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\title{Notes}

\begin{abstract}
%% Text of abstract
\end{abstract}


\section{Introduction}
In the kinetmatic regime (droplets are much smaller than the mean free pat
of the molecules, the effluent flux is given by
\EQ
\mathcal{F}=Nu/4
\EN
%where $N=C_{SiO2}-C_s$ is the molar consentration of the condensing gas,
where $N=C_{\Si\Ox_2}-C_s$ is the molar consentration of the condensing gas,
$C_{SiO2}$ is the molar concentration of SiO$_2$ gas, $C_{s}$ is
saturation concentration of the same gas and $u$
is the mean velocity of the gas molecules, which can be found from
the mean of the Maxwell-Botzmann distribution:
\EQ
u=\sqrt{8k_BT/(\pi m_{SiO2})},
\EN
where $k_B$ is Boltzmann's constant, $T$ is fluid temperature and $m_{SiO2}$ is
the molecular mass of SiO$_2$.
Hence, the units of $\mathcal{F}$ is mol/m$^2$/s. The total molar rate of
condensation on the surface of a particle with radius $r$ is then given by
\EQ
F=\mathcal{F}4\pi r^2.
\EN
From this it is clear that the volumetric change of the droplet becomes
\EQ
\frac{dV}{dt}=F M_{SiO2}/\rho_{ms},
\EN
where $M_{SiO2}=60$~kg/kmol is the molar mass of SiO$_2$ and
$\rho_{ms}\sim 2300$~kg/m$_3$ is the true density of a
microsilica particle/droplet.
For a spherical particle we know that
$dV=4\pi r^2dr$, such that
\EQA
\frac{dr}{dt}
&=&\frac{dV}{dt}\frac{1}{4\pi r^2}
=\mathcal{F}4\pi r^2 \frac{M_{SiO2}}{4\pi r^2\rho_{ms}} \nonumber \\
&=&(C_{SiO2}-C_s) \sqrt{8k_BT/(\pi m_{SiO2})}\frac{M_{SiO2}}{4\rho_{ms}} \nonumber \\
&=&A(C_{SiO2}-C_s) \sqrt{T},
\ENA
where $A=0.122$.



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