Mining Methods & Machinery

Belt Conveyor Capacity & Power

Volumetric and mass throughput of a troughed belt, and the power needed to lift the carried material.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A belt conveyor carries a continuous stream of broken rock whose volumetric capacity is simply the cross-sectional area of the load profile times the belt speed. Multiplying by the bulk density turns that into a mass rate (t/h). The power a conveyor draws is the force it must overcome times the belt speed; on an inclined run the dominant term is the rate at which the material is *lifted* — mass flow times gravity times lift height.

The Core Formula Matrix

Volumetric capacity:

Q_{v} = A v

[m^{3} / s]

(

A

= load cross-section,

v

= belt speed).

Mass capacity:

Q_{m} = A v ρ

[kg/s]

; in tonnes per hour

Q_{m} [t/h] = 3.6 A v ρ

(with

ρ

kg/m^{3}

).

Lift power:

P = \overset{m}{˙} g H

(

\overset{m}{˙}

= mass flow

[kg/s]

H

= lift height).

The ‘IIT Traps’

⚠
The 3.6 factor for t/h. $kg/s \times 3600/1000 = 3.6$ ; forgetting it leaves the answer 3.6× too small.
⚠
Bulk density, not solid density. Conveyor capacity uses the loose bulk density of broken rock, which is well below the intact density.
⚠
**Lift power uses vertical height $H$ , not belt length $L$ .** On an incline, only the rise lifts the material.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A belt carries a load cross-section of

0.2 m^{2}

at a speed of

2.5 m/s

. Its volumetric capacity is:

Q2MEDIUM2 Marks · NAT

A belt with load cross-section

0.15 m^{2}

runs at

2 m/s

carrying ore of bulk density

1600 kg/m^{3}

. Its mass capacity is ______ t/h. (Round off to the nearest whole number.)

t/h

Q3HARD2 Marks · NAT

An inclined conveyor lifts ore at a mass flow rate of

100 kg/s

through a vertical height of

20 m

. The power required to lift the material is ______ kW. (Round off to two decimal places.)