# shadding vortex

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 February 9, 2011, 12:48 shadding vortex #1 New Member   Join Date: Feb 2011 Posts: 9 Rep Power: 5 Hi! i'm a cfd newbie and i'm trying to solve a tipical cfd problem: catching shedding vortex (shown in the figure below) generated by a steady flow around a circle (2D) with Fluent i'm trying to solve it with a k-eps model (unsteady analisys). for the mesh i use a boundary layer around the circle and a triangular growing size mesh all over, (more thick in the tail) is this correct? and about the wall condition, do i have to use one? thank you very much!

 February 10, 2011, 03:01 #2 Super Moderator     Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 2,720 Rep Power: 27 here one example for laminar flow: Sans titre.png __________________ In memory of my friend Hervé: CFD engineer & freerider

 February 11, 2011, 06:51 #3 New Member   Join Date: Feb 2011 Posts: 9 Rep Power: 5 thanks, but is not i was looking for.... now i undertstand that i can obtain the phenomenou by chosing a different viscous turbolence model according to the reynolds number... is it correct? thanks

February 11, 2011, 07:05
#4
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Maxime Perelli
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Quote:
 Originally Posted by Lombo and about the wall condition, do i have to use one?
I answered to this question

Quote:
 Originally Posted by Lombo now i undertstand that i can obtain the phenomenou by chosing a different viscous turbolence model according to the reynolds number... is it correct?
No the phenomenon doesn't depend on the turbulence model, but on the Reynolds number
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In memory of my friend Hervé: CFD engineer & freerider

 February 11, 2011, 11:56 #5 New Member   Join Date: Feb 2011 Posts: 9 Rep Power: 5 yes! thank you very much! i resolved the case with k-eps model, and now i'm trying to change their values (k and eps) to understand how they affect the phenomenou... but i don't really understand... moreover i would know how to choose this parameters

 February 11, 2011, 12:25 #6 Super Moderator     Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 2,720 Rep Power: 27 k and eps are fixed (you impose a specific turbulent flowfield given by Re >> check litterature) So you are not supposed to change their values as you are not supposed to change the inlet velocity (for a given Re) __________________ In memory of my friend Hervé: CFD engineer & freerider

 February 11, 2011, 12:51 #7 New Member   Join Date: Feb 2011 Posts: 9 Rep Power: 5 i use fluent and with this software i have to specify the velocity at the inlet, k and eps (how choose them?)... i calculate the reynolds manually..

 February 14, 2011, 02:31 #8 Super Moderator     Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 2,720 Rep Power: 27 there are formulas for k and eps. Check online help chapter 7.2.2 "Determining Turbulence Parameters" __________________ In memory of my friend Hervé: CFD engineer & freerider

 February 15, 2011, 11:42 #9 New Member   Join Date: Feb 2011 Posts: 9 Rep Power: 5 what is the different between the steady and unsteady command? i tried to solve a simple 2-d problem of a beam section: with the steady solution a k-eps viscous model sometimes i cath the oscillation of the lift coefficient, but with the unsteady k-eps i don't reach the same results... i don't understand why this happens, because in general the flow should be unsteady....

February 16, 2011, 03:02
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Maxime Perelli
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Quote:
 Originally Posted by Lombo with the steady solution a k-eps viscous model sometimes i cath the oscillation of the lift coefficient, but with the unsteady k-eps i don't reach the same results...
If you get oscillations with a steady solver, then the oscillations are due to the convergence. Understand you solution is moving untill it reaches its converged value (if your model has a steady solution)
In unsteady solver, the oscillations are unsteady oscillations with respect of time (and not iterations as in steady solver)
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 March 3, 2011, 04:52 #11 Senior Member   Ugly Kid Joe Join Date: Aug 2010 Posts: 194 Rep Power: 5 Why are you trying a steady state solution ??????? Flow over a cylinder is not a steady state problem............ It is a transient case problem.

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