---
title: "Settling velocity of silica particle 0.02 cm at 20°C"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c69"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c69
site: BCT Engineering
---

# Settling velocity of silica particle 0.02 cm at 20°C

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2069 Chaitra, 8 marks
- Video: [Stokes law - settling velocity of the particle](https://www.youtube.com/watch?v=6jPTOKvs8x0) — JE Center By Ravi Kant Verma
- 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-c69
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Find the settling velocity of silica particle of size 0.02 cm with specific gravity 2.65 in water at 20°C. Take kinematic viscosity of water at 20°C as 1.007 centistokes.

## Short answer

Stokes' law gives 3.57 cm/s but Re=7.09. Transition-zone iteration converges to a settling velocity of about 2.63 cm/s (26.3 mm/s) at Re≈5.2.

## Answer

Settling velocity of a discrete spherical particle is first tried with Stokes' law, and the Reynolds number is checked. If $Re>1$ the transition formula must be used.

#### Data
$d=0.2$ mm $=0.02$ cm, $S_s=2.65$, $\nu=0.01007$ cm²/s (given: 1.007 centistokes), $g=981$ cm/s².

#### Step 1: Stokes' law (trial)
$$
v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.02^2}{18\times0.01007}=3.572\ \text{cm/s}
$$
$$
Re=\frac{v_s d}{\nu}=\frac{3.572\times0.02}{0.01007}=7.09
$$
Since $Re>1$, Stokes' law is not valid and the flow is in the transition zone ($1<Re<1000$).

#### Step 2: Transition zone (Fair and Geyer)
$$
C_d=\frac{24}{Re}+\frac{3}{\sqrt{Re}}+0.34,\qquad v_s=\sqrt{\frac{4g(S_s-1)d}{3C_d}}
$$
Trial and error:

| Trial | $v$ assumed (cm/s) | $Re=vd/\nu$ | $C_d$ | new $v$ (cm/s) |
|---|---|---|---|---|
| 1 | 3.572 | 7.09 | 4.849 | 2.983 |
| 2 | 2.983 | 5.93 | 5.623 | 2.771 |
| 3 | 2.771 | 5.50 | 5.980 | 2.687 |
| 4 | 2.687 | 5.34 | 6.137 | 2.652 |
| 5 | 2.652 | 5.27 | 6.203 | 2.638 |
| 6 | 2.638 | 5.24 | 6.232 | 2.632 |

The values converge to $v_s=2.63$ cm/s, $Re=5.2$, $C_d=6.25$.

**Answer: settling velocity $v_s\approx2.63$ cm/s ($26.3$ mm/s), with $Re=5.2$.** (Stokes' law would overestimate it as 3.57 cm/s.)

## Check yourself

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

1. Why is Stokes' law not valid for the 0.2 mm particle?
   A. Stokes trial gives Re = 7.09, above 1
   B. Re is below 1
   C. Temperature is too high
   D. Particle is too light
   Answer: A. Stokes' law applies only when Re < 1. (Key point: Stokes' law check)

2. Which method is then used?
   A. Transition zone with Cd = 24/Re + 3/sqrt(Re) + 0.34
   B. Newton's law with Cd = 0.4 only
   C. Stokes with a correction factor of 2
   D. Overflow rate method
   Answer: A. For 1 < Re < 1000 the Fair and Geyer Cd is used by trial and error. (Key point: Zone of flow)

3. What is the final vs, and how does it compare with Stokes?
   A. 2.63 cm/s, less than Stokes' 3.57 cm/s
   B. 3.57 cm/s, same as Stokes
   C. 26.3 cm/s
   D. 2.63 cm/s, more than Stokes
   Answer: A. Drag is higher than Stokes assumes, so the true velocity is lower: 2.63 cm/s with Re = 5.2. (Key point: Result)

## Related in Water Supply Engineering

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- [Settling velocity of discrete particle 0.17 mm at 21°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c70o) — Water Supply Engineering, chapter 6, asked 2×
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