---
title: "Circular combined sewer, 60 ha"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:combined-sewer-60ha"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/combined-sewer-60ha
site: BCT Engineering
---

# Circular combined sewer, 60 ha

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2071 Bhadra, 2+6 marks
- Video: [calculation of discharge of sewage in combined sewage sysytem | Environmental Engineering numerical](https://www.youtube.com/watch?v=dLrq2au_uLQ) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/combined-sewer-60ha
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3

## Question

Design a circular section of combined sewer from the following data:
Area to be served = 60 ha
Population = 65,000
Maximum permissible velocity = 3.2 m/sec
Time of entry = 5 minutes
Time of flow = 18 minutes
Rate of water supply = 235 lpcd
Overall runoff coefficient = 0.55
Assume suitably any other data required.

## Short answer

Total flow Q=2.535 m³/s. With n=0.013, choosing full-flow velocity 3.0 m/s gives area 0.845 m², D=1.037 m, adopted 1.05 m; actual v=2.93 m/s. Slope from Manning's is 0.00862 (1 in 116). Dry-weather velocity 1.58 m/s ensures self-cleansing.

## Answer

#### Part 1: Dry weather flow (sewage)  [2 marks]
Assume 80% of 235 lpcd becomes sewage and the peak factor is 3.

$$
Q_{sew}=\frac{65000\times235\times0.8\times3}{86400\times1000}=0.4243\ \text{m}^3/\text{s}
$$

#### Part 2: Design of the combined sewer  [6 marks]
**Storm water.** $t_c=t_e+t_f=5+18=23$ min. With the assumed formula $i=\dfrac{760}{t+10}$ mm/hr:

$$
\begin{aligned}
i&=\frac{760}{23+10}=23.03\ \text{mm/hr}\\
Q_{storm}&=\frac{0.55\times23.03\times60}{360}=2.111\ \text{m}^3/\text{s}\\
Q&=Q_{sew}+Q_{storm}=2.535\ \text{m}^3/\text{s}
\end{aligned}
$$

**Size.** The sewer is designed to run full at peak flow. Take a design velocity of 3.0 m/s (below the maximum permissible 3.2 m/s):

$$
A=\frac{Q}{v}=\frac{2.535}{3.0}=0.845\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=1.037\ \text{m}
$$

Adopt $D=1.05$ m. Then $v=\dfrac{Q}{\frac{\pi}{4}D^2}=2.93$ m/s (less than 3.2 m/s).

**Slope.** From Manning's formula with $R=D/4=0.263$ m, $n=0.013$:

$$
S=\left(\frac{v\,n}{R^{2/3}}\right)^2=\left(\frac{2.93\times0.013}{0.263^{2/3}}\right)^2=0.00862\approx\frac{1}{116}
$$

**Check at average dry weather flow** $=Q_{sew}/3=0.1414$ m³/s: $Q/Q_f=0.0558$, $d/D=0.160$, $v=1.58$ m/s. This is greater than 0.6 m/s, so the self-cleansing condition is satisfied.

**Answer: Circular combined sewer, D = 1.05 m, slope about 1 in 116, carrying 2.535 m³/s at 2.93 m/s (running full).**

## Check yourself

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

1. How is the time of concentration obtained here?
   A. t_c = time of entry + time of flow = 5 + 18 = 23 min
   B. t_c = 18 min only
   C. t_c = 18 - 5 = 13 min
   D. t_c = 5 min only
   Answer: A. Rainfall intensity is taken for t_c = t_e + t_f. (Key point: Time of concentration)

2. The design discharge for the combined sewer is
   A. Storm flow only, 2.111 m3/s
   B. Sewage peak (0.4243) plus storm flow (2.111) = 2.535 m3/s
   C. Sewage average flow 0.1414 m3/s
   D. Sewage flow multiplied by storm flow
   Answer: B. A combined sewer is designed for the sum of dry weather peak and storm runoff. (Key point: Combined flow)

3. Why is a design velocity of 3.0 m/s taken to get the area?
   A. It is the minimum self-cleansing velocity
   B. It is required by Manning's formula
   C. It is just below the maximum permissible 3.2 m/s, so A = Q/v gives the smallest safe pipe
   D. It equals the average dry weather velocity
   Answer: C. A = 2.535/3 = 0.845 m2 gives D = 1.037 m, adopt 1.05 m with v = 2.93 m/s < 3.2. (Key point: Size using velocity limit)

4. How is the slope found, and what does the dry weather check show?
   A. From Manning with R = D (S about 1 in 1000); v = 0.4 m/s so silting occurs
   B. Slope = Q/A; v at dry flow equals 2.93 m/s
   C. Slope is assumed 1 in 500; no dry flow check is needed
   D. From Manning with R = D/4 (S about 1 in 116); at average dry weather flow v = 1.58 m/s > 0.6, so self-cleansing is satisfied
   Answer: D. Full pipe has R = D/4; the dry-flow check confirms velocity at low flow remains above 0.6 m/s. (Key point: Slope and check at dry weather flow)

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