[k13113y]

Hello all,

I'm trying to run a multiphase simulation in a cube which has air on the top half and water on the bottom half.

When generating my mesh with blockMesh, how would I make the grading more fine around the interface (about half way up the vertical axis)?

[Antimony]

Hi,

If you are using upwards of OF2.4, you can do multigrading on your blockMesh to get your desired mesh.

Cheers,
Antimony

[akidess]

topoSet with fieldToCell between e.g. 0.1 and 0.9, then refineMesh on the resulting cellSet.

[k13113y]

Awesome thanks for getting back to me; I'd like to try both techniques.

Originally Posted by Antimony

Hi,

If you are using upwards of OF2.4, you can do multigrading on your blockMesh to get your desired mesh.

Cheers,
Antimony

For multigrading, I believe I would do something like this no?

simpleGrading
(
1
1
(a b c) (d e f) (g h i)
);

but I'm not sure what a-i should be because I don't understand the algorithm; for a domain with a dense middle in the z-dir, what do you think?

Originally Posted by akidess

topoSet with fieldToCell between e.g. 0.1 and 0.9, then refineMesh on the resulting cellSet.

This I'm actually going to try right now but I've never used refineMesh before; is it an action that goes in the topoSetDict?

[Antimony]

Hi,

Yup, you are right:

simpleGrading
(
1
1
(a b c) (d e f) (g h i)
);

This is perhaps best explained with an example. Assume that in z-direction your domain goes from say 0 to 100 (z-direction defined as pointing in direction of increasing z-coordinate) . For the first 20, let us say you want 20 cells with a size ratio of 0.25 (end cell is 1/4th start cell size). For the next 50, let us say you want 100 cells, all uniform. And for the remaining 30, let us say you go for 10 cells with size ratio of 4 (end cell is 4 times start cell).

So the simpleGrading now becomes:

simpleGrading
( 
   1
   1
   (
      (20 20 0.25)
      (50 100 1)
      (30 10 4)
   )
)

Personally I prefer to work in absolutes rather than in fractions. I prefer to use a (20), d (50) and g (30) as lengths that sum up to the total extent in that dimension (100). Also, I sum up the number of cells b (20), e (100), h (10) and put it in the number of cells (130) in the definition of the block: hex (....) (nx ny 130)

Hope this clarifies.

Cheers,
Antimony

[k13113y]

Ah its so simple now

Many thanks Antimony. If I can help you in the future, though unlikely, don't hesitate to ask.

[sandeep.pandey]

Hello all,

Being naive to OF, I am asking this question which you may find simple!

I have a small tube, and to create mesh, I used blockMesh with arcs with 5 hex as shown. Now I want to find mesh resolution in 'r' and 'theta' direction (as in z direction it will be same as defined in blockMesh). Is there any utility in ParaView or in OF? Also, any hint to calculate delta(r), delta(r.theta)?

[Antimony]

Hi,

When you specify the arcs, you end up specifying the mesh resolution in the radial and circumferential direction, no?

The x1 & x2 directions of the blockMesh are along the first and second edges used to define the blockMesh. So if your first edge is an arc, then the number of divisions would be equivalent to that of the circumferential direction. And if your second edge is one that connects the two arcs, then it would be equivalent to specifying the number of divisions along the radial direction. No?

Cheers,
Antimony

[sandeep.pandey]

Hi Antimony!

Thanks for quick reply, and sorry to ping here again.
In the blockMesh which I've received, they defined first 'arc' then defined the in 'hex' with number of division (as shown in attached image, just look at the lower five 'hex'). If I understand you correctly, resolution in r is 180, theta is ?


Thanks
Sandeep

[Antimony]

Hi,

Look at the block defined by hex (13 14 10 9 .... )

And look at how the edges 13-14, 14-10, 10-9 and 9-13 are defined. 13-14 and 9-10 are arcs an thus for that block they can be seen as the "circumferential" divisions.

14-10 and 13-9, nothing has been defined. So they are taken as straight arcs and thus can be seen as "radial" divisions.

Hope this clarifies at least some of it.

Cheers,
Antimony

[sandeep.pandey]

Originally Posted by Antimony

Hi,

Look at the block defined by hex (13 14 10 9 .... )

And look at how the edges 13-14, 14-10, 10-9 and 9-13 are defined. 13-14 and 9-10 are arcs an thus for that block they can be seen as the "circumferential" divisions.

14-10 and 13-9, nothing has been defined. So they are taken as straight arcs and thus can be seen as "radial" divisions.

Hope this clarifies at least some of it.

Cheers,
Antimony

Hey,

Thank you very much for the reply . I missed the latest reply from you because mail directly landed into spam . I figured it out with you previous message.

Thanks again
Sandeep