---
title: "Design pipes AB and BC, A RL 210 m, 44 and 18 lps"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c73"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c73
site: BCT Engineering
---

# Design pipes AB and BC, A RL 210 m, 44 and 18 lps

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2073 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-c73
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

In a part of water distribution system, the source is located at a point A with a RL of 210 m, a point B with RL of 154 m is at a distance of 700 m from point A and another point C with RL of 126 m is at a distance of 550 m from point B. Pipe line AB carries a discharge of 44 lps and pipe line BC carries a discharge of 18 lps. Taking minimum residual head as 10 m and Hazen William's coefficient as 100 for pipes, design pipe AB and BC.

## Short answer

Use Hazen-Williams equation with available head. For AB, calculated diameter 150.5 mm, adopt 200 mm. For BC, calculated 95.7 mm, adopt 100 mm. Velocities 1.40 and 2.29 m/s.

## Answer

C = 100, minimum residual head = 10 m. HGL at A = RL of A = 210 m. A to B: 700 m, 44 lps, RL of B = 154 m. B to C: 550 m, 18 lps, 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 = 210.00 m; ground RL at B = 154 m; residual head required = 10 m
- Available head $H = 210.00 - 154 - 10 = 46.00$ m
- Gradient $S = H/L = 46.00/700 = 0.06571$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 150.5$ mm, adopt commercial size **200 mm**
- Actual $h_f = 11.506$ m, HGL at B = 198.49 m, residual head = 44.49 m (≥ 10 m, OK)
- Velocity $v = Q/A = 1.40$ m/s

#### Pipe BC
- Discharge $Q = 18.000$ lps, length $L = 550$ m
- HGL at B = 198.49 m; ground RL at C = 126 m; residual head required = 10 m
- Available head $H = 198.49 - 126 - 10 = 62.49$ m
- Gradient $S = H/L = 62.49/550 = 0.11363$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 95.7$ mm, adopt commercial size **100 mm**
- Actual $h_f = 50.500$ m, HGL at C = 147.99 m, residual head = 21.99 m (≥ 10 m, OK)
- Velocity $v = Q/A = 2.29$ 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 | 150.5 | 200 | 11.51 | 44.49 | 1.40 |
| BC | 18 | 550 | 95.7 | 100 | 50.50 | 21.99 | 2.29 |

The head-based size of AB is 150.5 mm, which is slightly above 150 mm, so 200 mm is adopted (other option: 150 mm in part length and 200 mm in the rest). Velocity in BC is 2.29 m/s (< 3 m/s).

**Answer: AB = 200 mm and BC = 100 mm.**

## Check yourself

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

1. For AB (HGL at A = 210 m, RL of B = 154 m, residual 10 m), the available head is:
   A. 56 m
   B. 46 m
   C. 66 m
   D. 36 m
   Answer: B. H = 210 - 154 - 10 = 46 m, so S = 46/700 = 0.0657. (Key point: Available head for AB)

2. The calculated diameter of AB is 150.5 mm. Why is 200 mm adopted?
   A. It is slightly above 150 mm, so the next larger commercial size is taken
   B. Velocity must be above 3 m/s
   C. 200 mm is always used for mains
   D. 150 mm is not manufactured
   Answer: A. 150.5 mm exceeds the 150 mm size, so the next larger size 200 mm is adopted (or 150 and 200 mm in parts). (Key point: Why 200 mm for AB)

3. The HGL at B used to design BC is:
   A. 154 m
   B. 210 m
   C. 198.49 m
   D. 164 m
   Answer: C. HGL at B = 210 - h_f(AB) = 210 - 11.506 = 198.49 m. (Key point: Head at B for BC)

4. BC (18 lps, 100 mm) has v = 2.29 m/s. What does this show?
   A. It is too high; use 15 mm
   B. It is below the 3 m/s limit, so acceptable
   C. It must be below 0.6 m/s
   D. Velocity is not checked
   Answer: B. Velocity should stay below about 3 m/s; 2.29 m/s is acceptable. (Key point: Velocity check)

## Related in Water Supply Engineering

- [Design pipes AB and BC, source RL 180 m](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-as67) — 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, 20,000 population, f = 0.04](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-b66) — 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×
- [Design pipes AB, BC and AC, C = 110](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-j66) — Water Supply Engineering, chapter 7
- [Design pipes AB, BC; 160 lpcd, peak factor 3](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c71) — Water Supply Engineering, chapter 7

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