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November 24, 2005, 10:46 |
Entropy from Gibbs and N-S equations
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#1 |
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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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November 24, 2005, 23:39 |
Re: Entropy from Gibbs and N-S equations
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#2 |
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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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November 27, 2005, 13:30 |
Re: Entropy from Gibbs and N-S equations
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#3 |
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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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