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...

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