---
title: "Minimum sewer diameter, 80000 population"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:sewer-min-diameter-80000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-min-diameter-80000
site: BCT Engineering
---

# Minimum sewer diameter, 80000 population

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2070 Bhadra, 8 marks
- Video: [design of sewer, Quantity of sewage in combined sewer Problem, sewer size and velocity calculation](https://www.youtube.com/watch?v=oCX5F2VnK7Q) — 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/sewer-min-diameter-80000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3

## Question

The population of a town is 80,000 persons with a water supply rate of 145 lpcd. Assuming 80% of water supply contributes for sewage flow, taking Manning's N as 0.013, average slope as 1:400 and peak factor as 3, determine the minimum diameter of sewer required to carry the maximum discharge if it runs at 0.75 depths?

## Short answer

Peak sewage Q=0.3222 m³/s, d/D=0.75 gives Q/Q_f=0.9119, so Q_f=0.3534 m³/s. From Manning's with n=0.013, S=1/400, D=0.632 m; provide 675 mm. Velocity at d/D=0.75 is 1.33 m/s, well within permissible limits.

## Answer

#### Design flow
$$
Q_{max}=\frac{80000\times145\times0.8\times3}{86400\times1000}=0.3222\ \text{m}^3/\text{s}
$$

Data: $S=1/400$, $n=0.013$ (constant), depth ratio $d/D=0.75$.

For $d/D=0.75$: $\theta = 2\cos^{-1}(1-2d/D)=240.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.8045$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.2067$, so $v/v_f=(R/R_f)^{2/3}=1.1335$ and $Q/Q_f=(A/A_f)(v/v_f)=0.9119$ (n taken constant).

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

$$
Q_f=\frac{Q}{Q/Q_f}=\frac{0.3222}{0.9119}=0.3534\ \text{m}^3/\text{s}
$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00250$ and $n=0.013$:

$$
D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.3534\times0.013}{0.3117\times0.05000}\right)^{3/8}=0.632\ \text{m}
$$

Adopt the next standard size, **D = 675 mm**.

#### Check of velocity and capacity
For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.175$ m/s and $Q_f=0.4203$ m³/s.

$$
\begin{aligned}
v&=v_f\times(v/v_f)=1.175\times1.1335=1.33\ \text{m/s}\\
Q_{capacity}&=Q_f\times(Q/Q_f)=0.4203\times0.9119=0.3833\ \text{m}^3/\text{s}\ (\ge 0.3222)
\end{aligned}
$$

The velocity at maximum flow, 1.33 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: Minimum diameter required = 0.632 m; provide 675 mm.**

## Check yourself

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

1. Q_max for 80,000 persons, 145 lpcd, 80% to sewer, peak factor 3 is
   A. 0.1074 m3/s
   B. 0.3222 m3/s
   C. 0.4028 m3/s
   D. 0.9667 m3/s
   Answer: B. 80000 x 145 x 0.8 x 3 / 86400000 = 0.3222 m3/s. (Key point: Design flow)

2. At d/D = 0.75 (constant n), Q/Q_f is
   A. 0.9119
   B. 0.75
   C. 1.1335
   D. 0.8045
   Answer: A. Q/Q_f = 0.8045 x 1.1335 = 0.9119. (Key point: Discharge ratio at 0.75 depth)

3. The full discharge to be used in the D formula is
   A. 0.3222 x 0.9119
   B. 0.3222 m3/s
   C. 0.3222 / 0.9119 = 0.3534 m3/s
   D. 0.3222 / 0.75
   Answer: C. A pipe running 0.75 deep carries 91.2% of Q_f, so Q_f is larger than the design flow. (Key point: Full discharge)

4. The minimum calculated diameter and adopted size are
   A. 0.632 m and 600 mm
   B. 0.632 m and 675 mm
   C. 0.75 m and 750 mm
   D. 0.675 m and 632 mm
   Answer: B. Computed D is the minimum; the next standard size 675 mm is adopted (capacity 0.383 > 0.322). (Key point: Minimum diameter)

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