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-   -   Laminar vs Turbulent Navier-Stokes (http://www.cfd-online.com/Forums/main/85801-laminar-vs-turbulent-navier-stokes.html)

truman March 7, 2011 12:34

Laminar vs Turbulent Navier-Stokes
 
I am trying to wrap my head around the practical considerations of solving laminar vs turbulent Navier-Stokes. I understand that the plain Navier Stokes equations describe the most general case and that if they were solved at sufficient spatial and time resolution (as in DNS) they would describe the appropriate laminar or turbulent flow for whatever application that you were considering. I also understand that Reynolds (or Favre) averaging seeks to average out the turbulent fluctuating terms so that such high resolution is not necessary.

Referring to http://www.cfd-online.com/Wiki/Favre_averaged_Navier-Stokes_equations, the Favre averaged equations (30) - (32) look just like the original Navier-Stokes equations (1) - (3) with some additional turbulent terms that need to be modeled via a turbulent closure model like k-omega or k-epsilon. So, assuming laminar flow and neglecting the turbulence terms, it looks like we are back to the original Navier-Stokes except each instantaneous variable has been replaced with its averaged equivalent. This leads me to my point of confusion, if you attempt to solve the full Navier-Stokes equations without sufficient spatial and time resolution, are you just assuming laminar conditions and solving for the laminar solution? Will this solution approach the full turbulent solution as you refine in space and time? A related question: if you neglect all time derivatives in equations (1) - (3), what are you actually solving for some pseudo Reynolds or Favre averaged solution or will you arrive at a steady-state laminar solution? None of these conclusions feel right to me, but I can't quite put my finger on what is wrong.

Any help is appreciated.
Thank you.

truffaldino March 7, 2011 12:58

Reynolds averaged equations look like navier stokes if you forget that viscousity is now an unknown variable. Eddy viscousity changes from point to point and extra equations are needed to compute it.

truman March 7, 2011 13:27

But if you make the assumption that your flow is laminar, the eddy viscosity goes to zero and you are only left with the molecular viscosity and the original Navier-Stokes equations. I guess part of my question comes down to: what is the difference between assuming laminar flow and solving the full Navier-Stokes equations, aside from resolution?

truffaldino March 7, 2011 13:42

Sorry I have not read your question carefully.
Here is the answer:
Turbulence exists on large scales too, this is not the problem of resolution, rather it is problem of stability of solution of NS equation which can be broken at large scales too, depending on Reynolds number.

If you take NS equation on more coarse grid, solution will also become unstable after reynolds number will exceed critical value.

In other words at certain reynolds numbers laminar solution is not stable and chaotic solution you will get does not approach turbulent solution, as this is a chaotic systems where limit does not exists (unless you are going to DNS scale).

truman March 7, 2011 13:51

Ok, but then how do you solve a laminar flow problem at high Reynolds number, say a high-speed laminar boundary layer?

truffaldino March 7, 2011 14:22

Quote:

Originally Posted by truman (Post 298282)
Ok, but then how do you solve a laminar flow problem at high Reynolds number, say a high-speed laminar boundary layer?

High-speed laminar boundary layer does not exist at whole lenght (as Re exceeds certain limit there is a transition to turbulence along the stream). This is an essence of turbulence, that non-chaotic solution is not stable! This is why it is so difficult to account for it.

To get insight why it happens take a look at Orr-Sommerfeld theory.

truman March 7, 2011 14:44

All of my undergrad aerodynamics just came back to me. Thank you, it all makes a lot more sense now.


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