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How to determine viscous resistance and inertial resistance porous media

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Old   April 15, 2015, 14:37
Default How to determine viscous resistance and inertial resistance porous media
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Djilali
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I can't understand how i can enter these coeffiecents please look at the picture and i work with cavites porous media

-----

please i wanna your help
my problem is simulation enclosure porous media and i dont what does mean these coefficients look at the picture
this is the attached picture...

thanks
Djilali
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File Type: jpg help.jpg (83.5 KB, 19 views)

Last edited by wyldckat; April 18, 2015 at 15:17. Reason: merged posts since they were on the same topic
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Old   April 18, 2015, 15:42
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Quick question: Have tried clicking on the "Help" button?
I say this because the documentation provided by the application should explain what all of those parameters are, and how they should be calculated.
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Old   April 18, 2015, 16:13
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Quote:
Originally Posted by wyldckat View Post
Quick question: Have tried clicking on the "Help" button?
I say this because the documentation provided by the application should explain what all of those parameters are, and how they should be calculated.
first of all thanks
help doesn't work if you knows any things about it plz help me
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Old   April 18, 2015, 16:47
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I did just now a quick search with Google for
Code:
CFD porous media
and found the following threads:
Those threads have a lot of details on how this works.


I then searched for:
Code:
fluent "porous zone" direction-1
and found this page: http://jullio.pe.kr/fluent6.1/help/html/ug/node236.htm
Everything you're asking for seems to be explained on that page.


Good luck!
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Old   April 18, 2015, 16:50
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Quote:
Originally Posted by nelly View Post
Hi Oky, did you get to find the parameters?

If not, here is the explanation.

The formula is
Dp/L= [(viscosity/alpha)*velocity]+ [C2*1/2*density*velocity^2]
(dp/l =viscousr resistance + inertial resistance)

and


plot dp/l vs velocity in excel. The pressure drop can be obtained from cfd.

The values for velocity can be arbitrarily chosen say for ex. 0 to 20 m/s (even more if you want more points)

Plot Dp/L vs velocity based on above formula and examine the curve how it looks like. If it is a straight line then use only dp/l=viscous resistance.
If it is a quadratic equation use (dp/l =viscousr resistance + inertial resistance). Find K and C2 with these curves.

Hope this helps .

cheers
Nelly
from where i get dp/l vs velocity and what does mean c2 and what is the relation betwen c2 and the inertial resistance at the picture ?

Last edited by wyldckat; April 18, 2015 at 17:13. Reason: Added proper [QUOTE] markers
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Old   April 18, 2015, 17:14
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As I wrote before, everything is explained on this page:
Quote:
Originally Posted by wyldckat View Post
http://jullio.pe.kr/fluent6.1/help/html/ug/node236.htm
Everything you're asking for seems to be explained on that page.
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Old   April 18, 2015, 17:34
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Quote:
Originally Posted by wyldckat View Post
As I wrote before, everything is explained on this page:
i understand viscous resistance terme is 1/alpha (1/m^2)
but c2 how can i enter and what is the relation betwen c2 and inertial resistance Direction-1(1/m)
and thanks bro
Djilali
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Old   April 18, 2015, 18:28
Default Inertial Losses in Porous Media????????????????
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Quote:
Quote:
please look at the picture and tell me what relation betwen c and inertial Resistance Direction-1(1/m) in this picture




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Old   April 19, 2015, 11:04
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Quote:
Originally Posted by djodjo View Post
but c2 how can i enter and what is the relation betwen c2 and inertial resistance Direction-1(1/m)
I Googled just now for this:
Code:
fluent porous c2
And found this thread: FLUENT porous zone inputs - which explains where C2 is defined. As for what the direction is for, it's explained on the previous link that I've mentioned already twice.
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Old   April 20, 2015, 07:07
Default hey Bruno Santos
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hey Bruno Santos
i wanna ask you why we puts the values of viscous and inertial Resistances in direction-1 bigger than direction 2 i work with 2D and should i put the same way? and i work on heat transfer in enclosure porous media one side is has temperature heat the red one and other has temperature cold the blue one and the walls orizontale are adabatic. as you know my assumptions the medium is homogenous and isotropic
look at the picture you will understand what i mean
please bruno help me
thank you for everything you did with me before
Djilali



Which method suitable for my problem to derive viscous and inertial loss coefficients?

1. Method: Deriving Porous Media Inputs Based on Superficial Velocity, Using a known Pressure Loss.
2.Method: Using the Ergun Equation to Derive Porous Media Inputs for a Packed Bed
3.Method: Using an Empirical Equation to Derive Porous Media Inputs for Turbulent Flow Through a Perforated Plate.
Well my model is 2d and i have flow in both directions. why that the viscous and inertial resistance in the y-direction is 10 times larger than that in the x-direction.so does this mean that I have to set X=0.1 and Y=1
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Old   April 26, 2015, 14:31
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Hi Djilali,

Quote:
Originally Posted by djodjo View Post
i wanna ask you why we puts the values of viscous and inertial Resistances in direction-1 bigger than direction 2 i work with 2D and should i put the same way?
By what I can figure out from table 8.2, the Y and Z directions have a lot more porous resistance so that the problem becomes almost 1D, instead of 2D or 3D. In other words, the problem is almost only solved along X.

Quote:
Originally Posted by djodjo View Post
Which method suitable for my problem to derive viscous and inertial loss coefficients?

1. Method: Deriving Porous Media Inputs Based on Superficial Velocity, Using a known Pressure Loss.
2.Method: Using the Ergun Equation to Derive Porous Media Inputs for a Packed Bed
3.Method: Using an Empirical Equation to Derive Porous Media Inputs for Turbulent Flow Through a Perforated Plate.
I don't know. The image seems to describe a closed box. Method 2 might work, since it's designed for a "packed bed".

Quote:
Originally Posted by djodjo View Post
Well my model is 2d and i have flow in both directions. why that the viscous and inertial resistance in the y-direction is 10 times larger than that in the x-direction.so does this mean that I have to set X=0.1 and Y=1
Again, it seems the model is 3D, but the numerics are configured in such a way that the flow almost only acts along X (direction 1).
I don't know what you should do in this case. It seems to me that you should ask for more details from whomever gave you this task.

Good luck! Best regards,
Bruno
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Old   April 26, 2015, 17:58
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Quote:
Originally Posted by wyldckat View Post
Hi Djilali,


By what I can figure out from table 8.2, the Y and Z directions have a lot more porous resistance so that the problem becomes almost 1D, instead of 2D or 3D. In other words, the problem is almost only solved along X.


I don't know. The image seems to describe a closed box. Method 2 might work, since it's designed for a "packed bed".


Again, it seems the model is 3D, but the numerics are configured in such a way that the flow almost only acts along X (direction 1).
I don't know what you should do in this case. It seems to me that you should ask for more details from whomever gave you this task.

Good luck! Best regards,
Bruno
Thank you very much Bruno
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