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Sutherland's law

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:<math>C_1 = \frac{\mu_{ref}}{T_{ref}^{3/2}}(T_{ref} + S)</math>
:<math>C_1 = \frac{\mu_{ref}}{T_{ref}^{3/2}}(T_{ref} + S)</math>
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! Gas !! <math>\mu_0 [\frac{kg}{m s}]</math> !! <math>T_0 [K]</math> !! <math>S [K]</math> !! <math>C_1 [\frac{kg}{m s K ^ {0.5}}]</math>  
! Gas !! <math>\mu_0 [\frac{kg}{m s}]</math> !! <math>T_0 [K]</math> !! <math>S [K]</math> !! <math>C_1 [\frac{kg}{m s K ^ {0.5}}]</math>  
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Revision as of 17:20, 17 May 2007

In 1893 William Sutherland, an Australian physicist, published a relationship between the dynamic visocity, \mu, and the absolute temperature, T, of an ideal gas. This formula, often called Sutherland's law, is based on kinetic theory of ideal gases and an idealized intermolecular-force potential. Sutherland's law is still commonly used and most often gives fairly accurate results with an error less than a few percent over a wide range of temperatures. Sutherland's law can be expressed as:

\mu = \mu_{ref} \left( \frac{T}{T_{ref}} \right)^{3/2}\frac{T_{ref} + S}{T + S}
T_{ref} is a reference temperature.
\mu_{ref} is the viscosity at the T_{ref} reference temperature
S is the Sutherland temperature

Some authors instead express Sutherland's law in the following form:

\mu = \frac{C_1 T^{3/2}}{T + S}

Comparing the formulas above the C_1 constant can be written as:

C_1 = \frac{\mu_{ref}}{T_{ref}^{3/2}}(T_{ref} + S)
Sutherland's law coefficients:
Gas \mu_0 [\frac{kg}{m s}] T_0 [K] S [K] C_1 [\frac{kg}{m s K ^ {0.5}}]
Air 1.716 \times 10^{-5} 273.15 110.4 1.458 \times 10^{-6}

References

  • Sutherland, W. (1893), "The viscosity of gases and molecular force", Philosophical Magazine, S. 5, 36, pp. 507-531 (1893).
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