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#1 |
New Member
Join Date: Jan 2018
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I have a question regarding turbulence kinetic energy at the wall. I simulated compressible air flow in a pipe in Ansys CFX 18.0 (steady state). In k-epsilon turbulence model, near wall flow is modelled with wall function, no matter how fine the mesh is. As far as I know, the value of turbulence kinetic energy k at the wall should be always equall to 0. Instead of this, I obtained the highest value, which is then slowly decreasing with the distance from the wall. Moreover, despite no slip boundary condition, velocity at the wall is suprisingly high (approximately 120 m/s). If I use k-w SST model with coarser mesh (y+=27), wall function is switched on and the same problem occurs. K-w SST with fine mesh in boundary layer (y+=0.5) provides good results, i.e. k=0 on the wall. Figure in attachment presents distribution of k vs distance from the wall (purple line: k-w SST fine mesh, green line: k-epsilon fine mesh, blue line: k-w SST coarse mesh, red line: k-epsilon coarse mesh). Even if I set in turbulence option k=0, the value on the wall remains the highest and then rapidly decreases to 0. I looked to the different publications and turbulence kinetic energy at the wall is always equal to 0 for k-epsilon. Does anybody know what is going on? Why use of wall function provides so unphysical behaviour?
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#2 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
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Looks normal.
The turbulent kinetic energy is only 0 exactly on a no-slip wall. For boundary layers, k reaches a maximum very close to the wall (somewhere from y+ ~30 or ~100). k reaches a maximum as you approach the wall and then is zero at the wall. Very close to the wall k varies linearly with distance. This is what makes wall functions so annoying to implement for k. You'll only be able to see k approach 0 if you have a super-fine mesh and can completely resolve it and it will never be zero because you are doing FVM and you don't have cells "on the wall." |
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#3 | |
Senior Member
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Quote:
Just note that CFX is a node based finite volume solver, which means that your variables are defined at the mesh nodes, which include boundary ones. Now, I'm not an expert of this approach but consider the following: a) For classical cell centered FV solvers, you don't have variables defined at the walls, only bc. Which means that the solver can do whatever it takes at the bc level to solve the equations properly, while still showing you "expected" node values at the postprocessing stage (still, this is not always correct). Those accustomed with Fluent should know such difference between flagging or not the node values checkbox in the postprocessing panel. b) Node based solvers can't do this anymore, because the values at the wall are now part of the solution and cannot be faked (unless a stupid marketing approach is used in the company producing the solver). Thus, considering a wall function approach in CFX, they are obliged to put some value in the wall node that would correctly represent the wall function case. Such value is not 0, otherwise you would not be using wall functions at all. Actually, as stated by LuckyTran, in CFX you can't interpret node values at walls as actual node values. They are more representative of the near wall volume. Thus, even at the postprocessing stage, it makes sense to see something different from 0 if your cell at the wall is coarse. EDIT: You may want to give a look at this page https://www.sharcnet.ca/Software/Ans...fxTurbTreaWall |
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#4 |
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Thank you LuckyTran and sbaffini for replies. If I understood corectlly, when wall function is used, first node of the mesh is moved to the end of viscous sublayer, thereby value displayed in postprocessor comes from that point and not from the wall itself, although boundary condition k=0 at the wall is preserved and used in calculation. Am I rigth?
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#5 |
Senior Member
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Quite the opposite actually
![]() The mesh is not moved, but the bc used at the wall for k is relative to a point away from the wall. It should actually be a Neumann condition dk/dn = 0, so that it's not 0 unless it is also in the fluid. |
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#6 |
New Member
Join Date: Jan 2018
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Ok, I got it, thank you for detailed explanation.
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