Estimate the required drive power of a belt conveyor carrying bulk material (ore, gravel, grain and the like). Vary the mass flow, belt speed, length and lift height to see the friction and lift power breakdown, the total required power and the effective tension at the drive pulley update in real time.
Parameters
Mass flow rate
t/h
Mass of material conveyed per hour
Belt speed v
m/s
Horizontal length L
m
Conveyor length (tail pulley to head pulley)
Lift height H
m
Negative for downhill — gives negative lift power
Running-resistance coefficient f
Equivalent friction of idler bearings, belt flexing, etc.
Belt + rotating-parts mass per length
kg/m
Belt plus rotating mass of the idlers
Results
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Required power P (kW)
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Friction power (kW)
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Lift power (kW)
—
Effective tension Te (N)
—
Material mass on belt (kg/m)
—
Lift power fraction (%)
—
Inclined belt conveyor — conveying animation
Bulk material is carried up an inclined belt from a low loading point to a higher discharge point. The belt and lumps travel up the incline and tumble off the end. H is the lift height, v the belt speed.
Required power P [W] is the sum of the friction (running-resistance) power and the lift power. f: running-resistance coefficient, q_total: total moving mass on the belt [kg/m], L: conveyor length [m], v: belt speed [m/s], ṁ: material mass flow [kg/s], H: lift height [m], g = 9.81 m/s².
$$T_e=\frac{P}{v}$$
Effective tension Te [N] is the required power divided by the belt speed — the net circumferential force the drive pulley applies to the belt. It sets the belt strength, pulley friction and take-up device.
What is the Belt Conveyor Power Simulator?
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Those long conveyor belts you see moving things in factories and mines — how big a motor do you actually need to turn one?
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Good question. The required power of a belt conveyor is basically "two terms added together". One is the friction power — the power to keep dragging the belt and the material on top of it horizontally across the rollers. The other is the lift power — the power to raise the material to a higher place. Try raising the "lift height H" on the left, and you will see the lift-power card jump up.
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I see. So for a conveyor that only runs horizontally, the lift power is zero and I only need to worry about friction?
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Exactly. And the friction power is the tricky one. The formula is P_f = f·q_total·g·L·v, and it is proportional to the conveyor length L. On a mine conveyor several kilometres long, just dragging the belt itself and the rotating rollers eats an enormous power before you have moved any material. That is why q_total includes not only the material mass but also the mass of the belt and the idler rotating parts — the "belt + rotating-parts mass per length" slider on the left.
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What is the running-resistance coefficient f? Is it different from a friction coefficient?
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f is what is called an "apparent friction coefficient" — it is not pure sliding friction. The rolling resistance of the idler bearings, the internal loss as the material flexes up and down with the belt, the resistance of the belt riding over the rollers — f lumps all those small resistances into one number. For a standard installation it is around f = 0.020-0.025. Poor alignment, cold weather with stiff lubrication or a lot of dust all push f up. In standards like DIN 22101, this f is the heart of the power calculation.
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You can set the lift height H negative. Does a downhill conveyor need no motor?
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Nice thing to spot. Making H negative makes the lift power negative — the potential energy of the falling material "pushes" the conveyor along. When this negative lift power exceeds the friction power, the required power P itself goes negative. Now the conveyor is not a drive but a brake. You have to absorb the surplus energy with dynamic braking or regeneration, and the design problem becomes keeping it from running away. On steep downhill conveyors that bring ore down from a hill, some installations actually generate electricity while conveying.
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There is also an "effective tension Te" shown. How is that different from power?
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Power P is "work per unit time"; Te is "the net force on the belt". They are linked by Te = P/v. For the same power, the slower you run the belt, the larger Te becomes. And Te is the number that sets the belt's tensile strength, the friction margin against slip at the drive pulley (the Euler / capstan equation) and the weight of the take-up. You size the motor by power and the belt and pulleys by Te — you need both together to call it a conveyor design.
Frequently Asked Questions
The required power is the sum of the friction (running-resistance) power and the lift power. The friction power is P_f = f·q_total·g·L·v, proportional to the running-resistance coefficient f, the total moving mass per unit belt length q_total (material plus belt and rotating parts), the conveyor length L and the belt speed v. The lift power is P_h = ṁ·g·H, set only by the material mass flow ṁ and the height H it is raised. This tool adds the two to give the total required power P_total.
On a long, nearly horizontal conveyor the friction power dominates. Because the friction power is proportional to the conveyor length L, a run of several hundred metres consumes a large power just dragging the belt itself. On a short, steep incline, the lift power dominates instead. The "Lift power fraction" card shows the split. At the default settings (L=100 m, H=10 m) the lift term is about 74% of the total.
Setting the lift height H negative (downhill conveying) makes the lift power P_h negative — the potential energy of the falling material helps drive the conveyor. When this negative lift power exceeds the friction power, the total required power P_total itself goes negative and the conveyor becomes a brake rather than a drive. The surplus energy must then be handled by dynamic braking or regeneration, and the design focus shifts to preventing overspeed. This tool correctly shows the conditions where P_total goes negative.
