---
title: "Plain sedimentation tank for 4 MLD"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-b78"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-b78
site: BCT Engineering
---

# Plain sedimentation tank for 4 MLD

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2078 Bhadra, 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-b78
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Design the plain sedimentation tank for treating 4 MLD of water. Assume necessary data suitably. Sketch designed sedimentation tank with dimensions.

## Short answer

For 4 MLD, assuming 4 h detention and 3 m effective depth, volume is 666.7 m³ and plan area 222.2 m². With L:B=3:1, width is 8.7 m, length 26.0 m. SOR 17.7 m/day, horizontal velocity 0.106 m/min are acceptable. Dimensions: 26.0 m × 8.7 m × 3.0 m effective depth (4.0 m overall).

## Answer

#### Assumed data
Flow $Q=4$ MLD $=4000$ m³/day. Assume detention period $t=4$ h (3 to 4 h for plain sedimentation), effective depth $H=3$ m, $L:B=3:1$ (usual 3:1 to 5:1).

#### Step 1: Volume and area
$$
V=\frac{4000}{24}\times4=666.7\ \text{m}^3,\qquad A=\frac{V}{H}=\frac{666.7}{3}=222.2\ \text{m}^2
$$

#### Step 2: Dimensions
$$
3B^2=222.2\Rightarrow B=8.61\ \text{m},\quad L=3B=25.82\ \text{m}
$$
Adopt $B=8.7$ m, $L=26.0$ m.

#### Step 3: Checks
$$
t=8.7\times26.0\times3/4000\times24=4.07\ \text{h},\quad \text{SOR}=\frac{4000}{226.2}=17.7\ \text{m}^3/\text{m}^2/\text{day}\ (15\text{ to }30),\quad v=\frac{4000}{8.7\times3\times1440}=0.106\ \text{m/min}\ (<0.3)
$$
All values are within the usual limits. Provide a free board of 0.5 m and a sludge zone of 0.5 m with the floor sloping 1 in 10 towards the sludge pit, so the overall depth is 4.0 m.

```text
 PLAN
  +--------+--------------------------------+--------+
  | inlet  |          L = 26.0 m            | outlet |  B = 8.7 m
  | baffle |                                |  weir  |
  +--------+--------------------------------+--------+
 SECTION
  inlet -> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ FB 0.5 m
  baffle   |          H = 3.0 m               |-> outlet
           |__________________________________|
            \__ sludge zone 0.5 m (1:10) __/ -> drain
```

**Answer: rectangular plain sedimentation tank 26.0 m $\times$ 8.7 m $\times$ 3.0 m effective depth (4.0 m overall).**

## Check yourself

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

1. Which detention period and L:B are assumed for plain sedimentation?
   A. 3 to 4 h and 3:1 to 5:1
   B. 10 min and 1:1
   C. 12 h and 10:1
   D. 30 min and 2:1
   Answer: A. Typical values for plain tanks are 3 to 4 h and L:B of 3:1 to 5:1. (Key point: Assumed design values)

2. What plan area results for 4000 m3/day, t = 4 h and H = 3 m?
   A. 222.2 m2
   B. 666.7 m2
   C. 55.6 m2
   D. 1333 m2
   Answer: A. V = 666.7 m3, A = 666.7/3 = 222.2 m2. (Key point: Plan area)

3. Which limits are checked on the adopted tank?
   A. SOR 15 to 30 and horizontal velocity below 0.3 m/min
   B. SOR above 100 and velocity above 1 m/min
   C. Only chlorine dose
   D. Only depth
   Answer: A. SOR = 17.7 m3/m2/day and v = 0.106 m/min are both within limits. (Key point: Checks)

## Related in Water Supply Engineering

- [Plain sedimentation tank for 3.75 L/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as69b) — 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
- [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
- [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×
- [Sedimentation tank 6 m x 18 m x 3 m, 4 ML/day](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-a76) — 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

## Related in other subjects

- [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
- [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
- [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
- [Sludge digestion tank, 6.5 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-digestion-tank-6p5mld) — Sanitary Engineering, chapter 8