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 800 kg through a vertical distance of 36 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 2.1 m/s? 16.5 kW

(b) Assuming the motor is 90 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 14 cents per kilowatt-hour. Need help in this part!

Explanation / Answer

given

m = 800 kg

h = 36 m

a. load speed, v = 2.1 m/s

hence

power = mg*v = 16480.8 W = 16.4808 kW

b. for 90 % efifciency

electric energy needed, E = 16.4808/0.9 = 18.312 kW

hence

for 0.85 of full load

E = 0.85*18.312 = 15.5652 kW

time = 0.6*12 = 12*6/10 hours

hence

energy = E*t = 112.06944 kWh

now 1 kWh = 12 cents

hence

total cost per day = 1344.83328 cents = 13.4483328 dollars

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.