Motor Sizing for a Conveyor Start-Up
A 15 m inclined belt conveyor lifts aggregate 4 m at 0.8 m/s. Loaded belt mass is 900 kg, belt speed 0.8 m/s, and the drive drum is 400 mm diameter. Running friction and incline resistance total 4.2 kN at the belt. The conveyor starts fully loaded 30 times per shift, each start accelerating from rest to full speed in 2.5 s. A 7.5 kW, 4-pole induction motor (1440 rpm) with a direct-on-line starter is proposed. The gearbox ratio is fixed by the drum. The maintenance team reports nuisance tripping of the overload on start-up and asks whether to fit a larger motor or a soft starter.
1. Calculate the steady-state power and the acceleration torque, and determine whether the 7.5 kW motor is adequate.
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Steady state: belt speed 0.8 m/s, resistance 4.2 kN. P = Fv = 4200×0.8 = 3.36 kW. Plus the lift component is already in the 4.2 kN resistance, so steady power is 3.36 kW.
Acceleration: linear acceleration a = 0.8/2.5 = 0.32 m/s². Inertial force = ma = 900×0.32 = 288 N. Total start force = 4200 + 288 = 4488 N. Start power at full speed = 4488×0.8 = 3.59 kW - but the motor must supply peak torque during acceleration, not just power.
Drum torque: drum radius 0.2 m, so torque = 4488×0.2 = 898 N·m at the drum. Drum speed = 0.8/(2π×0.2) = 0.637 rev/s = 38 rpm. Motor at 1440 rpm, so gear ratio ≈ 38:1. Motor torque required = 898/38 = 23.6 N·m. A 7.5 kW motor at 1440 rpm produces rated torque = 7500/(2π×1440/60) = 49.7 N·m. So steady and average acceleration torque are within rating - the problem is the starting current and torque characteristic of a DOL induction motor, not steady capability.
2. Explain the tripping and recommend the fix, with reasoning about starting current and torque.
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A DOL induction motor draws 6-8 times rated current at locked rotor while producing only about 1.5-2.5 times rated torque. The overload relay sees this inrush for the 2.5 s acceleration and trips, even though the thermal capacity of the motor is not exceeded. The motor's rated torque (≈50 N·m) comfortably exceeds the 23.6 N·m required, so a bigger motor is the wrong fix - it would draw even more inrush current and may not solve the trip.
The correct fix is a soft starter or VSD to limit starting current and control acceleration torque. A soft starter ramps voltage so current is held to about 2-3 times rated and torque builds smoothly over the set ramp; a VSD gives full torque at low current and also allows the 2.5 s ramp to be tuned. Recommend a soft starter as the lower-cost solution, set the ramp to about 3 s and the current limit to 300% of rated, and verify the overload relay is set to the motor's service-factor current rather than a lower value. The judgement is to match the starting method to the load's inertia and duty cycle, not to oversize the motor.