---
title: "Sewer for 200 ha catchment at 0.6 depth"
subject: "Sanitary Engineering"
chapter: 3
key: "semester_6/sanitary_engineering:sewer-runoff-0p6-depth-200ha"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sewer-runoff-0p6-depth-200ha
site: BCT Engineering
---

# Sewer for 200 ha catchment at 0.6 depth

Sanitary Engineering · Chapter 3: Design and Construction of Sewers

- Asked in: 2080 Chaitra, 8 marks; 2075 Bhadra, 8 marks (asked 2 times)
- 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/sewer-runoff-0p6-depth-200ha
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-3

## Question

A sewer carries a runoff water to its 0.6 depth at maximum flow which is entering from the catchment area of 200 hectare having overall coefficient of runoff as 0.45 and time of concentration of 55 minutes. The velocity in the sewer is to be maintained as 1.5 m/s at peak flow. Determine the diameter and slope of the cement concrete sewer with the Manning's coefficient as 0.013.

## Short answer

Peak runoff computed by rational method with i=760/(55+10) mm/hr gives 2.923 m³/s. Using partial flow ratios at 0.6 depth (Q/Q_f=0.6718), the required full-flow area gives diameter 1.99 m (adopt 2.10 m). Slope from Manning's equation with v=1.5 m/s is 1 in 1192.

## Answer

#### Step 1: Peak runoff (rational method)
$t_c=55$ min. No rainfall formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed:

$$
\begin{aligned}
i&=\frac{760}{55+10}=11.69\ \text{mm/hr}\\
Q&=\frac{CiA}{360}=\frac{0.45\times11.69\times200}{360}=2.923\ \text{m}^3/\text{s}
\end{aligned}
$$

#### Step 2: Hydraulic elements at $d/D=0.6$
For $d/D=0.6$: $\theta = 2\cos^{-1}(1-2d/D)=203.07^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.6265$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1106$, so $v/v_f=(R/R_f)^{2/3}=1.0724$ and $Q/Q_f=(A/A_f)(v/v_f)=0.6718$ (n taken constant).

#### Step 3: Diameter
Here the velocity at peak flow is fixed ($v=1.5$ m/s), so the flow area is

$$
A_{p}=\frac{Q}{v}=\frac{2.923}{1.5}=1.949\ \text{m}^2
$$

$$
A_f=\frac{A_p}{A/A_f}=\frac{1.949}{0.6265}=3.111\ \text{m}^2\ \Rightarrow\ D=\sqrt{\frac{4A_f}{\pi}}=1.990\ \text{m}
$$

Adopt $D=2.10$ m.

#### Step 4: Slope
$R=\dfrac{R}{R_f}\times\dfrac{D}{4}=1.1106\times\dfrac{1.990}{4}=0.5525$ m. From Manning's formula $v=\dfrac{1}{n}R^{2/3}S^{1/2}$:

$$
S=\left(\frac{v\,n}{R^{2/3}}\right)^2=\left(\frac{1.5\times0.013}{0.5525^{2/3}}\right)^2=0.00084\approx\frac{1}{1192}
$$

(For the adopted $D=2.10$ m, the slope needed for the same 1.5 m/s is $1$ in 1281.)

**Answer: Diameter about 1.990 m (adopt 2.10 m), slope about 1 in 1192 (0.6 depth, v = 1.5 m/s, Q = 2.923 m³/s).**

## Check yourself

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

1. Which expression gives the peak runoff in m3/s used here (C, i in mm/hr, A in ha)?
   A. Q = C i A / 36
   B. Q = C i A / 360
   C. Q = C i A / 3600
   D. Q = C i / (360 A)
   Answer: B. In these units the rational formula is Q = CiA/360; here 0.45 x 11.69 x 200 / 360 = 2.923 m3/s. (Key point: Peak runoff by rational method)

2. No rainfall formula is given. What was assumed and which time was used?
   A. i = 760/(t+10) with t = 10 min
   B. A constant i of 50 mm/hr
   C. i = 760/(t+10) with t = time of concentration = 55 min
   D. i = 760/t with t = 55 min and no addition
   Answer: C. The assumed formula with t = 55 min gives i = 11.69 mm/hr. (Key point: Rainfall intensity assumption)

3. Why can the diameter be found from A = Q/v here?
   A. Velocity at peak flow is fixed at 1.5 m/s, so the flow area is Q/v and then A_f = A/(A/A_f)
   B. Because the sewer runs full
   C. Because Manning's formula is not valid at 0.6 depth
   D. Because the slope is already known
   Answer: A. A_p = 2.923/1.5 = 1.949 m2; dividing by A/A_f = 0.6265 gives A_f and D = 1.99 m. (Key point: Diameter from fixed velocity)

4. How is the slope found after D?
   A. Use S = (v n / D^(2/3))^2 with D directly
   B. Compute R = 1.1106 x D/4 and use S = (v n / R^(2/3))^2
   C. Use S = v / R
   D. Use the full-pipe slope S = (Q n / 0.3117 D^(8/3))^2 with Q full
   Answer: B. At d/D = 0.6 the hydraulic radius is 1.1106 times D/4; Manning then gives S about 0.00084 (1 in 1192). (Key point: Slope from Manning)

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