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October 29, 2010, 02:50 
Porous loss model coefficient query

#1 
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Stuart
Join Date: Jul 2009
Location: Portsmouth, England
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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 DupuitForchheimer equation (http://en.wikipedia.org/wiki/Darcy's...nonDarcy_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 DupuitForchheimer equation. So I just want to check if I should really be doubling this value in CFXPre 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 DupuitForchheimer equation in wiki and no actual fluids textbooks. Thanks. 

October 29, 2010, 07:02 

#2 
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Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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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 
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Hamed
Join Date: Jun 2010
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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 
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Glenn Horrocks
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What are you trying to do?


August 24, 2011, 00:34 

#5 
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Hamed
Join Date: Jun 2010
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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 
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Glenn Horrocks
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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 
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Hamed
Join Date: Jun 2010
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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 
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Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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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.


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