---
title: "Rectangular sedimentation tank, 0.2 mm/s, 3 hours"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c71"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c71
site: BCT Engineering
---

# Rectangular sedimentation tank, 0.2 mm/s, 3 hours

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2071 Chaitra, 8 marks
- Video: [Rectangular sedimentation tank design.](https://www.youtube.com/watch?v=cRtMpplW84I) — Engr Suliman Khan
- 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-c71
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Determine the size of a rectangular sedimentation tank having its length as twice of its width to settle the particles with settling velocity of 0.2 mm/sec with a settling period of 3 hours to treat water for population of 20000 with a peak allowance of 112 lpcd.

## Short answer

The design flow is 2.24 MLD. Using overflow rate equal to settling velocity (17.28 m/day) gives a surface area of 129.63 m². With L=2B, width is 8.05 m, length 16.10 m. From 3‑hour detention, effective depth is 2.16 m. Adopted effective size: 16.2 m × 8.1 m × 2.2 m (3.0 m total depth).

## Answer

#### Data
Population 20 000; peak allowance 112 lpcd; settling velocity $v_s=0.2$ mm/s; $L=2B$; settling (detention) period $t=3$ h.

#### Step 1: Flow
$$
Q=20\,000\times112=2\,240\,000\ \text{l/day}=2240\ \text{m}^3/\text{day}
$$

#### Step 2: Surface area (overflow rate = settling velocity)
$$
v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{Q}{v_0}=\frac{2240}{17.28}=129.63\ \text{m}^2
$$

#### Step 3: Width and length
$$
2B^2=129.63\Rightarrow B=8.05\ \text{m},\qquad L=2B=16.10\ \text{m}
$$

#### Step 4: Depth
$$
H=\frac{V}{A}=\frac{Q\,t}{A}=\frac{2240\times3/24}{129.63}=2.16\ \text{m}
$$
(Also $H=v_s\,t=0.2\times10^{-3}\times10\,800=2.16$ m.)

Adopt $B=8.1$ m, $L=16.2$ m and an effective depth of 2.2 m, plus 0.3 m free board and a 0.5 m sludge zone, giving a total depth of 3.0 m.

**Answer: required size 16.10 m $\times$ 8.05 m $\times$ 2.16 m effective depth; adopt a tank 16.2 m long $\times$ 8.1 m wide $\times$ 2.2 m effective depth (3.0 m overall).**

## Check yourself

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

1. Which relation gives the plan area?
   A. A = Q/vs
   B. A = Q x vs
   C. A = vs t
   D. A = Q t
   Answer: A. Setting overflow rate equal to vs gives A = 2240/17.28 = 129.63 m2. (Key point: Surface area)

2. What is vs = 0.2 mm/s in m3/m2/day?
   A. 17.28
   B. 172.8
   C. 1.728
   D. 0.2
   Answer: A. 0.2x10^-3 m/s x 86400 s/day = 17.28. (Key point: Unit conversion)

3. With L = 2B and A = 129.63 m2, what is B?
   A. 8.05 m
   B. 11.4 m
   C. 16.1 m
   D. 4.0 m
   Answer: A. 2B^2 = 129.63 gives B = 8.05 m and L = 16.1 m. (Key point: Width and length)

4. What is the effective depth?
   A. 2.16 m, equal to vs x t
   B. 3.5 m
   C. 0.72 m
   D. 6.5 m
   Answer: A. H = Q t/A = 2.16 m, equal to 0.2 mm/s x 10800 s. (Key point: Depth)

## Close variants set in other exams

- [Rectangular sedimentation tank, 0.2 mm/s, 25000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk79) — Water Supply Engineering, chapter 6, asked 2×

## Related in Water Supply Engineering

- [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, 9 million L peak demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c75) — Water Supply Engineering, chapter 6
- [Rectangular sedimentation tank for 10 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s71) — Water Supply Engineering, chapter 6
- [Settling tank 20 m x 2.5 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b66) — Water Supply Engineering, chapter 6, asked 2×
- [Settling tank 30 m x 3 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk68) — Water Supply Engineering, chapter 6, asked 2×
- [Rectangular sedimentation tank for 3000 persons](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-a75) — Water Supply Engineering, chapter 6

## Related in other subjects

- [Continuous flow sedimentation tank particle size](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-continuous-4m-60m) — Sanitary Engineering, chapter 7
- [Circular sedimentation tank, population 5500](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-industrial-5500) — Sanitary Engineering, chapter 7, asked 2×
- [Circular sedimentation tank, population 80000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-industrial-80000) — Sanitary Engineering, chapter 7, asked 2×