---
title: "Old tank 12 m x 5 m x 3 m as settling tank"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c68"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c68
site: BCT Engineering
---

# Old tank 12 m x 5 m x 3 m as settling tank

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2068 Chaitra, 10 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-c68
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

An old tank having dimension of 12 m x 5 m x 3 m is available in a village. It is proposed to use as a settling tank. At least 95 percentage of particles having diameter of 0.025 mm, specific gravity 2.65 is expected to remove on that tank at 20°C. What will be an overflow rate on using that tank? Does tank dimension is enough to remove 99 percentage of particles having diameter 0.04 mm at same conditions?

## Short answer

Required overflow rate is 0.0587 cm/s (50.8 m³/m²/day). The tank removes 100% of 0.04 mm particles, exceeding the 99% requirement.

## Answer

In an ideal settling tank, particles with $v_s\ge v_0$ are fully removed, and those with $v_s<v_0$ are removed in the ratio $v_s/v_0$, where $v_0=Q/A$ is the overflow rate.

#### Step 1: Settling velocity of the 0.025 mm particles
$d=0.0025$ cm, $S_s=2.65$, $\nu_{20}=0.01007$ cm²/s.
$$
v_s=\frac{981\times1.65\times0.0025^2}{18\times0.01007}=0.0558\ \text{cm/s}\quad(Re=0.014<1)
$$

#### Step 2: Overflow rate
For 95 % removal:
$$
v_0=\frac{v_s}{0.95}=\frac{0.0558}{0.95}=0.0587\ \text{cm/s}=50.8\ \text{m}^3/\text{m}^2/\text{day}\ (=2.11\ \text{m}^3/\text{m}^2/\text{h})
$$

Surface area $A=12\times5=60$ m², so the tank can treat
$$
Q=v_0A=0.000587\times60=0.03525\ \text{m}^3/\text{s}=3046\ \text{m}^3/\text{day}
$$
Horizontal velocity $=Q/(B\,H)=0.03525/(5\times3)=0.2350$ cm/s $=0.141$ m/min.

#### Step 3: Check for 99 % removal of 0.04 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 is greater than $v_0=0.0587$ cm/s ($v_s/v_0=2.43$), so these particles are removed **100 %**, which exceeds 99 %.

**Answer: overflow rate = 0.0587 cm/s (50.8 m³/m²/day); yes, the tank is sufficient to remove 99 % of 0.04 mm particles (it removes 100 %).**

## Check yourself

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

1. Given vs = 0.0558 cm/s for 0.025 mm particles, what is v0 for 95% removal?
   A. 0.0587 cm/s
   B. 0.0530 cm/s
   C. 0.0558 cm/s
   D. 0.0950 cm/s
   Answer: A. v0 = 0.0558/0.95 = 0.0587 cm/s. (Key point: Overflow rate for 95% removal)

2. Which dimensions are used to get the flow the tank can treat?
   A. Plan area 12 x 5 m
   B. Cross-section 5 x 3 m
   C. Volume 12 x 5 x 3 m
   D. Depth only
   Answer: A. Q = v0 A with A = 60 m2, giving 0.03525 m3/s. (Key point: Tank capacity)

3. Why is the tank enough for 99% removal of 0.04 mm particles?
   A. vs = 0.1429 cm/s exceeds v0 = 0.0587 cm/s
   B. vs is below v0
   C. They remain suspended
   D. Depth is 3 m
   Answer: A. vs/v0 = 2.43 > 1, so removal is 100%, above the 99% needed. (Key point: Larger particles)

## Close variants set in other exams

- [Old tank 11 m x 5 m x 3 m as settling tank](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c70) — Water Supply Engineering, chapter 6, asked 2×

## Related in Water Supply Engineering

- [Settling tank 30 m x 3 m, 0.03 mm at 25°C; 0.025 mm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k72) — Water Supply Engineering, chapter 6, asked 3×
- [Settling tank 15 m x 7 m x 3 m, 95% removal](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b81) — Water Supply Engineering, chapter 6
- [Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b79) — Water Supply Engineering, chapter 6, asked 3×
- [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 15 x 4 x 2.5 m, 96% of 0.015 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k78) — Water Supply Engineering, chapter 6
- [Settling tank 30 m x 3 m, 0.02 mm at 25°C; 0.01 mm at 10°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as70) — Water Supply Engineering, chapter 6, asked 3×

## Related in other subjects

- [Settling basin design for 0.2 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-2mm) — Hydro Power, chapter 5
- [Settling basin design for 0.15 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-15mm) — Hydro Power, chapter 5