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.

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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

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