---
title: "Rectangular sedimentation tank, 0.2 mm/s, 25000 population"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-bk79"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-bk79
site: BCT Engineering
---

# Rectangular sedimentation tank, 0.2 mm/s, 25000 population

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2079 Baisakh, 8 marks
- 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-bk79
- 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 a water for population of 25000 with a peak allowance of 110 lpcd. Draw neat sketch including all dimensions.

## Short answer

Flow is 2.75 MLD. Surface area from overflow rate (17.28 m/day) is 159.14 m². With L=2B, width is 8.92 m, length 17.84 m. Detention period of 3 h gives effective depth 2.16 m. Adopted dimensions: 18.0 m × 9.0 m × 2.2 m effective depth (3.0 m overall).

## Answer

#### Data
Population 25 000; peak allowance 110 lpcd; $v_s=0.2$ mm/s; $L=2B$; settling period $t=3$ h.

#### Step 1: Flow
$$
Q=25\,000\times110=2\,750\,000\ \text{l/day}=2750\ \text{m}^3/\text{day}
$$

#### Step 2: Surface area
$$
v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{2750}{17.28}=159.14\ \text{m}^2
$$

#### Step 3: Width and length
$$
2B^2=159.14\Rightarrow B=8.92\ \text{m},\qquad L=2B=17.84\ \text{m}
$$

#### Step 4: Depth
$$
H=\frac{Q\,t}{A}=\frac{2750\times3/24}{159.14}=2.16\ \text{m}
$$
Adopt $B=9.0$ m, $L=18.0$ m, effective depth 2.2 m, free board 0.3 m and sludge zone 0.5 m (total 3.0 m).

#### Sketch
```text
 PLAN
  +-------+------------------------------+--------+
  | inlet |         L = 18.0 m           | outlet |  B = 9.0 m
  +-------+------------------------------+--------+
 SECTION
  inlet ->  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~  FB 0.3 m
            |      H = 2.2 m           |-> outlet
            |__________________________|
             \__ sludge zone 0.5 m __/ -> drain
```

**Answer: tank 18.0 m $\times$ 9.0 m in plan with 2.2 m effective depth (3.0 m total).**

## Check yourself

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

1. What is the flow for 25000 persons at 110 lpcd?
   A. 2750 m3/day
   B. 275 m3/day
   C. 27500 m3/day
   D. 2250 m3/day
   Answer: A. 25000 x 110 = 2.75 x 10^6 l/day = 2750 m3/day. (Key point: Flow)

2. How is the plan area found?
   A. Flow divided by vs (overflow rate = vs)
   B. Flow times vs
   C. Flow divided by depth
   D. Volume times depth
   Answer: A. A = 2750/17.28 = 159.14 m2. (Key point: Area)

3. With L = 2B, which pair results?
   A. B = 8.92 m, L = 17.84 m
   B. B = 12.6 m, L = 12.6 m
   C. B = 17.84 m, L = 8.92 m
   D. B = 6 m, L = 12 m
   Answer: A. 2B^2 = 159.14, so B = 8.92 m. (Key point: Dimensions)

4. What effective depth is obtained from H = Q t/A?
   A. 2.16 m
   B. 3.0 m
   C. 1.08 m
   D. 4.32 m
   Answer: A. H = (2750 x 3/24)/159.14 = 2.16 m, equal to vs x t. (Key point: Depth)

## Close variants set in other exams

- [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×

## 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×