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Search took 0.00 seconds. Search: Posts Made By: ghorrocks
Forum: CFX January 26, 2012, 19:31
Replies: 16
Views: 2,531
Posted By ghorrocks
Let me give you a tip on the meshing - you need...

Let me give you a tip on the meshing - you need to define the two bodies as a single part and the interface surfaces will get a contiguous mesh. Look up multi-body parts in the tutorials or doco.
...
Forum: CFX January 26, 2012, 19:09
Replies: 16
Views: 2,531
Posted By ghorrocks
Good progress. You will probably need a...

Good progress.

You will probably need a source term linearisation coefficient. Read the documentation about source terms for this, but if you have defined a constant source then I think that goes...
Forum: CFX January 26, 2012, 05:59
Replies: 16
Views: 2,531
Posted By ghorrocks
You need to define an interface to connect them...

You need to define an interface to connect them with your current mesh. It can be a bit tricky selecting the correct surfaces but if you hide and unhide the bodies you can do it.

But a better...
Forum: CFX January 23, 2012, 16:39
Replies: 16
Views: 2,531
Posted By ghorrocks
You need to decide what approach you are going to...

You need to decide what approach you are going to use. If you remove the region then you use a inlet/outlet pair to produce the flow. While this is simple it does make it hard to link to a fan...
Forum: CFX January 23, 2012, 05:42
Replies: 16
Views: 2,531
Posted By ghorrocks
A subdomain is a 3D region. A boundary is a 2D...

A subdomain is a 3D region. A boundary is a 2D surface on the outside of a 3D region.
Forum: CFX January 22, 2012, 17:56
Replies: 16
Views: 2,531
Posted By ghorrocks
If you like you can use the porous flow equations...

If you like you can use the porous flow equations in your region. But because your region is a turbine and not a porous region then I cannot imagine why you would want to. A general momentum source...
Forum: CFX January 22, 2012, 06:32
Replies: 16
Views: 2,531
Posted By ghorrocks
Use a momentum source term to do this. They are...

Use a momentum source term to do this. They are move general and you can apply a resistance to the flow in any form you like that way.

And the source term can be applied as either a volume or a...
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