kpar: Fizzhead is correct.
There is one formula I found for the approximate time, t, a droplet takes to evaporate in stagnant surroundings:
t=([λ])([ρ]L)(D2)/[8 (kg)(LMTD)]
[λ]: latent heat of evaporation
[ρ]L: liquid density
D: original drop diameter
k
g: gas thermal conductivity
LMTD: logarithmic temperature difference
Let's work out an example with rough figures:
[λ]: 2,400 kJ/kg
[ρ]L: 1000 kg/m
3
D: 300 microns = 0.0003 m
k
g: 0.05 W/(m.K)
LMTD: 700 K
t=(2,400,000)(1000)(0.0003
2)/[8(0.05)(700)]~0.8 s
Estimates as this, may enable one to see whether at the prevailing gas velocities there is enough residence time to accomplish full evaporation and the desired gas quenching.
![[pipe] [pipe] [pipe]](/data/assets/smilies/pipe.gif)