Asked by Max
A grain bucket conveyor (grain leg) raises soybeans from the ground level up to the peak of its height, h= 150 ft. A bushel of soybean weight, W=60 lb/bu. The bucket elevator can deliver, Q= 120 bu/min, to the top spout. The motor shaft power to drive the elevator is 40 hp. Determine the power (hp) required to elevate the grain.
Answer: 32.7 hp
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Answer: 32.7 hp
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Answers
Answered by
Chanz
Bushels?? Really.
P = mgh/t
m/t is your converted Bu/min.
Then maybe convert the whole thing into hogsheads and find a different class.
P = mgh/t
m/t is your converted Bu/min.
Then maybe convert the whole thing into hogsheads and find a different class.
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