The supply pipe in a hydraulic ram installation is 60 mm in diameter and 4 m in length. The waste valve is 130 mm diameter having an effective lift of 6.5 mm and a weight of 13.25 N; it makes 125 beats per minute. Find the discharge through the delivery pipe per minute against a delivery head of 8 m above the water level in the supply reservoir. Neglect all losses.
Given: d = 0.06 m, L = 4 m, valve weight = 13.25 N, h = 6.5 mm
Waste valve diameter D = 130 mm
→ Base area of the waste valve = 0.7854 × 130² = 13273 mm²
Upward pressure required for closing the valve = weight of the valve/area
p = 13.25/13273 = 0.000998 N/mm²
→ pressure head = p/w = 0.000998 × 10^6/9810 = 0.102 m of water
∴ Flow velocity U through the waste valve =\sqrt{2gh}=\sqrt{2\times 9.81 \times 0.102 }=1.414 m/s
Area of flow through which the waste water flows = π × D × h = 2655 mm²
Area of the supply pipe = 0.7854 × 60² = 2827 mm²
∴ maximum velocity in the supply pipe U = 2655 × 1.414/2827 = 1.328 m/s
Retarding head =\frac{LU}{gt}=\frac{4\times 1.328}{9.81 \times t}=8 (difference of water level in the discharge and supply tanks)
→ t = 0.0677 s
During this time the water is supplied to the discharge line from the supply line.
However, the velocity changes and can be taken as the average velocity = U/2 = 0.664 m/s
∴ Quantity of water supplied/beat = 2827 × 10^{–6} × 0.664 = 1.27 × 10^{–4} m^3/s
quantity of water supplied per minute = 1.27 × 10^{–4} × 125 = 0.15625 m^3/\min = 15.625 l/\min