---
title: "Theory of settlement; Stoke's law"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:stokes-theory-settlement"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/stokes-theory-settlement
site: BCT Engineering
---

# Theory of settlement; Stoke's law

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2072 Chaitra, 8 marks; 2074 Asoj, 8 marks (asked 2 times)
- Video: [Sedimentation Analysis - Stokes Law](https://www.youtube.com/watch?v=U6qnDuZ0xn0) — Elementary Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/stokes-theory-settlement
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Explain briefly the theory of settlement of discrete particles through quiescent liquids (derive Stoke's law and state the conditions under which it is suitable for design of a sedimentation tank). How do you modify the theory to consider the temperature effect?

## Short answer

Stokes' law: v_s = g(S_s-1)d^2/(18ν) for discrete spherical particles at Re<1. It applies to laminar settling of non-flocculent particles without hindrance. Temperature effect: kinematic viscosity ν = 0.01775/(1+0.0337T+0.000221T²) cm²/s; settling at 20°C is about 1.3 times faster than at 10°C, so design uses the lowest water temperature.

## Answer

#### Theory of settlement of discrete particles
A **discrete particle** is a particle that keeps its size, shape and weight while settling (e.g. sand, silt). In a quiescent liquid, it first accelerates under gravity, then very quickly reaches a **terminal (settling) velocity** when the net downward force is balanced by drag.

#### Stokes' law
For a spherical particle falling freely at constant (terminal) velocity $v_s$, the weight minus buoyancy equals the fluid drag.

$$
\begin{aligned} \text{Net weight} &= \frac{\pi d^3}{6}(\rho_s-\rho)\,g \\ \text{Drag (viscous, } Re<1) &= 3\pi\mu d\,v_s \end{aligned}
$$

Equating the two:

$$
v_s = \frac{g\,(\rho_s-\rho)\,d^2}{18\,\mu} = \frac{g\,(S_s-1)\,d^2}{18\,\nu}
$$

where $d$ = particle diameter, $S_s$ = specific gravity of the particle, $\nu=\mu/\rho$ = kinematic viscosity.

**Conditions of application in design**
- The particle must be spherical, discrete (non-flocculent) and settle freely without being hindered by neighbours.
- The flow around it must be laminar, with Reynolds number $Re=\dfrac{v_s d}{\nu}<1$ (roughly particle diameter below about 0.1 mm for silt of $S_s=2.65$).
- The liquid is at rest or flows at very low velocity (no turbulence, no short-circuiting).
- For larger particles ($1<Re<1000$) the drag coefficient is taken as $C_d=\dfrac{24}{Re}+\dfrac{3}{\sqrt{Re}}+0.34$ and $v_s=\sqrt{\dfrac{4g(S_s-1)d}{3C_d}}$ is found by trial. For $Re>1000$, $v_s=1.8\sqrt{g d (S_s-1)}$ (Newton).

In an ideal rectangular tank the removal depends only on the overflow rate $v_0=Q/A=H/t$: a particle with $v_s\ge v_0$ is entirely removed.

#### Effect of temperature
Temperature changes the viscosity, and so the settling velocity ($v_s\propto 1/\nu$). The kinematic viscosity at temperature $T$ (°C) can be calculated from

$$
\nu = \frac{0.01775}{1+0.0337\,T+0.000221\,T^2}\ \text{cm}^2/\text{s}
$$

and for the same particle:

$$
\frac{v_{s,T_2}}{v_{s,T_1}} = \frac{\nu_{T_1}}{\nu_{T_2}}
$$

Typical values:

| Temp (°C) | 10 | 15 | 20 | 21 | 25 |
|---|---|---|---|---|---|
| $\nu$ (cm$^2$/s) | 0.01306 | 0.01139 | 0.01007 | 0.00978 | 0.00893 |

Hence the settling velocity at 20 °C is about 1.3 times that at 10 °C. For design, the lowest expected temperature is used, so the tank is also effective in warm weather.

## Check yourself

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

1. How is Stokes' law derived for a settling sphere?
   A. Net weight (weight minus buoyancy) is equated to viscous drag 3 pi mu d vs at terminal velocity
   B. Weight is equated to buoyancy only
   C. Drag is equated to the tank velocity
   D. Kinetic energy is equated to potential energy
   Answer: A. At terminal velocity the net downward force pi d^3 (rho_s - rho) g / 6 equals drag 3 pi mu d vs. (Key point: Derivation basis)

2. Stokes' law is valid when:
   A. Reynolds number is above 1000
   B. Flow is turbulent
   C. Particles are flocculent and hindered
   D. Reynolds number is below 1 (laminar flow around a discrete spherical particle)
   Answer: D. The viscous drag expression holds only for laminar flow, Re = vs d / nu < 1, with free settling of discrete particles. (Key point: Validity condition)

3. How is the settling velocity corrected for temperature?
   A. vs2/vs1 = nu2/nu1
   B. vs2/vs1 = nu1/nu2, since vs is inversely proportional to kinematic viscosity
   C. vs is independent of nu
   D. vs is proportional to nu
   Answer: B. For the same particle, vs varies as 1/nu, so the ratio of velocities is the inverse ratio of viscosities. (Key point: Temperature correction)

4. For Re above 1000 the settling velocity follows:
   A. Stokes' law, d^2/18nu
   B. Newton's law, vs = 1.8 sqrt(g d (Ss - 1))
   C. vs = constant 1 cm/s
   D. vs proportional to 1/d
   Answer: B. At very high Reynolds number the Newton relation applies instead of Stokes' law. (Key point: Large particles)

## Related in Water Supply Engineering

- [Rectangular sedimentation tank, 20000 population, 0.4 mm/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s73) — Water Supply Engineering, chapter 6
- [Rectangular sedimentation tank, 0.2 mm/s, 3 hours](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c71) — Water Supply Engineering, chapter 6, asked 2×
- [Rectangular sedimentation tank, 0.2 mm/s, 25000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk79) — Water Supply Engineering, chapter 6, asked 2×
- [Rectangular sedimentation tank for 10 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s71) — Water Supply Engineering, chapter 6
- [Rectangular sedimentation tank, 9 million L peak demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c75) — Water Supply Engineering, chapter 6
- [Plain sedimentation tank for 4 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b78) — Water Supply Engineering, chapter 6

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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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