Motor Selection and Drive Sizing for a Conveyor
A horizontal belt conveyor 30 m long carries bulk material at 40 t/h. The belt speed is 1.2 m/s, the load is uniformly distributed, and the effective friction coefficient for the loaded belt is 0.035. The conveyor is driven by a three-phase induction motor through a gearbox with ratio 25:1 and efficiency 0.92. The motor is a 415 V, 50 Hz, 4-pole machine. The plant engineer proposes a 7.5 kW motor. The starting torque required is 1.8 times the running torque, and the supply allows a direct-on-line start with a starting current of 6 times full-load current. The motor nameplate gives full-load efficiency 88% and power factor 0.84.
1. Calculate the running power and torque at the conveyor drum and at the motor, and check whether 7.5 kW is adequate.
Read the worked answer
Belt load per metre
- Mass flow = 40 t/h = 40 000/3600 = 11.11 kg/s.
- Belt speed = 1.2 m/s, so load per metre = 11.11/1.2 = 9.26 kg/m.
- Total load on the 30 m belt = 9.26 × 30 = 277.8 kg, weight = 277.8 × 9.81 = 2725 N.
Tractive force and power at the drum
- F = μ × W = 0.035 × 2725 = 95.4 N.
- Drum power = F × v = 95.4 × 1.2 = 114.5 W.
This is very small, which suggests the dominant load is belt and idler friction, not the material. If the belt itself and idlers are included, the effective friction coefficient of 0.035 should be applied to the total moving mass. If we conservatively take the belt and idler mass as comparable to the load, the tractive force roughly doubles to ~190 N, giving ~230 W. Even allowing a factor of five for belt weight, idler friction and incline, the drum power remains well under 2 kW.
Motor shaft power
- Motor output = drum power / gearbox efficiency = 114.5 / 0.92 = 124 W (or ~250 W with belt mass included).
Torque at the drum
- Drum speed: belt speed / drum circumference. Without the drum diameter this cannot be fixed, but drum torque = F × r. For a 0.3 m radius drum, T = 95.4 × 0.3 = 28.6 N·m.
Torque at the motor
- Motor speed 1500 rpm (synchronous); at ~1440 rpm full load, and gearbox ratio 25, drum speed = 1440/25 = 57.6 rpm.
- Motor torque = P/ω = 124/(2π × 1440/60) = 124/150.8 = 0.82 N·m.
Adequacy check. The calculated running power is far below 7.5 kW. A 7.5 kW motor is grossly oversized for the calculated load. However, the calculation is sensitive to the assumed friction coefficient and to whether belt and idler masses are included. The engineer must confirm the total moving mass, the incline (zero here), and the starting condition (a loaded belt start requires the 1.8× torque).
Starting torque check: required starting torque = 1.8 × running torque. A standard 7.5 kW 4-pole motor has a starting torque of about 2.0–2.5 times full-load torque, so it would start the conveyor. But an oversized motor runs at low load, with poor power factor and efficiency.
2. Evaluate the electrical supply implications and recommend a correctly sized motor.
Read the worked answer
Full-load current of the proposed 7.5 kW motor
- Input power = output / efficiency = 7500/0.88 = 8523 W.
- Line current I = P/(√3 × V × cos φ) = 8523/(1.732 × 415 × 0.84) = 8523/603.7 = 14.1 A.
- Direct-on-line starting current = 6 × 14.1 = 84.7 A.
This starting current may cause unacceptable voltage dip on a weak supply, and the protection settings must accommodate it.
Correctly sized motor
- Required output ≈ 0.25 kW (allowing for belt and idler mass). Apply a service factor of 1.5–2.0 for starting and shock: select 0.55 kW or 0.75 kW, the next standard sizes.
- For a 0.75 kW motor: input = 750/0.88 = 852 W, I = 852/603.7 = 1.41 A, starting current ≈ 8.5 A. This is far kinder to the supply.
Recommendations
- Re-survey the conveyor: measure the actual running current of the existing drive or conduct a pull test to establish the true tractive force. The 0.035 friction coefficient should be confirmed against the idler type and belt condition.
- Select a motor with a high starting torque characteristic (e.g. a design with starting torque ≥ 2.0 pu) to meet the 1.8× requirement, or use a variable-frequency drive to provide controlled acceleration and eliminate the high DOL starting current.
- If a VFD is used, the motor can be started at low current and ramped to speed, and the oversizing penalty is reduced because the drive can operate the motor at a better load point.
- Check the gearbox: a 25:1 ratio with the selected motor must transmit the drum torque with adequate service factor. The gearbox efficiency of 0.92 is reasonable for a helical unit.
- Verify protection coordination: the overload relay must be set to the motor full-load current, and the short-circuit protection must ride through the starting current for the acceleration time.