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October 26, 2008, 07:16 
How to draw a E(k)k graph?

#1 
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I know it was discussed before, but I still don't understand it and there's also no way to reply to those threads. So I put it here again. I am using LES for a channel Flow, and I want to draw a E(k)k graph as shown in many textbooks, but how? Now, I have velocity field in each time. Say, at time "1000" which is fully established turbulence field, here I have a velocity field U(x,y,z), and then I pick out a plane velocity field U(x,z) which is at the position of y^+=8, if the meshing is 32*64*32 in each direction, then the U(x,z) field is of the dimension of 32*32, and then??????? Thank you in advance! Regards, \Daniel 

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October 27, 2008, 09:09 
Re: How to draw a E(k)k graph?

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The FFT is carried out over the space dimensions, not the time. You can either chose to do that at a given time or to average the results over a given period of time.
First of all with 32*64*32 there aren't that many wavelengths to draw from..., especially that the graph is often drawn on a logarithmic scale. In theory, assume that your grid points are equally spaced in the (say) x dimension and that you have 512 points in the x dimension. You then have the array v(i) with i=1,512 and you want to check the kinetic energy of the flow by looking at v^2. So you have E(i)=v(i)*v(i), with i=1,512. Now that E is realy E(x), as x=x_i. So what you have to do is to carry out a FFT from the spatial dimension X to the spectral dimension k and you will obtain E(k). There are libraries with FFTs and you need to find out how it is done. Now if you are in 3D, then you need to average E(x,y,z) so that you get only E(x), this is a space average so that you sum over all the y and z and divide by the number of grid points y*z (say y_j, z_l then j*l); or alterantively you carry out a discrete (numerical) integration and then divide by the surface Y*Z (dimension in the y z coordinates). then you obtain E(x) and you can carry out the FFT in that dimension. Similarly you can carry out FFT in y and z. You can also (sometimes) wish to only do FFT at a given location of y_o, z_o in x, so you don't need to average by chose your "line" of "x" by fixing the value of y and z and carry out the FFT over the x. Now that was for a given time, but you can also do that for an entire period of time by integrating (averaging) your function E over the period of time (summing over all E(t) for all t and divide by the number of individual time steps you averaged), then you obtain E(x,y,z) and you can do your FFT as explained above. 

October 27, 2008, 11:54 
Re: How to draw a E(k)k graph?

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Thank you very much, Patrick!
First, I want to express my difficulty, you know in Channel Flow, which is different from a box turbulence (isotropic), the energy spectrum varies with y+, that is to say, E(k) is statically the same at a given y+ plane, no matter which line you draw in this plane. Please correct me if I am wrong. To my knowledge, 64*64*64 is enough for a LES channelflow simulation (Re_tau=180, etc.), do you mean 64 (64*64*64 mesh) or 32 (32*64*32 mesh) is too small comparing with 1024 points needed for FFT? Or do you also mean 64 or 32 is too samll a number to capture those small scale eddies? But, what can I do now? Refine the mesh? but "64*64*64 is enough for a LES simulation (Re_tau=180, etc.)" Second, how about this, in a mesh 32*64*32(x*y*z), first I pick out a plane which has 32*32(x*z) points, that is 1024 points' velocities, then I treat them as a single line, then it would be sufficient for a FFT? 

October 28, 2008, 10:57 
Re: How to draw a E(k)k graph?

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OK, let us assume that you have enough resolution to capture your eddies.
WHen you "pick up a plane" in x*z, you don't treat the 32x32=1024 points as a line as such, you treat it as a line in the Y dimension, so you have only 64 points in Y and you do your FFT with the 64 points in Y, while you either average the plane (average the 1064 points into one point for each single value of Y) or you chose a given coordinates in x and z and consider the Y dimension for x and z constant  again that gives you a "line" in Y. There are 1024 such lines that you can consider to do an FFT on, and each of them has only 64 points, so again you do the FFT on 64 points, which is not much I must say. I hope what I wrote is clear, as I am not sure I understood you and even less sure you understood me. Technical communication is not that easy when using text messages. 

December 9, 2014, 18:08 

#5 
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wang
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This method of calculation of Energy spectrum is totally wrong. The Eii in wavenumber is no directly relationship with ui*ui in physical space, so Eii in kx isn't the fourier transfer of ui*ui. It is the fft of Rii in physical space or integration of power of ukx in wave number.


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