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March 30, 2020, 01:07 |
Dividing Objects Symmetry
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#1 |
New Member
Babal Cobelt
Join Date: Dec 2019
Posts: 6
Rep Power: 6 |
Hey,
I want to simulate the flow of air with the heat transfer process on the square pipe with a very small size (side width of 3 mm). Because of its small size, I can only make it with a small amount of mesh so the results are not accurate. To get more accurate results and shorten the simulation time, I divided the square pipe into 4 parts of symmetry and performed a simulation on one of these parts. I have tried it but the results are not like a part of a 1/4 square pipe, but like a part of a complete pipe but with the size of 1/4 its original size. How do you make the 1/4 part of the square pipe really represent the state of the real 1/4 part of the whole square pipe? |
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March 30, 2020, 02:07 |
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#2 |
New Member
ZT
Join Date: Nov 2019
Posts: 14
Rep Power: 7 |
You will need to put symmetries on the two sides that would be on the inside of the pipe. For example, if you wanted to model the top-right 1/4 of the pipe, the top and right bc must be a wall, however the bottom and left bc must be a symmetry.
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March 30, 2020, 03:37 |
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#3 |
New Member
Babal Cobelt
Join Date: Dec 2019
Posts: 6
Rep Power: 6 |
Thank you for your response
In this case I have tried it with the bottom-left 1/4 of the pipe, the bottom and left b.c. are the walls, and the top and right b.c. are symmetry with adiabatic condition. But the result remains like a whole square pipe with 1/4 size of the original, while the process in it does not describe its condition which is 1/4 part. For example: the air flow in 1/4 of the pipe is centered at the blue dot like the following picture, which should be in the top right corner like the original. Last edited by babalcobelt; March 30, 2020 at 04:53. |
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March 30, 2020, 04:51 |
Symmetry
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#4 |
Senior Member
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If you use symmetry, then you won't get an option to apply adiabatic condition or any condition for that matter. Symmetry automatically implies adiabatic condition. So, in case after applying symmetry you still get an option for thermal condition where you can apply adiabatic, then that is not symmetry condition. Other option is to use Wall with Specified Shear of 0 and adiabatic. As long as there is no field other than flow, turbulence, and thermal, this would behave like symmetry.
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March 30, 2020, 05:23 |
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#5 |
New Member
Babal Cobelt
Join Date: Dec 2019
Posts: 6
Rep Power: 6 |
Got it
Thank you |
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Tags |
heat-transfer, quarter-section, square-channel, square-pipe |
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