You operate a small grain elevator near Champaign, Illinois. One of your silos u

Question

You operate a small grain elevator near Champaign, Illinois. One of your silos uses a bucket elevator that carries a full load of 700 kg through a vertical distance of 45 m. (A bucket elevator works with a continuous belt, like a conveyor belt.) (a) What is the power provided by the electric motor powering the bucket elevator when the bucket elevator ascends with a full load at a speed of 2.2 m/s?
kW

(b) Assuming the motor is 82 percent efficient, how much does it cost you to run this elevator, per day, assuming it runs 60 percent of the time between 7:00 a.m. and 7:00 p.m. with an average load of 85 percent of a full load? Assume the cost of electric energy in your location is 16 cents per kilowatt hour. $

Explanation / Answer

a) Power = work/time

work done = 700*9.8*45 (mgh)

time = 45/2.2

Power = 700*9.8*45/(45/2.2) = 15092 W

b) Cost = 0.16*(.82)*(12*60*60*0.6)(.85*15.092) = 43624.91 cent or $436.24

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