# Kolomogorov length scale

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 November 21, 2002, 11:16 Kolomogorov length scale #1 Bobby Guest   Posts: n/a Hi, everyone, I have a small question in DNS. How can I estimate the necessary grid nodes in a special computation, i.e. the smallest length scale? I think it must be from the Kolomogorov length scale, but how to recon in detais? Thanks for your help.

 November 22, 2002, 09:20 Re: Kolomogorov length scale #2 student Guest   Posts: n/a Hai you have rleation ship betweeen number of nodes and reynolds number with kolmogorv length scale. eta=L/(Re)^3/4. Where L is the box dimension and equal to Ndeltax. where N is the number of nodes in each diemnsionas. I think I answered your question bye student

 November 22, 2002, 09:50 Re: Kolomogorov length scale #3 Tom Guest   Posts: n/a How many grid points you need in a DNS? There is no straight answer to this question. It depends on the numerical method you are using and the kind of statistics you want to compute. The general rule is that the grid spacing should be three times the Kolmogorov length scale. If you you are only interested in mean and rms values of the velocity the grid spacing can be larger but if you want to compute higher order statistics you need more grid points perhaps. If you want to calculate a homogeneous flow, channel flow or boundary layer flow you can also look into the literature to see what other people are using. Tom

 November 22, 2002, 10:37 Re: Kolomogorov length scale #4 Bobby Guest   Posts: n/a Thanks for your reply. I do know that the smallest grid scale need not be equal to Kolomorogov scale (some value about 10-15*Kscale is enough in Moin(1998)'s opinion). What I want to know is how to calculate this ruler scale in details? (I do DNS of free jets.) Thank you very much.

 November 22, 2002, 10:44 Re: Kolomogorov length scale #5 Bobby Guest   Posts: n/a Thanks for your information. I am wondering if this simple method can be evaluated in all DNS cases. and could you please tell me where this formula come from (the original paper discussing about this)? Thanks.

 November 22, 2002, 11:15 Re: Kolomogorov length scale #6 Tom Guest   Posts: n/a In the far field of a round jet the dissipation is self-similar, see the discussion on page 127-128 and the figure on page 130 in the book of Pope, Turbulent flows. Given the diameter of the jet, the initial velocity and viscosity you can then calculate the Kolmogorov length scale as a function of the distance to the orifice. There are also DNS of round jets reported in literature. These can give you an idea what kind of resolution you need. Tom

 November 22, 2002, 11:29 Re: Kolomogorov length scale #7 Bobby Guest   Posts: n/a Thanks for your help.

 November 22, 2002, 13:25 just check.. #8 Patrick Godon Guest   Posts: n/a You can also check after you performed the simulations (de facto) that your scale is correct, rather than checking before (ab initio). Just plot the energy as a function of the wave number k, and make sure that the end of the spectrum has the slope you expect for the Kolmogorov law (depends whether it is 2D or 3D) and that it covers at least one order of magnitude in k. Your artificial viscosity should kill the higher k without damaging the slope and the rest of the spectrum. You want to have enough resolution (grid points) to reach the part of the spectrum that shows nicely the Kolmorov slope, and a (artificial) visosity that reaches only the smallest scale (higher k). So you have to play with the number of grid points and the Reynold number numerically untill you reach these conditions. Depending on the numerical scheme you use and the order of the numerical method, these numbers can vary a lot for a specific flow problem. Patrick

 November 23, 2002, 04:54 Re: just check.. #9 Tom Guest   Posts: n/a OK, this works nicely when you have a spectral method or a higher order FDM because then you calculate a good spectrum. But when you use a lower (second) order FDM or FVM I doubt if you can calculate a clear spectrum.

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