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Simulating an unsteady phenomenon with a steady model 

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April 13, 2016, 07:12 
Simulating an unsteady phenomenon with a steady model

#1 
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Jochem Grietens
Join Date: Nov 2013
Location: Belgium
Posts: 16
Rep Power: 12 
Dear colleagues,
I have been struggling with a problem lately and am hoping you can help me. We have made a model of a (turbulent) flow and we are not able to obtain the results we would expect. We have used a wepsilon steady state model to capture a complex flow. Basically, any small change to the model,mesh or solver presents us with way different flow results. I have been wondering if this might be a case of trying to capture and unsteady flow field with a steady solution. Starting from this hunch i have some related questions: 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 ... Any help would be very helpful. I am planning on delving into the mathematics of CFD myself but unfortunately i need some answers on short notice so any help would be of great help. Thanks 

April 13, 2016, 08:14 

#2  
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Filippo Maria Denaro
Join Date: Jul 2010
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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 timedepende 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, 12:51 

#3  
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Alex
Join Date: Jun 2012
Location: Germany
Posts: 3,384
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Quote:
2) Using solutions from a steadystate solver as initial conditions for a timedependent 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 timedependent 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 steadystate solution: yes. Otherwise rather not. 

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