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dirk July 21, 2005 15:10

grid size for DNS

I did some calculations with RANS models of a jet in cross flow. Now I am thinking about how fine the mesh for a DNS should be. I have a jet Reynolds number based on jet velocity and hole diameter (=8mm) of about 4600 and the momentum loss thickness Reynolds number of the mainstream just upstream the jet is around 370. So firstly: which one is decisive for determining the grid size for a DNS (I suppose it is the first one) and secondly: how small will it be ?

Thanks in advance, Dirk

yuan July 21, 2005 21:02

Re: grid size for DNS
Estimate the Komogorov scale, say l, then use dx=(5~20)l. maybe the you need N=100~200 crossflow!

hhb8622078 July 21, 2005 21:38

Re: grid size for DNS

dirk July 25, 2005 14:43

kolmogorov scales

I found these formulas for estimating kolmogorov scales: length=(nu^3/eps)^0.25 and time=(nu/eps)^0.5 . Are these correct ? My minimum kolmogorov length from realizable k-eps computation is 0.04mm. I think, the grid size in these regions must not be bigger than this.

noName July 25, 2005 15:42

Re: kolmogorov scales
Yes, the formulae are correct.

In general, it is hard to accurately estimate the Kolmogorov scale from a simulation (using say k, epsilon and nu). It's best to use some experimental data for this.

That being said, a true DNS should be able to resolve the Kolmogorov-scale wavenumber, so you need a resolution of at least half the Kolmogorov scale. However, this is usually computationally very expensive, so published DNSs usually resolve only about 6 times the Kolmogorov scale. Pope's book on Turbulence, explains why this is a reasonable grid size.

yuan July 27, 2005 04:02

Re: kolmogorov scales
You may read Moin%Mahesh,1998. Ann Rev Fluid Mech,30,p539 for a practical estimate. The practical DNS is not so loyal to the resolution rule, i.e. a coarser mesh is adopted, or your mesh is too fine to work.

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