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March 6, 2017, 13:43 
kequation

#1 
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Here:
http://www.afs.enea.it/project/neptu...3.htm#eqrsmk it says that for the boundary conditions of the reynold stresses the kequation is solved globally and only the values at the boundaries are taken as input to the boundary conditions for the reynoldstresses. I'm wondering: Do I not need boundary conditions for this kequation at the wall again or are these velocityfluctuations/turbulence = zero since they vanish for the same reason as the "real" velocity (no slip)? Phyiscally is it because within the vicious sublayer any arising turbulence is immediately damped, because of the "strong" vicous role in it?? 

March 7, 2017, 03:09 

#2 
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Lucky Tran
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You still need boundary conditions for the k equation. That is why there are nearwall treatment options when you use RSM.
In equilibrium boundary layers yes. However, "any turbulence arising" implies that there can be turbulence (or at least production of k). Immediately damped means that dissipation is equal to the production rate. Hence, equilibrium boundary layers are ones where the production rate is equal to the dissipation rate. However, these two combined allows k to be finite near walls. 

March 7, 2017, 06:23 

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And what are the precise boundary conditions for k then? Do you have a link explaining it? 

March 7, 2017, 15:07 

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The boundary condition for k at walls should be the zerogradient type.


March 8, 2017, 06:04 

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why? Whats the reason for this? Last edited by Diger; March 8, 2017 at 09:14. 

March 9, 2017, 02:17 

#6 
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Lucky Tran
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See section 4.16.2.4 in the Fluent Theory Guide.
The zerogradient condition is simply, turbulence doesn't magically appear or disappear. It's like the adiabatic boundary condition for the energy (no spontaneous heat in or out). If you set k=0 or any number at the wall, that means the wall is a massive turbulent kinetic energy reservoir/sink. 

March 9, 2017, 06:08 

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