For AB the calculated diameter is 18.5 mm. What is done next?

Adopt 20 mm, the next larger commercial size, then check head loss and residual head

Adopt 15 mm, the next smaller size

Why is the HGL at B (974.85 m) and not 985 m used to design pipe BC?

Pipe BC is fed through AB, so the head lost in AB is already gone at B

2068 Chaitra · 8 marks

Design pipelines AB, BC, AD; 1 kg/cm2 at tap

Short answer

Use Hazen-Williams equation with C=100. Pipe AB (0.25 lps, 120 m) adopts 20 mm, pipe BC (0.1 lps, 150 m) adopts 15 mm, pipe AD (0.5 lps, 160 m) adopts 25 mm. Residual heads exceed 10 m at all taps.

Design pipelines AB, BC and AD for the following pipe network. A minimum pressure of 1 kg/cm $^2$ is required at the tap. Take Hazen William constant C = 100.

[Figure: Source RL = 985 m at A. AB: L = 120 m, 0.25 lps, to B with Tap no. 1 (RL = 960 m). BC: L = 150 m, 0.1 lps, to C with Tap no. 2 (RL = 920 m). AD: L = 160 m, 0.5 lps, to D with Tap no. 3 (RL = 940 m).]

Residual pressure = 1 kg/cm² = 10 m; C = 100. HGL at A = 985 m. Discharges as given: AB = 0.25 lps, BC = 0.10 lps, AD = 0.50 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 $Q = 0.250$ lps, length $L = 120$ m
HGL at A = 985.00 m; ground RL at B = 960 m; residual head required = 10 m
Available head $H = 985.00 - 960 - 10 = 15.00$ m
Gradient $S = H/L = 15.00/120 = 0.12500$
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 $h_f = 10.146$ m, HGL at B = 974.85 m, residual head = 14.85 m (≥ 10 m, OK)
Velocity $v = Q/A = 0.80$ m/s

Pipe BC

Discharge $Q = 0.100$ lps, length $L = 150$ m
HGL at B = 974.85 m; ground RL at C = 920 m; residual head required = 10 m
Available head $H = 974.85 - 920 - 10 = 44.85$ m
Gradient $S = H/L = 44.85/150 = 0.29902$
From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$ : $D_{calc} = 10.9$ mm, adopt commercial size 15 mm
Actual $h_f = 9.434$ m, HGL at C = 965.42 m, residual head = 45.42 m (≥ 10 m, OK)
Velocity $v = Q/A = 0.57$ m/s

Pipe AD

Discharge $Q = 0.500$ lps, length $L = 160$ m
HGL at A = 985.00 m; ground RL at D = 940 m; residual head required = 10 m
Available head $H = 985.00 - 940 - 10 = 35.00$ m
Gradient $S = H/L = 35.00/160 = 0.21875$
From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$ : $D_{calc} = 21.4$ mm, adopt commercial size 25 mm
Actual $h_f = 16.474$ m, HGL at D = 968.53 m, residual head = 28.53 m (≥ 10 m, OK)
Velocity $v = Q/A = 1.02$ m/s

Summary

Pipe	Q (lps)	L (m)	Calc. D (mm)	Adopted D (mm)	h_f (m)	Residual (m)	v (m/s)
AB	0.25	120	18.5	20	10.15	14.85	0.80
BC	0.1	150	10.9	15	9.43	45.42	0.57
AD	0.5	160	21.4	25	16.47	28.53	1.02

Static head is highest at C (985 − 920 = 65 m), so the pipes BC (and AB feeding it) must be of a class suitable for at least 65 m.

Answer: AB = 20 mm, BC = 15 mm, AD = 25 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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