---
title: "Design pipelines AB, BC, AD; 1 kg/cm2 at tap"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c68"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c68
site: BCT Engineering
---

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

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2068 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c68
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

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).]

## 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.

## Answer

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A minimum pressure of 1 kg/cm^2 at the tap is used in the design as what head?
   A. 1 m
   B. 10 m
   C. 100 m
   D. 0.1 m
   Answer: B. 1 kg/cm^2 equals 10 m of water, so 10 m residual head is required at each tap. (Key point: Residual pressure conversion)

2. How is the available head for pipe AB found?
   A. HGL at A + RL of B + 10 m
   B. HGL at A - RL of B - 10 m
   C. RL of A - HGL at B
   D. HGL at A - 10 m only
   Answer: B. Available head = upstream HGL - ground RL of the tap - required residual head = 985 - 960 - 10 = 15 m. (Key point: Available head)

3. For AB the calculated diameter is 18.5 mm. What is done next?
   A. Adopt 15 mm, the next smaller size
   B. Adopt 18.5 mm exactly
   C. Adopt 20 mm, the next larger commercial size, then check head loss and residual head
   D. Adopt 50 mm to be safe
   Answer: C. The next larger commercial size is adopted, then actual h_f and residual head are checked. (Key point: Choosing the commercial size)

4. Why is the HGL at B (974.85 m) and not 985 m used to design pipe BC?
   A. B is lower than A
   B. BC is shorter
   C. The flow in BC is smaller
   D. Pipe BC is fed through AB, so the head lost in AB is already gone at B
   Answer: D. Pipes in series: HGL at B = 985 - 10.146 = 974.85 m, which is the head available to start BC. (Key point: Upstream HGL for BC)

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