Porous loss model coefficient query

siw · October 29, 2010, 02:50

Hi,

The two Momentum Source loss models (isotropic and directional) shown in the CFX Theory Guide (eqs. 1.146 and 1.149) show that the loss coefficient is multiplied by density / 2.

When comparing these equations with the Dupuit-Forchheimer equation (http://en.wikipedia.org/wiki/Darcy's...non-Darcy_flow) the loss coefficient there (Greek letter beta) is multiplied only by density.

I have dervied my loss coefficient from experimental data as like in the Dupuit-Forchheimer equation. So I just want to check if I should really be doubling this value in CFX-Pre so that CFX halves it as shown in the Theory Guide. I'm not sure why CFX uses the 1 / 2. But I've only found reference to the Dupuit-Forchheimer equation in wiki and no actual fluids textbooks.

Thanks.

ghorrocks · October 29, 2010, 07:02

I don't know the answer to your question, but I know what I would do if I needed to know.

I would set up a simple simulation with a block of porous material and simulate a known flow velocity across it to get a pressure drop. I would then work out the analytical pressure drop and make sure that the loss coefficient I put in was generating the pressure drop expected.

It is always better to work these things out for yourself anyway.

enghamed · August 23, 2011, 00:58

If I don't the value for linear ressistance coefficient, can I not input any value? just select thethe isotropic loss to linear and quadratic resistance coefficients but then don't tick the box for entering a value. Is this ok?

thanks

ghorrocks · August 23, 2011, 06:38

What are you trying to do?

enghamed · August 24, 2011, 00:34

I am modeling a sloshing tank with mesh motion. Now I want to put in cylindrical baffel inside the tank which want to be 50% porous. I set it as isotropic loss but I dont know the linear and quadratic resistance coefficients. Can I not tick the linear and quadratic resistance coefficients options so I dont have to enter the value for it?

Another problem that i have faced is that in the porous setting I cant define mesh motion, but i simulate sloshing by mesh motion so what should i do?
thx

ghorrocks · August 24, 2011, 06:21

I am not sure you can have a porous region with mesh motion. If this cannot be done then you should impose the porous region yourself as a momentum source term.

enghamed · August 24, 2011, 06:51

thx. can you tell me how should I impose the porous region as momentum source? how is this done? what do i need to do?

thx a lot.

ghorrocks · August 24, 2011, 07:41

Determine the equation of your porous region (ie pressure drop versus velocity) and impose that as a momentum source. This is discussed in the documentation. You will probably need to include the source term linearisation term - again, see the doco.

October 29, 2010, 02:50	Porous loss model coefficient query	#1
siw Senior Member Stuart Join Date: Jul 2009 Location: Portsmouth, England Posts: 734 Rep Power: 25	Hi, The two Momentum Source loss models (isotropic and directional) shown in the CFX Theory Guide (eqs. 1.146 and 1.149) show that the loss coefficient is multiplied by density / 2. When comparing these equations with the Dupuit-Forchheimer equation (http://en.wikipedia.org/wiki/Darcy's...non-Darcy_flow) the loss coefficient there (Greek letter beta) is multiplied only by density. I have dervied my loss coefficient from experimental data as like in the Dupuit-Forchheimer equation. So I just want to check if I should really be doubling this value in CFX-Pre so that CFX halves it as shown in the Theory Guide. I'm not sure why CFX uses the 1 / 2. But I've only found reference to the Dupuit-Forchheimer equation in wiki and no actual fluids textbooks. Thanks.

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October 29, 2010, 07:02		#2
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,719 Rep Power: 143	I don't know the answer to your question, but I know what I would do if I needed to know. I would set up a simple simulation with a block of porous material and simulate a known flow velocity across it to get a pressure drop. I would then work out the analytical pressure drop and make sure that the loss coefficient I put in was generating the pressure drop expected. It is always better to work these things out for yourself anyway.

August 23, 2011, 00:58		#3
enghamed Member Hamed Join Date: Jun 2010 Posts: 43 Rep Power: 15	If I don't the value for linear ressistance coefficient, can I not input any value? just select thethe isotropic loss to linear and quadratic resistance coefficients but then don't tick the box for entering a value. Is this ok? thanks

August 23, 2011, 06:38		#4
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,719 Rep Power: 143	What are you trying to do?

August 24, 2011, 00:34		#5
enghamed Member Hamed Join Date: Jun 2010 Posts: 43 Rep Power: 15	I am modeling a sloshing tank with mesh motion. Now I want to put in cylindrical baffel inside the tank which want to be 50% porous. I set it as isotropic loss but I dont know the linear and quadratic resistance coefficients. Can I not tick the linear and quadratic resistance coefficients options so I dont have to enter the value for it? Another problem that i have faced is that in the porous setting I cant define mesh motion, but i simulate sloshing by mesh motion so what should i do? thx

August 24, 2011, 06:21		#6
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,719 Rep Power: 143	I am not sure you can have a porous region with mesh motion. If this cannot be done then you should impose the porous region yourself as a momentum source term.

August 24, 2011, 06:51		#7
enghamed Member Hamed Join Date: Jun 2010 Posts: 43 Rep Power: 15	thx. can you tell me how should I impose the porous region as momentum source? how is this done? what do i need to do? thx a lot.

August 24, 2011, 07:41		#8
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,719 Rep Power: 143	Determine the equation of your porous region (ie pressure drop versus velocity) and impose that as a momentum source. This is discussed in the documentation. You will probably need to include the source term linearisation term - again, see the doco.