Can CFX5 calculate porous media?
 

and porosity tensor? I didn't find such physical model in manual of it.Maybe it cannot.

Re: Can CFX5 calculate porous media?
 

The can be done using a momentum source term. There are fields for linear and quadratic isotropic resistances. If you need to create an anisotropic resistance, you can specify the appropriate momentum source for the specified direction.

For example, to add a linear resistance in the X direction, specify your X-momentum source by the following expression:

-Alpha*Speed

where: Alpha => linear resistance coefficient Speed = sqrt(u^2+v^2+w^2)

If you need an anisotropic quadratic resistance, you would enter the expression:

-Beta*Speed*u

where: Beta => quadratic resistance coefficient

Regards, Robin

Re: Can CFX5 calculate porous media?
 

That is momentum source (body force), not porous medium and porosity tensor. You see, they are different in CFX4.

Re: Can CFX5 calculate porous media?
 

Robin,

My experience is that this results in a very unstable situation because f.i. the momentum force for u depends on u itself. Any suggestions how to improve this?

Astrid

Re: Can CFX5 calculate porous media?
 

Hi Astrid,

If you have a non-linear term (i.e. u*u), you need to linearize it. Although there is no way to directly linearize a momentum source term in a single direction, there is a trick by which you can achieve this.

If you enter a quadratic resistance, the solver will automatically linearize it, but you will end up with a quadratic resistance in all directions. To make it anisotropic, negate the same source in the directions you do not want them.

For example, should you want a value beta as a quadratic resistance in the X direction, create an isotropic resistance of beta and negate the resistance term in the Y and Z with the following source terms:

X-mom = 0 [kg m^-1 s^-1] Y-mom = beta*Speed*v Z-mom = beta*Speed*w

What you are left with is a properly linearized quadratic resistance in the X direction.

Regards, Robin

Re: Can CFX5 calculate porous media?
 

Robin,

I am trying to represent a monolith by a porous medium. Over this monolith the pressure drop is 200 Pa. A simple calculation according to the linear representation dp/dx=c*u, results in c = 100. However, the resistance in the monolith is anisotropic. For my case this means that the flow in reality can only go in x-direction. Flow in y- and z- direction is impossible. Thus, for these directions I require a resistance of for example 1000 or (much) higher. So:linear resistance = 1000
x-momentum = (1000-100)*u
y-momentum = 0
z-momentum = 0



But no succes at all. The monolith is placed in a circulation zone. As a result, the u-velocity in the monolith is partially positive and partially negative.
This is sweeped up by the u momentum towards extreme high gas velocities.

Astrid



P.S. In CFX-4 it is child's play..........

Re: Can CFX5 calculate porous media?
 

use x-momentum = (1000-100)*speed

where

speed = sqrt(u^2+v^2+w^2)

"In CFX-4 it is child's play.........."

Yes, I agree this is painful. Perhaps in a later release it will be made better.

Robin

Re: Can CFX5 calculate porous media?
 

I presume you mean 'Definitely' and not 'Perhaps'.

Astrid

Re: Can CFX5 calculate porous media?
 

The linearization of directional source terms will be available in 5.6.

Robin

Re: Can CFX5 calculate porous media?
 

Porosity models have not seemed to be a priority in CFX-5 yet. I wonder how many people are using them in CFX-4? At any rate I don't think you'll see an improvment in this area until at least 5.7.

Neale.

Re: Can CFX5 calculate porous media?
 

Hmmmmm. I am performing the simulations in CFX-4 now. Of course this works fine but I have to get used to > 2000 iterations in cfx4 in stead of ±200 in cfx5. And parallelisation is ... poor, and .... well, never mind.

Please speed up the rate, Astrid