The effective tension Te is the net circumferential force the drive pulley applies to the belt, given by Te = P_total / v. For the same required power, a slower belt speed v gives a larger Te. Te is the reference value for selecting the belt tensile strength, for the friction transfer at the drive pulley (slip margin via the Euler / capstan equation), and for sizing the take-up weight. Checking Te as well as power tells you whether the belt will neither break nor slip.
Real-World Applications
Mines and crushing plants: Open-pit mines and quarries use long-distance belt conveyors to carry blasted ore and crushed stone to crushers and stockpiles. The run can reach several kilometres, and at that scale the friction power overwhelmingly dominates. Shaving a little off the running-resistance coefficient f — high-efficiency idlers, low-rolling-resistance belt cover rubber — changes the annual electricity bill substantially. On steep downhill conveyors bringing material down from a hill, the lift power goes negative, and some installations use regenerative braking to supply part of the plant's electricity.
Ports and bulk handling: At bulk terminals that transfer coal, iron ore, grain or cement clinker between ships and yards, multi-stage conveyor systems are built together with ship loaders, stackers and reclaimers. The required power and effective tension of each conveyor are summed to choose the drive arrangement (single- or multiple-pulley drive) and the take-up device. Because the throughput varies through the day, power is estimated for the peak mass flow.
Power stations and cement plants: Belt conveyors sit at the heart of the process — coal feed to the boilers of a coal-fired power station, raw-material and clinker handling in a cement plant. Because continuous operation is the rule, the required power feeds straight into the electricity cost, and a small efficiency gain makes a large long-term difference. The lift power of inclined sections and the friction power of horizontal sections are separated so the optimum speed and belt width can be chosen section by section.
Distribution centres and airports: Conveyors carrying unit loads (cases, parcels, baggage) have smaller throughput and lift than bulk conveyors, but the same reasoning applies directly. Because there are many conveyors, keeping the friction power of each one low matters to the whole facility's power draw. In airport baggage systems, the power of merge, divert and incline sections is estimated to select motors and inverters.
Common Misconceptions and Pitfalls
The biggest misconception is "you only need to consider the lift power". The work of raising material H metres is intuitive, so people sometimes size the motor on the lift power ṁgH alone. But on a long, nearly horizontal conveyor the friction power f·q_total·g·L·v is far larger. The friction power is proportional to the conveyor length L, and q_total includes not just the material but the mass of the belt and the idler rotating parts. The friction power does not fall to zero even when running empty — that is the decisive difference from the lift power. Stretch L in this tool and you will see the friction power overtake the lift power.
Next, the assumption that "the running-resistance coefficient f is a fixed value". The default f = 0.022 here is a representative figure for a standard installation, room temperature and a standard speed. The real f varies over the range 0.015-0.05 with installation accuracy (idler misalignment and steps), temperature (cold weather stiffens the grease and raises resistance), belt sag, the moisture and stickiness of the material, and the state of maintenance. On long conveyors, static friction and inertia add at start-up, so a momentarily larger power and tension are needed than in steady running. Taking f optimistically leads to field troubles where the conveyor cannot start or the belt slips.
Finally, "required power is not the same as motor rating". What this tool calculates is the required power at the drive-pulley shaft. The actual motor rating is found by dividing this by the transmission efficiency of the gearbox and coupling (roughly 0.85-0.95) and then multiplying by a margin factor (about 1.1-1.25) for start-up margin, load variation and future expansion. The effective tension Te is a separate axis of study. For the same required power, a slower belt speed gives a larger Te, which decides whether the belt's tensile strength and the friction transfer at the drive pulley (wrap angle and the Euler equation) hold up. Size the motor by power and the belt and pulleys by effective tension — always study both together.
How to Use
Enter the mass flow rate (qNum, tonnes/hour) and select material type (qRange: ore, gravel, grain) to define bulk load characteristics
Set belt velocity (vNum, m/s) and span length (lNum, meters) for your conveyor geometry
Input elevation gain (hNum, meters) if the belt runs inclined; the simulator calculates friction power and lift power separately
Click Calculate to obtain Required Power (kW), Effective Tension Te (N), and the lift power fraction (%)
Worked Example
Iron ore conveyor: 150 tonnes/hour, belt velocity 2.5 m/s, horizontal span 45 m, friction coefficient 0.5. Required power = 18.4 kW (friction 12.8 kW + lift 0 kW). Now add 12 m elevation: lift power rises to 5.6 kW, total becomes 23.9 kW, with lift fraction 23.4%. Material mass on belt reaches 22.5 kg/m at steady state.
Practical Notes
Friction power dominates in long horizontal runs; incline angle steeper than 15° typically triggers significant lift power demand and requires motor upsizing
Wet ore and fine gravel increase effective friction coefficient by 0.1–0.2; account for seasonal moisture when selecting drive capacity
Effective tension Te scales linearly with span length but non-linearly with belt speed—doubling speed roughly quadruples air drag and bearing losses
For grain (low density, ~0.8 tonnes/m³) same throughput needs wider belt or faster speed than iron ore; check idler spacing to prevent sagging