# Entropy from Gibbs and N-S equations

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 November 24, 2005, 10:46 Entropy from Gibbs and N-S equations #1 JF Guest   Posts: n/a Hi all, My question concerns the computation of the entropy by the Gibbs equation expressed in the following form : Tds = dh - dp/rho If the variables in the preceding equation have been computed by the Reynolds averaged Navier-Stokes equations, shouldn't we use a Reynolds averaged form of the Gibbs equation to compute the entropy ? If not, why is it valid to apply the Gibbs equation directly to the averaged variables ? Any thoughts on this ? Thanks. JF

 November 24, 2005, 23:39 Re: Entropy from Gibbs and N-S equations #2 versi Guest   Posts: n/a I did not have experience in calculating the entropy. But I think these variables (T,P,rho) are already averaged ones. One need not again and again applying the so-called but unclear "Average' process to the thermodynamic relations.

 November 27, 2005, 13:30 Re: Entropy from Gibbs and N-S equations #3 JF Guest   Posts: n/a Thanks for responding versi > But I think these variables (T,P,rho) are already averaged ones It is that statment that I don't understand. I think that the Gibbs equation can be primarly applied to the instantaneous Navier-Stokes equations. If I want to use it with an averaged form of the Navier-Stokes I should apply the same averaging process to the Gibbs equation as well. For example, when you use the law of perfect gas, you use an average form of it : (p)_bar = (R*rho*T)_bar ==> p_bar = R*rho_bar*T_tilde where _bar is the normal averaging operator and _tilde the density based averaging operator. In this case, formally, the averaged and non-averaged equations are identical. However, if you apply the averaging to the Gibbs equation you'll find additional terms as you find the Reynolds stresses in the N-S equations. Is it right ? What am I missing ?

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