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-   -   thermo settings for natural and forced convection (http://www.cfd-online.com/Forums/openfoam-solving/87598-thermo-settings-natural-forced-convection.html)

braennstroem April 24, 2011 14:07

thermo settings for natural and forced convection
 
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

Chris Lucas April 26, 2011 03:54

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

braennstroem April 27, 2011 14:55

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

Chris Lucas May 1, 2011 12:33

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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