interFoam --> FAVRE

kaaja · March 30, 2018, 10:46

Hi.
InterFoam has varying density.
Must one not then use the form

$\frac{\partial (\rho u_i}{\partial t} + \frac{\partial(\rho u_i u_j)}{\partial x_j} = ...$

instead of

$\rho \frac{\partial u_i}{\partial t} + \rho \frac{\partial( u_i u_j)}{\partial x_j} = ...$

since the density is non-constant?

And, if one must use the first form, isn't it really FANS (Favre averaged Navier Stokes) one solves when using interFoam and a RAS turbulence model?

LuckyTran · March 30, 2018, 14:19

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)
    );

The gov eqns for RANS & FANS are identical except for a tiny correlation term in the viscous stress that everyone always ignores anyway, so that it is not necessary to distinguish between RANS/FANS. If your density is not-constant then it is FANS. If your density is constant then it is FANS with a constant density (or RANS).

kaaja · March 30, 2018, 22:51

Quote:

Originally Posted by LuckyTran

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)
    );

The gov eqns for RANS & FANS are identical except for a tiny correlation term in the viscous stress that everyone always ignores anyway, so that it is not necessary to distinguish between RANS/FANS. If your density is not-constant then it is FANS. If your density is constant then it is FANS with a constant density (or RANS).

Thanks for replying!
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?

LuckyTran · March 30, 2018, 23:08

Quote:

Originally Posted by kaaja

Thanks for replying!
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?

I don't follow what you are trying to say. This here is the RANS equations AND also the FANS equations. What determines whether they are RANS/FANS is when you decide what U is. Is it the Reynolds-averaged velocity or the Favre-averaged velocity? But the code and solver does not care because it is still U. You can call U by any other letter, e.g. W if you like.

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?

kaaja · March 31, 2018, 00:32

Quote:

Originally Posted by LuckyTran

I don't follow what you are trying to say. This here is the RANS equations AND also the FANS equations. What determines whether they are RANS/FANS is when you decide what U is. Is it the Reynolds-averaged velocity or the Favre-averaged velocity? But the code and solver does not care because it is still U. You can call U by any other letter, e.g. W if you like.

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?

I think I understand what you are saying: InterFoam doesn't care. It is a matter of interpretation of the U. That is OK.

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?

LuckyTran · March 31, 2018, 02:05

Quote:

Originally Posted by kaaja

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?

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?

kaaja · March 31, 2018, 02:15

Quote:

Originally Posted by LuckyTran

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?

So the coclusion is that if one wants to be a politcally correct ass, then one has to say FANS, not RANS.

Thanks for helping me sorting this out!

LuckyTran · March 31, 2018, 02:16

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.

Quote:

Originally Posted by kaaja

So the coclusion is that if one wants to be a politcally correct ass, then one has to say FANS, not RANS.

Thanks for helping me sorting this out!

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.

kaaja · March 31, 2018, 03:25

Quote:

Originally Posted by LuckyTran

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 is very interesting! I have not thought about the possible different origins of density variations (turbulence and variation in the mean itself)...

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?

sheaker · March 31, 2018, 03:50

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

March 30, 2018, 10:46	interFoam --> FAVRE	#1
kaaja Member Join Date: Apr 2017 Posts: 68 Rep Power: 9	Hi. InterFoam has varying density. Must one not then use the form $\frac{\partial (\rho u_i}{\partial t} + \frac{\partial(\rho u_i u_j)}{\partial x_j} = ...$ instead of $\rho \frac{\partial u_i}{\partial t} + \rho \frac{\partial( u_i u_j)}{\partial x_j} = ...$ since the density is non-constant? And, if one must use the first form, isn't it really FANS (Favre averaged Navier Stokes) one solves when using interFoam and a RAS turbulence model?

March 30, 2018, 14:19		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	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) ); The gov eqns for RANS & FANS are identical except for a tiny correlation term in the viscous stress that everyone always ignores anyway, so that it is not necessary to distinguish between RANS/FANS. If your density is not-constant then it is FANS. If your density is constant then it is FANS with a constant density (or RANS).

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