Register Blogs Members List Search Today's Posts Mark Forums Read

April 13, 2016, 07:14
#2
Senior Member

Filippo Maria Denaro
Join Date: Jul 2010
Posts: 3,745
Rep Power: 41
Quote:

I try to give some infos:

- Turbulence is always unsteady, what is called "steady" has only statistical meaning.
- If you are using RANS, you assume your flow problem has a statistical steady state. That depends on the type of flow problem.
- If you solve RANS formulation, you implicitly assume that any computed variable is steady in the sense of

<f>(x) = lim t->+Inf (1/t) Int [t0,t] f(x,t) dt

- for example, if you solve the flow over a cylinder, you cannot see the unsteady vortex shedding but only the steady statistical part of the separated flow. This is not a bug/error of the solution, is exactly what you are asking for from RANS.
- The unsteady version of the RANS formulation is called URANS. It has sense if the unsteadiness is due to an external time-depende forcing (for example the piston moving in a cylinder)
- Using RANS/URANS the solution is strongly dependenent on the turbulence model, less on the mesh and type of discretization.

April 13, 2016, 11:51
#3
Senior Member

Alex
Join Date: Jun 2012
Location: Germany
Posts: 1,648
Rep Power: 26
Quote:
 Originally Posted by JochemGrietens 1) How could one check if we are actually dealing with an unsteady phenomenon? 2)I was thinking of using the steady solution and using this as the starting conditions for a transient computation. I would thing that a clear change in velocity field would be visible if the phenomenon was actually steady and only small time dependence would occur if it is in fact steady. 3)What happens if you solve a unsteady phenomenon ( like a Von Karman street ) with a steady model using a turbulence model. Would you get total bogus results or would you get some kind of time average? 4)When performing a steady model to unsteady conditions, would you get a different steady solution every time you run the model or would it be the same every time ? 5) following question 4: would the solution become very dependent on the starting conditions ? like the start value of k or the mesh size ...
1) Preliminary knowledge about the physics of your problem. Poor convergence of a steady-state approach. Actually running an unsteady simulation to see the time-dependency
2) Using solutions from a steady-state solver as initial conditions for a time-dependent simulation is one possible approach. Especially if you are not interested in the transition from a known initial condition to the final solution.
3) With this approach you can get some correct results, e.g. the frequency of the vortex shedding in your example. However, do not expect the results to be as good as a LES. As FMDenaro already pointed out, URANS turbulence models are better suited for time-dependent flows where the unsteadiness arises from the boundary conditions.
4) For some cases you can get a unique converged solution in this case. But in many cases, you will get a poorly converged solution that lacks basic plausibility, e.g. solution not symmetrical for the flow around a cylinder- Depends on the flow, the modeling and the solver settings.
5) in the case of a poorly converged steady-state solution: yes. Otherwise rather not.
__________________
Please do not send me CFD-related questions via PM

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post neetu kumari OpenFOAM 4 June 14, 2016 18:16 Attesz CFX 7 January 5, 2013 04:32 Lisa Main CFD Forum 11 July 5, 2000 14:37 J Roued Main CFD Forum 13 August 31, 1999 13:32 Jens Thurso Main CFD Forum 1 May 4, 1999 09:50

All times are GMT -4. The time now is 06:56.