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[blockMesh] defining and connecting two circular pipes using blockMesh |
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#1 |
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Hale
Join Date: May 2013
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Hi,
I'm having trouble in defining two circular pipes that are connected with an angle of 90 degrees using blockMesh. Is there any tutorials or any example that shows how to define such a L-shaped circular pipe in openfoam? thanks alot |
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#2 |
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Yosmcer Mocktai
Join Date: Apr 2013
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Posts: 50
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You can make this with a block and some edges.
Here are two examples I just made. You can copy it and adapt to your needs: Code:
/*--------------------------------*- C++ -*----------------------------------*\ | ========= | | | \\ / F ield | OpenFOAM: The Open Source CFD Toolbox | | \\ / O peration | Version: 2.2.0 | | \\ / A nd | Web: www.OpenFOAM.org | | \\/ M anipulation | | \*---------------------------------------------------------------------------*/ FoamFile { version 2.0; format ascii; class dictionary; object blockMeshDict; } // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * // convertToMeters 1; vertices ( (0 1.292893219 0.707106781)//0 (0 2.707106781 0.707106781)//1 (0 2.707106781 -0.707106781)//2 (0 1.292893219 -0.707106781)//3 (2.707106781 0 0.707106781)//4 (1.292893219 0 0.707106781)//5 (1.292893219 0 -0.707106781)//6 (2.707106781 0 -0.707106781)//7 ); blocks ( hex (5 4 7 6 0 1 2 3) (50 50 50) simpleGrading (1 1 1) ); edges ( arc 0 1 (0 2 1) arc 1 2 (0 3 0) arc 2 3 (0 2 -1) arc 3 0 (0 1 0) arc 4 5 (2 0 1) arc 5 6 (1 0 0) arc 6 7 (2 0 -1) arc 7 4 (3 0 0) arc 0 5 (0.914213563 0.914213563 0.707106781) arc 3 6 (0.914213563 0.914213563 -0.707106781) arc 1 4 (1.914213562 1.914213562 0.707106781) arc 2 7 (1.914213562 1.914213562 -0.707106781) ); boundary ( ); mergePatchPairs ( ); // ************************************************** *********************** / Code:
/*--------------------------------*- C++ -*----------------------------------*\ | ========= | | | \\ / F ield | OpenFOAM: The Open Source CFD Toolbox | | \\ / O peration | Version: 2.2.0 | | \\ / A nd | Web: www.OpenFOAM.org | | \\/ M anipulation | | \*---------------------------------------------------------------------------*/ FoamFile { version 2.0; format ascii; class dictionary; object blockMeshDict; } // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * // convertToMeters 1; vertices ( (0 1 0)//0 (0 2 1)//1 (0 3 0)//2 (0 2 -1)//3 (3 0 0)//4 (2 0 1)//5 (1 0 0)//6 (2 0 -1)//7 ); blocks ( hex (5 4 7 6 1 2 3 0) (50 50 50) simpleGrading (1 1 1) ); edges ( arc 0 1 (0 1.292893219 0.707106781) arc 1 2 (0 2.707106781 0.707106781) arc 2 3 (0 2.707106781 -0.707106781) arc 3 0 (0 1.292893219 -0.707106781) arc 4 5 (2.707106781 0 0.707106781) arc 5 6 (1.292893219 0 0.707106781) arc 6 7 (1.292893219 0 -0.707106781) arc 7 4 (2.707106781 0 -0.707106781) arc 0 6 (0.707106781 0.707106781 0) arc 1 5 (1.414213562 1.414213562 1) arc 2 4 (2.121320344 2.121320344 0) arc 3 7 (1.414213562 1.414213562 -1) ); boundary ( ); mergePatchPairs ( ); // ************************************************** *********************** // Theses case are quite general with pipes of radius 1 and small curve radius to 1 too (you may need to not have this curve). Last edited by Yosmcer; July 30, 2013 at 05:48. |
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#3 |
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Thank you very much Yosmcer. It was a great help
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#4 |
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Yosmcer Mocktai
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I used the Pythagoras theorem with the simplified fact that the triangles were isosceles.
So basically, for a radius circle (hypotenuse) of 1, I have sqrt(2)/2. For more general shapes, you can use trigonometry. ![]() |
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#5 |
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Hale
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Thanks you very much for the explanation.
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