---
title: "Circular sedimentation tank for 58 MLD"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:sed-tank-58mld-circular"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-58mld-circular
site: BCT Engineering
---

# Circular sedimentation tank for 58 MLD

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2076 Baisakh, 8 marks
- 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-58mld-circular
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a circular sedimentation tank for 58 MLD of sewage of a temperature of 20°C. Assume specific gravity of organic matters and inorganic matters are 1.12 and 2.65 respectively.

## Short answer

Provide two circular tanks, each 32.5 m diameter, 3.0 m side water depth (total 4.0 m), detention 2.0 h, surface loading 35.0 m³/m²/day.

## Answer

Assumptions: detention 2 h at the design flow of 58 MLD, side water depth 3 m. Two tanks are provided. The particle sizes are not given, so the sizes removed are only checked at the end (20°C, $\nu=1.01\times 10^{-6}$ m²/s).

#### Step 1: Design flow

$$
Q = 58\times 10^6/1000 = 58000\ \text{m}^3/\text{day}
$$

#### 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).

$$
V = Q\,t = \frac{58000}{24}\times 2.0 = 4833.3\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{4833.3}{3.0} = 1611.1\ \text{m}^2
$$

#### Step 3: Diameter

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

$$
D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 805.6}{\pi}} = 32.03\ \text{m}\ \Rightarrow\ \text{adopt } D = 32.5\ \text{m}
$$

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

$$
\begin{aligned}
\text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{29000}{829.6} = 35.0\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\
\text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{29000}{\pi\times 32.5} = 284.0\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\
\text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.06\ \text{h}
\end{aligned}
$$

**Size of particles removed**: the tank removes all particles whose settling velocity is $\ge$ the overflow rate $v_0 = 35.0/86400 = 4.05\times 10^{-4}$ m/s. From Stokes' law, $d=\sqrt{18\nu v_0/(g(G-1))}$:

- inorganic particles ($G=2.65$): $d = 0.021$ mm
- organic particles ($G=1.12$): $d = 0.079$ mm

Particles larger than these are fully removed (Stokes' law is valid, since $Re<1$).

```text
   Section (circular, radial flow)
      inlet well        weir/launder
   ->  |__|~~~~~~~~~~~~~~~~~~~~~|_
       |  \                  /  | H = 3.0 m
       |   \______    ______/   | + sludge zone
       +----------\__/----------+
                  sludge hopper -> sludge pipe
        <---------- D = 32.5 m ---------->
```
**Answer: 2 circular sedimentation tanks, each of diameter 32.5 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.0 m³/m²/day.**

## Check yourself

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

1. How is the tank area obtained for 58000 m^3/day, t = 2 h, H = 3 m?
   A. A = Q x H
   B. A = V / H with V = (58000/24) x 2 = 4833 m^3, so A = 1611 m^2
   C. A = Q / t
   D. A = V x H
   Answer: B. Volume is Q t, and area is volume divided by side water depth. (Key point: Capacity from detention)

2. Why are 2 tanks provided?
   A. To halve the detention time
   B. To double the surface loading
   C. To avoid a central inlet
   D. So that one can be shut down for cleaning
   Answer: D. Each tank has area A/2, so one can be taken out of service for cleaning. (Key point: Why two tanks)

3. How is D obtained for each tank (A1 = 805.6 m^2)?
   A. D = A1 / pi
   B. D = sqrt(4 A1 / pi) = 32 m, adopted 32.5 m
   C. D = 2 A1
   D. D = sqrt(A1 / 4)
   Answer: B. Circular area is pi D^2/4, so D = sqrt(4A/pi). (Key point: Diameter)

4. What is checked against the usual 25-40 m^3/m^2/day?
   A. Flow per tank divided by surface area, about 35
   B. Depth divided by detention
   C. Weir length
   D. Tank volume
   Answer: A. Surface loading = (Q/N)/(pi D^2/4) = 29000/829.6 = 35 m^3/m^2/day. (Key point: Check of surface loading)

5. What principle gives the size of particles removed?
   A. Particles with settling velocity at least the overflow rate are removed; Stokes' law gives d
   B. Only the largest grit is removed
   C. All particles are removed regardless of size
   D. Particles with settling velocity below overflow rate are fully removed
   Answer: A. A particle is fully removed if v_s >= v_0 = surface loading; Stokes' law converts this to diameter. (Key point: Particle size removed)

## Related in Sanitary Engineering

- [Grit chamber for 58 MLD at 25°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-58mld-25c) — 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
- [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×
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- [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

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- [Old tank 12 m x 5 m x 3 m as settling tank](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c68) — Water Supply Engineering, chapter 6, asked 2×