin ϴ + μ mg cos ϴ)

8.464 = m(9.8 sin 15 + (0.25) (9.8) cos 15)

8.464 = m(2.536 + 2.366)

8464 = m(4.9025)

m = (8464)/(4.9025)

m = 1726 

The quantity of rock transported per tonne is 1726/1000 = 1.726 kg per ton.