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Shear stresses from spectral energy density - channel flow |
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April 11, 2018, 19:17 |
Shear stresses from spectral energy density - channel flow
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#1 |
New Member
Carlos
Join Date: Nov 2016
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I am trying to compute the Reynolds shear stresses from the spectral energy density tensor , for the case of a 3D channel flow where x and z are homogeneous and y is spectrally discretized. Assuming a zero streamwise wavenumber , from the definition:
My question is: is a N X N matrix with N being the number of points in y, so, how can I compute at a concrete point in y? Is it the corresponding value of the diagonal of ? (I mean, summing over all ) |
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April 12, 2018, 03:52 |
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#2 |
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Filippo Maria Denaro
Join Date: Jul 2010
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I just wonder why do not compute directly the term <u'v'> in physical space...
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April 12, 2018, 04:31 |
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#3 |
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Carlos
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In this case, I don't have a DNS but a linearised model where I am able to construct E_{uv}, but I don't have u' and v' so that I can average. Actually my goal is to compare with a real DNS, where that info is readily available.
How would you compute <u'v'> from the matrix E_{uv} for this case? |
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April 12, 2018, 07:07 |
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#4 | |
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Filippo Maria Denaro
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Quote:
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April 12, 2018, 08:02 |
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#5 |
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Carlos
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The code solves a linear version of NS discretising spectrally in x and z (homogeneous) and using FD in y. So, E_{uv} is a function of the stream- and spanwise wavenumbers \alpha, \beta and y.
Focusing on the large structures (\alpha=0), <u'v'>(y) is given by the expression above, the integral of E_{uv} over all \beta. However, this is a N X N matrix (N is the number of points in y). So, my question, how to extract <u'v'> for a particular y from the integral of E_{uv} over all \beta. Is it the diagonal? the column? |
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April 12, 2018, 08:32 |
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#6 | |
Senior Member
Filippo Maria Denaro
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Quote:
<f>(y) = 1/(T*Lx*Lz) Int [0,T] Int [0,Lx] Int [0,Lz]f(x,y,z,t) dt dx dz This is what we actually compute from the DNS fields. On the other hand we can define for example E_uu(kx,y,z,t) = squared modulus of Fourier coefficient in the 1D FFT along x. E_uu(x,y,kz,t) = squared modulus of Fourier coefficient in the 1D FFT along z. You could also perform a 2D FFT in the x,z plane and getting the Fourier coefficients as functions of (kx,y,kz,t). Then, to get a function of the only y position you need to make the statistical averaging in kx,kz,t. Of course, in general the transformation from Fourier wavenumber and physical space involves the inverse Fourier transform. |
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April 12, 2018, 08:41 |
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#7 |
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Carlos
Join Date: Nov 2016
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OK, maybe my question is not on turbulence but on how we solve the problem. The thing is that I have a matrix for E_{uv}, (for each \alpha and \beta). Why a matrix? Because of the discretization in y.
If, for example, I would like to compute the variance of the velocity tensor, I just need to integrate E_{uv} for all \beta and then take the trace of the resulting matrix. However, what I want to compute is <u'v'> at a certain y position, and I don't know to what entries of the velocity tensor this corresponds. |
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April 12, 2018, 08:55 |
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#8 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
I assume you have a matrix (Nx X Nz) corresponding to the plane of the wavenumbers kx and kz. But you have one matrix for each of the discrete position y. Therefore, you need to fix the value of the same y where the <u'v'> are to be evaluated. Then you proceed in the same way along all the nodes in the y-direction. Of course, that is still at a certain t, the statistical averaging in time being still to be performed. |
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