---
title: "Design pipes AB and BC, source RL 180 m"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-as67"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-as67
site: BCT Engineering
---

# Design pipes AB and BC, source RL 180 m

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2067 Asar (old course)
- 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-as67
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipeline AB and BC in the pipe network as shown in figure.

[Figure: Source at A, RL = 180 m; A to B: 44 lps, L = 700 m, RL of B = 154 m; B to C: 18 lps, L = 550 m, RL of C = 126 m.] Minimum pressure in pipe line should be 1.5 kg/cm$^2$. Take Hazen William's coefficient as 100.

## Short answer

Use Hazen-Williams equation. For AB, calculated 201.9 mm, adopt 250 mm. For BC, calculated 107.7 mm, adopt 125 mm. Velocities 0.90 and 1.47 m/s.

## Answer

Source A has RL 180 m; minimum pressure 1.5 kg/cm² = 15 m; C = 100. A to B: 44 lps, 700 m, RL of B = 154 m. B to C: 18 lps, 550 m, RL of C = 126 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.

#### Pipe AB
- Discharge $Q = 44.000$ lps, length $L = 700$ m
- HGL at A = 180.00 m; ground RL at B = 154 m; residual head required = 15 m
- Available head $H = 180.00 - 154 - 15 = 11.00$ m
- Gradient $S = H/L = 11.00/700 = 0.01571$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 201.9$ mm, adopt commercial size **250 mm**
- Actual $h_f = 3.881$ m, HGL at B = 176.12 m, residual head = 22.12 m (≥ 15 m, OK)
- Velocity $v = Q/A = 0.90$ m/s

#### Pipe BC
- Discharge $Q = 18.000$ lps, length $L = 550$ m
- HGL at B = 176.12 m; ground RL at C = 126 m; residual head required = 15 m
- Available head $H = 176.12 - 126 - 15 = 35.12$ m
- Gradient $S = H/L = 35.12/550 = 0.06385$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 107.7$ mm, adopt commercial size **125 mm**
- Actual $h_f = 17.035$ m, HGL at C = 159.08 m, residual head = 33.08 m (≥ 15 m, OK)
- Velocity $v = Q/A = 1.47$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| AB | 44 | 700 | 201.9 | 250 | 3.88 | 22.12 | 0.90 |
| BC | 18 | 550 | 107.7 | 125 | 17.03 | 33.08 | 1.47 |

The small available head (11 m) for the 700 m pipe AB needs a large pipe. A 250 mm pipe gives a velocity of 0.9 m/s, which is acceptable.

**Answer: AB = 250 mm and BC = 125 mm.**

## Check yourself

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

1. A minimum pressure of 1.5 kg/cm^2 corresponds to a residual head of:
   A. 1.5 m
   B. 10 m
   C. 150 m
   D. 15 m
   Answer: D. 1 kg/cm^2 = 10 m, so 1.5 kg/cm^2 = 15 m. (Key point: Residual head conversion)

2. With source RL 180 m, RL of B 154 m, the head available for AB is:
   A. 11 m
   B. 26 m
   C. 41 m
   D. 15 m
   Answer: A. H = 180 - 154 - 15 = 11 m. (Key point: Head available for AB)

3. Why does AB need a large diameter (250 mm)?
   A. The head available is small for a long 700 m pipe carrying 44 lps
   B. The pressure needed is very high
   C. The pipe is vertical
   D. Because C is low
   Answer: A. Small available head over a long length gives a small gradient (0.0157), which needs a big diameter. (Key point: Large pipe for small head)

4. The head available for BC is 35.12 m. This is obtained from:
   A. 180 - 126 - 15
   B. HGL at B after losses in AB (176.12) - 126 - 15
   C. 154 - 126 - 15
   D. 176.12 - 154
   Answer: B. HGL at B = 180 - 3.881 = 176.12 m; then 176.12 - 126 - 15 = 35.12 m. (Key point: Series design of BC)

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