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Old   June 8, 2007, 05:49
Default Evoapration of droplet
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Hi! friends,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&infin/(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&infin;(T&infin;=60℃)=RH*ps&infin;(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?

Many thanks!

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Old   June 8, 2007, 15:05
Default Re: Evoapration of droplet
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In the initial period during which the pressure difference is negative in your formulae, the droplet (may) undergo the coagulating process, i.e., the dropet diameter getting bigger, as shown by Eq. (1). Do you account for this process?
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