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Entropy from Gibbs and N-S equations

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Old   November 24, 2005, 10:46
Default Entropy from Gibbs and N-S equations
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JF
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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
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Old   November 24, 2005, 23:39
Default Re: Entropy from Gibbs and N-S equations
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versi
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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.
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Old   November 27, 2005, 13:30
Default Re: Entropy from Gibbs and N-S equations
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JF
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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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