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Modified simpleFOAM using given Reynolds stress field

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Old   February 24, 2016, 11:55
Default Modified simpleFOAM using given Reynolds stress field
  #1
K62
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Martin Schmelzer
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Dear OpenFOAM-users,

My aim is to run a RANS simulation with a given Reynolds-stress tensor field R_{ij}. This R_{ij} will come from a previous RANS simulation of the same case but will be (slightly) modified, i.e. R_{ij} \rightarrow R^*_{ij}. In order to write a solver, which is able to do that, I used simpleFoam (OF2.3.0). The UEqn.C file contains the following lines

Code:
tmp<fvVectorMatrix> UEqn
    (
        fvm::div(phi, U)
      + turbulence->divDevReff(U)
      ==
        fvOptions(U)
    );

The pointer
Code:
turbulence->divDevReff(U)
is defined e.g. in LRR.C
Code:
tmp<fvVectorMatrix> LRR::divDevReff(volVectorField& U) const
{
...

return
        (
            fvc::div(R_)
          + fvc::laplacian(nut_, U, "laplacian(nuEff,U)")
          - fvm::laplacian(nuEff(), U)
        );
}

This leads to the following implementation of my new solver
Code:
tmp<fvVectorMatrix> UEqn
    (
        fvm::div(phi, U)
      + fvc::div(R)
      + fvc::laplacian(nut, U, "laplacian(nuEff,U)")
      - fvm::laplacian(nu+nut, U)
      ==
        fvOptions(U)
    );
Also the line
Code:
turbulence->correct();
is commented, since it is not necessary anymore.

My workflow is like that.

STEP 1: I run a simulation using the un-modified simpleFoam solver until the residuals go to machine accuracy or drop a value of 1e-10 or so. I then run the utility R, which computes the Reynolds stresses using the Boussinesq approximation

Code:
tmp<volSymmTensorField> kEpsilon::R() const
{
    return tmp<volSymmTensorField>
    (
        new volSymmTensorField
        (
            IOobject
            (
                "R",
                runTime_.timeName(),
                mesh_,
                IOobject::NO_READ,
                IOobject::NO_WRITE
            ),
            ((2.0/3.0)*I)*k_ - nut_*twoSymm(fvc::grad(U_)),
            k_.boundaryField().types()
        )
    );
}
STEP 2: The output field R is used as an input for the modified solver described above, so that no additional transport equation for R_{ij} is solved and it stays (so far) unmodified. Since, the field is exactly the one which is the output of the step 1, I naively (?) assumed that using it as an input for the modified solver will lead to the same U and p fields as in STEP 1.

For the cases I am looking at I observe that for a smooth geometry (2D hill), the modified solver converges but gives velocity and pressure fields which show differences at the wall of up to 5%. I guess this has to do with the fact how boundaries are treated by the different differential operators (laplacian, div and grad) in OpenFOAM, but I wasn't able to get to the core of this problem. Does anybody have an idea about that?

Best wishes,
Martin
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Old   March 23, 2017, 03:14
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  #2
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Yangmo Zhu
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Thanks you Martin, your thought is exactly the same as mine. Regarding the 5% difference, did you use the same system files in both case? For example, fvscheme.
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Old   March 24, 2017, 03:41
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Yangmo Zhu
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Hi Martin, although I didn't work out your method, I found another way to deal with it. I want to share it with you:

1. Calculate nut according to R;
2. Add nut to nu as an internal field;
3. Solve the problem as laminar flow;
4. Done.
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