Belt pull: what sizes a conveyor drive

The crane of Chapter 4 handles one load at a time. This chapter turns to the opposite case: a continuous, uninterrupted flow of material or people. Belt conveyors, escalators, cableways and the huge machines of open-cast mining move product at a steady rate along a fixed path. Their control problems are no longer about one suspended mass. They are about keeping many drives in step, staying reliable and safe over long unattended runs, and coordinating several high-power mechanisms into one productive system. This lesson is course sections 5.1.1 to 5.1.3.

Conveyor types

The belt conveyor is the workhorse, carrying bulk and unit loads over a few metres to many kilometres. A roller conveyor moves discrete items on driven or gravity rollers, common in warehouses and assembly lines. A chain (apron) conveyor carries heavy, hot or abrasive material where a belt would not survive.

The mechanics differ, but all three present a mainly constant-torque load: the torque needed to move the material against friction and gravity hardly depends on speed, so power rises in direct proportion to belt speed. That is exactly the load a variable-frequency drive handles best: change the speed to match throughput, with the motor giving constant torque across the range.

Effective belt pull

Sizing starts from the effective belt pull $F_e$, the total resistance the belt must overcome at the drive pulley. For a belt of length $L$ carrying $m'_\text{load}$ kg/m of material through a rise $H$:

$$F_e = f g L\,(m'_\text{belt} + m'_\text{load}) + m'_\text{load}\, g H \tag{5.1}$$

where $f$ is an equivalent friction coefficient for the idlers and belt. The first term is friction on everything that moves; the second lifts the material. Only the material is lifted: the belt goes up on the carrying strand and comes back down on the return. Drive torque and motor power follow:

$$T_\text{drive} = F_e R_\text{pulley}, \qquad P_\text{motor} = \frac{F_e v}{\eta} \tag{5.2}$$

with $v$ the belt speed and $\eta$ the drive-train efficiency.

The belt must not slip

The tensions on either side of the drive pulley, $T_1$ on the tight side and $T_2$ on the slack side, differ by exactly the effective pull, $F_e = T_1 - T_2$. Their ratio is bounded by the capstan (Euler) relation

$$\frac{T_1}{T_2} \le e^{\mu\theta}$$

with $\mu$ the friction between belt and pulley lagging and $\theta$ the wrap angle in radians. Beyond that ratio the belt slips on the pulley. So the take-up must hold at least $T_2 = F_e / (e^{\mu\theta} - 1)$ on the slack side. For the Tutorial 5.1 belt, with 180° of wrap and $\mu = 0.35$, $e^{\mu\theta} \approx 3.0$ and the take-up must hold about 5.0 kN. A larger wrap (a snub pulley) or a better lagging lowers it.

Where the resistance comes from: DIN 22101

The lumped eq. 5.1 is a quick first estimate. A conveyor that is actually going to be built is specified by the fuller model of DIN 22101 (or the equivalent CEMA method), which splits the resistance into four physically distinct terms. This matters because the terms scale differently with length, load and speed, so which one dominates, and therefore which design change actually saves power, depends on the machine.

This is a more complete model, not a rearrangement of eq. 5.1. The simple form counts the belt mass once and ignores the idlers; the full model counts the belt on both strands and adds the rotating idler mass. For the inclined example above, the simple friction term gives 1.99 kN while the DIN primary resistance, with a typical 20 kg/m of idlers, gives 3.06 kN: more than half as much again.

Primary resistance $F_H$ is the rolling and flexing resistance along both strands: idler bearings, the belt cover indenting over each idler, belt and material flexing between idler sets. It takes one artificial friction coefficient on the whole moving mass:

$$F_H = f L g\,\left[m'_R + (2 m'_B + m'_L)\cos\delta\right] \tag{5.3}$$

where $m'_R$ is the rotating idler mass per metre, $m'_B$ the belt (counted twice, carrying and return), $m'_L$ the material and $\delta$ the incline. $f$ runs from 0.016 for a well-aligned, well-kept conveyor to 0.030 for a dusty, poorly aligned one: almost a factor of two on the main term of a long conveyor. Alignment and idler condition are energy issues, not only maintenance ones.

Secondary resistance $F_N = (C - 1) F_H$ covers the losses concentrated at the ends: accelerating the material at the loading point, skirtboard friction, belt cleaners, pulley bearings. The length coefficient $C$ falls from about 3 on a 20 m conveyor to about 1.05 on a 2 km one. On a short conveyor the secondary resistance is about twice the primary; on a long overland conveyor it is a few percent. That is why short conveyors are disproportionately power-hungry per tonne.

Slope resistance $F_{St} = m'_L L g \sin\delta = m'_L g H$ is the useful lifting work. It depends only on the rise and the mass lifted, not on length, speed or friction, and it is the only term that can be recovered: on a decline it becomes a driving force and the machine regenerates.

Special resistances $F_S$ collect whatever the standard model leaves out: tilted idlers set to steer the belt, ploughs, extra cleaners.

The total is $F_e = F_H + F_N + F_{St} + F_S$, and again $P = F_e v / \eta$.

Tutorial 5.1 (400 m, flat)Course example (120 m, 14 m rise)Overland (2 km, 6°)

Quick formula (eq. 5.1)DIN 22101 (eq. 5.3)

Length L

400 m

Incline δ

0.0° · H = 0.0 m

Material m′_load

90 kg/m

Belt m′_belt

25 kg/m

Friction coefficient f

0.022

0.016 for a well-aligned, well-kept conveyor; 0.030 for a dusty, poorly aligned one.

Belt speed v

2.5 m/s

Friction dominates: idler quality, belt cover and alignment are where the power goes.

Effective pull F_e

9.93 kN

Motor power F_e v / η

27.6 kW

Standard motor, if one drive

30 kW

F_e = f g L (m'_belt + m'_load) + m'_load g H ; P = F_e v / η (η = 0.9)

Predict first

Switch the Tutorial 5.1 conveyor from the quick formula to the DIN model (20 kg/m of idlers). What happens to the effective pull?

It stays near 9.9 kNIt rises by about 10 %It rises by about 70 %

The decomposition is a diagnosis. If $F_{St}$ dominates, the only real savings are in lifting less or recovering energy on the decline. If $F_H$ dominates, look at idler quality, the belt cover compound and alignment. If $F_N$ dominates, the conveyor is short and the loading arrangement is where to look.

FoundationStart here if this is new to you

Pulling a loaded sledge across flat snow, the effort is all friction, and it grows with the weight of the sledge and the load. Pulling it up a slope, you also lift the load, and that part does not care how far you walk, only how high you climb. A conveyor is a sledge that never stops: the friction part depends on its length, the lift part only on its rise.

ExplorerGo deeper: derivations and open questions

Decline. Set the incline to −8° on the overland preset. Which term changes sign, and what must the drive do for the whole of its running life? What would you fit instead of a braking resistor?

Short belts. With the DIN model, reduce the length to 20 m. What fraction of the pull is now secondary resistance, and why does a short feeder belt cost more power per tonne than a long one?