CFD Online Discussion Forums - LiquidEvaporationBoil & LiquidEvaporation

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LiquidEvaporationBoil & LiquidEvaporation

Hi,

I'm running a test case with sprayFoam. I have used two evaporation models "LiquidEvaporationBoil" and "LiquidEvaporation". The results are different. After some investigations, I figured out, if boiling does not occurs, theoretically two models should behave the same.

dMassPC[lid] in LiquidEvaporation is clear

Code:

dMassPC[lid] +=kc*(Cs - Cinf) *pi*sqr(d)*liquids_.properties()[lid].W()*dt

where kc is mass transfer coefficient and Cs is vapour concentration at surface of droplet

However I can not understand how dMassPC[lid] is calculated in LiquidEvaporationBoil (when there is not boiling)

Code:

         else

            {

                // evaporation



                // surface molar fraction - Raoult's Law

                const scalar Xs = X[lid]*pSat/pc;



                // molar ratio

                const scalar Xr = (Xs - Xc)/max(SMALL, 1.0 - Xs);



                if (Xr > 0)

                {

                    // mass transfer [kg]

                    dMassPC[lid] += pi*d*Sh*Dab*rhos*log(1.0 + Xr)*dt;

                }

            }

I found some information baout the evaporation part in the "LiquidEvaporationBoil" model. It is called "The Spalding evaporation model" (Spalding (1953)). you can find more information in the following link

Quote:

http://ethesis.inp-toulouse.fr/archi...a_sanchez2.pdf

Spalding number in LiquidEvaporationBoil

I know the post is quite old but I have a question concerning the use of Xr in the expression:

dMassPC[lid] += pi*d*Sh*Dab*rhos*log(1.0 + Xr)*dt;

In literature this expression is always written using Bm, the Spalding mass number:
#mass ratio
const scalar Bm=(Ys-Yc)/max(SMALL,1.0-Ys)

#Molar ratio
const scalar Xr = (Xs - Xc)/max(SMALL, 1.0 - Xs)

Does anybody know the rationale behind this choice in OpenFOAM?