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No 3Dimensionality in Flow Past a Cylinder Using RANS in STAR-CCM+ |
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December 9, 2015, 17:34 |
No 3Dimensionality in Flow Past a Cylinder Using RANS in STAR-CCM+
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#1 |
New Member
A E
Join Date: Dec 2015
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Hi,
I am trying to simulate flow past a circular cylinder for Re = 4,000 . For now, I am using RANS models in Star-CCM+. I have noticed that any k-epsilon based models in Star-CCM+ (i.e. SKE, RKE, RSM) do not have any 3 dimensionality. Meaning, there is no amplitude modulations in forces when the phenomenon is model in a 3D configuration and the result of 3D and 2D simulations are almost identical. In general, the results of k-epsilon models( drag, lift, and Strouhal number) are closer to the experiment compared to k-omega, although the model shows no 3D effects and the values of velocities and vorticity do not change in depth. On the other hand, using 3D simulations with k-omega results in lower values of drag and lift and amplitude modulation is observed in results. I do not think it is due to mesh refinement as I have tested different mesh settings and sizes. CFL number is kept less than 1.0 through out the simulation and y+ values are also less than unity. Anyone has any idea why this happens? Any help is appreciated. Thanks, A E |
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December 10, 2015, 10:45 |
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#2 |
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robo
Join Date: May 2013
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Are the sidewalls of the domain symmetry or no-slip? If they're symmetry, then what you're simulating is an infinite cylinder, which should be the same as the 2D result. If no-slip, then that should introduce some spanwise variance, and if you're not seeing it, then I would check things like the y+ on the sidewall. A 2-eq model is also going to be isotropic, which I believe will work to suppress spanwise variation. You could try using an RSM model to see if the anisotropy helps. I'm also unclear if you're looking at steady state or transient. If transient, then LES might be the better formulation for the problem. In all cases, the Re is a little low; I'm not sure what the critical Re should be for this problem off the top of my head but I doubt you'd be far past transition if you've even hit it, and I hesitate to make turbulence model recommendations for flow around transition, as that is really not my area.
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December 10, 2015, 18:39 |
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#3 |
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A E
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For this range of Re, vortex shedding should be 3D and quantities vary along the cylinder length. I also check the RSM model and as it is based on k-epsilon, the 3D and 2D models' response are the same.
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December 10, 2015, 19:02 |
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#4 |
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robo
Join Date: May 2013
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I'm not sure what you checked, but RSM models are not based on k-epsilon. Some of them do have an epsilon equation, although it is usually formulated differently. If you're studying vortex shedding, then you're running transient, in which case I would recommend LES as it is better formulated for unsteady problems. Although again, at that low Reynold's number, I am not sure you'd even see much turbulence in the flow as I think you'd be close to transition.
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December 10, 2015, 19:19 |
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#5 |
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A E
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Thanks for your reply.
Yeah, I think the RMS model in Star-CCM+ has the epsilon equation. And as I said there is no difference between 2D and 3D simulations when I use RSM (like k-e and unlike k-w) I will eventually use DES for my simulations but I just like to know why k-epsilon does behave like this and k-w does not. |
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December 10, 2015, 19:39 |
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#6 |
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robo
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I can't say for certain, but again, I feel like the low Reynold's number is gonna be a problem. None of those turbulence models are formulated for Reynold's numbers that low, and I am not sure how they'd behave. 4000 may not even be turbulent for that flow. What happens when you run with no turbulence model?
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December 10, 2015, 19:42 |
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#7 |
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A E
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For this range the boundary layer is laminar and wake is turbulent. I ran a 2D simulation with no turbulence and the result was not good at all, way higher than k-e and k-w results.
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December 29, 2015, 12:31 |
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#8 |
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Reza
Join Date: Mar 2009
Location: Appleton, WI
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A while back I remember trying this. I was using another package back then, and I was using a structured mesh, so my grid was symmetric with respect to the centerline and also uniform along the cylinder, so I had trouble getting the vortex street. What I had to do was to use an asymmetric inlet velocity for the first few iterations, to disturb the symmetry for the first few time steps, and that did it. Something like:
($Time < 1) ? ($inletVel + $somefactor * ($$Position[0] + $$Position[1])) : $inletVel should do it. This means, for the first second, the inlet velocity is not symmetric and is a bit disturbed (you can control the amplitude with the value of $somefactor). Just remember, you need to run the simulation for a bit longer to make sure the results are not because of the imposed disruption but because of the inherent unsteadiness. |
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December 29, 2015, 13:32 |
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#9 |
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A E
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Thanks for your comments. I think it is only because of the type of turbulence model I am using. k-e needs a lot of modifications for near the wall region and low Re number and at the end of the day it is an isotropic model, so there should be no 3D ness in it. But, so is k-w; k-w also needs modifications for low Re # and it is isotropic.
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June 9, 2017, 10:03 |
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#10 |
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abubakar izhar
Join Date: Oct 2015
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As you have said that symmetry condition introduces two dimensionality the SHOULD We use periodic conditions on side Walls or can i use just Wall bc. What will be the effect of periodic size Walls, wouldnt it will be introduces two dimensionality effect in 3D Simulation.
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June 9, 2017, 20:59 |
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#11 |
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A E
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Thanks for your reply. It appeared that the reason lies behind using damping functions for k-e models. All models w/o damping functions seem to produce 3D ness + No turbulence model case
These damping functions seem to literally damp out any 3D structures in the wake. A E |
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Tags |
k-epsilon, k-omega, rans modelling, star-ccm+, vortex shedding |
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