Method and interpretation

Schmidt Number

See how to enter values, calculate a result, and check the limits of its interpretation.

The two diffusivities the Schmidt number compares

Sc compares momentum diffusivity (kinematic viscosity ν) with the molecular diffusivity D of a species. The MIT 3.185 notes call ν/D the mass transfer Prandtl number, also known as the Schmidt number: it is Pr = ν/α with D in place of α.

The Michigan Tech CM3120 notes group Pr, Sc and Le as ratios of the diffusivities ν, α and D, with Sc = Le·Pr and mass Péclet number Pe = Re·Sc. Sc is therefore fixed by species, solvent and temperature, not by flow speed or size. MIT 3.185 adds that mass transfer Prandtl numbers in liquids are uniformly much greater than one, because solute diffusion is far slower than viscous momentum transfer.

Inputs: solvent properties and a species-pair diffusivity

Sc = μ / (ρD) = ν / D

μ is the solvent viscosity (Pa·s), ρ the solvent density (kg/m³) and D the molecular diffusivity of the species in that solvent (m²/s). D belongs to a species-solvent pair, not to one substance. Multiply cm²/s by 10⁻⁴ for m²/s: Cornell's 0.0000197 cm²/s is 1.97×10⁻⁹ m²/s.

The calculator defaults μ = 0.001 Pa·s, ρ = 1000 kg/m³ and D = 1×10⁻⁹ m²/s give Sc = 0.001 ÷ (1000 × 1×10⁻⁹) = 1000, a rough small solute in water. Once species and temperature are known, replace them with sourced properties as in the examples.

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Sc and Pr magnitudes in water and air

ν and Pr: IAEA; D: Cornell; Sc = ν/D: our calculation; seawater CO₂: CRAN marelac example output.

Diffusivity and Schmidt number by species-solvent pair, with the solvent Prandtl number
Species, solvent, temperatureν (m²/s)D (m²/s)Sc = ν/DSolvent Pr
Oxygen, water, 20 °C1.0020 × 10⁻⁶1.97 × 10⁻⁹508.66.981
Oxygen, water, 60 °C4.757 × 10⁻⁷4.82 × 10⁻⁹98.693.005
Oxygen, nitrogen (air approximation), 20 °C, 1 atm1.5104 × 10⁻⁵2.19 × 10⁻⁵0.690about 0.71
CO₂, seawater (salinity 35), 20 °Cnot listednot listedabout 660not listed

Water: Pr about 7, oxygen Sc above 500; heat spreads fast, dissolved gas slowly. In gases D ≈ ν in order, so Sc stays near 1.

Examples with dissolved and gaseous oxygen

Example 1. Dissolved oxygen in water at 20 °C

Given: Water at 20 °C: μ = 1.0000×10⁻³ Pa·s, ρ = 998.2 kg/m³, ν = 1.0020×10⁻⁶ m²/s, Pr = 6.981 (IAEA Table 3.3, saturated liquid used as a low-pressure approximation). Oxygen D = 0.0000197 cm²/s = 1.97×10⁻⁹ m²/s (Cornell, Wilke–Chang estimate).

Calculation: With μ, ρ and D in the calculator, Sc = 1.0000×10⁻³ ÷ (998.2 × 1.97×10⁻⁹) = 508.5. Using ν directly, 1.0020×10⁻⁶ ÷ 1.97×10⁻⁹ = 508.6.

Against Pr: Sc/Pr = 508.5 / 6.981 = 72.84, so thermal diffusivity is about 73 times the oxygen diffusivity. MIT 3.185 gives δu/δT = 0.975 Pr1/3 for a high-Pr laminar flat-plate boundary layer. Applied to mass transfer, δu/δc = 0.975 × 508.51/3 = 7.78, against 0.975 × 6.9811/3 = 1.86 for heat: the concentration layer is about 4.2 times thinner than the thermal layer.

Uncertainty: Wilke–Chang is usually within ±10% for dilute nondissociating solutes (Cornell), so Sc lies between 462 and 565.

