[J.J.Jameson]

Hello,

This is my first post, so I'm sorry if I'm doing something wrong! I have searched through the forum, but there was no answer given.

I am having trouble with Star-CCM's function to convert a mesh to 2D. I am modelling an contraction in a pipe. The model is effectively a rotated solid with the axis of revolution being the x-axis.

When I choose to convert the generated mesh to 2D, I get an error message saying "there are no boundaries in the z=0 plane". Considering the z=0 plane runs through the centre of the pipe, I do not understand why I'm getting this error message. Any ideas?

[azt]

hi

you probably do not have your geometry on the xy plane at z=0.
Put the centre at 0,0,0 and create a block (say) from 0,0,-1 to 0,0,1 and 0,1,-1 to 0,1,1. Extrude this 'sketch' in the x direction to create a block, say 20 long. Mesh this block then use the command create 2d grid from the drop down menu. Once you have your 2d planer grid, create cyclic interface along the line at y=0.

You should end up with a 2d plane with corner points 0,0 0,1 20,1 20,0 this is a pipe of radius 1 and 20 m long.

When creating the 3d block keep the +/- z size approximately the same as the +x dimension. This will aid meshing, but is a real pain if your geometry is large.

azt

[J.J.Jameson]

Thank you for your answer, but I am certain I have geometry at the plane. I am attaching an image of one of the meshes with the z=0 plane highlighted.
http://imgur.com/CeN6E3q

[cwl]

You should half your model so - that its flat part would lie in z=0 plane.

[DaveyBaby]

Hi,
I have an idea, not sure about it but here goes:
As you are solving in 2D, maybe you don't need to generate a cylinder at all. You could make a 2d sketch of half of the central plane section and extrude it normal to the sketch plane. You could then convert the mesh to 2d in the same way as the tutorial:
Tutorial Guide > Incompressible Flow > Steady Flow: Backward Facing Step > Setting Up the Backward Facing Step Case > Generating the Volume Mesh and Converting From 3D to 2D
It will look exactly the same as if you converted it from a cylinder by "un-revolving". More to the point, I think the 2d physics and the way Star solves will be equivalent. This last point is the part I'm not 100% on, but it seems reasonable to me.
Given how for 0 degrees of revolution the NS equations for a pipe become the 2d ones for a rectangle, and for a cuboid with 0 distance extrusion they also become the same? The question remains if for Star's 2D (one cell thick) there is something additional about the boundary conditions or something else that you need to consider.
Perhaps someone else could help with that?

[DaveyBaby]

Actually, I just saw that azt suggested this! Maybe you and I both skipped over it as he started with saying that you may not be on the central plane...