---
title: "Design pipelines AB, BC and CD, C = 120"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c74"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c74
site: BCT Engineering
---

# Design pipelines AB, BC and CD, C = 120

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2074 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-c74
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

A layout of water distribution is as shown below. Design pipelines AB, BC and CD considering Hazen-William's constant of 120. Minimum pressure required at B, C and D is 12 m of water.

[Figure: Source at A; AB: discharge = 0.25 lps, length = 700 m; BC: discharge = 0.15 lps, length = 500 m; CD: discharge = 0.1 lps, length = 600 m. Reduced levels: A = 500 m, B = 480 m, C = 465 m, D = 440 m.]

## Short answer

AB = 32 mm, BC = 20 mm, CD = 15 mm. Residual heads: 15.7 m at B, 19.0 m at C, 17.1 m at D, all exceeding the 12 m minimum. Velocities range 0.31–0.57 m/s.

## Answer

Hazen-Williams constant C = 120; the residual pressure at B, C and D must be at least 12 m. HGL at A = RL of A = 500 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 = 120.

#### Pipe AB
- Discharge $Q = 0.250$ lps, length $L = 700$ m
- HGL at A = 500.00 m; ground RL at B = 480 m; residual head required = 12 m
- Available head $H = 500.00 - 480 - 12 = 8.00$ m
- Gradient $S = H/L = 8.00/700 = 0.01143$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 28.1$ mm, adopt commercial size **32 mm**
- Actual $h_f = 4.281$ m, HGL at B = 495.72 m, residual head = 15.72 m (≥ 12 m, OK)
- Velocity $v = Q/A = 0.31$ m/s

#### Pipe BC
- Discharge $Q = 0.150$ lps, length $L = 500$ m
- HGL at B = 495.72 m; ground RL at C = 465 m; residual head required = 12 m
- Available head $H = 495.72 - 465 - 12 = 18.72$ m
- Gradient $S = H/L = 18.72/500 = 0.03744$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.2$ mm, adopt commercial size **20 mm**
- Actual $h_f = 11.711$ m, HGL at C = 484.01 m, residual head = 19.01 m (≥ 12 m, OK)
- Velocity $v = Q/A = 0.48$ m/s

#### Pipe CD
- Discharge $Q = 0.100$ lps, length $L = 600$ m
- HGL at C = 484.01 m; ground RL at D = 440 m; residual head required = 12 m
- Available head $H = 484.01 - 440 - 12 = 32.01$ m
- Gradient $S = H/L = 32.01/600 = 0.05335$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 14.5$ mm, adopt commercial size **15 mm**
- Actual $h_f = 26.922$ m, HGL at D = 457.09 m, residual head = 17.09 m (≥ 12 m, OK)
- Velocity $v = Q/A = 0.57$ 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 | 700 | 28.1 | 32 | 4.28 | 15.72 | 0.31 |
| BC | 0.15 | 500 | 18.2 | 20 | 11.71 | 19.01 | 0.48 |
| CD | 0.1 | 600 | 14.5 | 15 | 26.92 | 17.09 | 0.57 |

The pipe sizes are small because the flows are only a few tenths of a litre per second, and the velocities (0.3–0.6 m/s) are low; they are acceptable for a rural system with the limited head available. Maximum static head is at D: 500 − 440 = 60 m, so a pipe class for 60 m (about 0.6 MPa) is needed.

**Answer: AB = 32 mm, BC = 20 mm, CD = 15 mm.**

## Check yourself

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

1. What is the HGL at the source A?
   A. 500 m (RL of A)
   B. 512 m
   C. 488 m
   D. 480 m
   Answer: A. With no tank height given, HGL at A equals its RL = 500 m. (Key point: Starting HGL)

2. After AB, how is the available head in BC found?
   A. Using HGL at A again
   B. Using the HGL at B after losses (495.72 m) minus RL of C minus 12 m
   C. Using RL of B
   D. Using static head
   Answer: B. Each downstream pipe starts from the HGL left after the upstream head loss. (Key point: Next pipe head)

3. How is the pipe class for this layout chosen?
   A. From maximum static head, D: 500 - 440 = 60 m
   B. From the head loss in AB
   C. From diameter
   D. From C value
   Answer: A. Pipe class follows maximum static head, which is at the lowest point D. (Key point: Pipe class)

## Related in Water Supply Engineering

- [Design pipelines AB and BC, C = 100, 10 m head](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c70) — Water Supply Engineering, chapter 7
- [Design pipelines AB, BC, AD; 1 kg/cm2 at tap](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c68) — Water Supply Engineering, chapter 7
- [Design pipes AB and BC, source RL 1050 m](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c76) — Water Supply Engineering, chapter 7
- [Design pipes AB and BC, A RL 210 m, 44 and 18 lps](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c73) — Water Supply Engineering, chapter 7
- [Hardy-Cross network discharge AB, BC, AD, CD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardy-cross-as69) — Water Supply Engineering, chapter 7
- [Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apipe-design-ab-bc-ad-110) — Water Supply Engineering, chapter 7, asked 2×

## Related in other subjects

- [Pipe outlet minimum water level, 50 lps](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Apipe-outlet-min-water-level-50lps) — Irrigation and Drainage Engineering, chapter 7
- [Verify suggested discharge distribution (Hardy Cross)](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahardy-cross-verify-suggested-distribution) — Hydraulics, chapter 3
- [Hose siphon discharge and minimum pressure](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahose-siphon-flow-and-minimum-pressure) — Hydraulics, chapter 2
- [Force on 120° bend, 300 lps](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipe-bend-120-force-009) — Fluid Mechanics, chapter 7, asked 2×