Stokes' Law Calculator

Stokes' law diagram for drag force — sphere of radius r moving through a fluid at velocity v with drag force F_d
Drag force equals six times pi times viscosity times radius times velocity

Solution

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How It Works

Stokes' Law gives the drag force on a small sphere moving slowly through a viscous fluid: Fd = 6πμrv. It applies when the Reynolds number is low (Re < 1), meaning viscous forces dominate over inertial forces. The drag is proportional to velocity, viscosity, and sphere radius. The terminal velocity equation Vt = gd²(ρp−ρm)/(18μ) calculates the constant speed a spherical particle reaches when drag force balances the net gravitational force. This is widely used in sedimentation analysis and particle settling studies.

Example Problem

A 0.5 mm radius sphere falls through oil (μ = 0.1 Pa·s) at 0.01 m/s. What is the drag force?

  1. Identify the knowns. Sphere radius r = 0.5 mm = 5 × 10⁻⁴ m, fluid dynamic viscosity μ = 0.1 Pa·s, and velocity v = 0.01 m/s. The Reynolds number is well below 1, so Stokes flow applies.
  2. Identify what we're solving for. We want the viscous drag force Fd acting on the sphere.
  3. Write Stokes' Law in symbols: Fd = 6π × μ × r × v.
  4. Substitute the known values: Fd = 6π × 0.1 × (5 × 10⁻⁴) × 0.01.
  5. Simplify the arithmetic: Fd = 6π × 5 × 10⁻⁷ = 3π × 10⁻⁶.
  6. State the result with units: **Fd ≈ 9.42 × 10⁻⁶ N** (about 9.42 micronewtons) of viscous drag opposing the motion.

When to Use Each Variable

  • Solve for Drag Force — when you know the viscosity, sphere radius, and velocity, e.g., calculating resistance on a falling-ball viscometer sphere.
  • Solve for Viscosity (Drag) — when you know drag force, radius, and velocity, e.g., measuring fluid viscosity with a falling-ball or rising-bubble apparatus.
  • Solve for Radius (Drag) — when you know drag force, viscosity, and velocity, e.g., estimating particle size from measured drag in a sedimentation experiment.
  • Solve for Terminal Velocity — when you know particle and fluid properties, e.g., predicting settling rates for sediment particles in a clarifier.
  • Solve for Particle Diameter — when you know terminal velocity and fluid properties, e.g., determining the size of particles that settle at a measured rate.
  • Solve for Viscosity (Terminal) — when you know particle properties and terminal velocity, e.g., back-calculating viscosity from sedimentation data.

Key Concepts

Stokes' Law describes the drag force on a small sphere moving slowly through a viscous fluid, valid when the Reynolds number is below 1 (creeping flow regime). Drag is directly proportional to velocity, viscosity, and sphere radius — doubling any one doubles the drag. Terminal velocity occurs when drag equals the net gravitational force on the particle, resulting in constant settling speed. This is the foundation for sedimentation analysis, particle sizing, and viscometry.

Applications

  • Sedimentation analysis: determining soil particle size distribution by measuring settling times in a hydrometer test
  • Viscometry: measuring fluid viscosity using falling-ball and rising-bubble viscometers
  • Air quality: modeling the settling and suspension of aerosol particles, fog droplets, and dust
  • Wastewater treatment: designing clarifiers and settling tanks by predicting particle settling velocities
  • Pharmaceutical manufacturing: controlling particle size in suspension formulations for consistent drug delivery

Common Mistakes

  • Applying Stokes' Law above Re = 1 — at higher Reynolds numbers, inertial effects become significant and the linear drag relationship breaks down
  • Using diameter instead of radius in the drag equation — Fd = 6 pi mu r v uses radius, not diameter
  • Forgetting the density difference in terminal velocity — buoyancy reduces the effective gravitational force, and omitting fluid density overestimates settling speed
  • Ignoring particle shape — Stokes' Law applies to perfect spheres; irregular particles require correction factors

Frequently Asked Questions

When does Stokes' Law apply?

Stokes' Law is valid at low Reynolds numbers (Re < 1), which means small, slow-moving spheres in viscous fluids. Examples include sediment settling, fog droplets, and blood cells.

What is terminal velocity in Stokes flow?

Terminal velocity occurs when drag equals the net gravitational force on the sphere. At that point the sphere falls at constant speed. For a dense sphere in a light fluid, Vt = gd²(ρp − ρm) / (18μ).

How is Stokes' Law used in practice?

It is used to measure fluid viscosity (falling-ball viscometer), determine particle size in sedimentation analysis, and model aerosol behavior in air quality studies.

What equation does Stokes' Law use for drag force?

Fd = 6π × μ × r × v, where μ is the fluid dynamic viscosity, r is the sphere radius, and v is the sphere's velocity relative to the fluid. Drag scales linearly with all three — doubling any one doubles the drag force.

What is the difference between Stokes drag and Newton drag?

Stokes drag (Fd ∝ v) applies at Re < 1 where viscous forces dominate. Newton drag (Fd ∝ v²) applies at Re > ~1000 where inertial forces dominate. Between these regimes, drag follows a more complex empirical curve and neither simple formula is accurate.

