A power plant has a conveyor that carries coal from a barge on the Ohio River to a storage pile. The beginning of the conveyor starts 10 ft below ground level and carries material up about 40 ft above the ground.The conveyor is powered by an electric motor located in a pit, which was prone to flooding every time it rained. Maintenance workers decided to replace the electric-motor drive with hydraulics because a hydraulic motor would tolerate being under water occasionally.
The electric motor was rated at 15 hp at 1,800 rpm and supplied with 440 Vac, three phase, 60 Hz. Input speed to the conveyor’s gearbox was 1,800 rpm, so the electric motor used a belt drive with a 1:1 speed ratio. At full pressure, the hydraulic motor transmitted 18 hp at 900 rpm, so it used a 2:1 belt drive in order to provide the 1,800-rpm input speed to the conveyor. This gave a 20% safety factor for efficiency loss.
The conveyor ran well with the 15-hp electric motor, and the hydraulic motor was able to run the conveyor when it was empty. However, when the conveyor was loaded and stopped at the end of the work shift, the hydraulic motor could not start and run the conveyor.
All the math was checked and no error was found. So why was the electric motor able to power the conveyor, but the hydraulic motor wasn’t?
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Congratulations to Paul Lopez, CFPS, of Michaelson Fluid Power Inc., Smithfield, R.I., whose name was drawn at random from those who correctly answered May’s “Troubleshooting Challenge.”