Design main transmission pipes PQ and QR

Average demand 60 lpcd. The peak factor is not stated; 3 is assumed. C = 100 and residual head = 15 m at Q and R. HGL at P = 325 m. Pipe PQ carries the demand of Q and R: (2500 + 21000) = 23,500 people, and pipe QR carries the 21,000 people at R.

• $Q_{PQ} = 23500 \times 60 \times 3/86400 = 48.96$ lps
• $Q_{QR} = 21000 \times 60 \times 3/86400 = 43.75$ 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 PQ

• Discharge $Q = 48.958$ lps, length $L = 600$ m
• HGL at P = 325.00 m; ground RL at Q = 185 m; residual head required = 15 m
• Available head $H = 325.00 - 185 - 15 = 125.00$ m
• Gradient $S = H/L = 125.00/600 = 0.20833$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 123.6$ mm, adopt commercial size 150 mm (raised from the head-based size because velocity would exceed 3 m/s)
• Actual $h_f = 48.788$ m, HGL at Q = 276.21 m, residual head = 91.21 m (≥ 15 m, OK)
• Velocity $v = Q/A = 2.77$ m/s

Pipe QR

• Discharge $Q = 43.750$ lps, length $L = 250$ m
• HGL at Q = 276.21 m; ground RL at R = 45 m; residual head required = 15 m
• Available head $H = 276.21 - 45 - 15 = 216.21$ m
• Gradient $S = H/L = 216.21/250 = 0.86485$
• From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 88.4$ mm, adopt commercial size 150 mm (raised from the head-based size because velocity would exceed 3 m/s)
• Actual $h_f = 16.506$ m, HGL at R = 259.71 m, residual head = 214.71 m (≥ 15 m, OK)
• Velocity $v = Q/A = 2.48$ m/s

Summary

The static head at Q is 325 − 185 = 140 m and at R is 325 − 45 = 280 m, which is too high for ordinary pipes (and the residual heads of 91 m and 215 m are far more than needed). A break pressure tank at Q (RL 185 m) should be provided: the pressure is reduced to zero there, QR then has a static head of 140 m only, and the head available for QR becomes about 185 − 45 − 15 = 125 m, but the velocity limit of 3 m/s (125 mm gives 3.6 m/s) still requires 150 mm.

Answer: PQ = 150 mm and QR = 150 mm (with a BPT recommended at Q).