---
title: "Rectangular sedimentation tank, 20000 population, 0.4 mm/s"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-s73"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-s73
site: BCT Engineering
---

# Rectangular sedimentation tank, 20000 population, 0.4 mm/s

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2073 Shrawan, 8 marks
- Video: [CE405 - Module 3 - Design of Rectangular sedimentation tank-Plain Sedimentation](https://www.youtube.com/watch?v=Kdfp60mW6n4) — Splash Environs
- 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-s73
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Design a rectangular sedimentation tank for water treatment in a city with population of 20000. Considering the settling velocity of particles 0.4 mm/sec, Length = 2 x Width and detention period of 2 hours.

## Short answer

Tank is 12 m long, 6 m wide, with 3.0 m effective depth and 3.8 m total depth. Overflow rate equals the settling velocity of 34.56 m³/m²/day, and horizontal velocity is 0.093 m/min.

## Answer

#### Data and assumptions
Population 20 000; per-capita demand is not given, so 120 lpcd is assumed. Settling velocity $v_s=0.4$ mm/s; $L=2B$; detention period $t=2$ h.

#### Step 1: Flow
$$
Q=20\,000\times120=2\,400\,000\ \text{l/day}=2400\ \text{m}^3/\text{day}
$$

#### Step 2: Surface area
For all particles with $v_s=0.4$ mm/s to be removed, the overflow rate must equal $v_s$:
$$
v_0=v_s=0.4\times10^{-3}\times86400=34.56\ \text{m}^3/\text{m}^2/\text{day}
$$
$$
A=\frac{Q}{v_0}=\frac{2400}{34.56}=69.44\ \text{m}^2
$$

#### Step 3: Width and length
$$
2B^2=69.44\Rightarrow B=5.89\ \text{m},\qquad L=2B=11.79\ \text{m}
$$
Adopt $B=6$ m, $L=12$ m ($A=72$ m²).

#### Step 4: Depth
$$
V=Q\,t=\frac{2400}{24}\times2=200\ \text{m}^3,\qquad H=\frac{V}{A}=\frac{200}{69.44}=2.88\ \text{m}
$$
(The same as $H=v_st=0.4\times10^{-3}\times7200=2.88$ m.) Adopt effective depth 3.0 m, so the detention time $=72\times3/2400\times24=2.16$ h.

Add free board 0.3 m and a sludge zone 0.5 m: total depth 3.8 m. Horizontal velocity $=2400/(6\times3\times1440)=0.093$ m/min (OK).

**Answer: tank 12 m long $\times$ 6 m wide $\times$ 3.0 m effective depth (3.8 m overall).**

## Check yourself

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

1. To remove all particles with vs = 0.4 mm/s, what must the overflow rate be?
   A. Equal to vs (or less)
   B. Twice vs
   C. Half of vs
   D. Independent of vs
   Answer: A. Particles are fully removed when v0 = Q/A <= vs, so A = Q/vs. (Key point: Surface area from settling velocity)

2. What is the surface area for Q = 2400 m3/day and vs = 34.56 m/day?
   A. 69.44 m2
   B. 86.4 m2
   C. 14.4 m2
   D. 138.9 m2
   Answer: A. A = 2400/34.56 = 69.44 m2. (Key point: Area value)

3. With L = 2B, what are B and L?
   A. B = 5.89 m, L = 11.79 m
   B. B = 8.3 m, L = 8.3 m
   C. B = 11.8 m, L = 5.9 m
   D. B = 3.5 m, L = 7 m
   Answer: A. 2B^2 = 69.44 gives B = 5.89 m, L = 11.79 m. (Key point: Width and length)

4. How is the depth obtained?
   A. H = V/A, equal to vs x t
   B. H = Q/A
   C. H = L/2
   D. H = vs/t
   Answer: A. H = 200/69.44 = 2.88 m, equal to 0.4 mm/s x 7200 s. (Key point: Depth)

## 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, 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×
- [Settling tank for 41,600 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c76) — 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
- [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×

## Related in other subjects

- [Settling basin design for 0.2 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-2mm) — Hydro Power, chapter 5
- [Settling basin design for 0.15 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-15mm) — Hydro Power, chapter 5
- [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×