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April 19, 2010, 05:02 |
SST and K-omega worst than laminar, why?
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#1 |
New Member
Join Date: Mar 2009
Posts: 4
Rep Power: 17 |
Hello,
I'm simulating a cavity heated on the bottom and filled of water in transient state (free convection, low Ra number apparently). I'm trying to validate my case with a PIV system. I used an sst and k-omega models (I used the same mesh with yplus<1) but my results with this two models are not so good. After, I tried without turbulent models (laminar with the same mesh) and the results are quite good. Some one can explain me why, because even if my case is laminar, the turbulent models must to give me the same results or close, no? I’m asking this, because for the duration that I study maybe I’ll pass to the transient turbulent zone, and which model I’ll use? thanks Last edited by Jervds; April 19, 2010 at 05:22. |
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April 19, 2010, 07:39 |
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#2 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 17,870
Rep Power: 144 |
You are not alone:
In my PhD thesis, written almost a decade ago (wow! time flies) I found that for internal combustion engine modelling, using no turbulence model in an LES type setup proved to be just as accurate, and in some respects more accurate than traditional turbulence models. http://hdl.handle.net/2100/248 If the flow is low Re/Ra number then the turbulence is not strong and you may well be better off using a laminar flow model. Generally this is because of the simplifications (ie some cross terms get ignored in turbulent flow models but are included in laminar) and additional dissipation of the turbulence model (which is what turbulence models do). Looks like you have found another example. |
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Tags |
k-omega, laminar, sst |
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