---
title: "Design pipes AB and BC, 75 lpcd, tank 6 m high"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-b81"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-b81
site: BCT Engineering
---

# Design pipes AB and BC, 75 lpcd, tank 6 m high

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2081 Bhadra, 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-b81
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipes; AB and BC having 400 m and 500 m length respectively. Assume Hazen Williams coefficient C as 100 and maximum demand as 3 times the average demand. The average water consumption is 75 lpcd and population is distributed within the two blocks of 850 in AB section and 480 in BC section. The elevated storage tank is fixed at point A above 6 m height. The R.L. of points A, B and C are 2160, 2125 and 2115 meters respectively. The minimum pressure head of water is to be 5.0 m. Check velocity in the pipes also.

## Short answer

Pipe AB: 65 mm, residual head at B 26.9 m, velocity 1.04 m/s. Pipe BC: 40 mm, residual at C 8.4 m, velocity 0.99 m/s. Both meet minimum 5 m head, velocities acceptable.

## Answer

Data: C = 100, peak (maximum) demand = 3 × average, 75 lpcd, minimum residual head 5.0 m. The tank at A stands 6 m high, so the HGL at A = 2160 + 6 = 2166 m. The population of each section is taken as drawn at its end point (B for AB, C for BC), so pipe AB carries the demand of 850 + 480 = 1330 people and BC carries 480 people.

- $Q_{AB} = 1330 \times 75 \times 3/86400 = 3.464$ lps
- $Q_{BC} = 480 \times 75 \times 3/86400 = 1.250$ 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 $Q = 3.464$ lps, length $L = 400$ m
- HGL at A = 2166.00 m; ground RL at B = 2125 m; residual head required = 5 m
- Available head $H = 2166.00 - 2125 - 5 = 36.00$ m
- Gradient $S = H/L = 36.00/400 = 0.09000$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 53.7$ mm, adopt commercial size **65 mm**
- Actual $h_f = 14.143$ m, HGL at B = 2151.86 m, residual head = 26.86 m (≥ 5 m, OK)
- Velocity $v = Q/A = 1.04$ m/s

#### Pipe BC
- Discharge $Q = 1.250$ lps, length $L = 500$ m
- HGL at B = 2151.86 m; ground RL at C = 2115 m; residual head required = 5 m
- Available head $H = 2151.86 - 2115 - 5 = 31.86$ m
- Gradient $S = H/L = 31.86/500 = 0.06371$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 39.1$ mm, adopt commercial size **40 mm**
- Actual $h_f = 28.483$ m, HGL at C = 2123.37 m, residual head = 8.37 m (≥ 5 m, OK)
- Velocity $v = Q/A = 0.99$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| AB | 3.46354 | 400 | 53.7 | 65 | 14.14 | 26.86 | 1.04 |
| BC | 1.25 | 500 | 39.1 | 40 | 28.48 | 8.37 | 0.99 |

Velocities are between 0.6 m/s and 3 m/s, so the sizes are acceptable.

**Answer: AB = 65 mm and BC = 40 mm; residual heads 26.9 m at B and 8.4 m at C (both more than 5 m); velocities 1.04 m/s and 0.99 m/s.**

## Check yourself

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

1. Pipe AB serves 850 people, BC serves 480. Why does AB carry 1330 people's demand?
   A. AB is longer
   B. AB must also carry the flow that continues on to BC
   C. Peak factor applies twice
   D. The tank is at A
   Answer: B. The upstream pipe carries its own demand plus that passed downstream to BC. (Key point: Pipe discharge)

2. How is the available head for pipe AB found?
   A. HGL at A - RL of B - residual head = 2166 - 2125 - 5 = 36 m
   B. RL of A - RL of B = 35 m
   C. Tank height only = 6 m
   D. RL of B + 5 m
   Answer: A. HGL at A includes the 6 m tank height: 2166. Subtract ground RL at B and the required residual head. (Key point: Available head)

3. How is the diameter finally chosen?
   A. Round the calculated diameter down
   B. Adopt the next larger commercial size, then check head loss, residual head and velocity (0.6-3 m/s)
   C. Use the smallest pipe
   D. Use the same size for both pipes
   Answer: B. Calculated D (53.7 mm) is raised to 65 mm, then residual head and velocity are checked. (Key point: Diameter selection)

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