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Old   May 9, 2012, 07:33
Default symmetry plane error
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Dear all,
I am trying to model a free surface problem by replacing the free surface by a symmetry boundary condition. However, when I start the solver I am receiving a very strange error which says:

"When buoyancy is active, all symmetry plane normals must be perpendicular to the gravity vector"

I read some papers where the authors have used this type of boundary and the symmetry planes normals were not perpendicular to the gravity vector...so itīs quite strange for me...Does anyone has experienced the same problem?
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Old   May 9, 2012, 18:43
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Have a think about it - your geometry is not symmetric if gravity is not parallel to the symmetry plane. I have no idea what the papers you quote are talking about as the error message is correct.
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Old   May 9, 2012, 21:12
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Erik
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Some CFD programs (Star CCM for one) allow you to use a symmetry boundary condition on any surface not perpendicular to the gravity vector; CFX does not, but you can apply a free slip wall to get the same effect.
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Old   May 9, 2012, 21:14
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Sure, but it is not physically possible so I do not know why you would want to do it. But as you say, using a slip wall might be an acceptable option.
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Old   May 10, 2012, 05:16
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Dear all,
the problem is not the error message itself (which by the way is very clear), the problem is the fact that some points are not clear for me. Hence I would be very appreciated if someone could highlight me these 2 points:
1) In these 2 papers (the authors have used cfx) gravity is not parallel to the symmetry plane and they all say they have used a symmetry boundary condition to replace the free surface:
"Particle Image Velocimetry Measurements and Numerical Modeling of a Saline Density Current"
"Control of Turbidity Currents in Reservoirs by Solid and Permeable Obstacles"
According to Glennīs answer this is not possible, so i must conclude that they didnīt use it (which is quite surprising to me...)
2) I read somewhere that a typical way of simulating the free surface is making use of the rig lid assumption, which in the case of fluent/cfx is the same of using a free slip or symmetry boundary condition, which is the same of saying that the normal flux and the gradients of all variables are specified to be zero. So, why saying that " but it is not physically possible"?
Sorry if insist in these topics...my ideia is just to promote a healthy conversation between all.
Regards.
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Old   May 10, 2012, 06:40
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No problem, your questions are a good contribution to the forum.

What you describe is to use a symmetry plane to simply reproduce a slip wall. The flow has no symmetry, but as symmetry and slip walls aer mathematically similar they can be interchangeable.
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