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thermo settings for natural and forced convection |
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April 24, 2011, 15:07 |
thermo settings for natural and forced convection
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#1 |
Senior Member
Fabian Braennstroem
Join Date: Mar 2009
Posts: 407
Rep Power: 19 |
Hi,
I am still having a bit trouble understanding the thermoPhysicalProperties settings. I mainly have two different challenges, a natural convection, for which I am using buoyantSimpleFoam and a forced convection at Ma-number around 0.4, for which I plan to use rhoSimpleFoam (or buoyantSimpleFoam as well?). For the solvers hRhoThermo and hPsiThermo are used; as far as I know this means, that: - hPsiThermo (buoyantSimpleFoam), calculation based on enthalpy 'h' and compressibility 'psi' - hRhoThermo (rhoSimpleFoam), calculation based on enthalpy or sensible enthalpy What is the reason (maybe numerically!?) to have different thermo models for these kind of problems? Even more confusing for me is that the buoyantPimpleFoam solver is based hRhoThermo... why should it be different to buoyantSimpleFoam!? Thanks in advance! Fabian |
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April 26, 2011, 04:54 |
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#2 |
Senior Member
Christian Lucas
Join Date: Aug 2009
Location: Braunschweig, Germany
Posts: 202
Rep Power: 18 |
Hi Fabian,
both hPsiThermo and hRhoThermo more or less do the same. In hPsiThermo, the density is calculate by rho=psi*p with psi=1/(RT). In hRhoThermo, the density is calculated (assuming you use a perfect gas) rho=p/(RT). The main advantage of hPsiThermo is that the pressure correction of the density (see solver) is very simple because psi is a constant (for perfect gases). The main disadvantage is that the equation already assumes perfect gas behavior. So, the reason (as I understand it) of hRhoThermo is to avoid this disadvantage and to make it possible to use other equation of state like the polynomial by introducing an internal rho field in the thermo library. Regards, Christian |
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April 27, 2011, 15:55 |
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#3 |
Senior Member
Fabian Braennstroem
Join Date: Mar 2009
Posts: 407
Rep Power: 19 |
Hi Christian,
thanks for your help! This would mean, that if I expect a perfect gas the hPsi approach would be more stable from the numerics point of view!? Regards! Fabian |
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May 1, 2011, 13:33 |
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#4 |
Senior Member
Christian Lucas
Join Date: Aug 2009
Location: Braunschweig, Germany
Posts: 202
Rep Power: 18 |
Hi,
no, it wouldn't be more stable (if you code it correctly), but it would be faster ( assuming you code to correctly). Regards, Christian |
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