---
title: "Old tank 11 m x 5 m x 3 m as settling tank"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c70"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c70
site: BCT Engineering
---

# Old tank 11 m x 5 m x 3 m as settling tank

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2070 Chaitra, 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-c70
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

An old tank having dimension of 11 m x 5 m x 3 m is available in a village. It is proposed to use as a settling tank. At least 93 percent of particles having diameter of 0.025 mm, specific gravity 2.65 is expected to remove on the tank at 20°C. What will be an overflow rate on using that tank? Does the tank dimension is enough to remove 99 percentages of particles having diameter 0.05 mm at same conditions?

## Short answer

Required overflow rate is 0.0600 cm/s (51.9 m³/m²/day). The tank provides 100% removal of 0.05 mm particles, which easily meets the 99% target.

## 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 93 % removal:
$$
v_0=\frac{v_s}{0.93}=\frac{0.0558}{0.93}=0.0600\ \text{cm/s}=51.9\ \text{m}^3/\text{m}^2/\text{day}\ (=2.16\ \text{m}^3/\text{m}^2/\text{h})
$$

Surface area $A=11\times5=55$ m², so the tank can treat
$$
Q=v_0A=0.000600\times55=0.03301\ \text{m}^3/\text{s}=2852\ \text{m}^3/\text{day}
$$
Horizontal velocity $=Q/(B\,H)=0.03301/(5\times3)=0.2200$ cm/s $=0.132$ m/min.

#### Step 3: Check for 99 % removal of 0.05 mm particles
$d=0.005$ cm:
$$
v_s=\frac{981\times1.65\times0.005^2}{18\times0.01007}=0.2232\ \text{cm/s}\quad(Re=0.111<1)
$$
$v_s=0.2232$ cm/s is greater than $v_0=0.0600$ cm/s ($v_s/v_0=3.72$), so these particles are removed **100 %**, which exceeds 99 %.

**Answer: overflow rate = 0.0600 cm/s (51.9 m³/m²/day); yes, the tank is sufficient to remove 99 % of 0.05 mm particles (it removes 100 %).**

## Check yourself

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

1. What is the overflow rate v0 of a tank?
   A. Q/A, flow divided by plan area
   B. Q/(B H)
   C. V/Q
   D. H x L
   Answer: A. Overflow rate is flow per unit surface area. (Key point: Overflow rate definition)

2. Given vs = 0.0558 cm/s, what is v0 for 93% removal?
   A. 0.0600 cm/s
   B. 0.0519 cm/s
   C. 0.0558 cm/s
   D. 0.93 cm/s
   Answer: A. v0 = 0.0558/0.93 = 0.0600 cm/s. (Key point: Overflow rate for 93% removal)

3. How is the flow the 11 m x 5 m tank can treat found?
   A. Q = v0 x A, with A = 55 m2
   B. Q = v0 x volume
   C. Q = vs x depth
   D. Q = A/v0
   Answer: A. Q = 0.000600 x 55 = 0.03301 m3/s. (Key point: Flow capacity)

4. Why are 0.05 mm particles removed 100%?
   A. vs = 0.2232 cm/s is greater than v0
   B. vs is less than v0
   C. Stokes' law fails
   D. They are lighter
   Answer: A. When vs >= v0 the particle is removed completely. (Key point: Check of larger particles)

## Close variants set in other exams

- [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×

## 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