---
title: "Oxidation pond for Terai town, 12,000"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:oxidation-pond-terai-12000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/oxidation-pond-terai-12000
site: BCT Engineering
---

# Oxidation pond for Terai town, 12,000

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2077 Chaitra, 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-terai-12000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Calculate detention time and dimension of an oxidation pond for a town in Terai Region of Nepal with the following data. Draw neat sketch with necessary components showing buffer zone and detail dimensions as designed. Population = 12,000; Sewage flow = 100 lpcd; BOD of incoming sewage = 250 mg/l; Assume operational depth at 1.1 m.

## Short answer

For 12,000 people at 100 lpcd and 250 mg/l BOD, Q=1200 m³/d, BOD load=300 kg/d. Using 300 kg/ha/day, area=1 ha, volume=11000 m³, detention=9.17 days. Adopt 71 m × 142 m at 1.1 m depth, total depth 1.6 m with free board, and provide a buffer zone around the embankment.

## Answer

#### Assumptions
Terai has a warm climate, so a surface organic loading of $\lambda_s=300$ kg BOD/ha/day (usual range 200-400) is adopted. Length : width $=2:1$. Operational depth $=1.1$ m (given). Free board 0.5 m, embankment slopes 1 : 2 (inside) and 1 : 1.5 (outside), crest width 3 m.

#### Step 1: Flow and load

$$
\begin{aligned}
Q &= 12000\times100/1000 = 1200\ \text{m}^3/\text{day}\\
\text{BOD load} &= 1200\times250/1000 = 300\ \text{kg/day}
\end{aligned}
$$

#### Step 2: Area, volume and detention time

$$
\begin{aligned}
A &= \frac{300}{300}=1.00\ \text{ha}=10000\ \text{m}^2\\
V &= A\times d = 10000\times1.1=11000\ \text{m}^3\\
t &= \frac{V}{Q}=\frac{11000}{1200}=9.17\ \text{days}\ (\text{within 5-30 days: OK})
\end{aligned}
$$

#### Step 3: Dimensions

$$
W=\sqrt{\frac{A}{2}}=\sqrt{\frac{10000}{2}}=70.7\ \text{m (adopt 71 m)},\qquad L=2W=141.4\ \text{m (adopt 142 m)}
$$

Pond water surface (at mean depth) $=71\ \text{m}\times142\ \text{m}=10082$ m². Total depth $=1.1+0.5=1.6$ m. If possible, provide two ponds in parallel (each about half the area), so that one can be desludged.

#### Sketch (plan)
```text
 +--------------------------------------------+
 |  buffer zone (trees), 50 m (assumed)       |
 |   +------------------------------------+   |
 |   | crest 3 m                          |   |
 |   |  +------------------------------+  |   |
 |   |  |     POND 71 m x 142 m        |  |   |
 |   |  | inlet ->  d = 1.1 m -> outlet|  |   |
 |   |  +------------------------------+  |   |
 |   +------------------------------------+   |
 +--------------------------------------------+
  Section: crest 3m, slope 1:2, FB 0.5 m, d 1.1 m
```
**Answer: detention time $=9.17$ days; pond area $=10000$ m² ($\approx71\ \text{m}\times142\ \text{m}$ at mean depth), depth 1.1 m, with a buffer zone around the embankment.**

## Check yourself

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

1. For 12,000 people at 100 lpcd and BOD 250 mg/l, what are the flow and BOD load?
   A. 1200 m3/day and 300 kg/day
   B. 1200 m3/day and 30 kg/day
   C. 120 m3/day and 300 kg/day
   D. 12000 m3/day and 3000 kg/day
   Answer: A. Q = 12000 x 100 / 1000 = 1200 m3/day; load = 1200 x 250 / 1000 = 300 kg/day. (Key point: Flow and load)

2. With a loading of 300 kg BOD/ha/day, the pond area is
   A. 0.1 ha
   B. 10 ha
   C. 3.6 ha
   D. 1.0 ha (10,000 m2)
   Answer: D. A = 300/300 = 1 ha = 10,000 m2. (Key point: Area)

3. What is the detention time at depth 1.1 m?
   A. 9.17 days
   B. 5.5 days
   C. 13.6 days
   D. 1.1 days
   Answer: A. V = 10000 x 1.1 = 11000 m3; t = 11000/1200 = 9.17 days, within 5-30 days. (Key point: Detention time)

4. For L : W = 2 : 1 and area 10,000 m2, the adopted dimensions are about
   A. 100 m x 100 m
   B. 50 m x 200 m
   C. 35 m x 70 m
   D. 71 m x 142 m
   Answer: D. W = sqrt(10000/2) = 70.7 m, L = 2W = 141.4 m. (Key point: Dimensions)

## Related in Sanitary Engineering

- [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 theory; 2500 persons](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-2500-persons) — 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
- [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
- [Conventional ASP design, population 50,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-50000-80lpcd) — Sanitary Engineering, chapter 7, asked 2×
- [Oxidation pond and area procedure](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aoxidation-pond-area-procedure) — Sanitary Engineering, chapter 7

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