Calculate Stokes settling velocity, reverse-size particle diameter, or estimate settling time with unit conversion and Reynolds-regime checks.

Enter dynamic viscosity, not kinematic viscosity, and particle diameter, not radius. Blank gravity uses 9.80665 m/s².

Stokes model for an isolated sphere in a quiet Newtonian fluid, after terminal speed is reached. Downward settling only; particle density must exceed fluid density.

Gravitational acceleration

Settling Velocity Formula

Settling velocity here is the modeled downward terminal speed of an isolated spherical particle in a quiet Newtonian fluid when gravity and buoyancy balance viscous drag. The calculator uses the Stokes-law form below with dynamic viscosity μ, not kinematic viscosity. It also reverses the equation to find diameter or divides distance by terminal speed to estimate time. Time excludes initial acceleration.

Vₛ = (g(ρₛ - ρf)d²) / (18μ)
Symbol Meaning Typical SI Unit Role in the Calculation
Vs Settling velocity m/s Calculated output
ρs Density of the solid particle kg/m³ Higher values increase settling speed
ρf Density of the fluid kg/m³ Higher values reduce the density contrast
d Particle diameter m Strongest input effect because diameter is squared
μ Dynamic viscosity Pa·s Higher viscosity slows settling
g Gravitational acceleration m/s² Usually taken as 9.81 m/s²

Because d is squared, small diameter changes can produce large differences in the final answer.

How to Calculate Settling Velocity

  1. Enter the density of the solid particle.
  2. Enter the density of the surrounding fluid.
  3. Enter the particle diameter, not the radius.
  4. Enter the fluid’s dynamic viscosity in the selected unit: 1 cP = 1 mPa·s = 0.001 Pa·s.
  5. Calculate the settling velocity and confirm the result is physically reasonable.

Unit note: if your data gives kinematic viscosity ν, convert to dynamic viscosity first using μ = ρfν with consistent SI units. Equivalently, ν = μ/ρf. The calculator’s Reynolds check is Re = ρfvd/μ = vd/ν. See NASA’s viscosity and Reynolds definitions.

μ = ρfnu
Input Change Effect on Settling Velocity
Larger diameter Increases strongly; doubling the diameter makes the settling velocity about 4 times larger.
Higher solid density Increases settling speed.
Higher fluid density Decreases the relative density difference and slows settling.
Higher viscosity Decreases settling speed.

When This Estimate Is Most Reliable

Condition Why It Matters
Small, nearly spherical particles The Stokes form is derived for spherical-particle settling behavior.
Very low Reynolds number settling Accuracy is best in the Stokes or creeping-flow range.
Particle Reynolds number less than about 1 Re < 1 is a common screening convention, not an accuracy guarantee; creeping-flow accuracy improves as Re becomes much smaller than 1.
Relatively calm fluid Strong turbulence can change actual settling behavior and particle paths.

At Re ≥ 1, treat the output as a formal Stokes calculation rather than a reliable terminal-speed or design prediction; use an appropriate drag correlation. Even low Re does not remove particle-shape, concentration, wall or fluid-motion effects. MIT’s settling chapter derives the spherical-particle Stokes model.

Common Input Mistakes

  • Using radius instead of diameter.
  • Entering kinematic viscosity when the calculator expects dynamic viscosity.
  • Mixing units while checking the math by hand.
  • Using bulk material density instead of the actual particle density.
  • Applying the equation to particles that are too large for Stokes settling.

Interpreting the Result

Result What It Means
Positive value The particle tends to settle downward through the fluid.
Near zero Settling is very slow, so particles may remain suspended longer.
Negative value In a signed Stokes relation this indicates rising. This downward-only calculator instead rejects particle density at or below fluid density.
Unexpectedly large value Recheck diameter and viscosity units first, then confirm the Stokes approximation is appropriate.

Example

For a particle with ρs = 2650 kg/m³, ρf = 1000 kg/m³, d = 0.0002 m, and ν = 1.0 × 10-6 m²/s:

Vₛ = (9.81((2650) / (1000) - 1)(0.0002)²) / (18(1.0 × 10⁻⁶)) ≈ 0.036 m / s

The equation gives 0.035970 m/s, about 3.6 cm/s downward, but Re = 7.194. This example is outside the usual Stokes range and is not a reliable terminal-speed prediction. Its kinematic viscosity corresponds to μ = 1000 × 10−6 = 0.001 Pa·s = 1 cP for entry in the widget.

Low-Re check: with the same densities, μ = 1 cP, diameter 10 µm (0.01 mm), and standard gravity 9.80665 m/s², the model gives 0.0000898942917 m/s = 0.0898942917 mm/s and Re ≈ 0.000898943. Travel over 1 m at that terminal speed takes about 185.403 minutes.