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k-omega-SST: what happens when nut is set to 'calculated' at the wall for y+<1 |
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May 11, 2020, 17:09 |
k-omega-SST: what happens when nut is set to 'calculated' at the wall for y+<1
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Join Date: Feb 2020
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Hello everyone,
I'm trying to model a heat transfer problem and therefore want to pay special attention to my boundary layer treatment. The plan is to resolve the boundary layer with prisms so that y+<1 and therefore not rely on wall functions. I want to use the k-omega SST turbulence model as the literature shows that it performs well in my problem and because it's capable of fully resolving the boundary layer. My original plan was to set the following BC at the wall: k: kLowReWallFunction omega: omegaWallFunction nut: nutLowReWallFunctions However, I am quite unsure about the BC for nut. From the OF BC Guide, I know that nutLowReWallFunction simply sets nut to zero. Intuitively, this makes sense to me, as there are no turbulent fluctuations at the wall and therefore zero turbulent viscosity. Please correct me though if I'm wrong here. A colleague of mine told me that he usually uses 'calculated' for the nut wall BC. I don't really understand the point behind this though: if this BC is used, how exactly is nut calculated and how is this result going to look like? Is it simply calculated as ? (Equation for nut fromk-omega-SST model) If yes, it looks like it does go to zero because the turbulent kinetic energy k should be zero at the wall (at least according to my intuition). I hope my question is somewhat clear, it would be really cool if someone could help me out there. Thanks! |
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