Question 10.4.2: It is required to transmit 36.75 kW power from an accumulato......
It is required to transmit 36.75 kW power from an accumulator through a pipeline of 10 cm diameter and 1500 m long. The ram is loaded with a weight of 1250 kN and the friction loss in the pipeline equals 2.5% of the total power being transmitted. Work out the diameter of the ram if the friction factor f = 0.01 for the pipeline.
Given:
P = 36.75 kW, d = 0.1 m, L = 1500 m, W = 1250 kN
power loss = 0.025 × 36.75 = 0.9187 kW
Friction head loss h_f =\frac{4fl}{d}\frac{U^2}{2g}=\frac{4\times 0.01 \times 1500 }{2\times 9.81 \times 0.1} U^2 =30.58 U^2
Power loss in friction = w Q h_f = 9810 × 0.7854 × 0.1^2 × U × 30.58U^2 = 2353 U^3 = 918.7
→ U, flow velocity in pipe = 0.731 m/s and Q = 0.7854 × 0.12 × U = 5.738 × 10^{–3} m^3/s
This discharge comes form the accumulator. If H is the pressure head of water in the accumulator then
Power developed by accumulator = w Q H = 36.75 × 1000 (given)
or, 9810 × 5.738 × 10^{–3} × H = 36.75 × 1000 → H = 652.87 m of water
Pressure intensity = w H = 9810 × 652.87 = 6.4 MN/m²
But pressure intensity = W/ram area
or,
6.4 \times 10^6=\frac{125 \times 1000}{\pi/4 \times D^2} \rightarrow D=0.499 m