---
title: "Oxidation pond design, 250 m³/d; theory"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:oxidation-pond-250m3d"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-250m3d
site: BCT Engineering
---

# Oxidation pond design, 250 m³/d; theory

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2070 Bhadra, 8 marks
- Video: [Design of oxidation pond | volume and retention time of oxidation pond | oxidation pond problem](https://www.youtube.com/watch?v=GdBM8beuUc0) — 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/oxidation-pond-250m3d
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design an oxidation pond to treat 250 m³/d of sewage from a community with permissible organic loading of 450 kg/ha/d. The influent BOD is 250 mg/l and the efficiency of the pond is maintained to be 90%. Describe the theory of oxidation pond.

## Short answer

BOD load = 250 × 250/1000 = 62.5 kg/d; at 450 kg/ha/day, area = 0.139 ha (1389 m²). With 1.2 m depth, volume = 1667 m³, detention = 6.67 days, so adopt about 27 m × 53 m. Effluent BOD at 90% efficiency is 25 mg/l.

## Answer

#### Theory of oxidation pond
An oxidation pond is a shallow earthen pond in which sewage is treated by the combined action of bacteria and algae in sunlight.
1. **Aerobic bacteria** in the upper layer oxidise organic matter: $\text{organics}+O_2\rightarrow CO_2+H_2O+\text{new cells}+NH_3+PO_4$.
2. **Algae** take the CO₂, ammonia and phosphates and, with sunlight, produce oxygen: $CO_2+H_2O+\text{light}\rightarrow\text{algae}+O_2$; this oxygen is used by the bacteria (symbiosis), with additional aeration by the wind.
3. **Anaerobic bacteria** at the bottom digest the settled sludge to CH₄, CO₂ and H₂S.
4. Sunlight, high pH and long detention destroy pathogens.

```text
  sunlight -> algae --O2--> bacteria
              <--CO2,NH3,PO4--
  ~~~~~~~ surface ~~~~~~~
  anaerobic sludge at bottom
```

#### Design
**Assumption:** operating depth $d=1.2$ m; length : width $=2:1$.

$$
\begin{aligned}
\text{BOD load} &= 250\times250/1000=62.5\ \text{kg/day}\\
\text{Area} &= \frac{62.5}{450}=0.139\ \text{ha}=1389\ \text{m}^2\\
\text{Volume} &= 1389\times1.2=1667\ \text{m}^3\\
\text{Detention time} &= \frac{1667}{250}=6.67\ \text{days}\\
W &= \sqrt{1389/2}=26.4\ \text{m},\quad L=2W=52.7\ \text{m}\\
\text{Effluent BOD} &= 250\times(1-0.9)=25\ \text{mg/l}
\end{aligned}
$$

**Answer: pond area $1389$ m² (about 27 m x 53 m), depth 1.2 m (plus 0.5 m free board), detention 6.67 days, effluent BOD about 25 mg/l.**

## Check yourself

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

1. For 250 m3/d of sewage with BOD 250 mg/l, the BOD load is
   A. 625 kg/day
   B. 6.25 kg/day
   C. 62.5 kg/day
   D. 250 kg/day
   Answer: C. Load = 250 x 250 / 1000 = 62.5 kg/day. (Key point: BOD load)

2. With permissible loading 450 kg/ha/day, the area is
   A. 1389 m2
   B. 13,890 m2
   C. 139 m2
   D. 2813 m2
   Answer: A. A = 62.5/450 = 0.139 ha = 1389 m2. (Key point: Pond area)

3. At depth 1.2 m, the detention time is
   A. 1.2 days
   B. 12 days
   C. 3.3 days
   D. 6.67 days
   Answer: D. V = 1389 x 1.2 = 1667 m3; t = 1667/250 = 6.67 days. (Key point: Detention time)

4. With 90% efficiency, the effluent BOD is
   A. 225 mg/l
   B. 25 mg/l
   C. 90 mg/l
   D. 2.5 mg/l
   Answer: B. Effluent = 250 x (1 - 0.9) = 25 mg/l. (Key point: Effluent BOD)

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