---
title: "Sewer diameter for a 12 sq km area"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:sewer-diameter-12-sqkm"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-diameter-12-sqkm
site: BCT Engineering
---

# Sewer diameter for a 12 sq km area

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2073 Magh, 8 marks
- Video: [design of sewer, Quantity of sewage in combined sewer, capacity of sewer, rational formula rainfall](https://www.youtube.com/watch?v=OgLXaYwVC70) — 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-diameter-12-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 12 sq.km with a population density of 250 persons per hectare. The average rate of sewage flow is 235 lpcd. The coefficient of runoff for 50% farthest area is 0.3 and rest of the area is 0.75. Time taken to reach storm water inlet from the farthest point of the catchment is 25 min. Assume storm duration = 20 min and n = 0.013.

## Short answer

Total design discharge 46.78 m³/s. At slope 1/500 the required diameter is 5.343 m (5400 mm) but full velocity 4.20 m/s exceeds scouring limit. Reducing slope to 1/1500 yields acceptable diameter about 6.6 m with velocity 2.77 m/s.

## Answer

The sewer is taken as a combined sewer. Missing data are assumed: slope 1 in 500, peak factor 3, sewer running half full. The 235 lpcd given is already the sewage flow, so no further reduction is applied.

#### Step 1: Sewage flow
Area = 12 km² = 1200 ha, population $=250\times1200=300000$.

$$
Q_{sew}=\frac{300000\times235\times3}{86400\times1000}=2.4479\ \text{m}^3/\text{s}
$$

#### Step 2: Storm water
$C=0.5(0.3)+0.5(0.75)=0.525$. Time of concentration = 25 min; storm duration = 20 min. The intensity for the given 20 min storm is taken from the assumed formula $i=\dfrac{760}{t+10}$ mm/hr:

$$
\begin{aligned}
i&=\frac{760}{20+10}=25.33\ \text{mm/hr}\\
Q_{storm}&=\frac{0.525\times25.33\times1200}{360}=44.333\ \text{m}^3/\text{s}
\end{aligned}
$$

(Because the 20 min storm is shorter than $t_c$ = 25 min, the farthest part of the area has not yet contributed. Using the full area is on the safe side.)

#### Step 3: Total discharge
$$
Q=2.4479+44.333=46.781\ \text{m}^3/\text{s}
$$

#### Step 4: Sewer design ($S=1/500$, $n=0.013$, half full)
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:

$$
Q_f=\frac{Q}{Q/Q_f}=\frac{46.781}{0.5000}=93.562\ \text{m}^3/\text{s}
$$

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 $n=0.013$:

$$
D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{93.562\times0.013}{0.3117\times0.04472}\right)^{3/8}=5.343\ \text{m}
$$

Adopt the next standard size, **D = 5400 mm**.

#### Check of velocity and capacity
For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=4.203$ m/s and $Q_f=96.248$ m³/s.

$$
\begin{aligned}
v&=v_f\times(v/v_f)=4.203\times1.0000=4.20\ \text{m/s}\\
Q_{capacity}&=Q_f\times(Q/Q_f)=96.248\times0.5000=48.124\ \text{m}^3/\text{s}\ (\ge 46.781)
\end{aligned}
$$

The velocity at maximum flow, 4.20 m/s, is above the maximum non-scouring limit of 3.0 m/s, so this design is **not safe against scouring**.

#### Step 5: Revised slope
Since the slope was only assumed, a flatter slope is chosen. With $S=1/1500$:

$$
D=\left(\frac{93.562\times0.013}{0.3117\times0.02582}\right)^{3/8}=6.565\ \text{m}
$$

Adopt $D=6.6$ m. Then $v_f=2.77$ m/s, and the half-full capacity is $47.45$ m³/s ($\ge 46.781$). The velocity of 2.77 m/s lies between 0.6 and 3.0 m/s, so this is satisfactory.

**Answer: Design discharge = 46.78 m³/s. At 1 in 500 the required diameter is 5.343 m (5400 mm), but the velocity of 4.20 m/s is too high; at 1 in 1500 provide about 6.6 m diameter (v = 2.77 m/s).** For such a large flow, two or more sewers in parallel or a box/arch section would be used in practice.

## Check yourself

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

1. 12 km2 at 250 persons/ha gives
   A. 30,000
   B. 300,000 persons
   C. 3,000,000
   D. 120,000
   Answer: B. 12 km2 = 1200 ha; 250 x 1200 = 300,000. (Key point: Population)

2. 50% of area with C = 0.3 and rest with C = 0.75 gives overall C
   A. 0.375
   B. 0.75
   C. 0.525
   D. 1.05
   Answer: C. C = 0.5(0.3) + 0.5(0.75) = 0.525. (Key point: Weighted runoff coefficient)

3. How is total discharge obtained?
   A. Q = Q_sewage + Q_storm = 1.958 + 44.33 = 46.29 m3/s
   B. Q = Q_storm only
   C. Q = Q_sewage x Q_storm
   D. Q = Q_storm / peak factor
   Answer: A. The combined sewer must carry both dry weather peak flow and storm runoff. (Key point: Combined discharge)

4. The 5400 mm pipe gives 4.20 m/s. What does that indicate?
   A. It is satisfactory since above 0.6 m/s
   B. It exceeds the 3.0 m/s limit, so it is not within limits; parallel sewers or other sections may be used
   C. It is satisfactory since the pipe is large
   D. It shows the slope should be flatter than 1 in 500 for self-cleansing
   Answer: B. Velocity above about 3 m/s risks scour, so the huge flow is split or another section is used. (Key point: Final velocity check)

## Related in Sanitary Engineering

- [Sewer diameter for a 2.5 sq km area](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asewer-diameter-2p5-sqkm) — Sanitary Engineering, chapter 3
- [Minimum sewer diameter, 80000 population](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asewer-min-diameter-80000) — Sanitary Engineering, chapter 3