Question

A belt conveyor (7 ply, 900 mm wide) is to be used to convey ore on a roadway whose gradient is 1 in 10 against the load. If the peak-loading rate is 150 t/h calculate the power required and the maximum permissible length of the conveyor. Belt speed is 1.5 m/s, angle of wrap 2400 , coefficient of friction 0.25, drive efficiency 85% and the mass of moving parts of the conveyor is 75 kg/m. Allow for a maximum belt tension of 5.25 kN/ply/m and an idler friction factor of 0.03

Answer 1

The maximum load that the conveyor can handle is given by:

Load = Belt tension x belt width x coefficient of friction

= 5.25 x 900 x 0.25

= 1143.75 N/m

The power required to move the conveyor is given by:

Power = Load x speed x gradient x idler friction factor

= 1143.75 x 1.5 x 0.1 x 0.03

= 17.15625 W/m

The total power required to move the conveyor is given by:

Power = Power per unit length x conveyor length

= 17.15625 x L

The maximum permissible length of the conveyor can be found by rearranging the equation above:

L = Power / Power per unit length

= 150 / (17.15625 x 0.85)

= 96.8 m