Audited 31 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Kinematic Viscosity of Air Calculator

Kinematic viscosity of dry air at any absolute temperature and pressure, from Sutherland's law and the ideal-gas density: ν = μ ⁄ ρ in m²/s.

Kinematic Viscosity of Air Calculator

K
Pa
Kinematic viscosity
1.569 × 10⁻⁵
The kinematic viscosity of dry air at the entered temperature and pressure, in square metres per second.
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.

Water is about 55 times more viscous than air at 20 °C, yet by the measure that matters for most flow problems the order flips: air's kinematic viscosity, about 1.5 × 10⁻⁵ m²/s, is fifteen times water's. The reason is density. Kinematic viscosity divides a fluid's resistance to shearing, its dynamic viscosity μ, by how much mass there is to move, its density ρ. Air is very light, so the little viscosity it has goes a long way.

That quotient, ν = μ ⁄ ρ, is the number that sits in the denominator of a Reynolds number, Re = vL ⁄ ν, and so it decides whether the air over a wing, through a duct or around a chimney flows smoothly or turns turbulent. It is also not a constant. It rises steeply with temperature and is inversely proportional to pressure, so a value copied from a sea-level table can be badly off for a hot exhaust, a pressurised line or air at altitude.

The calculator builds ν from two textbook models. Sutherland's 1893 formula gives the dynamic viscosity of air from its absolute temperature alone, and the ideal-gas law gives its density from temperature and pressure. Enter the absolute temperature in kelvin and the absolute pressure in pascals, and it returns the kinematic viscosity in square metres per second, together with a note on what the model covers.

The model describes dry air behaving as an ideal gas. It does not adjust for humidity, for the departures from ideal behaviour that appear at high pressure, for changes in composition at extreme temperatures, or for gas so thin that it stops acting as a continuous fluid. Within ordinary engineering conditions, from cold winter air to hot ducts, at pressures from high altitude up to a few atmospheres, it is the standard way to get the value a Reynolds number needs.

What is kinematic viscosity of air calculator?

Kinematic viscosity ν is dynamic viscosity divided by density. Dynamic viscosity, in pascal-seconds, measures the shear stress needed to keep neighbouring layers of fluid sliding past each other at a given rate. Dividing by density, in kilograms per cubic metre, gives a quantity in square metres per second, the same units as a diffusion coefficient, and that is the best way to picture it: ν is how quickly momentum spreads sideways through the fluid, just as heat or a dissolved dye would. Where ν is large, a disturbance in velocity is smoothed out quickly; where it is small, inertia dominates and flows break up into turbulence more easily. The older unit is the stokes, 1 cm²/s or 10⁻⁴ m²/s, and lubricant data often use the centistokes, 10⁻⁶ m²/s. For air near room temperature ν is about 15 to 16 centistokes; at 15 °C and standard sea-level pressure it is about 1.4607 × 10⁻⁵ m²/s.

Which of the two viscosities a calculation needs depends on the question. Forces and stresses use the dynamic viscosity: the shear stress on a surface, the drag on a slowly settling particle, the pressure drop that Hagen and Poiseuille's law gives for laminar pipe flow. Questions about flow regime, how far momentum diffuses and when inertia overwhelms friction, use the kinematic viscosity, which is why it is the one inside the Reynolds number.

How to use this calculator.

  1. Enter the absolute temperature in kelvin: add 273.15 to a Celsius reading, so a 25 °C room is 298.15 K. For Fahrenheit, add 459.67 and multiply by 5 ⁄ 9.
  2. Enter the absolute pressure in pascals. Standard sea level is 101,325 Pa. A gauge reading must have atmospheric pressure added, so a line at 2 bar gauge holds about 301,325 Pa absolute.
  3. For a site at altitude, use the actual local pressure, not the sea-level-corrected figure a weather report gives. At the 11,000 m of the standard atmosphere's tropopause, pressure is about 22,632 Pa and temperature 216.65 K.
  4. Read ν in m²/s. Multiply by 10⁶ for centistokes or by 10.7639 for square feet per second.
  5. Use the value in Re = vL ⁄ ν, with the velocity in m/s and the characteristic length in metres, both measured for the same flow.

