---
title: "Design pipes AB, BC; 160 lpcd, peak factor 3"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c71"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c71
site: BCT Engineering
---

# Design pipes AB, BC; 160 lpcd, peak factor 3

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2071 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-c71
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipelines AB and BC for the water distribution network shown below. Take per capita demand of water 160 lpcd. Assume peak factor = 3, Hazen William's Constant C = 100. The residual head at any point in the distribution system should not be less than 10 m.

[Figure: Service reservoir at A, RL = 1000 m. A to B: L = 200 m, B has RL = 950 m, population = 3600 supplied along this section. B to C: L = 500 m, population = 4400, C has RL = 900 m.]

## Short answer

Use Hazen-Williams equation with C=100 and per capita demand 160 lpcd, peak factor 3. Pipe AB (44.44 lps, 200 m) adopts 150 mm and pipe BC (24.44 lps, 500 m) adopts 125 mm. Velocities are 2.52 m/s and 1.99 m/s.

## Answer

Per capita demand 160 lpcd, peak factor 3, C = 100, minimum residual head 10 m. HGL at A = 1000 m. The population along each section is taken as supplied at its end node: AB carries the demand of 3600 + 4400 = 8000 people, and BC carries 4400 people.

- $Q_{AB} = 8000 \times 160 \times 3/86400 = 44.44$ lps
- $Q_{BC} = 4400 \times 160 \times 3/86400 = 24.44$ 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 $Q = 44.444$ lps, length $L = 200$ m
- HGL at A = 1000.00 m; ground RL at B = 950 m; residual head required = 10 m
- Available head $H = 1000.00 - 950 - 10 = 40.00$ m
- Gradient $S = H/L = 40.00/200 = 0.20000$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 120.2$ mm, adopt commercial size **150 mm** (raised from the head-based size because velocity would exceed 3 m/s)
- Actual $h_f = 13.595$ m, HGL at B = 986.40 m, residual head = 36.40 m (≥ 10 m, OK)
- Velocity $v = Q/A = 2.52$ m/s

#### Pipe BC
- Discharge $Q = 24.444$ lps, length $L = 500$ m
- HGL at B = 986.40 m; ground RL at C = 900 m; residual head required = 10 m
- Available head $H = 986.40 - 900 - 10 = 76.40$ m
- Gradient $S = H/L = 76.40/500 = 0.15281$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 101.2$ mm, adopt commercial size **125 mm**
- Actual $h_f = 27.296$ m, HGL at C = 959.11 m, residual head = 59.11 m (≥ 10 m, OK)
- Velocity $v = Q/A = 1.99$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| AB | 44.4444 | 200 | 120.2 | 150 | 13.60 | 36.40 | 2.52 |
| BC | 24.4444 | 500 | 101.2 | 125 | 27.30 | 59.11 | 1.99 |

**Answer: AB = 150 mm and BC = 125 mm.** For AB the head-based size is 125 mm, but its velocity would be 3.6 m/s, which is too high, so 150 mm is adopted (v = 2.5 m/s). Residual heads are above 10 m at B and C.

## Check yourself

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

1. Why does AB carry 8000 people's demand?
   A. Peak factor 8
   B. It supplies B (3600) and passes on the demand for C (4400)
   C. Population doubles
   D. It is the longest
   Answer: B. Upstream pipe carries all downstream demand. (Key point: Pipe AB discharge)

2. Q = 8000 x 160 x 3/86400 gives about?
   A. 44.4 lps
   B. 14.8 lps
   C. 133 lps
   D. 4.4 lps
   Answer: A. Peak flow = 3 x average: 3,840,000/86,400 = 44.4 lps. (Key point: Discharge)

3. Why is 150 mm adopted for AB instead of 125 mm?
   A. Velocity in 125 mm would be about 3.6 m/s, above 3 m/s
   B. Cost
   C. Head loss too low
   D. Residual head too high
   Answer: A. The head-based size gives excessive velocity, so a larger pipe is used. (Key point: Velocity limit)

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