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 October 26, 2010, 10:02 Quadratic Resistance Coefficient when energy source added. #1 New Member   JoFFe JoFF Join Date: Oct 2010 Posts: 7 Rep Power: 9 Hi all. I am modeling a porous domain with an energy source added to it. Compressible helium is flowing through this porous domain. I can analytically calculate the linear and quadratic resistance coefficients for when energy is not added. In that case it works just fine and the pressure drop is coherent with my analytic calculation. It also works if I use incompressible helium and heat is added. As soon as i use helium as ideal gas and heat is added, my equations for the resistance coefficients don't work any more. Does anybody know how to calculate resistance coefficients when an energy source(heat) is added to the porous domain? Would really appreciate some help! Thanks

 October 26, 2010, 11:17 #2 Senior Member   Join Date: Apr 2009 Posts: 532 Rep Power: 14 Apply the loss using the permeability and loss coefficient instead, since the loss coefficient is multipled by density so will the momentum loss will automatically adjust as the density changes. The quadratic resistance is related to the loss coefficient as: Quadratic Resistance Coeff = Loss Coeff * (Density/2) So if you've calculated your quadratic loss for a given density, you can back out the Loss Coeff.

 October 26, 2010, 13:25 #3 New Member   JoFFe JoFF Join Date: Oct 2010 Posts: 7 Rep Power: 9 Thanks for the tip, I will try that soon. But do you mean that i should add that as an expresseion or calculate a value and enter that in the filed for loss coefficient? By the way: I have been using expressions as resistance coefficients before with the density included, but for some reson the pressure drop goes against the same value as I set as inlet pressure. Do you se any problem with using these boundary conditions? Inlet: Total pressure: 20 bar Outlet: Mass flov: 0.78294 kg/s Reference pressure: 0 atm I'm using this because the maximum allowed pressure in the system is 20 bar. Any suggestions or ideas if this is okey or if there is a better way? Last edited by JoFFe; October 26, 2010 at 14:36.

 October 27, 2010, 08:42 #4 Senior Member   Bharath kumar Join Date: Apr 2009 Posts: 169 Rep Power: 10 i think the solution you obtain is right.porous domain is an imaginary thing, inside the domain.in real condition also it will happen like that only or if you want force the drop you can adjust the coefficient little on either side, give you the exact thing you need. stumpy says the same thing

 Tags energy, linear, porous, quadratic, resistance

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