# Kolmogorov/Energy spectrum for turbulent boundary layer

Previously, I have calculated energy spectrum for 3D isotropic turbulent flow data which is equally spaced in all three directions and then to compute the energy spectrum, one performs Fourier transform and then accumulates energy located in different wavenumber bins and then gets the ($k^{−5/3}$) slope and it all works fine.

Now if I want to extend the same for an anisotropic case like turbulent boundary layer case, how should I modify the technique? I cannot take a Fourier transform in the wall normal direction anymore and I am not sure how to compute the energy spectrum as such. There is no problem if I compute the spectrum for streamwise and spanwise plane since they are both defined using Fourier basis.

So my question is two-fold:

1. How do we compute energy spectrum for a 3D turbulent boundary layer?
2. If we cannot compute it for a 3D case i.e. if the argument is valid only for a 2D planar cut at various wall normal locations, then how do we define a scale for turbulent boundary layer cases, particularly along wall normal direction?
• What kind of data do you have? I thought a time series on a probe position would be enough? – Bernhard May 14 '14 at 20:41
• I have DNS data wherein I have a fully resolved isotropic case for a single time step. For an experiment, one uses a probe to get the spectrum, but for simulation its done differently as I explained above. – Sidhha May 15 '14 at 15:04
• Why can you not do the same thing with your DNS? You encountered the trick with the Fourier transforms does not work, right? – Bernhard May 15 '14 at 17:41
• For the turbulent boundary layer which is anisotropic, the wall normal direction is aperiodic and non-uniform. So unlike the isotropic case, I cannot take a Fourier transform in the wall normal direction. I am not sure how people obtain a 3D energy spectrum for boundary layer. In literature, I have only seen 2D planar results along streamwise and spanwise directions. – Sidhha May 15 '14 at 18:22
• You can probe a point, right? Or is there an objection against that? – Bernhard May 15 '14 at 18:50