# k-equation

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 March 6, 2017, 13:43 k-equation #1 Senior Member   Join Date: Nov 2015 Posts: 109 Rep Power: 3 Here: http://www.afs.enea.it/project/neptu...3.htm#eq-rsm-k it says that for the boundary conditions of the reynold stresses the k-equation is solved globally and only the values at the boundaries are taken as input to the boundary conditions for the reynold-stresses. I'm wondering: Do I not need boundary conditions for this k-equation at the wall again or are these velocity-fluctuations/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
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Lucky Tran
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You still need boundary conditions for the k equation. That is why there are near-wall treatment options when you use RSM.

Quote:
 Originally Posted by Diger Phyiscally is it because within the vicious sublayer any arising turbulence is immediately damped, because of the "strong" vicous role in it??
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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Quote:
 Originally Posted by LuckyTran You still need boundary conditions for the k equation. That is why there are near-wall 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.

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

 March 7, 2017, 15:07 #4 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,854 Rep Power: 26 The boundary condition for k at walls should be the zero-gradient type.

March 8, 2017, 06:04
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Quote:
 Originally Posted by LuckyTran The boundary condition for k at walls should be the zero-gradient type.
pure neumann?
why? Whats the reason for this?

Last edited by Diger; March 8, 2017 at 09:14.

 March 9, 2017, 02:17 #6 Senior Member   Lucky Tran Join Date: Apr 2011 Location: Orlando, FL USA Posts: 1,854 Rep Power: 26 See section 4.16.2.4 in the Fluent Theory Guide. The zero-gradient 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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Quote:
 Originally Posted by LuckyTran See section 4.16.2.4 in the Fluent Theory Guide. The zero-gradient 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.
Ah yeah, I see...Like a diffusing species in a box which doesn't react with the walls and therefore doesn't get lost. (which is actually the same as the temperature thing)

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