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May 30, 2003, 03:43 |
Symmetry
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#1 |
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Hi everybody!
Please can you give me a answer to my question: I have a 3D-Volume with a velocity inlet (calculated by given massflow) Now I split the volume and use only a quarter to do calculations. I also use symmetry-BC Are the calculations done for the hole volume or only for a quarter? With best regards Tobi |
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May 30, 2003, 08:13 |
Re: Symmetry
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#2 |
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Hi!
If you use the simmetry bc it means that conditions are expected to be the same on the other side of bc (in stedy case). So calculations can be made only for that quater of volume - this speeds up he calculation. The flow parameters (velocity, pressure field...) in the other 3/4 of the volume are simmetric to the calculated ones. regards MATEUS |
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May 30, 2003, 09:37 |
Re: Symmetry
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#3 |
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What about non-steady case?
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May 30, 2003, 10:00 |
Re: Symmetry
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#4 |
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The calculation is done only on 1/4 of the volume.
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May 30, 2003, 10:08 |
Re: Symmetry
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#5 |
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If the flow you're modeling has unsteady characteristics, like gas flow in a liquid-gas column or granular solid flow in a gas-solid flow, you shouldn't use symmetry condition.
You can apply symmetry condition for example to a flow of water in a straight pipe, even if you do an unsteady calculation, because you expect the flow behaviour to be symmetrical. If you impose a symmetry condition to a flow with unsteady behaviour you risk to obtain non-physical results. |
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May 30, 2003, 10:56 |
Re: Symmetry
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#6 |
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But it is correct to use symmetry to model a free-slip wall or free surface (that does not change shape)even in highly transient flows?
I want to use symmetry to model a wall that has free-slip. In this way I want to get rid of a large portion of domain that is far from the region of interest, and I believe the flow beyond the placement of the symmetry condition has little effect on the remainder of flow, even though the flow is highly time dependent everywhere. (The velocity normal to the placement of the free-slip wall is approximately 0, even in the true flow situation) |
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May 30, 2003, 16:14 |
Re: Symmetry
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#7 |
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I think you can use symmetry.
If I'm not wrong, your flow is transient but also symmetrical, so if you use symmetry, you don't impose an unreal condition. Hi ap |
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May 30, 2003, 17:17 |
Re: Symmetry
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#8 |
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Hi
From experiments, I have ever seen any periodical flow that presents symetry. I guess you could made this asumption. However, you should be aware of the reality! Regards Alex Munoz |
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June 4, 2003, 09:43 |
Re: Symmetry
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#9 |
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maybe symmetry show 1/2, but not show pipe or cube
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