---
title: "Oxidation pond theory; colony of 15000"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:oxidation-pond-15000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-15000
site: BCT Engineering
---

# Oxidation pond theory; colony of 15000

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2076 Bhadra, 8 marks
- 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-15000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Describe the theory of oxidation pond. A new colony of 15000 populations is supplied with water at 200 lpcd. The sewage from colony is to be treated in an oxidation pond. The BOD and suspended solids are each 340 mg/l. Assuming the organic loading of 300 kg/ha/day, calculate the dimension of the pond. If the daily flow is for 9 hours and the average velocity of sewage is 0.9 m/s, what is the diameter of the inlet pipe required for the oxidation pond?

## Short answer

Assume 80% of 200 lpcd reaches the pond: Q=2400 m³/d, BOD load=816 kg/d. At 300 kg/ha/day, area=2.72 ha; with depth 1.2 m, volume=32640 m³, detention=13.6 days, adopt about 117 m × 234 m. For 9-hour flow at 0.9 m/s, pipe area=0.0823 m², so diameter ≈ 325 mm.

## Answer

An **oxidation pond** (waste stabilisation pond, facultative pond) is a shallow earthen basin, 1-1.5 m deep, in which sewage is purified naturally by the mutual action of bacteria and algae in the presence of sunlight.

- **Aerobic zone (top):** aerobic bacteria oxidise organic matter using oxygen produced by algae and by wind aeration: $\text{organics}+O_2\rightarrow CO_2+H_2O+\text{new cells}+NH_3+PO_4$.
- Algae use sunlight, the CO₂, nitrogen and phosphate released by the bacteria to grow and release oxygen: $CO_2+H_2O+\text{light}\rightarrow\text{algae}+O_2$.
- **Anaerobic zone (bottom):** settled solids are digested to CH₄, CO₂ and H₂S.
- Pathogens die by sunlight, high pH and long detention. Typical detention is 5-30 days, loading 200-400 kg BOD/ha/day, and BOD removal 70-90%.

```text
   sunlight
   \ | /
  ~~~~~~~~~~~~~~~~~~~~~~~~~~ surface (wind aeration)
   Algae --O2--> Bacteria (aerobic zone)
   Algae <--CO2,NH3,PO4-- Bacteria
  - - - - - - - - - - - - - - - - -
   Anaerobic zone: sludge -> CH4, CO2
  ==================== bottom
```

#### Design
**Assumptions:** 80% of the water supplied becomes sewage; depth of pond $=1.2$ m; length : width $=2:1$.

$$
\begin{aligned}
Q &= 15000\times200\times0.8/1000=2400\ \text{m}^3/\text{day}\\
\text{BOD load} &= 2400\times340/1000=816\ \text{kg/day}\\
\text{Area} &= \frac{816}{300}=2.72\ \text{ha}=27200\ \text{m}^2\\
W &= \sqrt{27200/2}=116.6\ \text{m},\quad L=2W=233.2\ \text{m}\\
\text{Volume} &= 27200\times1.2=32640\ \text{m}^3,\quad t=\frac{32640}{2400}=13.6\ \text{days}
\end{aligned}
$$

Adopt a pond of about $117\ \text{m}\times234\ \text{m}\times1.2$ m (preferably as two ponds in parallel).

#### Inlet pipe diameter
The daily flow arrives in 9 hours, so

$$
\begin{aligned}
q &= \frac{2400}{9\times3600}=0.0741\ \text{m}^3/\text{s}\\
A_{pipe} &= \frac{q}{v}=\frac{0.0741}{0.9}=0.0823\ \text{m}^2\\
d &= \sqrt{\frac{4A_{pipe}}{\pi}}=0.324\ \text{m}=324\ \text{mm}
\end{aligned}
$$

**Answer: pond area $27200$ m² ($\approx 117\ \text{m}\times234\ \text{m}$, depth 1.2 m, detention 13.6 days); inlet pipe diameter about 325 mm.**

## Check yourself

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

1. Why is the flow taken as 2400 m3/day for 15,000 people at 200 lpcd?
   A. 80% of the water supplied is assumed to become sewage
   B. All supplied water becomes sewage
   C. Only 20% becomes sewage
   D. Flow equals population x 200 x 8
   Answer: A. Q = 15000 x 200 x 0.8 / 1000 = 2400 m3/day. (Key point: Sewage flow assumption)

2. With BOD 340 mg/l and loading 300 kg/ha/day, the area is
   A. 8.16 ha
   B. 0.272 ha
   C. 27.2 ha
   D. 2.72 ha (27,200 m2)
   Answer: D. Load = 2400 x 340 / 1000 = 816 kg/day; A = 816/300 = 2.72 ha. (Key point: Pond area)

3. The daily flow arrives in 9 hours. The inlet pipe discharge is
   A. 2400/(24 x 3600) = 0.0278 m3/s
   B. 2400/(9 x 3600) = 0.0741 m3/s
   C. 2400/9 = 267 m3/s
   D. 2400 x 9/3600 = 6 m3/s
   Answer: B. The flow is concentrated in 9 hours, so q = 2400/(9 x 3600) = 0.0741 m3/s. (Key point: Inlet discharge)

4. With velocity 0.9 m/s, the inlet pipe diameter is about
   A. 32 mm
   B. 1.2 m
   C. 325 mm
   D. 160 mm
   Answer: C. Area = 0.0741/0.9 = 0.0823 m2; d = sqrt(4A/pi) = 0.324 m. (Key point: Pipe diameter)

## Related in Sanitary Engineering

- [Oxidation pond theory; 2500 persons](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-2500-persons) — 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
- [Sedimentation tank and oxidation pond design](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-and-oxidation-pond-10000) — Sanitary Engineering, chapter 7
- [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
- [1-day BOD 300 mg/l to 5-day at 30°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Abod-1day-300-to-5day-30c) — Sanitary Engineering, chapter 5
- [Oxidation pond: theory and commissioning](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-mechanism-commissioning) — Sanitary Engineering, chapter 7, asked 3×

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