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\title{The Theory Concerning Droplet Generator}
\author{}
\institute{}
\begin{document}
\begin{CJK}{GBK}{song}
\begin{frame} % Cover slide
\titlepage
\end{frame}
\frame{\frametitle{Table of Contents}\tableofcontents}
\section{Introduction to the Mechanism of Break-up}
\subsection{Young-Laplace Equation}
\begin{frame}
\frametitle{Young-Laplace Equation}
The Young-Laplace equation describes the capillary pressure difference sustained across the interface between two fluids.
\begin{block}
\begin{equation}
\Delta p = T \bigl{(}\frac{1}{R_1} + \frac{1}{R_2} \bigr{)}
\end{equation}
\end{block}
$\Delta p$: The capillary pressure difference between two liquids. \\
$T$: The surface tension between two fluids, determined by the types of the fluids. \\
$R_1$ \& $R_2$: The principle radii of curvature for the interface. \\
\begin{figure}
{
\includegraphics[width=1.5in]{R1R2.png}
}
\end{figure}
\end{frame}
\subsection{The Mechanism for the Break-up of Liquid Jet}
\begin{frame}
\frametitle{The Mechanism for the Break-up of Liquid Jet}
\begin{columns}[c]
\column{0.5\textwidth}
\begin{block}{}
\begin{figure}
{
\includegraphics[width=1.5in]{Rayleigh.png}
%\caption{Intermediate stage of a jet breaking into drops}
% \label{fig:Rayleigh}
}
\end{figure}
\end{block}
\column{0.5\textwidth}
\begin{block}{The Young-Laplace Equation for $R_1$}
\begin{equation*}
\Delta p_{R_1} = T \bigl{(}\frac{1}{R_1} \bigr{)}
\end{equation*}
\end{block}
\begin{itemize}
\item We suppose $R_1$ is the radius of the liquid jet.
\item Because $R_1$ at the trough is smaller, the pressure due to surface tension is increased.
\item Because $R_1$ at the peak is bigger, the pressure due to surface tension is decreased.
\end{itemize}
%The Plateau-Rayleigh instability explains why and how a falling stream of fluid breaks up into smaller droplets. The Plateau-Rayleigh instability is derived based on the Rayleigh-Taylor instability. For our experiment, we utilize this method theory to decide the amplitude and frequency of the vibration, which perturbs the liquid jet generated by our droplet generator.
\end{columns}
\end{frame}
\begin{frame}
\frametitle{The Mechanism for the Break-up of Liquid Jet}
\begin{columns}[c]
\column{0.5\textwidth}
\begin{block}{}
\begin{figure}
{
\includegraphics[width=1.5in]{Rayleigh.png}
}
\end{figure}
\end{block}
\column{0.5\textwidth}
\begin{block}{The Young-Laplace Equation for $R_2$}
\begin{equation*}
\Delta p_{R_2} = T \bigl{(}\frac{1}{R_2} \bigr{)}
\end{equation*}
\end{block}
\begin{itemize}
\item We suppose $R_2$ is the radius of curvature for the arc.
\item Because $R_2$ at the trough is negative, the pressure due to surface tension is negative.
\item Because $R_2$ at the peak is positive, the pressure due to surface tension is positive.
\end{itemize}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{The Mechanism for the Break-up of Liquid Jet}
\begin{block}{}
\begin{equation}
\Delta p = \Delta p_{R_1} + \Delta p_{R_2} = T \bigl{(}\frac{1}{R_1} + \frac{1}{R_2} \bigr{)}
\end{equation}
\end{block}
From the statement above, we could know that the values of $R_1$ and $R_2$ both determine the pressure difference.
\begin{itemize}
\item If $R_1$ dominates, the pressure at the trough is greater than that of the peak. The liquid will accumulate towards the peak and the liquid will break up.
\item If $R_2$ dominates, the pressure at the peak is greater than that of the trough. The liquid will accumulate towards the trough and the liquid will not break up.
\end{itemize}
\bf What we are doing with droplet generator is to determine which radius of curvature will dominate, $R_1$ or $R_2$.
\end{frame}
\section{The Principle Formulae for Our Design}
\begin{frame}
\frametitle{The Principle Formulae for Our Design}
The principle formulae composed of the factors crucial to our design is:
\begin{block}{}
\begin{equation}
d_{d} = 1.145 \cdot \bigl{(}\frac{v_j \cdot d_j^2}{f_{opt}}\bigr{)}^{\frac{1}{3}}
\end{equation}
\end{block}
$d_{d}$: The diameter of the droplets broken up. \\
$v_j$: The velocity of the liquid jet at the outlet of the nozzle.\\
$d_j$: The diameter of the jet, which is not identical to the inner diameter of the nozzle.\\
$f_{opt}$: The optimum frequency of the disturbance, which could be represented as a plane wave. \\
\end{frame}
\subsection{Derivation of the Disturbance}
\begin{frame}
\frametitle{Derivation of the Disturbance}
\begin{columns}[c]
\column{0.5\textwidth}
\begin{block}{}
\begin{figure}
{
\includegraphics[height=0.8\textheight]{Wave.png}
}
\end{figure}
\end{block}
\column{0.5\textwidth}
We suppose that $x$ direction is vertical and $y$ direction is perpendicular to $x$ direction. We could derived our equation of motion in $y$ direction:
\begin{block}{Equation of Motion}
\begin{equation}
\rho \frac{\partial v}{\partial t} = \mu \frac{\partial^2 v}{\partial x^2}
\end{equation}
\end{block}
$\rho$: The density of the liquid. \\
$\mu$: The dynamic viscosity between the liquid and the air. \\
$v$: The velocity of the wave in the direction of $y$.\\
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Derivation of the Disturbance}
\end{frame}
\subsection{Derivation of the Jet Velocity at the Outlet of the Nozzle}
\begin{frame}
\frametitle{Derivation of the Jet Velocity at the Outlet of the Nozzle}
According to the Bernoulli Equation,
\begin{block}{The Bernoulli Equation}
\begin{equation}
H = z + \frac{p_1}{\rho g} + \frac{v_1^2}{2g} = h + \frac{p_2}{\rho g} + \frac{v_2^2}{2g}
\end{equation}
\end{block}
$H$: The \textbf{hydraulic head} (压力水头) of the Bernoulli Equation, which is a constant. \\
$p_1$ \& $p_2$: The pressures at two positions in the fluid. \\
$v_1$ \& $v_2$: The velocities at two positions in the fluid. \\
$z$: The elevation of the point above a reference plane. \\
$h$: The energy loss of the liquid, which could be represented as:
\begin{block}{The Energy Loss}
\begin{equation}
h = \gamma \frac{v_2^2}{2g} + \frac{32 D \mu v_2}{\rho g d_0^2}
\end{equation}
\end{block}
\end{frame}
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\end{document}
