---
title: "Stokes flow and settling velocity"
subject: "Applied Computational Fluid Dynamics"
chapter: 2
key: "semester_7/applied_computational_fluid_dynamics:stokes-flow-settling-particle"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/stokes-flow-settling-particle
site: BCT Engineering
---

# Stokes flow and settling velocity

Applied Computational Fluid Dynamics · Chapter 2: Governing Equations and Numerical Foundations

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/stokes-flow-settling-particle
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-2

## Question

What is Stokes flow? State the conditions and give the drag force on a sphere. A sand particle of diameter 0.1 mm and density 2650 kg/m³ settles in still water at 20 °C ($\rho = 998$ kg/m³, $\mu = 1.002\times10^{-3}$ Pa·s). Find its terminal velocity and check that Stokes' law is valid.

## Short answer

Stokes flow occurs at Re ≪ 1, where inertia is negligible and ∇p = μ∇²u holds. Drag on a sphere: F_D = 3πμdV. Terminal velocity V_t = (ρ_p−ρ_f)gd²/(18μ) = 8.99×10⁻³ m/s; particle Re = 0.89, so Stokes law is valid.

## Answer

#### Stokes flow
**Stokes (creeping) flow** is flow at very low Reynolds number ($Re\ll1$, in practice $Re<1$) in which inertia is negligible compared with viscous forces. The Navier-Stokes equation reduces to the linear form
$$
\nabla p=\mu\nabla^2\vec V,\qquad \nabla\cdot\vec V=0
$$
Conditions: small particle or length scale, high viscosity, or very low velocity (e.g. micro-fluidics, sedimentation, lubrication films). The solution is linear and time-reversible.

For a sphere of diameter $d$ moving at velocity $V$ in an infinite fluid, **Stokes' drag** is
$$
F_D=3\pi\mu dV,\qquad C_D=\frac{24}{Re}
$$

#### Terminal velocity
At terminal velocity: weight $-$ buoyancy $=$ drag
$$
\frac{\pi d^3}{6}(\rho_p-\rho_f)g=3\pi\mu dV_t\ \Rightarrow\ V_t=\frac{(\rho_p-\rho_f)\,g\,d^2}{18\mu}
$$
Substituting ($d=1\times10^{-4}$ m):
$$
V_t=\frac{(2650-998)\times9.81\times(1\times10^{-4})^2}{18\times1.002\times10^{-3}}=\frac{1652\times9.81\times10^{-8}}{0.018036}=8.99\times10^{-3}\ \text{m/s}
$$

#### Check of validity
$$
Re=\frac{\rho_fV_td}{\mu}=\frac{998\times8.99\times10^{-3}\times10^{-4}}{1.002\times10^{-3}}=0.895<1
$$
The particle Reynolds number is below 1, so Stokes' law is acceptable (slightly at the upper limit).

The drag force is $F_D=3\pi\mu dV_t=8.5\times10^{-9}$ N, equal to the net weight.

**Answer:** $V_t=8.99$ mm/s (about 9 mm/s), $Re=0.89$, so Stokes' law is valid.

## Check yourself

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

1. Stokes (creeping) flow occurs when:
   A. Re << 1 and inertia is negligible compared with viscous forces
   B. Re >> 1000
   C. The fluid is inviscid
   D. The flow is turbulent
   Answer: A. At very low Reynolds number the Navier-Stokes equation reduces to the linear form grad p = mu del^2 V. (Key point: Stokes flow)

2. The Stokes drag on a sphere of diameter d at velocity V is:
   A. F = 6 pi mu d V
   B. F = 3 pi mu d V
   C. F = 3 pi mu V
   D. F = 1/2 rho V^2 A
   Answer: B. F_D = 3 pi mu d V, equivalent to C_D = 24/Re. (Key point: Stokes drag)

3. At terminal velocity:
   A. Drag equals the weight only
   B. Buoyancy equals the weight
   C. Weight minus buoyancy equals drag
   D. Drag is zero
   Answer: C. Net weight balances drag, giving V_t = (rho_p - rho_f) g d^2 / (18 mu). (Key point: Terminal velocity balance)

4. For the sand particle V_t = 8.99 mm/s. The check of Stokes' law gives:
   A. Re = 89.5, invalid
   B. Re = 8.95, invalid
   C. Re = 0.0895, far too low
   D. Re = 0.895 < 1, so valid
   Answer: D. Re = 998 x 8.99e-3 x 1e-4 /1.002e-3 = 0.895, below 1. (Key point: Validity check)

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