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July 19, 2013, 04:22 
steady for unsteady flow?

#1 
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Shenren Xu
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Location: London, U.K.
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Hello,
I have a quite fundamental question about using steady flow solver on (physically) unsteady flow. Let's take the example of laminar flow around cylinder, and I want to find the bifurcation point, i.e., the Reynold's number above which flow will be unsteady with periodic vortex shedding. I run the steady flow solver with gradually increasing Reynold's number until the steady flow solver does not fully converge any more, presumably it goes to limit cycle oscillation. My question is, how much would the demarcation point I would find through this numerical example reflect the physical phenomenon. And how would using an explicit or implicit steady solver affect the demarcation point I find? And finally, my numerical experience tells me that demarcation point, i.e., the critical Reynold's number using implicit solver is higher than the explicit solver, why is there a difference in the demarcation point I find using explicit and implicit solvers? Thanks in advance for any shared thoughts on this. Cheers, Shenren 

July 19, 2013, 04:27 

#2  
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Filippo Maria Denaro
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Quote:
I would consider a different stategy: use always timemarching scheme, changing the Re number you will find that the steady solution is no longer reached. Implicit or explicit schemes of the same accuracy order must reach the same solution, provided that you are using the same timestep. Some difference can be due to the fact that implicit schemes require some particular BC at the new update and this task can introduce some approximation that does not appear for explicit scheme 

July 19, 2013, 04:34 

#3  
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Shenren Xu
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Hi Filippo.
I think what I did is the same as what you proposed. I used a (pseudo) timemarching scheme and change the Re number until it does not converge. My explicit solver uses RungeKutta+Multigrid+blockJacobi preconditioner, so the timeaccuracy is not preserved. By "timemarching scheme", are you actually suggesting to use a unsteady solver with physical time step? As for implicit solver, for me, it's just a matrix preconditioner on the LHS, should it not affect the steady solution that's reached, if ever? Please let me know if I didn't explain clearly. Shenren Quote:


July 19, 2013, 04:53 

#4 
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Filippo Maria Denaro
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yes, I mean a timemarching integration using a physical timestep.
Are you sure that in your implicit method the BCs for Un+1 is never required? However, implicit and explicit integration of the same accuracy, using the same time step, have the same magnitude order in the LTE, differences appearing in the sign of the coefficients... 

July 19, 2013, 05:06 

#5  
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Shenren Xu
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Okay, then it's a different approach, and essentially you are saying I should use an unsteady solver. Maybe there is a fundamental problem with my approach of using only steady solver, I will think about it carefully.
I'm not using timedependent boundary condition, so no, the BCs for Un+1 is never required. And by 'LTE', you mean leading truncation error? Quote:


July 19, 2013, 05:17 

#6 
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Filippo Maria Denaro
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yes, LTE = local truncation error
Note that when the solution becomes unsteady at some Re number, you have to prescribe a suitable BC Un+1 at the outlet, otherwise you somehow will force the steady condition (due to elliptical part of the solution). 

July 19, 2013, 05:26 

#7 
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Shenren Xu
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Thanks. I see. This approach probably will give me a better estimate
of the ctirical Re number. Then again, why using the explicit solver and the Re number found by checking whether it converges or not does not give me a satisfying answer? I'm more interested in this more 'theoretical' problem than the more practical problem of predicting the critical Re #. Any thought on that? :P 

July 19, 2013, 05:32 

#8  
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Filippo Maria Denaro
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Quote:
Many issues can be involved...it is difficult for me thinking about without checking details ... also the chosen tolerance for the residuals can cause differences... 

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