Question

An electrical lift make 12 double journey per hour. A load of 5 tonnes is raised by it through a height 50 m and it returns empty. The lift takes 65 seconds to go up and 48 seconds to return. The weight of the cage is 1/2 tonne and that of the counterweight is 2.5 tonne. The efficiency of the hoist is 80 per cent that of the motor is 85 %. Calculate the hourly consumption in kWh.

Answer 1

The lift is shown in Fig..

Weight raised during upward journey

= 5 + 1/2 − 2.5 = 3 tonne = 3000 kg-wt

Distance travelled = 50 m

Work done during upward journey = 3000 × 50 = 15 × 104 m-kg

Weight raised during downward journey

= 2.5 − 0.5 = 2 tonne = 2000 kg

Similarly, work done during downward journey

= 2000 × 50 = 10 × 10−4 m-kg.

Total work done per double journey

= 15 × 104 + 10 × 104 = 25 × 104 m-kg

Now, 1, m-kg = 9.8 joules

∴ Work done per double journey

= 9.8 × 25 × 104 J = 245 × 104 J

No. of double journey made per hour = 12

∴ work done per hour = 12 × 245 × 104 = 294 × 105 J

Energy drawn from supply = 294 × 105 /0.8 × 0.85 = 432.3 × 105 J

Now, 1 kWh = 36 × 105 J

∴ Energy consumption per hour = 432.3 × 105 /36 × 105 = 12 kWh