I have to select a motor for a mixing tank. The tank is 7-ft in diameter and 4-ft high. Two-thirds of the tank will be filled with polymer slurry. I expect the impeller diameter to be 3 ft with a rotational speed of about 300 rpm. Can you please tell me how to calculate the horsepower requirement of the motor? I would appreciate if you could quote a formula.
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Tom Bishop, P.E. Forum Moderator 17 Posts
Re: Motor horsepower equation25 June 2007 at 1:29pmThe answer to the question is that the impeller size and rotational speed are incompatible with the vessel size, regardless of the contents. Any 3-ft diameter impeller operating at 300 rpm will send the tank contents flying. The least power input would be with a hydrofoil impeller, but at 40 hp, the contents would remain in the tank for only seconds, unless the vessel were completely enclosed. A pitched-blade turbine would require a 250-hp motor and a straight-blade impeller would require a 500-hp motor. Even at 50 rpm, a straight-blade turbine would provide intense mixing, with a 3-hp motor. Either the impeller size or the rotational speed must be drastically reduced to make this design feasible.