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November 2, 2017, 03:29 |
Two dimensional cut
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#1 |
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Hello everyone,
I have a problem, I can not wrap my head around. I have an 2D-cut of a 3D isotropic turbulence DNS. Now I would like to get the energy spectrum from this. But it does not seem so easy, since I can just FT in the x-yplane and not in the z-direction. So I just have something like v'(kx,ky,z), with v' being the FT of v. So I can not simply set E=(v'^2)/2 . Does anyone have an idea? |
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November 2, 2017, 03:45 |
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#2 | |
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Filippo Maria Denaro
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Quote:
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November 2, 2017, 03:59 |
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#3 |
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Hey FMDenaro,
thanks for the reply. Yes, the boundary condition is periodic. My problem is, that i just have this one slice. I do not have the full 3D information, just this one 2D slice. In the 3D case I would do a 3d FFT and of the velocity field: v'=FT(v) and then calculate the energy spectrum as E=<v'v'>/2. But now I can not do this, because i don't have the information of the third direction, so I can just FT in x and y but not in z. I hope this makes it clearer. |
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November 2, 2017, 04:03 |
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#4 | |
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Filippo Maria Denaro
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Well, you can do the 1D energy spectra analysis, your case is homogeneous and isotropic. In the (x,y) plane you can do the FFT fixing y and working at several x stations, then performing averaging. Viceversa for the FFT along y. |
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November 2, 2017, 04:14 |
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#5 |
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I calculated the 1D energy spectrum from the correlation function like they suggest in Pope (6.206), as the FT of the velocity correlation fucntion.
Is this what you had in mind? With this I could get the full energy spectrum function E(k) via Pope(6.217). BUT I ran into the problem, that I can just determine E11(k1) but for calculating the full energy spectrum function I need E11(k). And I dont know how to calculate that. Is it clear what I meant, since I am referring to the book a lot and using there notation? Also thanks again for the quick and helpful responses. |
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November 2, 2017, 04:39 |
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#6 |
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Filippo Maria Denaro
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Usually I perform the FFT of the velocity field and plot the squared modulus of the Fourier coefficients, it is just a slight difference.
You can only do the best with what you have. Owing to the isotropy, in the (x,y) plane you can consider the slice of the shell, working on the radius in the 2D plane. I suppose that in the plane you have also the measure of the third velocity component along z, that is w(x,y). |
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November 2, 2017, 05:08 |
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#7 |
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Hey, yes, I have the third component as well.
I am not sure, if I understood correctly, but as far as I understood, you are suggesting , that I take the FT of vx -> v'x and plot the squared. I think this leads me back to my first question. I can only do the FT for the x and y direction, so the fourier component of vx would be v'x(kx, ky, z=0). This is not the same as v'x(kx, ky, kz=0). If i had the latter I would agree, that E11(kx,ky,kz=0)= [v'x(kx, ky, kz=0)]^2. But I don't see why this should hold: E11(kx,ky,kz=0) = [v'x(kx, ky, z=0)]^2. If this were correct, I could do it for the total energy spectrum as well, couldn't I: E(kx, ky, kz=0) = [v'x(kx, ky, z=0)]^2 +[v'y(kx, ky, z=0)]^2 + [v'y(kx, ky, z=0)]^2 . Maybe I am missing a crucial point about isotropy, but in my mind I have to FT in the z-direction as well (which I can not) if I want to do it this way. Thanks again for your time, I really appreciate it. |
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November 2, 2017, 06:09 |
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#8 |
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Filippo Maria Denaro
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The best you can do is computing at the cut plane zc=constant the components Euu(kx,ky,zc), Evv(kx,ky,zc), Eww(kx,ky,zc) by a 2D FFT.
Howevere, I would start from the 1D spectra Eii(kx,y,zc), averaging them along y. |
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November 2, 2017, 06:19 |
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#9 |
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Hmm ok thanks.
Do you see any way to get the total energy spectrum function E(k) from this ? |
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November 2, 2017, 07:29 |
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#10 | |
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Filippo Maria Denaro
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have a look to Sec.4.3 https://www.io-warnemuende.de/tl_fil...Chap4_WS08.pdf |
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November 2, 2017, 08:41 |
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#11 |
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Hmmm I read the associated Pope chapter but there I ran into the problem I wrote in comment #5. But maybe I have to read it more carefully.
Thanks anyways |
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November 2, 2017, 10:30 |
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#12 |
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Filippo Maria Denaro
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