Conveyor Speed Increase: Sprocket Change or Motor Overload?
An existing pallet conveyor is driven by a 1.5 kW four-pole motor at 1450 rpm through a worm gearbox with ratio 30:1, a 15-tooth driving sprocket, a 45-tooth driven sprocket and a chain with 19.05 mm pitch. The driven sprocket is mounted on a 160 mm diameter head drum. Production requires the belt speed to rise from about 0.135 m/s to 0.150 m/s. A maintenance technician proposes replacing the 45-tooth driven sprocket with a 40-tooth sprocket, arguing that the motor has 1.5 kW and 'should have enough margin'. After the change, operators report a 12 Hz vibration, chain noise and occasional motor overload trips when the conveyor starts fully loaded. The chain has no automatic tensioner, the centre distance is fixed at 800 mm, and the motor is fed by a direct-on-line starter. The conveyor must start with a full load of 400 kg pallets, and the gearbox has a rated output torque of 950 Nm.
1. Calculate the original and new belt speeds, the required conveyor power and the approximate chain tension. Show why the motor overloads even though the gearbox torque rating is not exceeded.
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Gearbox output speed is 1450/30 = 48.33 rpm.
Original: n_drum = 48.33 × 15/45 = 16.11 rpm. v = πDn/60 = π × 0.160 × 16.11/60 = 0.135 m/s. Drum torque T = P/ω = 1500/(2π × 16.11/60) = 890 N·m. Drum force F = T/r = 890/0.080 = 11.1 kN.
New: n_drum = 48.33 × 15/40 = 18.13 rpm. v = π × 0.160 × 18.13/60 = 0.152 m/s. If the conveyor force stays at 11.1 kN, required power is Fv = 11.1×10³ × 0.152 = 1.69 kW. That is 12.5% above the 1.5 kW motor rating, so the motor overloads on a full-load start.
Chain tension: driven sprocket pitch radius r ≈ pz/(2π). Original r = 0.01905 × 45/(2π) = 0.1365 m, so chain tension ≈ 890/0.1365 = 6.5 kN. New r = 0.01905 × 40/(2π) = 0.1213 m, so tension ≈ 890/0.1213 = 7.3 kN, about 12% higher. The gearbox output torque is only 890/(40/15) = 334 N·m, well below 950 N·m; the limiting component is motor power, not gearbox torque.
2. The 12 Hz vibration is measured after the change. What is the likely source, and what checks and mechanical changes should be made before accepting the modification?
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The driving sprocket rotates at 48.33 rpm with 15 teeth, giving a tooth-mesh frequency of 15 × 48.33/60 = 12.1 Hz. The 12 Hz peak is therefore at the sprocket mesh frequency, not a random bearing frequency. The modification also raised chain tension by about 12% and the chain has no automatic tensioner, so mesh forces and slack-side vibration are worse. A structural resonance near 12 Hz is possible.
Check chain slack against the manufacturer's allowance, sprocket alignment, sprocket wear, lubrication and the fixed 800 mm centre distance. Measure vibration at the bearing housing and on the frame with the chain removed to separate structural resonance from chain excitation. Fit an automatic tensioner or idler if the layout allows, correct alignment, and replace worn sprockets as a set. If a resonance exists, change stiffness or add damping rather than simply over-tensioning the chain. Do not accept the change until the motor overload and chain tension are solved.
3. Recommend a correct engineering solution to reach 0.150 m/s, including the starting problem. Explain why a larger motor alone is not sufficient.
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The conveyor needs about 1.69 kW at 0.152 m/s if the load force is unchanged. A 2.2 kW motor gives margin, but the gearbox, chain, bearings and drum must be checked for the higher speed and the higher chain tension. Keeping the 45-tooth sprocket and using a VFD to raise motor speed is one option: required drum speed is 0.150 × 60/(π × 0.160) = 17.9 rpm, so gearbox output must be 17.9 × 45/15 = 53.8 rpm and motor speed 53.8 × 30 = 1614 rpm. That is within a normal 4-pole motor's overspeed range, but the load still requires about 1.69 kW, so a 1.5 kW motor is not adequate.
A larger motor alone does not solve the direct-on-line start or the chain vibration. Add a soft starter or VFD to reduce start current and shock, fit a chain tensioner, align the sprockets, and verify the gearbox thermal rating at the new speed. If the load can be reduced to about 1.3 kW, the existing motor may be retained; otherwise use a 2.2 kW motor and review the entire drive train.