Design pipelines AB and BC, C = 100, 10 m head

C = 100; minimum residual pressure at B and C = 10 m. HGL at A = 950 m. The discharge in BC is not given in the figure; it is assumed that half of the 1.3 lps is used at B, so BC carries 0.65 lps (if the actual discharge is known, substitute it in the same steps).

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.

Pipe AB

• Discharge $Q = 1.300$ lps, length $L = 2750$ m
• HGL at A = 950.00 m; ground RL at B = 920 m; residual head required = 10 m
• Available head $H = 950.00 - 920 - 10 = 20.00$ m
• Gradient $S = H/L = 20.00/2750 = 0.00727$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 62.0$ mm, adopt commercial size 65 mm
• Actual $h_f = 15.836$ m, HGL at B = 934.16 m, residual head = 14.16 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.39$ m/s

Pipe BC

• Discharge $Q = 0.650$ lps, length $L = 1700$ m
• HGL at B = 934.16 m; ground RL at C = 894 m; residual head required = 10 m
• Available head $H = 934.16 - 894 - 10 = 30.16$ m
• Gradient $S = H/L = 30.16/1700 = 0.01774$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 39.6$ mm, adopt commercial size 40 mm
• Actual $h_f = 28.847$ m, HGL at C = 905.32 m, residual head = 11.32 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.52$ m/s

Summary

Answer: AB = 65 mm and BC = 40 mm (GI), with residual heads of 14.2 m at B and 11.3 m at C.