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LES Compressible Smagorinsky Model

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Old   September 26, 2006, 06:30
Default The les Smagorinsky Model inco
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iyer_arvind
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The les Smagorinsky Model incorporated in OpenFOAM, for compressible flows, add some more terms, which i have not been able to understand, can anyone give me a link to what method has been followed for compressible LES Smagorinsky Model?
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Old   October 9, 2006, 00:43
Default The Smagorinsky model coded is
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The Smagorinsky model coded is as follows

volTensorField D = symm(gradU);
volScalarField a = ce_/delta();
volScalarField b = (2.0/3.0)*tr(D);
volScalarField c = 2*ck_*delta()*(dev(D) && D);
k_ = sqr((2*b + sqrt(sqr(b) +4*a*c))/(2*a));
muSgs_ = ck_*rho()*delta()*sqrt(k_);

while the model commonly documented is much simpler. This looks like a solution of a quadratic equation, but i am still in search of a link for interpreting the above model.

the incompressible Smagorinsly model is simply

nuSgs_ = ck_*delta()*sqrt(k(gradU));
nuSgs_.correctBoundaryConditions();

i would be very thankful if someone could point out a link to the compressible model.
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Old   June 24, 2008, 10:13
Default Is there any reference paper f
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Is there any reference paper for the implementation of compressible smagorinsky model???
if there is any link either in the forum please provide it to me.

cheers!

Nishant
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Thanks and regards,

Nishant
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Old   June 24, 2008, 12:02
Default Hi The following title sug
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Hi

The following title suggest that it might be interesting. Haven't read it myself:

C. Fureby
On Sub-Grid Scale Modeling in Large Eddy Simulations of Compressible Fluid Flow
Phys. Fluids, 8, 1301, 1996.

Best regards,

Niels
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Old   July 21, 2010, 02:51
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Quote:
Originally Posted by iyer_arvind View Post
The Smagorinsky model coded is as follows

volTensorField D = symm(gradU);
volScalarField a = ce_/delta();
volScalarField b = (2.0/3.0)*tr(D);
volScalarField c = 2*ck_*delta()*(dev(D) && D);
k_ = sqr((2*b + sqrt(sqr(b) +4*a*c))/(2*a));
muSgs_ = ck_*rho()*delta()*sqrt(k_);

while the model commonly documented is much simpler. This looks like a solution of a quadratic equation, but i am still in search of a link for interpreting the above model.

the incompressible Smagorinsly model is simply

nuSgs_ = ck_*delta()*sqrt(k(gradU));
nuSgs_.correctBoundaryConditions();

i would be very thankful if someone could point out a link to the compressible model.
Hi iyer_arvind

Have you got the reference for the dynSmagorinsky model used in OpenFOAM by any chance? I am stuck at the same place trying to understand it as it is different from the original dynamic Smagoribsky from Lilly 1992.

Thanks
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Old   December 1, 2010, 11:42
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Hello everybody,

I have another question regarding the compressible Smagorinsky model. Why does the model ask for the file k within start directory? I use OpenFOAM-1.6 with rhoPisoFoam. I thought, Smagorinsky works without it. The incompressible model does not ask.

Best regards,
Georg
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Old   December 3, 2010, 14:05
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Hello Georg,

Looking at the code, I guess that the compressible version of the Smagorinsky model requests the "k" file through inheritance..

The Smagorinsky LES model is a child of the general base class GenEddyVisc. In the GenEddyVisc.C file you can see:
Code:
00073     k_
00074     (
00075         IOobject
00076         (
00077             "k",
00078             runTime_.timeName(),
00079             mesh_,
00080             IOobject::MUST_READ,
00081             IOobject::AUTO_WRITE
00082         ),
00083         mesh_
00084     ),
Which means that the implementation of the compressible Smagorinsky model (and as a matter of fact, all the models having GenEddyEq as a base class) will require k to be in your start directory, and that whether the turbulence model uses it, or not..

I guess this is inconsistent with the incompressible implementation of the turbulent models, where the request for "k" is programmed in the turbulence model itself, meaning that it will not be used/asked for if not needed by the model. Consequently, the incompressible Smagorinsky model does not complaining about "k" not being present in the start directory..

I don't know why it was implemented that way, but I guess you could ask the developers themselves via the openfoam/bugs site. I would be curious to know what the reason is.

Kind regards,

Francois.
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Old   December 8, 2010, 05:37
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Hi Francois,

thank you for your quick answer. I have reported that issue to OpenCFD. I will keep you informed.

