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#1 |
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Join Date: Apr 2017
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#2 |
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Lucky
Join Date: Apr 2011
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It does use the first. I copy-pasted a part of UEqn.H from interFoam.
Code:
fvVectorMatrix UEqn ( fvm::ddt(rho, U) + fvm::div(rhoPhi, U) + turbulence->divDevRhoReff(rho, U) == fvOptions(rho, U) ); |
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#3 | |
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Quote:
I agree. This should have the implication that one make a writeup of the governing equations for a case that uses interFoam, one should not write up the RANS equations, one should write the FANS equations (which has the difference you said), agree? |
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#4 | |
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Lucky
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Quote:
Code:
fvVectorMatrix UEqn ( fvm::ddt(rho, U) + fvm::div(rhoPhi, U) + turbulence->divDevRhoReff(rho, U) == fvOptions(rho, U) ); The 2nd equation you provided is not the equation used in interFoam. So what is your point? |
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#5 | |
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Quote:
My point is that if one in a study apply interFoam, and one has varying denisty, than it is more correct to say that the U one gets out is the Favre-averaged U than saying that it is the Reynolds averaged U, agree? |
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#6 |
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Lucky
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Yes but no. The arguments for either are pedantic. Trying to be politically correct might turn you into an ass. Are penguins flightless birds or are most birds flying penguins?
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#7 | |
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Quote:
Thanks for helping me sorting this out! |
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#8 |
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Lucky
Join Date: Apr 2011
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Also, the Reynolds decomposition is compatible with variations in mean density. What is missing in RANS is the contribution from turbulent density fluctuations. For many practical flows the density changed induced by turbulence is quite small. So it is not enough to say varying density.
That's still not correct because the Favre variables are chosen to satisfy the Reynolds conditions, which means that FANS is actually RANS. This should not be a surprise because the governing equations are the same for both. |
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#9 | |
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Quote:
That's still not correct because the Favre variables are chosen to satisfy the Reynolds conditions, which means that FANS is actually RANS. This should not be a surprise because the governing equations are the same for both.[/QUOTE] I think I mix up notation a bit here. When I said RANS I was thinking of the normal time averaged equations. When Favre is applied the name of the equations is still RANS, if I get you correct? RANS is a general description that applies independent of averaging type? |
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#10 |
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Oskar
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Hello.
If I haven't made a mistake in my thesis, Favre Average Navier Stokes equation are presented in attachment. Accent '-' means Reynolds averaging, accent '~' means Favre averaging. The conclusion is that only velocity and energy are Favre averaged. That's why those equations looks the same in RANS and in FANS. Please, let me know if I am wrong! Have a nice day! Oskar |
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interfoam favre fans |
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