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April 28, 2007, 05:12 
Symmetry breaking : Vortex Shedding

#1 
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Hi,
For the flow past a Cylinder at Low Reynolds numbers say around 100, Ive observed the standard periodic asymmetric vortex shedding after a particular time (tc). For T < tc, the pair of vortices are symmetric and stable. My question is, what breaks the symmetry after T=tc ? If one runs the simulation for the entire run, one would notice that one of the vortices (either the top or the bottom one) starts to stretch and finally induce asymmetry.. Any one has insight why this happens to only one vortex and not both initially ? Regs, Dominic 

May 1, 2007, 02:43 
Re: Symmetry breaking : Vortex Shedding

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The symmetry breaks due to perturbations in your simulation, caused by machine precision issues as well as algorithmic issues (implementation and/or asymmetries that may be inherent in the algo itself). If a flow is fundamentally stable these perturbations don't lead to asymmetric flow. However, if the flow is in an unstable regime, then the perturbations will eventually lead to what you observe in your simulations. Given the above explanation, it is easy to understand why one vortex may stretch longer than the other  as a minimum, check, for example, just the grid data to verify that indeed the grid points are not _exactly_ symmetric with respect to the axis of symmetry (check the last few digits of the coord values)
Adrin 

May 1, 2007, 03:48 
Re: Symmetry breaking : Vortex Shedding

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Thanks for your comments. I do agree on the numerical part. But at this Re = 100, experiments also do show vortex shedding and my question is in this fundamental aspect.. not necessary from the numerical point of view. Vortex shedding is a result of symmetry breaking as observed by many people. But why such a symmetry breaking ? I read a couple of papers that state that a hopf bifurcation takes place. My question is more fundamental.. What are the origins of this physical perturbation ?


May 1, 2007, 08:28 
Re: Symmetry breaking : Vortex Shedding

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"Hopf bifurcation"
That's exactly the point. As Adrin points out, the symmetric solution (which is stable at a Reynolds number of 40) is unstable at Reynolds numbers of more than 50. That's the case in reality/experiment, and with any flow model able to predict the instability (obviously including yours). However, even an unstable solution might theroretically persist if there were no perturbation to trigger the instability. For example, carefully designed experiments have resulted in laminar pipe flow up to Reynolds numbers of 40000, although such flows are unstable above Re=2300 (laminar to turbulent transition is also a stability problem, like your vortex shedding phenomenon). But that's like trying to make a pencil stand on its tip... virtually impossible. To increase your understanding, you first need to make a distinction between the instability and a perturbation. It's not the perturbation that causes the instability. Perturbations only serve as triggers to let a prevalent instability come to play. Your papers on bifurcation will give you insight on the origins of the instability. The question of stability, in a nutshell, is this: If you take the symmetric solution and you slightly displace one of the vortices and let it go, will it jump back to the original position (stable), or will it drift away (unstable)? The behavior depends on Reynolds number. In your case, the asymmetry (and eventually shedding) is *caused* by the instability of the symmetric solution, and *initiated* by numerical perturbations (any roundoff error will do, even on a perfect grid). 

May 1, 2007, 14:19 
Re: Symmetry breaking : Vortex Shedding

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Thanks Mani, Just to add more: Does it mean that all numerical solutions will result in this _numerical_initiated asymmetry that will eventually lead to vortex shedding ?


May 1, 2007, 16:16 
Re: Symmetry breaking : Vortex Shedding

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Yes, any timeaccurate numerical algorithm that correctly predicts the flow instability can show this behavior. That should include any CFD scheme that yields discretized equations consistent with the unsteady NS equations. Of course, if a solution is very diffusive (because of the scheme, coarse grid, insufficient temporal resolution...) the instability can be artficially suppressed or delayed to higher Reynolds numbers.


May 1, 2007, 16:17 
Re: Symmetry breaking : Vortex Shedding

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Any numerical solution that is _not_ highly diffusive will do. If you have a highly diffusive scheme and/or use a coarse grid you lower the effective Reynolds number potentially below the critical Reynolds number and end up having a symmetric solution. Try a Reynolds number that is slightly above the critical value, say 55 in your case, using a coarse grid and see what happens. If you have a perfect grid and wellimplemented code, the solution should either remain symmetric forever or at least delayed to a very long time...
Conside the numerical perturbations (caused by any "imperfections") as the analogue to perturbations in your "imperfect" experiments, as Mani pointed out. Adrin 

October 31, 2016, 21:26 
Implicit VS Explicit Solver effect on Vortex Shedding

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Nima Jedari Attari
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Hi
Does using Implicit Solver have effect on not seeing numerical asymmetry and so not seeing vortex shedding? because in an Explicit solver, we must scan the grid from top to down (or down to top) and this makes some kind of asymmetry but in an Implicit Solver, we solve all the equation in whole of the grid, at the same time (in each time step or computational step).this means that : for example top of the grid has no privilege over the bottom of the grid or viceversa 

October 31, 2016, 21:33 

#9 
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A properly coded implicit solver will also capture vortex shedding.


November 1, 2016, 04:47 

#10 
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Filippo Maria Denaro
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I agree, no matter implicit or explicit time integration is used. However, I suggest to use a small time step if you want to describe properly the shedding.


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