Design pipes AB and BC, source RL 180 m

Source A has RL 180 m; minimum pressure 1.5 kg/cm² = 15 m; C = 100. A to B: 44 lps, 700 m, RL of B = 154 m. B to C: 18 lps, 550 m, 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 = 180.00 m; ground RL at B = 154 m; residual head required = 15 m
• Available head $H = 180.00 - 154 - 15 = 11.00$ m
• Gradient $S = H/L = 11.00/700 = 0.01571$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 201.9$ mm, adopt commercial size 250 mm
• Actual $h_f = 3.881$ m, HGL at B = 176.12 m, residual head = 22.12 m (≥ 15 m, OK)
• Velocity $v = Q/A = 0.90$ m/s

Pipe BC

• Discharge $Q = 18.000$ lps, length $L = 550$ m
• HGL at B = 176.12 m; ground RL at C = 126 m; residual head required = 15 m
• Available head $H = 176.12 - 126 - 15 = 35.12$ m
• Gradient $S = H/L = 35.12/550 = 0.06385$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 107.7$ mm, adopt commercial size 125 mm
• Actual $h_f = 17.035$ m, HGL at C = 159.08 m, residual head = 33.08 m (≥ 15 m, OK)
• Velocity $v = Q/A = 1.47$ m/s

Summary

The small available head (11 m) for the 700 m pipe AB needs a large pipe. A 250 mm pipe gives a velocity of 0.9 m/s, which is acceptable.

Answer: AB = 250 mm and BC = 125 mm.