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November 27, 2006, 07:39 
porous domain problem

#1 
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hello everyone
i needed to model a porous domain. velocity  pressure loss 0.1333 m/s  500 pascals 0.21667 m/s  1000 pascals 0.4 m/s  2000 pascals the above was available about the porous media to be modelled. the velocity range will always be within 0.1 m/s to 0.4 m/s. so with these three data, i fitted a curve Y = a X + b X^2. and found the values for a and b. then i used the value of 'a' to be the linear resistance coefficient and value of 'b' to be the quadratic resistance coefficient. is this methodology correct? if the methodology is correct, why the following is not matching? i modelled a pipe with the porous media and applied the velocity to be 0.133m/s and calculated the pressure drop across the porous domain. the pressure drop computed from CFX was 1.7 pascal. (but from the given data, it should have been 500 pascals). thanks in advance 

November 27, 2006, 14:06 
Re: porous domain problem

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The "Y" on LHS of that curve fitting equation is actually delta_p/delta_L, where delta_L is the characteristic length scale of porous domain. You will probably find out that 1.7 [pa] ~ 500 [pa]*thickness of porous domain.


November 28, 2006, 03:06 
Re: porous domain problem

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hi longbow thanks for your immediate response. yes my thickness of the porous domain was 3.2mm. so 500[Pa]*0.0032 [m] = 1.6 [Pa/m]. so this 500 pascals of pressure drop is for unit length of the porous media. right? but now i have a question. the experimental data has been given for 3.2mm thick polypropylene material. typical air permeabilities @ 3.2mm thick velocity  pressure loss 0.1333 m/s  500 pascals 0.21667 m/s  1000 pascals 0.4 m/s  2000 pascals doesn't this mean that the pressure loss of 500 pascals is for velocity of 0.13m/s. probably i may be wrong in what i think.... can anyone explain me the above? also can anyone tell how experiment will be conducted or any links where i could find some more details.
thanks in advance sudhakar 

November 28, 2006, 08:20 
Re: porous domain problem

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You state:
500[Pa]*0.0032 [m] = 1.6 [Pa/m] Shouldn't it be 500[Pa]/0.0032 [m] = 156250 [Pa/m] ? 

November 28, 2006, 08:39 
Re: porous domain problem

#5 
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sorry it is 500[Pa]*0.0032 [m] = 1.6 [Pam]
thanks sudhakar 

November 28, 2006, 10:19 
Re: porous domain problem

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The curve fitting equation should be delta_p/thickness=a*U+b*U^2. Therefore, to have 500[pa] pressure drop for 0.13[m/s], you need to multiply the "a" and "b" you have got with thickness of porous media domain.


November 29, 2006, 10:48 
Re: porous domain problem

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In case of a quadratic resistance only, the specified resistance coeff. can be calculated using the following formula:
Kq = dP / ( dX * U^2) [kg/m^4]  Michael 

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