---
title: "Circular sedimentation tank, population 80000"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:sed-tank-industrial-80000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-industrial-80000
site: BCT Engineering
---

# Circular sedimentation tank, population 80000

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2070 Bhadra, 8 marks
- Video: [Sedimentation tank Solved design dimension of tank | design of sedimentation tank | waste water engg](https://www.youtube.com/watch?v=P3phfraI2-g) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-industrial-80000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

What will be the circular sewage sedimentation tank dimension for an industrial area having population of 80,000? The average water demand is 135 lpcd. Assume that 78% water reaches at treatment plant and maximum demand is 2.5 times average demand. Dimension of the suspended silica particles available in influent water more than 0.12 mm.

## Short answer

Two circular tanks, each 19.5 m diameter, 3.0 m side water depth (total 4.0 m), detention 2.0 h, surface loading 35.3 m³/m²/day.

## Answer

**Assumptions:** sewage reaching the plant $=0.78\times135 = 105.3$ lpcd, designed for the maximum rate ($2.5\times$ average): $Q_{avg}=80000\times 105.3/1000 = 8424.0$ m³/day. Detention 2 h and depth 3 m are adopted; two tanks are provided. Silica $d=0.12$ mm ($G=2.65$) at 20°C.

#### Step 1: Design flow

$$
Q = 80000\times 135\times 0.78\times 2.5/1000 = 21060\ \text{m}^3/\text{day}
$$

#### Step 2: Capacity

Adopt detention period $t = 2.0$ h and side water depth $H = 3.0$ m (typical for plain sedimentation: 2-3 h, 2.5-4 m).

$$
V = Q\,t = \frac{21060}{24}\times 2.0 = 1755.0\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{1755.0}{3.0} = 585.0\ \text{m}^2
$$

#### Step 3: Diameter

Provide 2 circular tanks, each with area $A_1 = A/2$ (one tank can then be shut down for cleaning).

$$
D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 292.5}{\pi}} = 19.30\ \text{m}\ \Rightarrow\ \text{adopt } D = 19.5\ \text{m}
$$

Adopt a total depth of $H + 0.5\ (\text{free board}) + 0.5\ (\text{sludge zone}) = 4.0$ m, with a central inlet well and a peripheral weir.

#### Step 4: Checks

$$
\begin{aligned}
\text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{10530}{298.6} = 35.3\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\
\text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{10530}{\pi\times 19.5} = 171.9\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\
\text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.04\ \text{h}
\end{aligned}
$$

**Check:** for $0.1<d<1$ mm, Hazen's transition formula at 20°C gives $v_s = 60.6\times1.65\times0.12\times1.3 = 15.6$ mm/s $\approx 1348$ m/day ($d$ in mm). This is far greater than the overflow rate of the tank, so silica of 0.12 mm and larger is removed.

```text
   Section (circular, radial flow)
      inlet well        weir/launder
   ->  |__|~~~~~~~~~~~~~~~~~~~~~|_
       |  \                  /  | H = 3.0 m
       |   \______    ______/   | + sludge zone
       +----------\__/----------+
                  sludge hopper -> sludge pipe
        <---------- D = 19.5 m ---------->
```
**Answer: 2 circular sedimentation tanks, each of diameter 19.5 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.3 m³/m²/day.**

## Check yourself

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

1. What is the design flow for 80,000 people?
   A. 80000 x 135 x 0.78 / 1000 = 8424 m^3/day
   B. 80000 x 135 x 2.5 / 1000
   C. 80000 x 135 x 0.78 x 2.5 / 1000 = 21060 m^3/day
   D. 80000 x 135 / 0.78 x 2.5
   Answer: C. Includes both the 78% sewage fraction and maximum factor 2.5. (Key point: Design flow)

2. With 2 tanks, t = 2 h, H = 3 m, what is the area of each?
   A. 585 m^2
   B. 1755 m^2
   C. 292.5 m^2
   D. 146 m^2
   Answer: C. Total V = 1755 m^3 gives A = 585 m^2, so each of two has 292.5 m^2. (Key point: Area per tank)

3. What diameter follows from 292.5 m^2?
   A. D = sqrt(292.5) = 17 m
   B. D = sqrt(4 x 292.5/pi) = 19.3 m, adopted 19.5 m
   C. D = 292.5 / pi = 93 m
   D. D = 585 / 3 = 195 m
   Answer: B. D = sqrt(4A/pi). (Key point: Diameter)

4. What does the weir loading check ensure?
   A. Detention above 5 h
   B. Flow per metre of weir below about 300 m^3/m/day, here 171.9
   C. Depth below 1 m
   D. Surface loading above 100
   Answer: B. Weir loading = (Q/N)/(pi D) must be below about 300 to avoid carrying solids over the weir. (Key point: Weir loading check)

## Close variants set in other exams

- [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×

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## Related in other subjects

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