---
title: "Rectangular sedimentation tank for 3000 persons"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-a75"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-a75
site: BCT Engineering
---

# Rectangular sedimentation tank for 3000 persons

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2075 Asoj, 8 marks
- Video: [Sedimentation tank Solved design dimension of tank | design of sedimentation tank | waste water engg](https://www.youtube.com/watch?v=P3phfraI2-g) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-a75
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A rectangular sedimentation tank is to be provided to treat water for 3000 persons at per capita daily allowance of 120 liters. Propose the dimensions of the sedimentation tank assuming detention period of 6 hours.

## Short answer

Plan dimensions are 9.5 m × 3.2 m with 3 m effective depth and total depth 3.8 m. Detention time is 6.08 h, SOR 11.8 m³/m²/day, and horizontal velocity 0.026 m/min, all satisfactory.

## Answer

#### Data and assumptions
Population 3000; 120 lpcd; $t=6$ h. Assumed: effective depth $H=3$ m, $L:B=3:1$ (no peak factor is given).

#### Step 1: Flow and volume
$$
Q=3000\times120=360\,000\ \text{l/day}=360\ \text{m}^3/\text{day}
$$
$$
V=\frac{360}{24}\times6=90.0\ \text{m}^3
$$

#### Step 2: Plan area and size
$$
A=\frac{V}{H}=\frac{90.0}{3}=30.0\ \text{m}^2,\qquad 3B^2=30.0\Rightarrow B=3.16\ \text{m},\ L=3B=9.49\ \text{m}
$$
Adopt $B=3.2$ m and $L=9.5$ m.

#### Step 3: Checks
$$
t=\frac{3.2\times9.5\times3}{360}\times24=6.08\ \text{h},\quad \text{SOR}=\frac{360}{30.4}=11.8\ \text{m}^3/\text{m}^2/\text{day},\quad v=\frac{360}{3.2\times3\times1440}=0.026\ \text{m/min}
$$
The velocity is far below 0.3 m/min. The SOR is low (safe).

**Answer: provide a tank 9.5 m $\times$ 3.2 m in plan with 3 m effective depth (add 0.3 m free board and 0.5 m sludge zone, so 3.8 m total).**

## Check yourself

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

1. For 3000 persons at 120 lpcd and t = 6 h, what is the tank volume?
   A. 90 m3
   B. 360 m3
   C. 15 m3
   D. 540 m3
   Answer: A. Q = 360 m3/day, V = 360/24 x 6 = 90 m3. (Key point: Flow and volume)

2. Why must depth and L:B be assumed?
   A. The question gives only flow and detention time, so volume alone does not fix dimensions
   B. They are fixed by law
   C. To reduce detention time
   D. To change the flow
   Answer: A. V fixes only the product; H = 3 m and L:B = 3:1 are assumed to get plan size. (Key point: Assumed data)

3. With A = 30 m2 and L = 3B, what are B and L?
   A. B = 3.16 m, L = 9.49 m
   B. B = 5.5 m, L = 5.5 m
   C. B = 10 m, L = 3 m
   D. B = 1.9 m, L = 15.8 m
   Answer: A. 3B^2 = 30 gives B = 3.16 m, L = 9.49 m. (Key point: Dimensions)

## Related in Water Supply Engineering

- [Rectangular sedimentation tank, 0.2 mm/s, 3 hours](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c71) — Water Supply Engineering, chapter 6, asked 2×
- [Rectangular sedimentation tank, 9 million L peak demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c75) — Water Supply Engineering, chapter 6
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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
- [Rectangular sedimentation tank, 8x10^6 L/day](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as67) — Water Supply Engineering, chapter 6

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