---
title: "Separate sewer, check self-cleansing"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:separate-sewer-0p5-self-cleansing"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/separate-sewer-0p5-self-cleansing
site: BCT Engineering
---

# Separate sewer, check self-cleansing

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2072 Magh, 8 marks
- Video: [Design of Sewers & Sewerage systems | Numericals on Sewer Design | GATE 2022 | Environmental Engg |](https://www.youtube.com/watch?v=_1CRUHPE5lI) — TechnoCivil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/separate-sewer-0p5-self-cleansing
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3

## Question

Design a sewer for separate system to carry peak flow 0.5 m³/sec at a slope 10 in 10000. Sewer should run 0.7 times depth at peak flow. The value of n in Manning's formula is 0.012. Will the self cleansing velocity be maintained in the sewer during dry weather flow? Take peak factor = 3.

## Short answer

The sewer is designed as 900 mm at 1 in 1000. At average dry weather flow (0.1667 m³/s), the velocity is 0.83 m/s, which exceeds the self-cleansing velocity of 0.6 m/s, so self-cleansing is maintained.

## Answer

#### Given
Peak flow $Q_{max}=0.5$ m³/s, $S=10/10000=0.001$, $n=0.012$, $d/D=0.7$ at peak flow, peak factor = 3.

#### Design for peak flow
For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant).

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

$$
Q_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \text{m}^3/\text{s}
$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$:

$$
D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m}
$$

Adopt the next standard size, **D = 900 mm**.

#### Check of velocity and capacity
For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.975$ m/s and $Q_f=0.6202$ m³/s.

$$
\begin{aligned}
v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\
Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000)
\end{aligned}
$$

The velocity at maximum flow, 1.09 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

#### Check in dry weather flow
Average (dry weather) flow $=\dfrac{Q_{max}}{\text{peak factor}}=\dfrac{0.5}{3}=0.1667$ m³/s. For the adopted pipe ($D=0.900$ m, $Q_f=0.6202$ m³/s, $v_f=0.975$ m/s):

$$
\frac{Q}{Q_f}=\frac{0.1667}{0.6202}=0.2687\ \Rightarrow\ \frac{d}{D}=0.354,\ \ \frac{v}{v_f}=0.848
$$

$$
v=0.848\times0.975=0.83\ \text{m/s}
$$

At the minimum flow (about one-third of the average, $0.0556$ m³/s), $d/D=0.202$ and $v=0.60$ m/s.

The self-cleansing velocity is 0.6 m/s (0.75 m/s is preferred for sewage). The velocity at the average dry weather flow, 0.83 m/s, is well above 0.6 m/s, and even at the minimum flow the velocity is just about 0.6 m/s.

**Answer: Provide a 900 mm sewer at 1 in 1000. Velocity at peak flow = 1.09 m/s; at dry weather (average) flow = 0.83 m/s, so the self-cleansing velocity is maintained.**

## Check yourself

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

1. Peak flow is 0.5 m3/s with peak factor 3. What is the average dry weather flow?
   A. 1.5 m3/s
   B. 0.1667 m3/s
   C. 0.25 m3/s
   D. 0.0556 m3/s
   Answer: B. Average flow = Q_max / peak factor = 0.5/3 = 0.1667 m3/s. (Key point: Dry weather flow from peak factor)

2. How is the 900 mm diameter obtained for the peak flow?
   A. Q_f = 0.5 x 0.8372, then round up
   B. Use the dry weather flow in Manning's formula
   C. Q_f = 0.5/0.8372 = 0.597 m3/s, then Manning's full-pipe formula gives D = 0.887 m, adopt 900 mm
   D. Pick 900 mm since 0.9 is the maximum for separate sewers
   Answer: C. Design uses the peak flow at d/D = 0.7 to find the full-flow discharge and then the next standard size. (Key point: Sizing at 0.7 depth)

3. For the 900 mm sewer, Q/Q_f = 0.2687 at average flow. How is the velocity found?
   A. From the proportionate chart d/D = 0.354, v/v_f = 0.848, so v = 0.848 x 0.975 = 0.83 m/s
   B. v = v_f = 0.975 m/s because the pipe is the same
   C. v = 0.2687 x 0.975 m/s
   D. v = 1.09 m/s from the peak flow check
   Answer: A. Velocity depends on depth; the ratio v/v_f is applied to the full velocity. (Key point: Velocity at dry weather flow)

4. What is concluded about self-cleansing?
   A. 0.83 m/s is below 0.6 m/s so the sewer silts
   B. 0.83 m/s at average flow is above 0.6 m/s, so self-cleansing is maintained at average flow
   C. Self-cleansing cannot be judged without the peak flow
   D. The sewer fails because 0.75 m/s is needed at peak flow
   Answer: B. The velocity at average dry weather flow exceeds the minimum 0.6 m/s. (Key point: Conclusion on self-cleansing)

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