---
title: "Settling tank 15 m x 7 m x 3 m, 95% removal"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-b81"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-b81
site: BCT Engineering
---

# Settling tank 15 m x 7 m x 3 m, 95% removal

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2081 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_5/water_supply_engineering/sed-b81
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A tank with dimensions of 15 m x 7 m x 3 m is proposed to be used as a settling tank. It is expected to remove at least 95 percent of particles with a diameter of 0.035 mm and a specific gravity of 2.65 at 20°C. What will be the overflow rate and flow velocity of the tank at 20°C? Additionally, will the tank's dimensions be sufficient to remove 100 percent of particles with a diameter of 0.040 mm under the same conditions?

## Short answer

Settling velocity vs for 0.035 mm particles at 20°C is 0.1094 cm/s. Overflow rate v0 = vs / 0.95 = 0.1151 cm/s (99.5 m³/m²/day). Flow velocity = Q/(B H) = 0.5757 cm/s. For 0.040 mm particles vs = 0.1429 cm/s > v0, so 100% removal; tank dimensions adequate.

## Answer

In an ideal tank the fraction of particles removed is $v_s/v_0$ for $v_s<v_0$, and 100 % for $v_s\ge v_0$, where $v_0=Q/A$ is the overflow rate.

#### Step 1: Settling velocity of 0.035 mm particles at 20 °C
$d=0.0035$ cm, $S_s=2.65$, $\nu=0.01007$ cm²/s.
$$
v_s=\frac{981\times1.65\times0.0035^2}{18\times0.01007}=0.1094\ \text{cm/s}\quad(Re=0.038<1)
$$

#### Step 2: Overflow rate for 95 % removal
$$
\frac{v_s}{v_0}=0.95\Rightarrow v_0=\frac{0.1094}{0.95}=0.1151\ \text{cm/s}=99.5\ \text{m}^3/\text{m}^2/\text{day}
$$

#### Step 3: Flow and flow velocity
Surface area $A=15\times7=105$ m²; cross-section $=7\times3=21$ m².
$$
Q=v_0A=0.001151\times105=0.12091\ \text{m}^3/\text{s}\ (=10446\ \text{m}^3/\text{day})
$$
$$
\text{Flow velocity}=\frac{Q}{B\,H}=\frac{0.12091}{21}=0.00576\ \text{m/s}=0.5757\ \text{cm/s} =0.345\ \text{m/min}
$$
This is slightly above the usual guideline of about 0.3 m/min for avoiding scour of settled solids. The question only asks for the value, but in practice the tank would be made wider or deeper to reduce the velocity.

#### Step 4: Check for 100 % removal of 0.040 mm particles
$d=0.004$ cm:
$$
v_s=\frac{981\times1.65\times0.004^2}{18\times0.01007}=0.1429\ \text{cm/s}\quad(Re=0.057<1)
$$
$v_s=0.1429$ cm/s $>v_0=0.1151$ cm/s ($v_s/v_0=1.24>1$), so these particles are removed completely.

**Answer: overflow rate = 0.1151 cm/s (99.5 m³/m²/day); flow velocity = 0.5757 cm/s (0.345 m/min); yes, the tank is sufficient to remove 100 % of 0.040 mm particles.**

## Check yourself

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

1. For particles with vs < v0 in an ideal tank, what fraction is removed?
   A. vs/v0
   B. v0/vs
   C. 1 - vs/v0
   D. 100%
   Answer: A. Removal fraction is vs/v0 for particles settling slower than the overflow rate. (Key point: Removal in an ideal tank)

2. If vs = 0.1094 cm/s for the design particle, what overflow rate gives 95% removal?
   A. 0.1094 cm/s
   B. 0.1151 cm/s
   C. 0.1039 cm/s
   D. 0.95 cm/s
   Answer: B. v0 = vs/0.95 = 0.1151 cm/s. (Key point: Overflow rate for 95% removal)

3. How is the horizontal flow velocity found?
   A. Q divided by surface area
   B. Q divided by cross-section B x H
   C. Q divided by volume
   D. vs times L/H
   Answer: B. Velocity = Q/(B H) = 0.12091/21 = 0.00576 m/s. (Key point: Flow velocity)

4. Why is the tank sufficient for 100% removal of 0.040 mm particles?
   A. Their vs = 0.1429 cm/s exceeds v0 = 0.1151 cm/s
   B. Their vs is below v0
   C. The tank is deep enough
   D. Stokes' law fails for them
   Answer: A. Particles with vs >= v0 are fully removed; here vs/v0 = 1.24. (Key point: Check for larger particles)

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