---
title: "Settling tank 15 x 4 x 2.5 m, 96% of 0.015 mm"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-k78"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-k78
site: BCT Engineering
---

# Settling tank 15 x 4 x 2.5 m, 96% of 0.015 mm

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2078 Kartik, 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-k78
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

In continuous flow settling tank of 15 m length, 4 m width and 2.5 m effective depth, what detention time would you recommend to remove 96% of particles having diameter of 0.015 mm and specific gravity 2.65 at 15°C? Does the tank is enough to remove 99% of particles having size 0.020 mm at same conditions?

## Short answer

Settling velocity vs for 0.015 mm at 15°C = 0.01776 cm/s. For 96% removal, v0 = vs/0.96 = 0.01850 cm/s; detention time = H/v0 = 250/0.01850 = 3.75 h. vs for 0.020 mm is 0.03158 cm/s > v0, so 100% removal exceeds 99% requirement; tank is sufficient.

## Answer

In an ideal settling tank the fraction of particles removed is $v_s/v_0$ (for $v_s<v_0$), where $v_0=H/t$ is the overflow rate; larger particles are removed completely.

#### (a) Detention time for 96 % removal of 0.015 mm particles at 15 °C
$d=0.0015$ cm, $S_s=2.65$, $\nu_{15}=0.01139$ cm²/s, $H=250$ cm.
$$
v_s=\frac{981\times1.65\times0.0015^2}{18\times0.01139}=0.01776\ \text{cm/s}\quad(Re=0.0023<1)
$$
$$
\frac{v_s}{v_0}=0.96\Rightarrow v_0=\frac{0.01776}{0.96}=0.01850\ \text{cm/s}
$$
$$
t=\frac{H}{v_0}=\frac{250}{0.01850}=13511\ \text{s}=3.75\ \text{h}
$$
For this detention time the tank (15 m $\times$ 4 m $\times$ 2.5 m, volume 150 m³) can pass a flow $Q=V/t=40.0$ m³/h $=959$ m³/day (surface overflow rate $v_0=16.0$ m³/m²/day).

#### (b) Removal of 99 % of 0.020 mm particles
$d=0.002$ cm at the same temperature:
$$
v_s=\frac{981\times1.65\times0.002^2}{18\times0.01139}=0.03158\ \text{cm/s}\quad(Re=0.0055<1)
$$
Compare with the overflow rate $v_0=0.01850$ cm/s: $v_s/v_0=1.71>1$, so these particles are removed **100 %**, which is more than 99 %.

**Answer: detention time = 3.75 h (about 225 min); the tank is sufficient, since 0.020 mm particles are removed 100 % (more than the 99 % required).**

## Check yourself

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

1. To remove 96% of 0.015 mm particles, how is v0 found?
   A. v0 = vs/0.96
   B. v0 = vs x 0.96
   C. v0 = 0.96 vs^2
   D. v0 = vs
   Answer: A. Removal = vs/v0 = 0.96, so v0 = vs/0.96 = 0.0185 cm/s. (Key point: Overflow rate for partial removal)

2. With v0 = 0.0185 cm/s and H = 250 cm, what is the detention time?
   A. About 3.75 h
   B. About 0.375 h
   C. About 1.0 h
   D. About 37.5 h
   Answer: A. t = H/v0 = 250/0.0185 = 13511 s = 3.75 h. (Key point: Detention time)

3. Why is the tank enough for 99% removal of 0.020 mm particles?
   A. vs = 0.0316 cm/s exceeds v0, so removal is 100%
   B. vs equals v0
   C. Their vs is smaller than v0 but close
   D. Temperature rises
   Answer: A. A larger particle settles faster than v0 (ratio 1.71), so it is removed completely. (Key point: Larger particles)

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