Design pipes AB and BC, A RL 210 m, 44 and 18 lps

C = 100, minimum residual head = 10 m. HGL at A = RL of A = 210 m. A to B: 700 m, 44 lps, RL of B = 154 m. B to C: 550 m, 18 lps, RL of C = 126 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.

Pipe AB

• Discharge $Q = 44.000$ lps, length $L = 700$ m
• HGL at A = 210.00 m; ground RL at B = 154 m; residual head required = 10 m
• Available head $H = 210.00 - 154 - 10 = 46.00$ m
• Gradient $S = H/L = 46.00/700 = 0.06571$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 150.5$ mm, adopt commercial size 200 mm
• Actual $h_f = 11.506$ m, HGL at B = 198.49 m, residual head = 44.49 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.40$ m/s

Pipe BC

• Discharge $Q = 18.000$ lps, length $L = 550$ m
• HGL at B = 198.49 m; ground RL at C = 126 m; residual head required = 10 m
• Available head $H = 198.49 - 126 - 10 = 62.49$ m
• Gradient $S = H/L = 62.49/550 = 0.11363$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 95.7$ mm, adopt commercial size 100 mm
• Actual $h_f = 50.500$ m, HGL at C = 147.99 m, residual head = 21.99 m (≥ 10 m, OK)
• Velocity $v = Q/A = 2.29$ m/s

Summary

The head-based size of AB is 150.5 mm, which is slightly above 150 mm, so 200 mm is adopted (other option: 150 mm in part length and 200 mm in the rest). Velocity in BC is 2.29 m/s (< 3 m/s).

Answer: AB = 200 mm and BC = 100 mm.