An electric lift makes 14 double journeys per hour. A load of 5 tonne is raised by it through a height 50m and it returns empty.

The weight of the cage is half a tonne and its counterweight is 2 tonne. The efficiency of the hoist is 90% and that of the motor is 87%.

Calculate the hourly consumption in kWh.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

When the lift is moving upwards, the weight being raised is 5 ton + 0.5 ton and a weight of 2 ton which is the counterweight is being lowered. The load is raised through a height of 50 m.

The work to be done is m*g*h,

m*g*h = (5.5 - 2)* 9.8 * 50 KJ

=> 3.5*490 = 1715 KJ

The efficiency of the hoist is 90% and that of the electric motor is 87%.

So the work that actually has to be done is 1715 / (0.9 * 0.87)

=> 2190.2 KJ

When the lift comes down, the counterweight of 2 ton is going up and the cage of 0.5 ton is coming down.

Work to be done is 1.5*9.8*50 = 735 KJ

Including the loss due to the efficiency being less than 1, the work to be done is 735/ (0.9*0.87) = 938.6 KJ.

The lift makes 14 journeys in a hour. The total energy consumed in an hour is (2190.2 + 938.6)*14 = 43803.2 KJ.

In terms of KWh it is 12. 167 KWh

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial