---
title: "Design pipes AB and BC, source RL 1050 m"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c76"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c76
site: BCT Engineering
---

# Design pipes AB and BC, source RL 1050 m

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2076 Chaitra, 8 marks
- 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/pd-c76
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipelines AB and BC for the following pipe network. A minimum pressure of 1 kg/cm$^2$ is required at the tap. Take Hazen William constant as 100. Which pipe class has to be designed? Is there any option to reduce high level pipe class and how?

[Figure: Source A (RL = 1050 m); pipe AB, L = 320 m, carries 0.35 lps to B, Tap no. 1 (RL = 970 m); pipe BC, L = 250 m, carries 0.2 lps to C, Tap no. 2 (RL = 910 m).]

## Short answer

Without break pressure, AB = 20 mm, BC = 15 mm. The highest class must withstand 140 m static head in BC. Installing a break pressure tank at B reduces the static head to 60 m, allowing a lower class and BC redesigned to 20 mm.

## Answer

Residual pressure at each tap = 1 kg/cm² = 10 m. Source A has RL 1050 m, so HGL at A = 1050 m. Flows: AB = 0.35 lps, BC = 0.20 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 $Q = 0.350$ lps, length $L = 320$ m
- HGL at A = 1050.00 m; ground RL at B = 970 m; residual head required = 10 m
- Available head $H = 1050.00 - 970 - 10 = 70.00$ m
- Gradient $S = H/L = 70.00/320 = 0.21875$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.7$ mm, adopt commercial size **20 mm**
- Actual $h_f = 50.456$ m, HGL at B = 999.54 m, residual head = 29.54 m (≥ 10 m, OK)
- Velocity $v = Q/A = 1.11$ m/s

#### Pipe BC
- Discharge $Q = 0.200$ lps, length $L = 250$ m
- HGL at B = 999.54 m; ground RL at C = 910 m; residual head required = 10 m
- Available head $H = 999.54 - 910 - 10 = 79.54$ m
- Gradient $S = H/L = 79.54/250 = 0.31818$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 14.0$ mm, adopt commercial size **15 mm**
- Actual $h_f = 56.761$ m, HGL at C = 942.78 m, residual head = 32.78 m (≥ 10 m, OK)
- Velocity $v = Q/A = 1.13$ 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 | 320 | 18.7 | 20 | 50.46 | 29.54 | 1.11 |
| BC | 0.2 | 250 | 14.0 | 15 | 56.76 | 32.78 | 1.13 |

#### Pipe class
The pipe class must resist the maximum static (zero-flow) head:
- Section AB: static head at B = 1050 − 970 = 80 m
- Section BC: static head at C = 1050 − 910 = **140 m** (14 kg/cm² = 1.4 MPa)

So pipe BC must be of a high class that can withstand at least 140 m (for example GI heavy or HDPE PN 16 / PN 20), while AB needs only a class for 80 m (GI medium or PN 10). The high-class pipe is costly.

#### Reducing the high class
Provide a **break pressure tank (BPT) at B** (RL 970 m). The tank breaks the pressure to zero (atmospheric) and starts a new hydraulic line from RL 970 m, so the static head in BC becomes 970 − 910 = 60 m only, and the whole scheme can use a pipe class for 80 m (PN 10 / GI medium).

Redesign of BC with BPT at B: available head $H = 970 - 910 - 10 = 50$ m, $S = 0.200$, $D_{calc} = 15.4$ mm, adopt **20 mm**, $h_f = 13.98$ m, residual head = 46.02 m, $v = 0.64$ m/s. A pressure-reducing valve at B could also be used.

**Answer: AB = 20 mm, BC = 15 mm (without BPT). Highest class needed is for 140 m static head in BC; with a BPT at B the BC static head falls to 60 m and a lower class (for 80 m) is enough, with BC = 20 mm.**

## Check yourself

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

1. What decides the pipe class?
   A. Maximum static (zero-flow) head
   B. Maximum velocity
   C. Pipe length
   D. Residual head only
   Answer: A. Pipes must resist the highest static pressure; BC has 1050 - 910 = 140 m. (Key point: Pipe class)

2. How can the high pipe class be avoided?
   A. Use a smaller diameter
   B. Provide a break pressure tank at B to reset the static head
   C. Increase flow
   D. Use a longer pipe
   Answer: B. A BPT at B (RL 970) breaks pressure to atmospheric; BC static head becomes 60 m. (Key point: Reduce class)

3. 1 kg/cm2 minimum tap pressure equals what head?
   A. 1 m
   B. 10 m
   C. 100 m
   D. 0.1 m
   Answer: B. 1 kg/cm2 is about 10 m of water. (Key point: Residual head)

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