---
title: "Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pipe-design-ab-bc-ad-110"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pipe-design-ab-bc-ad-110
site: BCT Engineering
---

# Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2081 Baisakh, 8 marks; 2080 Bhadra, 8 marks (asked 2 times)
- Video: [Hardy Cross Method Explained: Environmental Engineering | GATE CE Numerical](https://www.youtube.com/watch?v=HCMZhXJmKHI) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pipe-design-ab-bc-ad-110
- 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 1.1 kg/cm$^2$ is required at the tap. Take Hazen Williams constant C = 100.

[Figure: reservoir at A with static water level, RL of A = 985 m. Pipe AB, L = 120 m, carries 0.35 lps and ends at B, where Tap no 1 (RL = 960 m) is located. Pipe BC, L = 150 m, carries 0.15 lps to C, where Tap no 2 (RL = 920 m) is located.]

## Short answer

Using the Hazen-Williams equation with C=100, available head, and required residual pressure of 11 m, the designed pipe diameters are AB: 25 mm, BC: 15 mm, and AD: 25 mm. All provide residual heads above 11 m and velocities below 1.1 m/s.

## Answer

Residual pressure at each tap = 1.1 kg/cm² = 11 m of water. Static water level at A = RL 985 m, so the HGL at A is 985 m. The figure gives only pipes AB and BC; for pipe AD the data of the companion problem is assumed (L = 160 m, Q = 0.5 lps, tap RL 940 m).

#### 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.

#### Design
#### Pipe AB
- Discharge $Q = 0.350$ lps, length $L = 120$ m
- HGL at A = 985.00 m; ground RL at B = 960 m; residual head required = 11 m
- Available head $H = 985.00 - 960 - 11 = 14.00$ m
- Gradient $S = H/L = 14.00/120 = 0.11667$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.3$ mm, adopt commercial size **25 mm**
- Actual $h_f = 6.383$ m, HGL at B = 978.62 m, residual head = 18.62 m (≥ 11 m, OK)
- Velocity $v = Q/A = 0.71$ m/s

#### Pipe BC
- Discharge $Q = 0.150$ lps, length $L = 150$ m
- HGL at B = 978.62 m; ground RL at C = 920 m; residual head required = 11 m
- Available head $H = 978.62 - 920 - 11 = 47.62$ m
- Gradient $S = H/L = 47.62/150 = 0.31745$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 12.6$ mm, adopt commercial size **15 mm**
- Actual $h_f = 19.990$ m, HGL at C = 958.63 m, residual head = 38.63 m (≥ 11 m, OK)
- Velocity $v = Q/A = 0.85$ 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 = 11 m
- Available head $H = 985.00 - 940 - 11 = 34.00$ m
- Gradient $S = H/L = 34.00/160 = 0.21250$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.5$ mm, adopt commercial size **25 mm**
- Actual $h_f = 16.474$ m, HGL at D = 968.53 m, residual head = 28.53 m (≥ 11 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 | 0.35 | 120 | 21.3 | 25 | 6.38 | 18.62 | 0.71 |
| BC | 0.15 | 150 | 12.6 | 15 | 19.99 | 38.63 | 0.85 |
| AD | 0.5 | 160 | 21.5 | 25 | 16.47 | 28.53 | 1.02 |

Maximum static head: B = 985 − 960 = 25 m, D = 45 m and C = 985 − 920 = 65 m. Pipe BC must therefore be of a class that can bear at least 65 m (about 0.65 MPa), e.g. GI medium or HDPE PN 10, while AB can be a lower class.

**Answer: AB = 25 mm, BC = 15 mm, AD = 25 mm (GI), all giving residual head above 11 m and velocity below 1.1 m/s.**

## Check yourself

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

1. A minimum pressure of 1.1 kg/cm2 at the tap corresponds to what head?
   A. 1.1 m
   B. 11 m
   C. 110 m
   D. 0.11 m
   Answer: B. 1 kg/cm2 is about 10 m of water, so 1.1 kg/cm2 is 11 m. (Key point: Residual head)

2. What is the available head for pipe AB?
   A. 985 - 960 - 11 = 14 m
   B. 985 - 960 = 25 m
   C. 960 - 11 = 949 m
   D. 985 + 960 - 11 m
   Answer: A. Available head = HGL at A - ground RL at B - required residual head = 14 m. (Key point: Available head for AB)

3. After computing the diameter from the gradient, what is done?
   A. Adopt the next larger commercial size and check head loss and residual head
   B. Adopt the next smaller size
   C. Use the calculated value exactly
   D. Double the diameter
   Answer: A. The next larger standard size is adopted, then actual head loss and residual head are verified. (Key point: Diameter selection)

4. What HGL is used at the start of pipe BC?
   A. 985 m, the reservoir level
   B. 978.62 m, the HGL at B after the loss in AB
   C. 960 m, the ground level at B
   D. 920 m
   Answer: B. BC starts from the HGL at B (985 - 6.38 = 978.62 m), not the reservoir level. (Key point: Pipe BC upstream head)

## Related in Water Supply Engineering

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