Kind regards,
Georg
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Old   January 11, 2011, 04:20
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Hello Francois,

me again. It seems, that there was no reason for the way the compressible model was implemented. The code was reorganized and will be part of the next release.

Regards,
Georg
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Old   January 11, 2011, 08:18
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Allright!
Good work, and thanks for the update.

Kind regards,
Francois.
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Old   July 19, 2011, 11:50
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Hi,
there is no new implementation of compressible Smagorinsky model in OF 2.0! Is there any other reference than Fureby because he doesn't use this kind of modelling? Where I can find more about the version of compressible Smagorinsky, which is implemented in OF? Why a separate calculation of k will be used?
Thx Dejan
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Old   January 25, 2012, 12:07
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Quote:
Originally Posted by morard View Post
Why a separate calculation of k will be used?
that was a bug see http://www.openfoam.com/mantisbt/view.php?id=389
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Old   January 25, 2012, 12:27
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Thanks Gregor!

This question was actually posed by my friend because he didn't have profile at that time. I can't wait to see his face when he find out that this was a bug all the time
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Old   January 25, 2012, 12:35
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And furthermore you can actually derive the quadratic k equation from the Fureby1996 paper. I just did't see how in the beginning.
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Old   July 11, 2012, 07:57
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As demanded:

To derive the quadratic k equation you take

-B*D = c_B*rho*k^(3/2)*delta^(-1) (1 equilibrium assumption)

and plug in

B = 2/3 rho*k*I-2*mu_t*D (2 subgrid scale stress tensor )

so you get:

- c_B*rho*k^(3/2)*Delta^(-1) = 2/3 rho*k*trace(D)-2*mu_t*D^2 (3)

with mu_t = c_k*Delta*k^(1/2) (4 Smagorinsky model)

you finally get the quadratic k equation used in OpenFOAM:

- c_B*rho*k^(3/2)*Delta^(-1) = 2/3 rho*k*trace(D)-2*c_k*Delta*k^(1/2)*D^2 (5)

So the basic difference between the Smagorinsky model in the compressible and incompressible case is (at least how it is implemented in OpenFOAM), whether you can neglect the rho*k*trace(D) term or not.

D = Strain rate tensor
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Last edited by gregor; July 11, 2012 at 07:57. Reason: typo
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Old   October 22, 2012, 11:53
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Gregor,

If I may ask this question: how is D is the strain rate tensor?
volTensorField D = symm(gradU);

Sij = 0.5(dUi/dxj + dUj/dxi) and it is clearly a symmetric tensor
But how is D= symm(fvc::grad(U)) = Sij???

Is the assumption that the cross-stream and span-wise derivatives are minute compared to the wall normal derivates?

Cheers,
Deji
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Old   October 22, 2012, 12:05
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D is actually the deformation rate tensor.
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Old   October 22, 2012, 12:10
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I think I've figured this out...
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Old   January 23, 2013, 18:30
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Quote:
Originally Posted by gregor View Post
As demanded:

To derive the quadratic k equation you take

-B*D = c_B*rho*k^(3/2)*delta^(-1) (1 equilibrium assumption)

and plug in

B = 2/3 rho*k*I-2*mu_t*D (2 subgrid scale stress tensor )

so you get:

- c_B*rho*k^(3/2)*Delta^(-1) = 2/3 rho*k*trace(D)-2*mu_t*D^2 (3)

with mu_t = c_k*Delta*k^(1/2) (4 Smagorinsky model)

you finally get the quadratic k equation used in OpenFOAM:

- c_B*rho*k^(3/2)*Delta^(-1) = 2/3 rho*k*trace(D)-2*c_k*Delta*k^(1/2)*D^2 (5)

So the basic difference between the Smagorinsky model in the compressible and incompressible case is (at least how it is implemented in OpenFOAM), whether you can neglect the rho*k*trace(D) term or not.

D = Strain rate tensor
Hi,

Can I ask you a question about Eq. (5)? Why is Eq. (5) a the quadratic k equation ? I know this is a simple question. I checked the source code and but that OPenfoam did solve it like a quadratic equation. Any suggestions are really appreciated.

best regards,
H
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Old   January 24, 2013, 07:51
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Hi H

take Eq. 5 and divide it by k^(1/2) and you get:

- c_B*rho*k*Delta^(-1) = 2/3 rho*k^(1/2)*trace(D)-2*c_k*Delta*D^2

now substitute k by k=K^2, this yields

- c_B*rho*K^2*Delta^(-1) = 2/3 rho*K*trace(D)-2*c_k*Delta*D^2

solve for K and resubstitute K=sqrt(k). That is at least my opinion of whats happening
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