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the Reynolds average of RANS and the filter function of LES

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Old   January 15, 2014, 04:33
Default the Reynolds average of RANS and the filter function of LES
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I am learning the CFD, and compiling the program by fortran.

After I learned a RANS program, there is a problem coming to me. The program seemly does not contain the function of Reynolds average, so I do not know how the Reynolds average influents the N-S of RANS.

Samely, I also do not know how the filter function influents the N-S of LES when I can not find where is the filter function in the whole program.

Any help will be greatly appreciated. Thanks
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Old   January 15, 2014, 05:08
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Filippo Maria Denaro
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Quote:
Originally Posted by yzeyue View Post
I am learning the CFD, and compiling the program by fortran.

After I learned a RANS program, there is a problem coming to me. The program seemly does not contain the function of Reynolds average, so I do not know how the Reynolds average influents the N-S of RANS.

Samely, I also do not know how the filter function influents the N-S of LES when I can not find where is the filter function in the whole program.

Any help will be greatly appreciated. Thanks
Hello,
your observation is correct, the answer is that you are right, no average in RANS and no filter in LES codes generally are present in subroutines.

The key is that the original NS equations are statistically averaged for RANS while being filtered for LES and such operation results in a new system formulated directly in terms of the RANS or LES variable.
Then, in RANS the unsteady terms vanishes while in LES is retained.
At the end, the difference appears in the unresolved term. For RANS is responsible for the residual as respect with the statistical averaged variable, for LES as respect with the filtered variable.
In practice, unlike the time derivative, RANS and LES equations could be confused, both RANS averaged and LES filter are implicitly assumed in a code, the difference being in the way the unresolved terms are modelled.
Finally, be careful that the computational grid size and the type of discretization of the equations act as implicit filtering for LES code, while in RANS the statistical averaging is much more relevant by means of the model and overwheelms the effect of the computational grid size
It is worthwhile remarking that some "explicit" filtered formulation for LES exist, but are not very common
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Old   January 15, 2014, 05:34
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Quote:
Originally Posted by FMDenaro View Post
Hello,
your observation is correct, the answer is that you are right, no average in RANS and no filter in LES codes generally are present in subroutines.

The key is that the original NS equations are statistically averaged for RANS while being filtered for LES and such operation results in a new system formulated directly in terms of the RANS or LES variable.
Then, in RANS the unsteady terms vanishes while in LES is retained.
At the end, the difference appears in the unresolved term. For RANS is responsible for the residual as respect with the statistical averaged variable, for LES as respect with the filtered variable.
In practice, unlike the time derivative, RANS and LES equations could be confused, both RANS averaged and LES filter are implicitly assumed in a code, the difference being in the way the unresolved terms are modelled.
Finally, be careful that the computational grid size and the type of discretization of the equations act as implicit filtering for LES code, while in RANS the statistical averaging is much more relevant by means of the model and overwheelms the effect of the computational grid size
It is worthwhile remarking that some "explicit" filtered formulation for LES exist, but are not very common
Thx very much!

This problem confuses me a long time.

But, I still do not know the sentence you had said----"the computational grid size and the type of discretization of the equations act as implicit filtering for LES code".

Does that mean that ---when we use LES, the small-scale fluctuation smaller the grid size will be filtered implicitly? and why does the discretization can also help in filtering?
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Old   January 15, 2014, 05:42
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for any type of method, FD, FV, SM, etc, the computational grid size introduce the cut-off filter at the Nyquist frequency Pi/h. All the scales smaller than h are filtered by the grid.
But there is much more than the grid filter, the choice of the method affects also the resolved scales greater than h. This happens for example while using second order central discretization that implicitly introduces a smooth transfer function (e.g. sin (kh)/kh) therefore, the wavenumbers components are smoothed approaching the Nyquist frequency.
Conversely, specral discretization are associated only to the sharp cut-off filter, no smoothing appears.

If you are interested in more details, you can read http://adsabs.harvard.edu/abs/2011JCoPh.230.3849D
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Old   January 15, 2014, 05:48
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Quote:
Originally Posted by FMDenaro View Post
for any type of method, FD, FV, SM, etc, the computational grid size introduce the cut-off filter at the Nyquist frequency Pi/h. All the scales smaller than h are filtered by the grid.
But there is much more than the grid filter, the choice of the method affects also the resolved scales greater than h. This happens for example while using second order central discretization that implicitly introduces a smooth transfer function (e.g. sin (kh)/kh) therefore, the wavenumbers components are smoothed approaching the Nyquist frequency.
Conversely, specral discretization are associated only to the sharp cut-off filter, no smoothing appears.

If you are interested in more details, you can read http://adsabs.harvard.edu/abs/2011JCoPh.230.3849D
Thx!U helped me a lot!
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Old   January 15, 2014, 05:53
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Quote:
Originally Posted by FMDenaro View Post
for any type of method, FD, FV, SM, etc, the computational grid size introduce the cut-off filter at the Nyquist frequency Pi/h. All the scales smaller than h are filtered by the grid.
But there is much more than the grid filter, the choice of the method affects also the resolved scales greater than h. This happens for example while using second order central discretization that implicitly introduces a smooth transfer function (e.g. sin (kh)/kh) therefore, the wavenumbers components are smoothed approaching the Nyquist frequency.
Conversely, specral discretization are associated only to the sharp cut-off filter, no smoothing appears.

If you are interested in more details, you can read http://adsabs.harvard.edu/abs/2011JCoPh.230.3849D
I also want to ask u a queation.

whether u programming by fortran, or use some open source software such as OpenFOAM?

I am a new LESer.
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