---
title: "Settling velocity of silica particle 0.05 cm at 10°C"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-as69a"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-as69a
site: BCT Engineering
---

# Settling velocity of silica particle 0.05 cm at 10°C

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2069 Asar, 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-as69a
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Find the settling velocity of silica particle of specific gravity 2.65 at 10°C if the diameter of the particle is 0.05 cm. Assume kinematic viscosity at 10°C is 0.91 centistokes.

## Short answer

Stokes' law gives 24.7 cm/s but Re=135.7 (>1) invalidates it. Using the transition-zone drag equation iteratively (Fair and Geyer) yields a settling velocity of about 9.40 cm/s (94.0 mm/s).

## 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
$d=0.5$ mm $=0.05$ cm, $S_s=2.65$, $\nu=0.0091$ cm²/s (given: 0.91 centistokes), $g=981$ cm/s².

#### Step 1: Stokes' law (trial)
$$
v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.05^2}{18\times0.0091}=24.705\ \text{cm/s}
$$
$$
Re=\frac{v_s d}{\nu}=\frac{24.705\times0.05}{0.0091}=135.74
$$
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 | 24.705 | 135.74 | 0.774 | 11.805 |
| 2 | 11.805 | 64.86 | 1.082 | 9.984 |
| 3 | 9.984 | 54.86 | 1.183 | 9.553 |
| 4 | 9.553 | 52.49 | 1.211 | 9.438 |
| 5 | 9.438 | 51.86 | 1.219 | 9.407 |
| 6 | 9.407 | 51.69 | 1.222 | 9.399 |

The values converge to $v_s=9.40$ cm/s, $Re=51.6$, $C_d=1.22$.

**Answer: settling velocity $v_s\approx9.40$ cm/s ($94.0$ mm/s), with $Re=51.6$.** (Stokes' law would overestimate it as 24.70 cm/s.)

## Check yourself

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

1. What is the first step to find the settling velocity?
   A. Try Stokes' law and check the Reynolds number
   B. Use Newton's law straight away
   C. Assume Cd = 1
   D. Use the overflow rate
   Answer: A. Stokes' law is tried first; Re then decides if it is valid. (Key point: First trial method)

2. Re = 135.74 from Stokes' trial. What does it mean?
   A. Stokes' law is invalid; use the transition zone formula
   B. Stokes' law is valid
   C. Particle will not settle
   D. Flow is laminar
   Answer: A. Stokes' law needs Re < 1; here 1 < Re < 1000 is the transition zone. (Key point: Validity of Stokes' law)

3. Which Cd expression is used in the transition zone?
   A. 24/Re + 3/sqrt(Re) + 0.34
   B. 24/Re
   C. 0.4
   D. 18 nu/d
   Answer: A. The Fair and Geyer expression is used in trial and error with vs = sqrt(4g(Ss-1)d/(3Cd)). (Key point: Drag coefficient)

4. What is the settling velocity?
   A. About 9.40 cm/s
   B. About 24.7 cm/s
   C. About 0.94 cm/s
   D. About 2.4 cm/s
   Answer: A. The trial and error converges to 9.40 cm/s; Stokes' 24.7 cm/s overestimates it. (Key point: Final result)

## Related in Water Supply Engineering

- [Settling velocity of silica particle 0.02 cm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c69) — Water Supply Engineering, chapter 6
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