---
title: "Design pipes RA and AB; 200 lpcd, peak factor 3"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-c69"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-c69
site: BCT Engineering
---

# Design pipes RA and AB; 200 lpcd, peak factor 3

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2069 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-c69
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipes RA and AB for the water distribution network shown below. Take per capita demand of water as 200 lpcd. Assume peak factor = 3 and Hazen Williams Constant C = 100. The residual pressure at any point in the distribution system should not be less than 15 m. Check velocity in the pipes also.

[Figure: Service reservoir R, RL 890 m; R to A: length = 200 m, RL of A = 840 m, population = 3600; A to B: length = 500 m, population = 4400, RL of B = 790 m.]

## Short answer

Use Hazen-Williams equation with C=100 and per capita demand 200 lpcd, peak factor 3. Pipe RA (55.56 lps, 200 m) adopts 200 mm and pipe AB (30.56 lps, 500 m) adopts 125 mm. Velocities are 1.77 m/s and 2.49 m/s.

## Answer

Per capita demand 200 lpcd, peak factor 3, C = 100, residual head ≥ 15 m. HGL at R = 890 m. The population along each section is treated as drawn at its end node (A for RA, B for AB): pipe RA carries 3600 + 4400 = 8000 people and AB carries 4400 people.

- $Q_{RA} = 8000 \times 200 \times 3/86400 = 55.56$ lps
- $Q_{AB} = 4400 \times 200 \times 3/86400 = 30.56$ 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 RA
- Discharge $Q = 55.556$ lps, length $L = 200$ m
- HGL at R = 890.00 m; ground RL at A = 840 m; residual head required = 15 m
- Available head $H = 890.00 - 840 - 15 = 35.00$ m
- Gradient $S = H/L = 35.00/200 = 0.17500$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 134.5$ mm, adopt commercial size **200 mm** (raised from the head-based size because velocity would exceed 3 m/s)
- Actual $h_f = 5.063$ m, HGL at A = 884.94 m, residual head = 44.94 m (≥ 15 m, OK)
- Velocity $v = Q/A = 1.77$ m/s

#### Pipe AB
- Discharge $Q = 30.556$ lps, length $L = 500$ m
- HGL at A = 884.94 m; ground RL at B = 790 m; residual head required = 15 m
- Available head $H = 884.94 - 790 - 15 = 79.94$ m
- Gradient $S = H/L = 79.94/500 = 0.15987$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 109.1$ mm, adopt commercial size **125 mm**
- Actual $h_f = 41.264$ m, HGL at B = 843.67 m, residual head = 53.67 m (≥ 15 m, OK)
- Velocity $v = Q/A = 2.49$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| RA | 55.5556 | 200 | 134.5 | 200 | 5.06 | 44.94 | 1.77 |
| AB | 30.5556 | 500 | 109.1 | 125 | 41.26 | 53.67 | 2.49 |

The head-based diameter of RA is 134.5 mm, but 150 mm would give a velocity of 3.14 m/s, which is above the usual limit of 3 m/s; 200 mm is therefore adopted (v = 1.77 m/s). All velocities are between 0.6 and 3 m/s and all residual heads are above 15 m.

**Answer: RA = 200 mm, AB = 125 mm.**

## Check yourself

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

1. Q_RA = 8000 x 200 x 3/86400 =?
   A. 55.56 lps
   B. 18.5 lps
   C. 30.56 lps
   D. 166.7 lps
   Answer: A. 4,800,000/86,400 = 55.56 lps. (Key point: Discharge RA)

2. Why is RA 200 mm rather than 150 mm?
   A. 150 mm gives v = 3.14 m/s, above 3 m/s
   B. Head loss too high
   C. Residual too low
   D. Cost
   Answer: A. Velocity limit of 3 m/s is exceeded with 150 mm. (Key point: Why 200 mm)

3. HGL at A 884.94, RL of B 790, residual 15. Available head?
   A. 79.94 m
   B. 94.94 m
   C. 100 m
   D. 49.94 m
   Answer: A. 884.94 - 790 - 15 = 79.94 m. (Key point: Available head AB)

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