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

Question

You operate a small grain elevator near Champaign, Iilinois. One of your silos uses a bucket elevator that carries a full load of 750 kg through a vertical distance of 36 m. (A bucket elevator works with a continuous beit, 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? 16.2 kw (b) Assuming the motor is 81 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 20 cents per killowatt hour Recall the definition of power

Explanation / Answer

a) Power = Work/time

Here Work = mgh = 750*9.8*36 = 264600 J

Time = distance/speed = 36/2.2 = 16.36 sec

Power = 264600/16.36= 16170 W = 16.2 kW

b) The net time that the elevator runs for, T = (60/100)*12 hours =7.2 hours.

the average power consumed when taking (the average) load assuming100% efficiency,

P = [(85/100)*m] gv

=> P = [(85/100)*750] *9.8*2.2

=> P = 13744.5 watts.

If P' is the the average power actually required given that we haveonly 81% efficiency, we can see that (81/100)

P' = P
=> 0.81 P' =13744.5 watts
=> P' = 13744.5 / 0.81 watts

         = 16968.52 watts

         = 16.97 kW

Thus, the average energy consumed in kWh,

E = average powerin kW * time in hr

    = P' * T

    = 16.97 kW * 7.2 hours

    = 122.173 kWh.

The net cost in $ = E * cost per kWh in $

                         = 122.173* $ (20 / 100)

                         = $ 24.43

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