Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)

Residual pressure at each tap = 1.1 kg/cm² = 11 m of water. Static water level at A = RL 985 m, so the HGL at A is 985 m. The figure gives only pipes AB and BC; for pipe AD the data of the companion problem is assumed (L = 160 m, Q = 0.5 lps, tap RL 940 m).

Method

Use the Hazen-Williams equation in SI units:

$$h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$

with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Design

Pipe AB

• Discharge $Q = 0.350$ lps, length $L = 120$ m
• HGL at A = 985.00 m; ground RL at B = 960 m; residual head required = 11 m
• Available head $H = 985.00 - 960 - 11 = 14.00$ m
• Gradient $S = H/L = 14.00/120 = 0.11667$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.3$ mm, adopt commercial size 25 mm
• Actual $h_f = 6.383$ m, HGL at B = 978.62 m, residual head = 18.62 m (≥ 11 m, OK)
• Velocity $v = Q/A = 0.71$ m/s

Pipe BC

• Discharge $Q = 0.150$ lps, length $L = 150$ m
• HGL at B = 978.62 m; ground RL at C = 920 m; residual head required = 11 m
• Available head $H = 978.62 - 920 - 11 = 47.62$ m
• Gradient $S = H/L = 47.62/150 = 0.31745$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 12.6$ mm, adopt commercial size 15 mm
• Actual $h_f = 19.990$ m, HGL at C = 958.63 m, residual head = 38.63 m (≥ 11 m, OK)
• Velocity $v = Q/A = 0.85$ m/s

Pipe AD

• Discharge $Q = 0.500$ lps, length $L = 160$ m
• HGL at A = 985.00 m; ground RL at D = 940 m; residual head required = 11 m
• Available head $H = 985.00 - 940 - 11 = 34.00$ m
• Gradient $S = H/L = 34.00/160 = 0.21250$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.5$ mm, adopt commercial size 25 mm
• Actual $h_f = 16.474$ m, HGL at D = 968.53 m, residual head = 28.53 m (≥ 11 m, OK)
• Velocity $v = Q/A = 1.02$ m/s

Summary

Maximum static head: B = 985 − 960 = 25 m, D = 45 m and C = 985 − 920 = 65 m. Pipe BC must therefore be of a class that can bear at least 65 m (about 0.65 MPa), e.g. GI medium or HDPE PN 10, while AB can be a lower class.

Answer: AB = 25 mm, BC = 15 mm, AD = 25 mm (GI), all giving residual head above 11 m and velocity below 1.1 m/s.