Audited 31 Jul 2026·Last updated 15 Sept 2026·5 citations·Tier 1·0 uses

Kinematic Viscosity of Air Calculator

Kinematic viscosity of air: Sutherland's law plus ideal-gas density give ν = μ/ρ at your temperature and pressure — the ν in every Reynolds number.

Kinematic Viscosity of Air Calculator

K
Pa
Kinematic viscosity
0
Result of nu = mu0 (T/T0)^(3/2) (T0+S)/(T+S) * R T / p using the entered coherent-SI magnitudes.
Model scope
Dry, dilute air modeled as an ideal gas with NASA's stated Sutherland constants; temperature and absolute pressure are explicit, while humidity, real-gas effects, composition changes, rarefaction, and a standards-grade property table are outside it.

Background.

Every Reynolds number computed for air — over a wing, through a duct, around a chimney — needs one property in its denominator: the kinematic viscosity ν = μ/ρ, the ratio of air's dynamic viscosity to its density. At room conditions it is about 1.57×10⁻⁵ m²/s, but it is nowhere near a constant, and this page computes it for the temperature and pressure you actually have.

The two ingredients move differently. Dynamic viscosity μ — the fluid's intrinsic stickiness — depends on temperature alone for a dilute gas and, counterintuitively, rises with heating: hotter molecules cross between flow layers faster and transfer more momentum. Sutherland's 1893 formula, μ = μ₀(T/T₀)^{3/2}(T₀+S)/(T+S) with S = 110.4 K for air, captures that rise to within about 2% from 100 to 1,900 K and remains the standard engineering correlation, used in NASA's compressible-flow references and most CFD codes.

Density brings in pressure through the ideal-gas law, ρ = p/(RT). Because ρ sits in the denominator of ν while μ ignores pressure, kinematic viscosity is strongly pressure- and altitude-dependent: halve the pressure and ν doubles. At airliner cruise altitude ν is roughly four times its sea-level value — one reason flight Reynolds numbers differ so much from wind-tunnel ones, and why tunnels are sometimes pressurised to compensate.

The model is dry, dilute air: Sutherland constants for the standard composition, ideal-gas density, no humidity correction (moist air is up to ≈1% less dense at summer conditions). For standards-grade property work there are dedicated tables; for the Reynolds numbers, settling times, and boundary-layer estimates of ordinary engineering, this correlation is the tool, and the scope note beside the result draws the line.

What is kinematic viscosity of air calculator?

Kinematic viscosity ν is dynamic viscosity divided by density, ν = μ/ρ, with units of m²/s — the diffusivity of momentum through the fluid. For air this page builds it from two standard models: Sutherland's law for μ(T), which rises with temperature as faster molecules exchange momentum between layers more effectively, and the ideal-gas law for ρ(T, p). The result is the ν that appears in the Reynolds number Re = vL/ν, in boundary-layer growth, and in Stokes settling — about 1.46×10⁻⁵ m²/s at 15 °C sea level, growing with altitude and temperature.

How to use this calculator.

  1. Enter the absolute temperature in kelvin — °C + 273.15, so a 25 °C lab is 298.15 K.
  2. Enter the absolute (not gauge) pressure in pascals: sea level 101,325 Pa, Denver ≈83,000 Pa, airliner cruise ≈23,000 Pa.
  3. Read ν in m²/s; multiply by 10⁴ if a reference quotes stokes, or by 10⁶ to compare with the ‘centistokes’-scale values of oils.
  4. Drop the result into your Reynolds number Re = vL/ν — a 0.1 m chord in a 10 m/s room-temperature stream gives Re ≈ 64,000, comfortably turbulent-transitional.
  5. For hot or high-altitude cases, resist the temptation to reuse the sea-level 1.5×10⁻⁵: at 600 K and 1 atm ν is roughly 5×10⁻⁵ m²/s, a factor-three shift that moves flow regimes.

