Question 10.6.2: An accumulator maintains a steady pressure of 8 Mpa in a 7.5......
An accumulator maintains a steady pressure of 8 Mpa in a 7.5 cm hydraulic main. A hydraulic lift is supplied with pressure water from the main and the point at which the supply to the lift is drawn off is at a distance of 600 m from the accumulator. The ram of the lift is 20 cm in diameter and the load on it, inclusive of its own weight, is 120 kN. Assuming the friction of the ram and cage etc. to be equivalent to an addition of 5% of the gross load on the ram, determine the speed with which the lift will ascend. Take f = 0.008 for the hydraulic main and neglect the minor losses.
Given: p = 8 × 10^6 N/m^2, d = 0.075 m, L = 600 m, D = 0.2 m, W = 120 kN, f = 0.008
Let the flow velocity in the main = U m/s. Then
h_f=\frac{4\times 0.008 \times U^2}{2\times 9.81 \times 0.075} =13.05 U^2
pressure intensity at the lift = p – w h_f = 8 × 10^6 – 13.05 U^2
and force on the ram = (8 × 10^6 – 13.05 U^2) × 0.7854 × 0.2^2 = (120 + 0.05 × 120) × 10^3
→
U = 5.581 m/s and Q = 0.7854 × 0.2² × U = 0.0246 m³/s
∴ volocity of lift\frac{Q}{\pi/4 \times 0.2^2}=0.783 m/s