You are designing a conveyor to run at a minimum velocity of 2.5 ft/sec. the req

Question

You are designing a conveyor to run at a minimum velocity of 2.5 ft/sec. the required minimum force to move product at this velocity is 10000 lbf. The diameter of the primary drive roller is 24in. This roller is mounted on a shaft passing through the roller and connected to a worm drive gear box used to turn it. Both the roller and gear box are locked to the shaft using sled runner keys. The gear box has a C-face style motor mount and an efficiency of 85%. What is the ratio of this gear box for an 1800 rpm motor? Adjust to commonly found ratio. What size motor [hp] is required to run the conveyor?

Explanation / Answer

Power = 10000 * 2.5 = 25000 lbf ft/sec

angular velocity of primary drive roller = 2.5 * 12 / 12 = 2.5 rad/s or 23.8732 rpm

for an 1800 rpm motor

gear ratio   =   23.8732 / 1800 = 0.013263

commonly found ratio   = 0.015

horse power of motor   = 25000 / .85 = 29411.765 lbf ft/s   or 53.476 hp

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