Design pipes PQ, PR and RS, C = 110

Residual pressure = 1 kg/cm² = 10 m at all taps; C = 110. HGL at source P = 950 m. The flows given for each pipe are taken as the discharge carried by that pipe (PR = 0.7 lps, RS = 0.3 lps, PQ = 0.25 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 = 110.

Pipe PR

• Discharge $Q = 0.700$ lps, length $L = 150$ m
• HGL at P = 950.00 m; ground RL at R = 860 m; residual head required = 10 m
• Available head $H = 950.00 - 860 - 10 = 80.00$ m
• Gradient $S = H/L = 80.00/150 = 0.53333$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 19.5$ mm, adopt commercial size 20 mm
• Actual $h_f = 71.565$ m, HGL at R = 878.43 m, residual head = 18.43 m (≥ 10 m, OK)
• Velocity $v = Q/A = 2.23$ m/s

Pipe RS

• Discharge $Q = 0.300$ lps, length $L = 160$ m
• HGL at R = 878.43 m; ground RL at S = 845 m; residual head required = 10 m
• Available head $H = 878.43 - 845 - 10 = 23.43$ m
• Gradient $S = H/L = 23.43/160 = 0.14647$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.5$ mm, adopt commercial size 20 mm
• Actual $h_f = 15.894$ m, HGL at S = 862.54 m, residual head = 17.54 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.95$ m/s

Pipe PQ

• Discharge $Q = 0.250$ lps, length $L = 180$ m
• HGL at P = 950.00 m; ground RL at Q = 930 m; residual head required = 10 m
• Available head $H = 950.00 - 930 - 10 = 10.00$ m
• Gradient $S = H/L = 10.00/180 = 0.05556$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.0$ mm, adopt commercial size 25 mm
• Actual $h_f = 4.303$ m, HGL at Q = 945.70 m, residual head = 15.70 m (≥ 10 m, OK)
• Velocity $v = Q/A = 0.51$ m/s

Summary

Static heads: Q = 950 − 930 = 20 m, R = 90 m, S = 105 m. Pipes PR and RS must therefore be of a class for at least 105 m; if the high pressure is a problem, a break pressure tank can be provided near R.

Answer: PQ = 25 mm, PR = 20 mm, RS = 20 mm. (Velocity in PR is 2.23 m/s, still within the 3 m/s limit.)