Question

An electric hoist makes 10 double journey per hour. In each journey, a load of 6 tonnes is raised to a height of 60 meters in 90 seconds. The hoist cage weighs 1/2 tonne and has a balance load of 3 tonnes. The efficiency of the hoist is 80 % and of the driving motor 88 %. Calculate (a) electric energy absorbed per double journey (b) hourly energy consumption in kWh (c) hp (British) rating of the motor required (d) cost of electric energy if hoist works for 4 hours/day for 30 days. Cost per kWh is 50 paise.

Answer 1

Wt. of cage when fully loaded = 6(1/2) tonne-wt.

Force exerted on upward journey = 6(1/2) - 3 = 3(1/2) tonne-wt.

= 3(1/2) x 1000 = 3,500 kg-wt.

Force exerted on downward journey = (3 - 1/2) = 2(1/2) tonnes-wt. = 2500 kg-wt

Distance moved = 60 m

Work done during upward journey = 3,500 × 60 m-kg

Work done during downward journey = 2,500 × 60 m-kg

Work done during each double journey = (3,500 + 2,500) × 60 = 36 × 104 m-kg

= 36 × 104 × 9.81 = 534 × 104 J

Overall η = 0.80 × 0.88

∴ Energy input per double journey

= 534 × 104 /0.8 × 0.88

= 505 × 104 J

(a) Electric energy absorbed per double journey = 505 × 104 /36 × 105 = 1.402 kWh

(b) Hourly consumption = 1.402 × 10 = 14.02 kWh

(c) Before calculating the rating of the motor, maximum rate of working should be found. It is seen that maximum rate of working is required in the upward journey.

Work done = 3,500 × 60 × 9.81 = 206 × 104 J

Time taken = 90 second

B.H.P of motor = (206 x 104)/(90 x 0.8 x 746) = 38.6

(d) Cost = 14.02 × (30 × 4) × 50/100 = Rs. 841.2