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January 5, 2014, 09:44 |
SStkOmega Initial Condition problem??
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#1 |
Senior Member
ArathoN
Join Date: Jul 2011
Posts: 137
Rep Power: 16 |
Hi I really hope some kind soul will help me After 2 months i couldn't find the solution by my self and i decided to ask the community (Hoping that this time some one will answer me).
I'm studying the Backward facing step and i'm using the Makiola Data to validate the code. Here the checkMesh output: Code:
Checking geometry... Overall domain bounding box (-1 -0.1 -0.0600333) (5 0.1 0.0600333) Mesh (non-empty, non-wedge) directions (1 1 0) Mesh (non-empty) directions (1 1 0) All edges aligned with or perpendicular to non-empty directions. Boundary openness (-9.79501e-18 6.68689e-16 9.46468e-20) OK. Max cell openness = 2.42701e-16 OK. Max aspect ratio = 180.003 OK. Minimum face area = 5.27601e-07. Maximum face area = 0.00749943. Face area magnitudes OK. Min volume = 6.33473e-08. Max volume = 1.33007e-05. Total volume = 0.130418. Cell volumes OK. Mesh non-orthogonality Max: 20.1903 average: 6.51521 Non-orthogonality check OK. Face pyramids OK. Max skewness = 0.444594 OK. Coupled point location match (average 0) OK. Code:
Reynold numbers: 15000, 64000 Aspect Ratio:2 Hinlet=Hstep=0.1 m I%= sqrt(0.1) * 100 (as in the makiola paper) nu=1.45e-5 U(Re1)=2.175 U(Re2)=9.28 l (mixing-length)= 10% Hstep K, Omega calculated with the formulas in the wiki (the ones with the mixing length) Omegawall= 60*nu/[ deltay^2 * Cmu] (i found this relation in the wilcox) To solve i searched a lot in the forum and changed in every aspect the boundary condition and also the schemes with no effect, i tried using the KEpsilon and it gives as expected a lesser value. Here some BC setting i've chosen for Re=15000: Code:
Omega Inlet: fixedValue 153.84 Outlet: zeroGradient upperWall: fixedValue 9700 lowerWall: fixedValue 9700 K Inlet: fixedValue 0.71 Outlet: zeroGradient upperWall: fixedValue 1e-16 lowerWall: fixedValue 1e-16 Code:
Omega Inlet: fixedValue 153.84 Outlet: zeroGradient upperWall: omegaWallFunction lowerWall: omegaWallFunction K Inlet: fixedValue 0.71 Outlet: zeroGradient upperWall: kqRWallFunction lowerWall: kqRWallFunction If it is needed i can also upload my scheme file and everything that is needed. P.S: I used the yPlusRAS to check the "yPlus" value and It is always less than the unit for the case Re=15000 Finally sorry for my bad English, i tried to write as correctly as i could. Damn i didn't notice the mistake in the title sorry for that. Last edited by ArathoN; January 7, 2014 at 21:02. Reason: some problemi were solved |
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January 7, 2014, 20:58 |
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#2 |
Senior Member
ArathoN
Join Date: Jul 2011
Posts: 137
Rep Power: 16 |
Please someone help me!!!! I changed again the mesh but i didn't get the Desired results, i even changed the geometry dimensions to the ones in the Makiola paper.
Right now I'm with no idea how to proceed. I noticed that if i use the zerogradient BC on the wall for the variables omega and k it will give some really strange results. If i plot the wallshearstress against the distance from the step the shearstress is always negative as if there is no recirculatinge zone. Does anyone know why? PS. The yplus is under 1 for both the walls. |
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January 9, 2014, 18:03 |
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#3 |
Senior Member
ArathoN
Join Date: Jul 2011
Posts: 137
Rep Power: 16 |
I've done more tests but with no success. I made even a fiber mesh but the results didn't change.
Now I have a doubt what if the turbolent length length was wrong? I considered it as 10% H, then I lowered it until 3% H. However the biggest turbolent scale length is is in the same order of the step height. I'll try this new setting then I'll comment hoping in some help meanwhile. |
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January 9, 2014, 21:16 |
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#4 |
Senior Member
ArathoN
Join Date: Jul 2011
Posts: 137
Rep Power: 16 |
Finally I have some good results, I choose there BC on the walls:
K-----> fixed value at 1e-16 Omega------->zerogradient Now I have a better reattachment length, xr/h=6.3 against the experimental value 5.8; at this point I decided to change the turbolent length scale and I considered it as 20℅ of Hstep giving me a xr/h=5.9 really close to the desired value. So I'll change again the turbulence length then I'll consider the other configurations; until now I focused one the 20° step; to validate my observation. P.S funny how this thread is turning into a monologue. I hope that my findings may help someone in the future :P |
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