+ -12 points Tipler6 7.P.090. My Notes + Ask Your Teacher You operate a small gr
ID: 1875138 • Letter: #
Question
+ -12 points Tipler6 7.P.090. My Notes + Ask Your Teacher You operate a small grain elevator near Champaign, Illinois. One of your silos uses a bucket elevator that carries a full load of 900 kg through a vertical distance of 44 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 2.4 m/s? kw (b) Assuming the motor is 85 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 13 cents per kilowatt hour. eBook Submit Answer Save Progress Practice Another VersionExplanation / Answer
a)
power = work done / time
= (force x distance)/time
= ([9.81x900] x 2.4)/1 (9.81, to overcome gravitational force of load)
= 21189.6 W = 21.2 kW
b)
load is reduced; power assuming full efficiency = work / time
= ([9.81x(900x0.85)] x 2.4) / 1
= 18011.16 W
real power we need to supply = 18011.16 x 100/85
= 21.2 kW
number of hours = 12 x 0.6 = 7.2 h
cost per day = power * number of hours * price per kilowatt hr
= 21.2 kW * 7.2h * $0.13/kWh
= $19.8
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.