Example 2. The same calculation in water heated to 60 °C

Given: Water at 60 °C: μ = 4.677×10⁻⁴ Pa·s, ρ = 983.2 kg/m³, ν = 4.757×10⁻⁷ m²/s, Pr = 3.005 (IAEA Table 3.3); oxygen D = 4.82×10⁻⁹ m²/s (Cornell).

Calculation: Sc = 4.677×10⁻⁴ ÷ (983.2 × 4.82×10⁻⁹) = 98.69, 5.15 times smaller than 508.5 at 20 °C. D rose 2.45 times and ν fell 2.11 times, and the two effects multiply.

Interpretation: δu/δc = 0.975 × 98.691/3 = 4.51, so the concentration layer is still thinner than the velocity layer. Do not freeze Sc at one temperature in a correlation; evaluate it at the boundary-layer temperature. With very different wall and bulk temperatures, compare Sc at both.

Example 3. Oxygen in air at 20 °C

Given: Cornell's gas-diffusion page lists D = 0.219 cm²/s for the oxygen-nitrogen pair at atmospheric pressure and 20 °C; treating air as nitrogen, D = 2.19×10⁻⁵ m²/s. Air ν = 1.5104×10⁻⁵ m²/s comes from IAEA Table 3.1 μ = 1.819×10⁻⁵ Pa·s and the 98.1 kPa density 1.166 kg/m³ scaled to 1 atm as an ideal gas, ρ = 1.2043 kg/m³.

Calculation: Sc = 1.5104×10⁻⁵ ÷ 2.19×10⁻⁵ = 0.690. At 20 °C, D in the gas is about 11,117 times D in water (2.19×10⁻⁵ / 1.97×10⁻⁹), which sets the liquid-gas gap in Sc.

Interpretation: Sc is close to air's Pr (about 0.71), so concentration and thermal layers are similar. The MIT ratio holds above about 0.6: 0.975 × 0.6901/3 = 0.862, a concentration layer slightly thicker than the velocity layer.

What Sc ≫ 1 means for concentration layers and meshes

With Sc in the hundreds, concentration changes crowd into a very thin wall layer, about 1/8 of the laminar velocity layer in Example 1. A CFD mesh sized for the velocity field may not resolve that gradient; if wall mass transfer is the key output, recheck the first cell height against the concentration layer.

Turbulence changes the picture. MIT 3.185 notes that eddies carry heat and solute about as well as momentum, so νt/Dt is very close to one. Molecular Sc matters little in the turbulent core but controls the resistance right at the wall, where turbulence dies out.

Oceanographers use Sc for air-sea gas exchange. The CRAN marelac function gas_schmidt estimates Sc for several gases in seawater at salinity 35 and 0 to 30 °C, returning about 660 for CO₂ at 20 °C, and gas_transfer combines Sc with the wind speed at 10 m to give a transfer velocity in m/s.

Checks before trusting a diffusivity value

Schmidt number questions

Why is Pr about 7 in water while Sc for oxygen exceeds 500?

Both use ν; only the denominators differ. In 20 °C water thermal diffusivity is about 73 times the oxygen diffusivity (Sc/Pr = 72.84 in Example 1). That ratio α/D is the Lewis number, matching Sc = Le·Pr in the Michigan Tech notes.

Why do gas mixtures give Sc near 1?

In a gas, molecular motion carries both momentum and species, so ν and D are of the same order: oxygen in air (Example 3) has ν = 1.5104×10⁻⁵ m²/s, D = 2.19×10⁻⁵ m²/s and Sc = 0.690. In liquids only molecular diffusion slows sharply, pushing Sc into the hundreds.

Is the turbulent Schmidt number in a CFD setup the same as this Sc?

No. This Sc is molecular viscosity over molecular diffusivity. The turbulent Schmidt number is the ratio of modelled turbulent viscosity to turbulent diffusivity, which MIT 3.185 describes as very close to one. Solvers take the two separately, and wall mass transfer is governed by the molecular Sc.

Why does air-sea gas exchange need Sc?

Transfer velocity depends on wind and on gas diffusion in water; marelac's gas_transfer takes the 10 m wind speed and Sc, and a supplied Sc overrides the gas and temperature arguments.

Once Sc fixes the diffusion properties of a species and solvent, these numbers carry it into an actual flow.

References