---
title: "Sewer diameter for a 2.5 sq km area"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:sewer-diameter-2p5-sqkm"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-diameter-2p5-sqkm
site: BCT Engineering
---

# Sewer diameter for a 2.5 sq km area

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2076 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-diameter-2p5-sqkm
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3

## Question

Calculate the diameter of a sewer to serve an area of 2.5 sq. km. with a population density of 400 persons per hectare. The average rate of water supply is 110 lpcd. The coefficient of runoff for 50% area is 0.5 and the remaining area is 0.6. The time taken to reach by the storm water at the considered point of sewer from the farthest point of the catchment area is 18 minute. Consider the storm duration of 21 minutes, values of Manning's n = 0.013 at all the depths of flow. The permissible slope is 1 in 500. Consider peak factor of 2.7. The sewer should run half full.

## Short answer

Computed sanitary flow 0.275 m³/s, storm flow 9.364 m³/s, total 9.639 m³/s. For half-full sewer, Q_full=19.28 m³/s; using Manning's with n=0.013, S=1/500 yields diameter 2.955 m. Provide 3000 mm circular sewer (runs half full, v=2.84 m/s within limits).

## Answer

The sewer is taken as a combined sewer (sewage + storm water), running half full at maximum flow.

#### Step 1: Sanitary sewage
Area = 2.5 km² = 250 ha, so population $=400\times250=100000$. Assume 80% of the supply becomes sewage; peak factor = 2.7.

$$
Q_{sew}=\frac{100000\times110\times0.8\times2.7}{86400\times1000}=0.2750\ \text{m}^3/\text{s}
$$

#### Step 2: Storm water
Weighted coefficient: $C=0.5(0.5)+0.5(0.6)=0.55$. No intensity formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed for the given storm duration of 21 min (the time of concentration is 18 min, so a 21 min storm covers the whole catchment).

$$
\begin{aligned}
i&=\frac{760}{21+10}=24.52\ \text{mm/hr}\\
Q_{storm}&=\frac{0.55\times24.52\times250}{360}=9.364\ \text{m}^3/\text{s}
\end{aligned}
$$

#### Step 3: Total design discharge
$$
Q=0.2750+9.364=9.639\ \text{m}^3/\text{s}
$$

#### Step 4: Sewer design (half full, $S=1/500$, $n=0.013$)
For $d/D=0.5$: $\theta = 2\cos^{-1}(1-2d/D)=180.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.5000$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.0000$, so $v/v_f=(R/R_f)^{2/3}=1.0000$ and $Q/Q_f=(A/A_f)(v/v_f)=0.5000$ (n taken constant).

#### Size of the sewer
Discharge when running full:

$$
Q_f=\frac{Q}{Q/Q_f}=\frac{9.6388}{0.5000}=19.2776\ \text{m}^3/\text{s}
$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00200$ and $n=0.013$:

$$
D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{19.2776\times0.013}{0.3117\times0.04472}\right)^{3/8}=2.955\ \text{m}
$$

Adopt the next standard size, **D = 3000 mm**.

#### Check of velocity and capacity
For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=2.840$ m/s and $Q_f=20.0739$ m³/s.

$$
\begin{aligned}
v&=v_f\times(v/v_f)=2.840\times1.0000=2.84\ \text{m/s}\\
Q_{capacity}&=Q_f\times(Q/Q_f)=20.0739\times0.5000=10.0370\ \text{m}^3/\text{s}\ (\ge 9.6388)
\end{aligned}
$$

The velocity at maximum flow, 2.84 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

**Answer: Diameter required = 2.955 m; provide 3000 mm circular sewer (running half full, v = 2.84 m/s).**

## Check yourself

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

1. Area 2.5 km2 and density 400 persons/ha give what population?
   A. 100,000
   B. 10,000
   C. 1,000,000
   D. 250,000
   Answer: A. 2.5 km2 = 250 ha, so 400 x 250 = 100,000. (Key point: Population from area)

2. How is the total design discharge found for this combined sewer?
   A. Storm water flow only
   B. Sewage flow only
   C. Sewage peak flow plus storm water flow (rational method)
   D. The larger of sewage and storm flow
   Answer: C. A combined sewer carries both: 0.275 + 9.364 = 9.639 m3/s. (Key point: Combined design discharge)

3. Half the area has C = 0.5 and half has 0.6. The overall C is
   A. 0.60, the larger value
   B. 0.55, the area-weighted average
   C. 0.50, the smaller value
   D. 1.10, the sum
   Answer: B. C = 0.5(0.5) + 0.5(0.6) = 0.55. (Key point: Weighted runoff coefficient)

4. Running half full, what full discharge is used for Manning's formula?
   A. Q_f = 9.639 m3/s
   B. Q_f = 9.639/2
   C. Q_f = 9.639 x 0.6718
   D. Q_f = 2 x 9.639 = 19.28 m3/s
   Answer: D. At half depth Q/Q_f = 0.5, giving D = 2.955 m, adopt 3000 mm. (Key point: Half-full sewer sizing)

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