# LiquidEvaporationBoil & LiquidEvaporation

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 August 26, 2016, 05:19 LiquidEvaporationBoil & LiquidEvaporation #1 New Member   Amir Join Date: Feb 2012 Posts: 12 Rep Power: 13 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; } }```

 August 26, 2016, 07:22 #2 New Member   Amir Join Date: Feb 2012 Posts: 12 Rep Power: 13 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 charryzzz likes this.

 July 31, 2019, 05:03 Spalding number in LiquidEvaporationBoil #3 New Member   Alessandro Join Date: Jul 2016 Posts: 11 Rep Power: 8 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? zhangyan likes this.