how to solve 3

Question

5. Calculate the quantity of rock, in tones per hour, which can be transported by a belt conveyor with the following particulars: Maximum tension in the belt=35,5kN Contact angle of the belt on the driving pulley=185 degrees Frictional force=2,5kN Belt speed= 95m/min Delivery height=35m Coefficient of friction=0,3

Answer 1

The Maximum tension in the belt T1 is 35.5 kN

Contact Angle 'a' is 185° = 185 * (π/180) = 3.2288 radians

Frictional Force = 2.5 kN

Belt speed = 95 m/min

Delivery Height = 35 m

Coefficient of Friction µ is 0.3

 

We know that

T2 = (35.5)/2.6343

T2 = 13.475 kN

Therefore the Tension in slight side is 13.475 kN

 

Effective Driving Force Fe = T1 - T2

Fe = 35.5 - 13.475

Fe = 22.02 kN

 

Therefore Total Force required to lift the load is

F  = Effective Driving force + Frictional Force

Since Effective Force and Frictional Force are in opposite direction,

F  = Effective Driving force - Frictional Force

F = 22.0 - 2.5 = 19.52 kN

 

From Newtons law of Motion, F= ma

ma = 19.52 kN

m * v/t = 19.52 kN           (Acceleration a = Speed / time)

m / t = 19.5*10^3/ 95/60

m / t = 12,330.90 kg/sec

m/t = 12,328.42 * 0.001 / (1/3600) tonnes/hour

m/t = 44,391.26 tonnes per hour

Therefore the Quantity of Rock in tonne per hour is 44,391.26 tonnes per hour.