How accurate is Stokes' Law for real particles?

Stokes' Law is exact for rigid spheres in unbounded Newtonian fluids at Re < 0.1, and within a few percent up to Re ≈ 1. Irregular particles, deformable droplets, near-wall effects, and non-Newtonian fluids all require correction factors. For Re > 1, use the full drag-coefficient curve instead.

Why do small particles settle so much slower than large ones?

Terminal velocity scales with the square of diameter (Vt ∝ d²). Halving the particle size cuts terminal velocity to one quarter. That's why clay (~2 μm) takes hours to settle in still water while sand (~200 μm) settles in seconds — even though their densities are similar.

Reference:

Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.

Worked Examples

Hematology

How fast do red blood cells sediment in plasma?

A red blood cell has an effective diameter of 7 μm, density of 1100 kg/m³ in plasma of density 1025 kg/m³, and plasma viscosity 0.0014 Pa·s. Use Stokes' law to estimate the terminal settling velocity behind the ESR (erythrocyte sedimentation rate) blood test.

  • Knowns: d = 7×10⁻⁶ m, ρp = 1100 kg/m³, ρm = 1025 kg/m³, μ = 0.0014 Pa·s, g = 9.81 m/s²
  • Vt = g × d² × (ρp − ρm) / (18 × μ)
  • Vt = 9.81 × (7×10⁻⁶)² × 75 / (18 × 0.0014)
  • Vt = 9.81 × 4.9×10⁻¹¹ × 75 / 0.0252
  • Vt = 3.605×10⁻⁸ / 0.0252

Vt ≈ 1.43×10⁻⁶ m/s (≈ 5.15 mm/hr)

Normal ESR is roughly 0-20 mm/hr; elevated values (>30 mm/hr) suggest inflammation. Stokes' law is the physical basis of the test, although in practice rouleaux formation and hematocrit shift the measured rate.

Environmental Engineering

How long does silt take to settle in a lake?

A 20 μm silt particle (ρp = 2650 kg/m³) settles in still lake water (ρm = 1000 kg/m³, μ = 0.001 Pa·s). What is the terminal settling velocity, and how long would it take to settle through a 10 m water column?

  • Knowns: d = 20×10⁻⁶ m, ρp = 2650 kg/m³, ρm = 1000 kg/m³, μ = 0.001 Pa·s, g = 9.81
  • Vt = g × d² × (ρp − ρm) / (18 × μ)
  • Vt = 9.81 × (2×10⁻⁵)² × 1650 / (18 × 0.001)
  • Vt = 9.81 × 4×10⁻¹⁰ × 1650 / 0.018
  • Vt = 6.475×10⁻⁶ / 0.018

Vt ≈ 3.60×10⁻⁴ m/s (≈ 31 m/day)

A 10 m water column would clear of suspended silt in about 7-8 hours under quiescent conditions — sediment ponds and clarifier basins use exactly this calculation to size their depth and residence time.

Viscometry

What viscous drag does a 5 mm steel ball feel falling through glycerin?

A falling-ball viscometer drops a 5 mm radius steel sphere through glycerin (μ = 1.4 Pa·s) at a measured velocity of 0.05 m/s. What is the Stokes drag force on the ball?

  • Knowns: μ = 1.4 Pa·s, r = 0.005 m, v = 0.05 m/s
  • Fd = 6 × π × μ × r × v
  • Fd = 6π × 1.4 × 0.005 × 0.05
  • Fd = 18.85 × 0.00035

Fd ≈ 6.60×10⁻³ N (about 6.6 mN)

Balanced against gravity minus buoyancy, this drag determines the terminal velocity used in the standard ASTM D1343 falling-ball viscosity method. The technique is accurate for Newtonian liquids in the laminar regime (Re < 1).

Stokes' Law Formulas

Stokes' Law describes the drag force on a small sphere moving slowly through a viscous fluid, and the terminal settling velocity it reaches when drag balances gravity:

Fd = 6π × μ × r × vViscous drag force on a sphere
Vt = g × d² × (ρp − ρm) / (18 × μ)Terminal velocity of a settling sphere

Where:

  • Fd — viscous drag force on the sphere (N)
  • μ — fluid dynamic viscosity (Pa·s or kg/(m·s))
  • r — sphere radius (m); diameter d = 2r
  • v — sphere's velocity relative to the fluid (m/s)
  • Vt — terminal settling velocity (m/s)
  • g — gravitational acceleration (m/s²); ≈ 9.81 on Earth
  • d — particle diameter (m)
  • ρp — particle (or sphere) density (kg/m³)
  • ρm — surrounding medium density (kg/m³)

These formulas are exact for rigid spheres in unbounded Newtonian fluids at Reynolds number Re < 0.1 and accurate within a few percent up to Re ≈ 1. At higher Re, inertial effects matter and the full drag-coefficient curve must be used. The terminal velocity equation comes from setting drag (6πμrVt) equal to net gravity (volume × density difference × g).

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