Design pipes AB and BC, source RL 1050 m

Residual pressure at each tap = 1 kg/cm² = 10 m. Source A has RL 1050 m, so HGL at A = 1050 m. Flows: AB = 0.35 lps, BC = 0.20 lps.

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 $Q = 0.350$ lps, length $L = 320$ m
• HGL at A = 1050.00 m; ground RL at B = 970 m; residual head required = 10 m
• Available head $H = 1050.00 - 970 - 10 = 70.00$ m
• Gradient $S = H/L = 70.00/320 = 0.21875$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.7$ mm, adopt commercial size 20 mm
• Actual $h_f = 50.456$ m, HGL at B = 999.54 m, residual head = 29.54 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.11$ m/s

Pipe BC

• Discharge $Q = 0.200$ lps, length $L = 250$ m
• HGL at B = 999.54 m; ground RL at C = 910 m; residual head required = 10 m
• Available head $H = 999.54 - 910 - 10 = 79.54$ m
• Gradient $S = H/L = 79.54/250 = 0.31818$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 14.0$ mm, adopt commercial size 15 mm
• Actual $h_f = 56.761$ m, HGL at C = 942.78 m, residual head = 32.78 m (≥ 10 m, OK)
• Velocity $v = Q/A = 1.13$ m/s

Summary

Pipe class

The pipe class must resist the maximum static (zero-flow) head:

• Section AB: static head at B = 1050 − 970 = 80 m
• Section BC: static head at C = 1050 − 910 = 140 m (14 kg/cm² = 1.4 MPa)

So pipe BC must be of a high class that can withstand at least 140 m (for example GI heavy or HDPE PN 16 / PN 20), while AB needs only a class for 80 m (GI medium or PN 10). The high-class pipe is costly.

Reducing the high class

Provide a break pressure tank (BPT) at B (RL 970 m). The tank breaks the pressure to zero (atmospheric) and starts a new hydraulic line from RL 970 m, so the static head in BC becomes 970 − 910 = 60 m only, and the whole scheme can use a pipe class for 80 m (PN 10 / GI medium).

Redesign of BC with BPT at B: available head $H = 970 - 910 - 10 = 50$ m, $S = 0.200$, $D_{calc} = 15.4$ mm, adopt 20 mm, $h_f = 13.98$ m, residual head = 46.02 m, $v = 0.64$ m/s. A pressure-reducing valve at B could also be used.

Answer: AB = 20 mm, BC = 15 mm (without BPT). Highest class needed is for 140 m static head in BC; with a BPT at B the BC static head falls to 60 m and a lower class (for 80 m) is enough, with BC = 20 mm.