# Laminar model the same as DNS?

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 February 20, 2003, 14:59 Laminar model the same as DNS? #1 Kerrin Guest   Posts: n/a Please forgive me for asking a question which may be obvious to many of you, but: Does a DNS simulation solve exactly the same equations as those used in laminar flow simulations? I have been trying to figure this out from the textbook but couldn't quite pin it down. Thanks, Kerrin bhanuday.sharma likes this.

 February 21, 2003, 23:05 Re: Laminar model the same as DNS? #2 versi Guest   Posts: n/a I did not do DNS. But I figure out the equations are the same, except that DNS use huge number of grids to resolve turbulent scales, and the solutions are not always steady state, yet you can make certain kind of statistics and average in time to obtain meaningful results.

 February 22, 2003, 07:15 Re: Laminar model the same as DNS? #3 gorka Guest   Posts: n/a I've never performed any DNS simulation but as far I know the solved equations are EXACTLY the same: Navier-Stokes equations and continuity equation (+ energy equation if needed). NO extra term is added, neither any physical modelling assumed. (In comparison with Reynolds Averaging approach) Hope this helps Gorka

 February 22, 2003, 09:10 Re: Laminar model the same as DNS? #4 Holidays Guest   Posts: n/a The main problem with DNS is that the number of nodes that is necessary is roughly proportional to the turbulent Re at the power of 3 (turbulent Re ~1% of Re) per turbulent recirculation length unit. Hence the cost in computer power, time (the time step needs be small on top of the grid size issue) and storage! Otherwise the equation are even, in principle, simpler to solve than the RANS.

 February 23, 2003, 14:59 Re: Laminar model the same as DNS? #5 Kerrin Guest   Posts: n/a Thanks for your replies. I had a suspician that the equations were the same but was not sure. If I was solving on a single workstation (ie limited computing power), I imagine that a normal RANS simulation would produce more accurate results than a DNS simulation (with far too few mesh points). I guess what I am asking is: Is there any point using a DNS simulation (with too few mesh points) to check the results of a turbulence model? Or is that just a waste of time? The reason I ask is that another CFD user said they were doing just this to check whether their RANS solutions were correct. This sounded wrong to me. Any thoughts? Kerrin

 February 24, 2003, 03:51 Re: Laminar model the same as DNS? #6 Tom Guest   Posts: n/a For a good DNS you need to solve the full three-dimensional (!) and time dependent Navier-Stokes equations (no terms neglected!). The time step has to be really small (CFL number around one) because you need to resolve all the rapid fluctuations in time. In general turbulent kinetic energy will be produced in the flow and this has to be taken away by dissipation otherwise it will accumulate. So need to resolve the small dissipative scales. In many practical applications the Reynolds number is so high (and the dissipative scales so small) that it is impossible to perform a full DNS, especially with a slow commercial CFD code. So I guess that 'they' are doing a very poorly resolved DNS and that the dissipative scales are absolutely not resolved. A good code would just blow up because of the accumulation of energy. That they can do a simulation means that the computations are somehow stabilized with a lot of numerical viscosity. You can say that they are doing some kind of large eddy simulation poorly defined and probably poorly resolved large eddy simulation (LES). The sub-grid model is the numerical viscosity added by the numerical scheme. I guess that you can use these kind of computation in the best case just to get a very global idea about the flow patterns but absolutely not for any kind of quantitave check of the RANS results.

 February 24, 2003, 15:12 Re: Laminar model the same as DNS? #7 Kerrin Guest   Posts: n/a Hi Tom, Thanks for helping to clarify this for me. I had suspicians along the line of what you had said but wanted to check it out with someone else. Kerrin