---
title: "Rectangular sedimentation tank for 10 MLD"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-s71"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-s71
site: BCT Engineering
---

# Rectangular sedimentation tank for 10 MLD

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2071 Shrawan, 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-s71
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A rectangular sedimentation tank is to treat 10 MLD of water. A detention basin of width to length ratio of 1/3 is proposed to trap all particles larger than 0.04 mm in size. Assuming a specific gravity of particles at 20°C is 2.65, compute the tank dimensions. If the depth of the tank is 3.5 m, calculate the detention time.

## Short answer

Settling velocity by Stokes' law is 0.1429 cm/s (123.45 m/day). For 10 MLD, required surface area is 81.01 m². With L=3B, dimensions are 15.59 m × 5.20 m. For depth 3.5 m, detention time is 0.68 h (about 41 min).

## Answer

#### Data
$Q=10$ MLD $=10\,000$ m³/day; trap all particles larger than 0.04 mm ($d=0.004$ cm); $S_s=2.65$; $T=20$ °C ($\nu=0.01007$ cm²/s); width : length $=1:3$; depth $=3.5$ m.

#### Step 1: Settling velocity
$$
v_s=\frac{981\times1.65\times0.004^2}{18\times0.01007}=0.1429\ \text{cm/s}=123.45\ \text{m/day}\quad(Re=0.057<1)
$$
To trap all particles of this size, the overflow rate must equal the settling velocity: $v_0=Q/A=v_s$.

#### Step 2: Surface area
$$
A=\frac{Q}{v_s}=\frac{10\,000}{123.45}=81.01\ \text{m}^2
$$

#### Step 3: Length and width
$L=3B$, so $3B^2=81.01$:
$$
B=5.20\ \text{m},\qquad L=3B=15.59\ \text{m}
$$

#### Step 4: Detention time for depth 3.5 m
$$
t=\frac{V}{Q}=\frac{81.01\times3.5}{10\,000}=0.0284\ \text{day}=0.68\ \text{h}=40.8\ \text{min}
$$
(This also equals $H/v_s=350/0.1429$ s $=40.8$ min.)

**Answer: tank 15.59 m long $\times$ 5.20 m wide ($A=81.0$ m²); detention time for H = 3.5 m is 0.68 h (41 min).** A longer detention (2 to 4 h) is normally adopted in practice by providing a larger tank to allow for inefficiencies such as short-circuiting.

## Check yourself

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

1. To trap all particles larger than 0.04 mm, what must hold?
   A. Overflow rate v0 = Q/A equals their settling velocity
   B. v0 is double vs
   C. Depth equals 3.5 m
   D. Re exceeds 1
   Answer: A. The smallest particle to be fully removed has vs = v0. (Key point: Condition to trap all particles)

2. With vs = 123.45 m/day and Q = 10000 m3/day, what is A?
   A. 81.01 m2
   B. 123.45 m2
   C. 12.3 m2
   D. 1234 m2
   Answer: A. A = Q/vs = 10000/123.45 = 81.01 m2. (Key point: Surface area)

3. With width : length = 1:3, which pair results?
   A. B = 5.20 m, L = 15.59 m
   B. B = 9 m, L = 9 m
   C. B = 15.59 m, L = 5.20 m
   D. B = 2.7 m, L = 30 m
   Answer: A. 3B^2 = 81.01 gives B = 5.20 m. (Key point: Width and length)

4. What is the detention time for H = 3.5 m?
   A. About 41 min (0.68 h)
   B. About 4 h
   C. About 6.8 h
   D. About 4.1 min
   Answer: A. t = V/Q = 81.01 x 3.5/10000 = 0.0284 day, equal to H/vs. (Key point: Detention time)

## 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×
- [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 for 3000 persons](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-a75) — 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 long, 3.5 m deep; 0.02 and 0.01 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b79) — Water Supply Engineering, chapter 6, asked 3×

## Related in other subjects

- [Sludge volume from sedimentation tank, 10 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-10mld-250) — Sanitary Engineering, chapter 8
- [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
- [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
- [Circular sedimentation tank for 58 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-58mld-circular) — Sanitary Engineering, chapter 7