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July 30, 2010, 13:09 
Symmetry Issue

#1 
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Dylan Kowalewski
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I'm having a problem with the symmetry boundary condition on a combustion chamber in fluent. When I use it on my model, there appears to be a wall where the symmetry plane is (ie. two chambers instead of one). Any help would be appreciated. 

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August 2, 2010, 01:24 

#2 
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Maxime Perelli
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does the plane appear with symmetry setup in the BC list (in fluent)?
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August 3, 2010, 03:19 

#3 
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Dylan Kowalewski
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Yes, the symmetry plane appears in the BC menu, but it seems to just calculate as if the half was one container, then just reflects that result over the plane.


August 3, 2010, 04:14 

#4 
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Maxime Perelli
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that's the definition of symmetry.
If your geometry is symmetric, and your BC are also symmetric. Then you split your geometry along its symmetry and you compute on one half, because you know that the result will be symmetric. Purpous of use of symmetry is to save computationnal effort. But maybe I didn't understand your problem
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August 3, 2010, 05:58 

#5 
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Dylan Kowalewski
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Yes, I'm splitting it in half to save computations, but there's a wall where the plane of symmetry is, so instead of having 1 chamber, I have 2. Is there any way to get rid of that?


August 3, 2010, 06:17 

#6 
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Maxime Perelli
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if you split your domain along your symmetry, then you delete the volume you don't need.
The closed volume remaining is "closed" by the symmetry plane (which is not considered as a wall... since it appears in fluent BC as symmetry). Can you post a picture?
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August 3, 2010, 09:16 

#7 
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Dylan Kowalewski
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So here's a quick render I did of the setup. The line that goes straight down the centre is the plane of symmetry. There are three entrances for gases, side facing you and on the left and right sides. The pathlines are clearly following the plane of symmetry as if it were a wall.
Thanks 

August 3, 2010, 09:32 

#8 
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Maxime Perelli
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Per definition, on symmetry plane you shouldn't have velocity normal to the plane.
In other words: on the symmetry you have V.n = 0 where n is the normal vector. So on symmetry you have only coplanar components and no normal component. If the pathlines go through the plane in the middle then your flowfield isn't symmetric (that's why I didn't call it symmetry since it is not mathematically correct)
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August 3, 2010, 09:35 

#9 
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Dylan Kowalewski
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I see, so would this be reasonable to do if I did not have those inlets pointing towards the plane?
Thanks for the help! 

August 4, 2010, 01:18 

#10 
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Maxime Perelli
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Are your Boundary Conditions also symmetric?
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In memory of my friend Hervé: CFD engineer & freerider 

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