---
title: "Oxidation pond theory; 2500 persons"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:oxidation-pond-2500-persons"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-2500-persons
site: BCT Engineering
---

# Oxidation pond theory; 2500 persons

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2071 Bhadra, 3+5 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-2500-persons
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Describe the theory of oxidation pond. Design an oxidation pond for treating domestic sewage of 2500 persons supplied with 225 lpcd of water. The BOD5 of the wastewater is 250 mg/l. Permissible organic loading for the pond is 550 kg/ha/day and the detention time is 12 days. Assume the width to length ratio of the pond as 1:2 and the operational depth as 1.25 m.

## Short answer

Q=450 m³/d (80% of supply), BOD load=112.5 kg/d. Loading gives 2045 m² but detention (12 d × 450 / 1.25 m) gives 4320 m², so use 4320 m². Adopt about 47 m × 93 m at 1.25 m operational depth, with 0.5 m free board; actual loading 260 kg/ha/day is safe.

## Answer

#### Theory of oxidation pond
An oxidation pond is a shallow earthen basin (1-1.5 m deep) in which sewage is stabilised naturally by bacteria and algae in sunlight.
- Bacteria in the upper aerobic zone oxidise organic matter with oxygen produced by algae and wind aeration.
- Algae use sunlight and the CO₂, ammonia and phosphates released by the bacteria, producing oxygen (symbiosis).
- Settled solids are digested anaerobically at the bottom.
- Pathogens die by sunlight, high pH and long detention. BOD removal is 70-90%.

#### Design
**Assumption:** 80% of the water supplied reaches the pond as sewage.

$$
\begin{aligned}
Q &= 2500\times225\times0.8/1000=450\ \text{m}^3/\text{day}\\
\text{BOD load} &= 450\times250/1000=112.5\ \text{kg/day}
\end{aligned}
$$

**Area from the permissible loading:** $A_1=\dfrac{112.5}{550}=0.205$ ha $=2045$ m².

**Area from the detention time (12 days, depth 1.25 m):**

$$
V=Q\,t=450\times12=5400\ \text{m}^3,\qquad A_2=\frac{V}{d}=\frac{5400}{1.25}=4320\ \text{m}^2
$$

The larger area governs: $A=4320$ m². The loading actually applied is $\dfrac{112.5}{0.432}=260$ kg/ha/day, which is below the permissible 550, so it is safe.

**Dimensions** (width : length $=1:2$):

$$
W=\sqrt{\frac{4320}{2}}=46.5\ \text{m (adopt 47 m)},\qquad L=2W=93.0\ \text{m (adopt 93 m)}
$$

Provide a pond of $47\ \text{m}\times93\ \text{m}\times1.25$ m operational depth, plus 0.5 m free board, with embankment slopes 1 : 2 and an inlet and outlet at opposite ends.

**Answer: pond area $4320$ m² ($\approx 47\ \text{m}\times93\ \text{m}$), depth 1.25 m (1.75 m including free board).**

## Check yourself

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

1. Why is the area taken as 4320 m2 rather than 2045 m2?
   A. The smaller area is always adopted
   B. The area from detention time is larger, so it governs and the loading is then below permissible
   C. The depth is ignored
   D. 2045 m2 gives a loading above 550 kg/ha/day
   Answer: B. A1 = 112.5/550 = 2045 m2 from loading; A2 = 5400/1.25 = 4320 m2 from 12 days detention. The larger governs. (Key point: Governing area)

2. With the adopted area, the applied loading is about
   A. 550 kg/ha/day
   B. 26 kg/ha/day
   C. 112 kg/ha/day
   D. 260 kg/ha/day
   Answer: D. 112.5 / 0.432 ha = 260 kg/ha/day, below the permissible 550, so safe. (Key point: Actual loading)

3. For W : L = 1 : 2 and area 4320 m2, the pond is about
   A. 47 m x 93 m
   B. 66 m x 66 m
   C. 93 m x 186 m
   D. 24 m x 47 m
   Answer: A. W = sqrt(4320/2) = 46.5 m, L = 93 m. (Key point: Dimensions)

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