---
title: "Sedimentation tank and oxidation pond design"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:sed-tank-and-oxidation-pond-10000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-and-oxidation-pond-10000
site: BCT Engineering
---

# Sedimentation tank and oxidation pond design

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2072 Magh, 8 marks
- Video: [Group 14 Design of Waste Stabilisation Ponds](https://www.youtube.com/watch?v=JYnTOiHP_dU) — Emma Nicolson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-and-oxidation-pond-10000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a sedimentation tank and oxidation pond for a town with the following data: Population = 10,000; Sewage flow = 100 lpcd; BOD of incoming sewage = 250 mg/l; BOD in the effluent of oxidation pond should be less than 30 mg/l.

## Short answer

Sedimentation tank: 7.5 m diameter, 2.5 m side water depth (total 3.5 m). Oxidation pond: 7000 m² area, 1.2 m depth, dimensions ~59.2 m × 118.3 m, detention 8.4 days.

## Answer

#### Data

$Q = 10000\times 100/1000 = 1000$ m³/day; raw BOD $=250$ mg/l; required effluent BOD $<30$ mg/l.

#### Part A: Sedimentation tank (circular)

Adopt detention $t=2.5$ h and side water depth $H=2.5$ m.

$$
\begin{aligned}
V &= \frac{1000}{24}\times 2.5 = 104.2\ \text{m}^3\\
A &= \frac{V}{H} = \frac{104.2}{2.5} = 41.7\ \text{m}^2\\
D &= \sqrt{\frac{4A}{\pi}} = 7.28\ \text{m}\ \Rightarrow\ \text{adopt } D=7.5\ \text{m}\\
\text{Surface loading} &= \frac{1000}{44.2} = 22.6\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$

The surface loading is a little below the usual 25-40 m³/m²/day. A lower overflow rate only means better settling, so the tank is on the safe side and acceptable. (If a loading within the range is wanted, adopt $t = 2$ h, giving $D \approx 6.5$ m and about 30 m³/m²/day.)

Provide total depth $= 2.5+0.5+0.5=3.5$ m. Primary sedimentation removes about 30% of the BOD, so the BOD leaving it is $250\times0.7=175$ mg/l.

#### Part B: Oxidation pond (facultative)

Adopt a surface organic loading $\lambda_s = 250$ kg BOD/ha/day (suitable for a warm climate; usual range 200-400) and depth 1.2 m.

$$
\begin{aligned}
\text{BOD load} &= Q\times BOD = 1000\times 175\times10^{-3} = 175.0\ \text{kg/day}\\
\text{Area} &= \frac{175.0}{250} = 0.70\ \text{ha} = 7000\ \text{m}^2\\
\text{Volume} &= 7000\times 1.2 = 8400\ \text{m}^3\\
\text{Detention time} &= \frac{V}{Q} = \frac{8400}{1000} = 8.4\ \text{days (usual 5-10 days)}
\end{aligned}
$$

Take length : width $=2:1$: $W=\sqrt{7000/2}=59.2$ m, $L=2W=118.3$ m. Provide 2 ponds in parallel (each half this area) so that one can be desludged. Embankment side slope 1 : 2 (inside) to 1 : 1.5 (outside), free board 0.5 m, top width 2.5-3 m.

#### Efficiency check

Required BOD removal in the pond $= \dfrac{175-30}{175}\times100 = 82.9\%$, which is within the 80-90% removal normally achieved at this loading and detention time, so the effluent will be below 30 mg/l.

**Answer: sedimentation tank D = 7.5 m, depth 2.5 m (+1 m); oxidation pond area 7000 m² (about 0.70 ha), depth 1.2 m, size about 59.2 m x 118.3 m, detention 8.4 days.**

## Check yourself

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

1. How is the circular tank diameter obtained for 1000 m^3/day, t = 2.5 h, H = 2.5 m?
   A. D = sqrt(4 x 1000 / pi)
   B. V = Q t = 104.2 m^3, A = V/H = 41.7 m^2, D = sqrt(4A/pi) = 7.28 m
   C. D = 1000 / 2.5
   D. A = Q x H
   Answer: B. Volume, then area, then diameter from the circle area. (Key point: Sedimentation tank sizing)

2. What BOD enters the oxidation pond?
   A. 250 mg/l
   B. 30 mg/l
   C. 175 mg/l, since primary settling removes about 30%
   D. 100 mg/l
   Answer: C. 250 x 0.7 = 175 mg/l. (Key point: BOD removed by sedimentation)

3. How is the pond area found?
   A. BOD load (kg/day) x loading rate
   B. BOD load = 175 kg/day divided by 250 kg/ha/day = 0.70 ha
   C. Q divided by depth
   D. Volume multiplied by depth
   Answer: B. Area = BOD load / surface organic loading = 0.7 ha = 7000 m^2. (Key point: Pond area from loading)

4. How is detention time checked and what removal is required?
   A. Detention = area / Q; removal 30%
   B. Detention = V/Q = 8400/1000 = 8.4 days; removal needed about 82.9%
   C. Detention = Q/V; removal 100%
   D. Detention = 1.2 days; removal 50%
   Answer: B. V = 7000 x 1.2 = 8400 m^3, giving 8.4 days; (175-30)/175 = 82.9%, within the 80-90% expected. (Key point: Detention and efficiency)

## Related in Sanitary Engineering

- [Oxidation pond for Terai town, 12,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-terai-12000) — Sanitary Engineering, chapter 7
- [Oxidation pond design, 250 m³/d; theory](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-250m3d) — Sanitary Engineering, chapter 7
- [Circular sedimentation tank, population 80000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-industrial-80000) — Sanitary Engineering, chapter 7, asked 2×
- [Circular sedimentation tank, population 5500](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-industrial-5500) — Sanitary Engineering, chapter 7, asked 2×
- [Conventional ASP design, population 1,00,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-100000-150lpcd) — Sanitary Engineering, chapter 7
- [Oxidation pond theory; colony of 15000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-15000) — Sanitary Engineering, chapter 7

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- [Rectangular sedimentation tank, 20000 population, 0.4 mm/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s73) — Water Supply Engineering, chapter 6
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