Evaopation of droplet
Hi!freinds,I was studying on the mathematical simulation of the droplet drying. The mathematical expression of the droplet's diameter is d(rd)/dt=-km*M1*(pvs-pv∞)/(R*ρl*Tav) (1) drying conditions: r0=0.00075mm, T∞(environment temporature)=60℃ and RH=80% km=Sh*Dg/(2*rd) (2) In equation(1): the molar mass of the water M1=0.0018kg/molar; R(gas constant) = 8.314; ρ1=1000kg/m3; Tav(temperature difference between the surface anfd the surroundings)=40℃ In equation(2): Dg(the diffusion coefficient of water vapor)=1.32e-5 Sh=2+2*(rd/r0)^0.5 (3) Questions are: The vapor pressure of the air pv∞(T∞=60℃)=RH*ps∞(saturated vapor pressure),=15936pa,but the saturated vapor pressure of droplet surface (T=20℃)=2331pa<15936pa. The temperature of droplet rise until the saturated vapor pressure of droplet surface is greater than the gas vapor pressure. Then the droplet starts evaporating. But the droplet diameter by the simulating is smaller than the experiment results, why?
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