Question 10.6.1: A hydraulic lift is required to lift a load of 12 kN through......
A hydraulic lift is required to lift a load of 12 kN through a height of 10 m, once in every 1.75 minutes. The speed of the lift is 0.75 m/s. During the working stroke of the lift, water from the accumulator and the pump at a pressure of 400 N/cm² is supplied to the lift. If the efficiency of the pump is 80% and that of the lift 75%, find the power required to drive the pump and the minimum capacity of the accumulator. Neglect frictional losses in the pipe.
Given: W = 12 kN, H = 10 m, t = 1.75 \min = 105 s, ν = \text{lift speed} = 0.75 m/s, p = 400 × 10^4 N/m^2, η_p = 0.8, η_l = 0.75
Total work done in the lift = useful work = W × ν = 9000 W or, 9 kW
Power supplied to lift per second = useful work/η_l = 9/0.75 = 12 kW
Time taken by lift to travel through 10 m = t′ = H/v = 10/0.75 = 13.33 s
Idle period of lift = t – t′ = 105 – 13.33 = 91.67 s.
The accumulator stores the energy in this idle period that is supplied during the working stroke. Let P_1 be the power output of the pump then power supplied by accumulator to the lift = 12 – P_1
Energy stored in the accumulator during idle period = 91.67 × P_1 kJ
= energy supplied by the accumulator in t^′ s = (12 – P_1) × 13.33 → P_1 = 1.523 kW
Input to the pump = P_1/η_p = 1.523/0.8 = 1.90375 kW
Energy stored in accumulator = accumulator capacity = P_1 × 91.67 = 13.96 kJ