The formula.

nu = mu0 (T/T0)^(3/2) (T0+S)/(T+S) * R T / p

The dynamic part comes from kinetic theory refined by Sutherland: an ideal hard-sphere gas would give μ ∝ √T, but real molecules attract weakly at a distance, making slow (cold) molecules act effectively larger. Sutherland modelled that with one extra constant S, giving μ = μ₀(T/T₀)^{3/2}·(T₀+S)/(T+S); for air μ₀ = 1.716×10⁻⁵ Pa·s at T₀ = 273.15 K with S = 110.4 K. Notice pressure is absent — for dilute gases, more molecules per volume also means proportionally shorter free paths, and the two effects cancel in μ. Density supplies the pressure dependence instead: ρ = p/(R_specific·T) with R = 287.05 J/(kg·K) for dry air. Dividing, ν = μRT/p: kinematic viscosity grows a little faster than T^{3/2} with temperature (both numerator effects align) and inversely with pressure. The engine evaluates the chain — Sutherland factor, gas density, quotient — in Decimal arithmetic and rounds once to twelve significant digits.

A worked example.

Example

What is the kinematic viscosity of air at 300 K and standard pressure (101,325 Pa)? Two quantities combine here. First, Sutherland's formula gives the dynamic viscosity: starting from the reference value 1.716×10⁻⁵ Pa·s at 273.15 K, the temperature ratio term (300/273.15)^1.5 = 1.151 and the Sutherland correction (273.15+110.4)/(300+110.4) = 0.935 multiply out to μ ≈ 1.846×10⁻⁵ Pa·s. Second, the ideal-gas density at these conditions is ρ = p/(RT) = 101,325/(287.05 × 300) ≈ 1.177 kg/m³. Kinematic viscosity is their quotient: ν = μ/ρ = 1.846×10⁻⁵ / 1.177 ≈ 1.569×10⁻⁵ m²/s — the 15.69 centistokes familiar from room-temperature aerodynamics tables. Note what pressure does: it never touches μ, but doubling it doubles ρ and therefore halves ν, which is why quoting a kinematic viscosity without its pressure is meaningless.

absolute Pressure Pa101,325
absolute Temperature K300

Frequently asked questions.

Why does air's viscosity increase with temperature when liquids get thinner?
Different momentum-transfer mechanisms. A liquid's viscosity comes from intermolecular cohesion, which heat loosens — hence thinner hot oil. A gas's comes from molecules wandering between flow layers and carrying their momentum along; heating speeds the wandering, so μ rises, roughly as T^{3/2}/(T+S). Kinematic ν rises even faster because heating also thins the air, shrinking the ρ in the denominator.
Why does pressure change ν but not μ?
Compressing a dilute gas packs in more momentum carriers, but each travels a proportionally shorter distance before colliding — the two effects cancel exactly in μ, a classic kinetic-theory result that holds well below ≈10% of the critical pressure. Density, though, is directly proportional to pressure, so ν = μ/ρ inherits a clean 1/p: at 5,500 m altitude, where pressure has halved, ν has doubled.
What value should I use for ‘standard’ air?
Depends on whose standard: 1.46×10⁻⁵ m²/s at the ICAO sea-level atmosphere (15 °C, 101,325 Pa), 1.57×10⁻⁵ at the 300 K of this page's example, about 1.51×10⁻⁵ at a 20 °C lab. The 7% spread among ‘room temperature’ conventions is why quoting the temperature with any ν matters — and why a calculator beats a memorised constant once precision matters at all.
How accurate is Sutherland's formula, and when does the model break?
For dry air it tracks reference data within about 2% from roughly 100 K to 1,900 K — ample for Reynolds-number work, where regime boundaries are order-of-magnitude affairs. It degrades where its dilute-gas assumptions do: very high pressures (real-gas effects on density arrive first), very high temperatures (dissociation changes the gas itself, relevant to re-entry), and rarefied conditions where the continuum picture fails. Humidity is a separate ≈1% effect this dry-air model ignores.
Where does kinematic viscosity actually enter engineering calculations?
Anywhere momentum diffusion competes with transport. The Reynolds number Re = vL/ν decides laminar versus turbulent in every pipe, duct, and boundary layer; laminar boundary-layer thickness grows as √(νx/v); Stokes settling of dust and droplets scales with ν through the drag on small particles. Since ν quadruples from sea level to cruise altitude, aircraft Reynolds numbers fall correspondingly — a wing model matched in a sea-level tunnel is quietly mismatched in flight unless the tunnel compensates with pressure or scale.

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nu = mu0 (T/T0)^(3/2) (T0+S)/(T+S) * R T / p
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