The formula.

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

ν = μ₀ × (T ⁄ T₀)^1.5 × (T₀ + S) ⁄ (T + S) × R × T ⁄ p

The first part is Sutherland's law for the dynamic viscosity μ. Simple kinetic theory treats gas molecules as hard spheres and predicts that viscosity grows with the square root of temperature, because faster molecules carry momentum between layers more quickly. Real molecules also attract each other weakly, which makes slow, cold molecules behave as if they were larger. William Sutherland captured that with a single extra constant, S. The constants used here are μ₀ = 1.716 × 10⁻⁵ Pa·s at the reference temperature T₀ = 273.15 K and S = 110.4 K, the same S that NASA's Glenn Research Center gives in imperial form as 198.72 °R.

Pressure does not appear in μ. Compressing a dilute gas puts more molecules in each layer to carry momentum, but each travels a proportionally shorter distance between collisions, and the two effects cancel. Maxwell predicted this from kinetic theory, and experiment bore it out over a wide range of ordinary pressures.

The second part is the ideal-gas density, ρ = p ⁄ (R × T), with the specific gas constant for dry air R = 287.058 J ⁄ (kg·K). Dividing μ by ρ is the same as multiplying by R × T ⁄ p, which is how both pieces end up in one line.

The result scales exactly as 1 ⁄ p: halve the pressure and ν doubles. Temperature acts twice, raising μ and lowering ρ together, so near room temperature ν grows roughly as T^1.77. Between 300 K and 600 K at one atmosphere, μ rises by a factor of about 1.63 while ν rises by about 3.27.

The equation will return a number for any positive temperature and pressure, including values far outside what it describes. It has no meaning near or below the temperatures at which air's nitrogen and oxygen liquefy, about 77 K and 90 K at atmospheric pressure; at pressures high enough for air to depart from ideal-gas behaviour; at the very high temperatures where molecules start to dissociate; or in gas so rarefied that molecules travel distances comparable to the object in the flow.

A worked example.

Example

Take air at 300 K and standard atmospheric pressure, 101,325 Pa. Start with the dynamic viscosity. The temperature ratio raised to the power 1.5 is (300 ⁄ 273.15)^1.5 = 1.1510. Sutherland's correction is (273.15 + 110.4) ⁄ (300 + 110.4) = 383.55 ⁄ 410.4 = 0.9346. Multiplying, μ = 1.716 × 10⁻⁵ × 1.1510 × 0.9346 = 1.8459 × 10⁻⁵ Pa·s. Next the density: ρ = 101,325 ⁄ (287.058 × 300) = 1.1766 kg/m³. The quotient is ν = 1.8459 × 10⁻⁵ ⁄ 1.1766 = 1.5689 × 10⁻⁵ m²/s, which the calculator reports as 1.56887 × 10⁻⁵ m²/s, or about 15.69 centistokes. For a 0.1 m model wing chord in a 10 m/s airstream at these conditions, the Reynolds number is 10 × 0.1 ⁄ (1.56887 × 10⁻⁵) ≈ 63,740. The same air in a 50 mm pipe at 2 m/s gives Re ≈ 6,374, well above the roughly 2,300 at which pipe flow stops being reliably laminar. Fly the same 0.1 m chord at 10 m/s at 11,000 m in the standard atmosphere, where ν is 3.9063 × 10⁻⁵ m²/s, and the Reynolds number falls to about 25,600, less than half its value in the 300 K test.

absolute Pressure Pa101,325
absolute Temperature K300

Frequently asked questions.

Why does air get more viscous when it is heated, when oil gets thinner?
The mechanisms differ. A liquid resists flow through cohesion between closely packed molecules, and heat loosens that grip. In a gas, viscosity comes from molecules wandering between layers and carrying momentum with them, and heat makes them wander faster. So μ rises with temperature, and ν rises faster still because the heated gas also becomes less dense.
Why does pressure change kinematic viscosity but not dynamic viscosity?
In a dilute gas, a higher pressure packs more momentum carriers into each layer but shortens the distance each one travels before colliding, and the two effects cancel in μ. Density, though, is directly proportional to pressure. Because ν = μ ⁄ ρ, it falls in exact proportion as pressure rises: at 2 bar gauge, about 301,325 Pa absolute, air at 300 K has ν ≈ 5.2755 × 10⁻⁶ m²/s, about a third of its value at one atmosphere.
What value should I use for standard air?
It depends on which standard. At the International Standard Atmosphere's sea level, 15 °C and 101,325 Pa, this model gives 1.4607 × 10⁻⁵ m²/s. At 20 °C it gives 1.5060 × 10⁻⁵, at 25 °C 1.5518 × 10⁻⁵ and at 300 K 1.5689 × 10⁻⁵. The spread between 15 °C and 300 K is about 7%, so always quote the temperature with the value.
How much does kinematic viscosity change with altitude?
Less than the pressure drop alone suggests, because the air also gets colder. In the standard atmosphere at 5,500 m, pressure is about half its sea-level value but ν is only about 1.58 times higher, at 2.3116 × 10⁻⁵ m²/s. At 11,000 m, with 22,632 Pa and 216.65 K, it is 3.9063 × 10⁻⁵ m²/s, about 2.67 times the sea-level figure.
How does air compare with water?
Water's dynamic viscosity at 20 °C is about 1.0 × 10⁻³ Pa·s, roughly 55 times air's. Its kinematic viscosity is about 1.0 × 10⁻⁶ m²/s, roughly fifteen times smaller than air's 1.5060 × 10⁻⁵, because water is about 830 times denser. A model in water at the same speed and size therefore reaches a Reynolds number about fifteen times higher, which is why water tunnels are useful for studying air flows.
Where does kinematic viscosity appear in engineering calculations?
Mainly in the Reynolds number Re = vL ⁄ ν, which decides between laminar and turbulent flow in pipes, ducts and boundary layers. It also sets laminar boundary-layer growth: by the Blasius solution the layer is about 5 × √(νx ⁄ U) thick, roughly 6.1 mm at 0.5 m along a plate in 5 m/s air at 20 °C. Stokes settling of dust uses the dynamic viscosity instead.
Does humidity change the kinematic viscosity of air?
A little. Water vapour is lighter than the nitrogen and oxygen it displaces, so humid air is slightly less dense than dry air at the same temperature and pressure, and its viscosity differs slightly too. The calculator models dry air only. For most Reynolds-number estimates the difference is small next to the uncertainty in velocities and lengths, but precise property work needs a moist-air model.
How do I convert the result to stokes, centistokes or square feet per second?
One stokes is 10⁻⁴ m²/s and one centistokes is 10⁻⁶ m²/s, so multiply the result by 10,000 for stokes or by 1,000,000 for centistokes: 1.56887 × 10⁻⁵ m²/s is 0.156887 St or 15.6887 cSt. For square feet per second multiply by 10.7639, giving about 1.6887 × 10⁻⁴ ft²/s at 300 K.
Why do some wind tunnels pressurise or chill their air?
To raise the Reynolds number of a small model toward that of the full-size aircraft. Re = vL ⁄ ν, and a scale model has a much smaller L. Raising the pressure cuts ν in direct proportion, and cooling the gas cuts it further by lowering μ and raising the density, so pressurised and cryogenic tunnels recover the Reynolds number that model scale gives away.
Can I use this for gases other than air?
No. Both constants are specific to air: S = 110.4 K and μ₀ = 1.716 × 10⁻⁵ Pa·s in Sutherland's law, and R = 287.058 J ⁄ (kg·K) in the density. Other gases have their own Sutherland constants and gas constants. Light gases such as hydrogen and helium have densities far below air's and kinematic viscosities several times higher, so applying air's values would give a wrong viscosity and a wrong density.

How this page was produced

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