\documentstyle[11pt]{article}

\begin{document}
\begin{titlepage}
\vspace*{1in}
\begin{center}
\LARGE Performance Optimization of a Pressure-driven Rocket
\end{center}
\vspace*{1in}
\begin{center}
Michael D. Mendyke\vspace*{2em}\\
16.621 Proposal\vspace*{1em}\\
\today\vspace*{1in}\\
Partner: Eric Shukan\hfill Advisor: Prof. Von Flotow

\end{center}

\end{titlepage}

\begin{center}
\Large \bf ABSTRACT
\end{center}
\vspace{2in}

There exists for any propulsion system a set of parameters which
characterize its performance. Using fluid mechanics and propulsion
theory, my partner and I have derived the equations relating the
pressure and fuel level of a pressure-driven, liquid-fueled rocket to
its resultant velocity.

This paper will discuss our theory for determining the optimum initial
fuel level for a given pressure.  Included is a description of an
apparatus we propose to use to verify the theory.  The results of the
experiment will be reported in a subsequent paper.  We propose this
project to fulfill the requirements of the Aeronautics and
Astronautics courses $16.621/.622$.

\newpage

\tableofcontents
\newpage

\begin{center}
{\LARGE Performance Optimization of a Pressure-driven Rocket}
\end{center}

\section{Introduction}

Consider a plastic toy rocket. It has a hollow interior, stabilizing
fins, and a hole at the bottom. When this rocket is pressurized with
air and released, the compressed air flows out of the hole, causing
a momentum change and providing thrust for the rocket until pressure
has been equilibrated. Typical heights that can be reached:
$5$--$15$~ft.

If we now fill some portion of the rocket with water, pressurize the
remaining air and release the rocket, the water will be forced
out of the rocket by the air inside; hence, the water takes the place
of the air as the exhaust. Typical heights that can be reached:
$75$--$100$~ft.

This result is the opposite of what might be expected at first glance;
the rocket launched with water has a smaller volume of pressurized air
and hence, less total stored energy than the rocket launched with air
alone. Also, the water propelled rocket weighs more.  Nonetheless, the
thrust of the rocket (and thus the height attained) is much greater
when water is used. Why is this the case? How can we predict the
height the rocket will reach? These are the questions we hope to
answer.

\section{Purpose}

The purpose of this project is to optimize the height produced by a
pressure-driven propulsion system for a given pressure. This will be
done in two steps: first, the relevant theory will be derived; then,
the theory will be verified with a dynamic experiment.

\section{Description of Project}

\subsection{General Procedure}

After a hypothesis has been formed based on the relevant theory, an
apparatus will be constructed to conduct experiments, providing
measured data to compare with predicted values. These results will be
analyzed and written up in a final paper.

\subsection{Theory}

In general, we would like to determine the conditions of optimum
performance of the rocket. To do this, we must characterize the
performance of the rocket as a function of all relevant parameters. If
we assume that we will use only air as the pressurized gas within the
chamber, and water as the only fuel, the rocket performance can be
predicted knowing only two key parameters: initial air pressure within
the rocket $(P_o)$; and initial volume fraction of water $(V_o)$ (When
$V_o=0$, there is no water and the equations will degenerate to the
air-only case).

Note that while $V_o$ varies from $0$--$1$, $P_o$ can go to infinity.
This implies that there is no limit to thrust or height, regardless of
the initial volume fraction. In reality we are limited by the
structural integrity of the rocket. We believe that {\em for any fixed
$P_o$,} $V_o$ is optimized when the water runs out at the same time
that the air equilibrates. If $V_o$ is greater, for example, there
will be leftover water when the pressure equilibrates.  This water did
not provide any thrust and acted only as dead weight, which is not
desirable.  If $V_o$ is less then the suggested optimum , then the
fuel runs out before it finishes depressurizing.  The rocket must
therefore finish its thrusting using only air, and it is clear from
the introduction of this paper that using water is better than using
air.

It is possible to verify this theory in two ways: statically and
dynamically. In both cases the thrust as a function of time $T(t)$ is
required. $T(t)$ can be found as follows:

\begin{enumerate}

\item Starting with Newton's Law:

\begin{equation}
F=Ma \rightarrow T=\dot{M}u_e
\end{equation}

\item From continuity:

\begin{equation}
 \dot{M} = \rho u_e A
\end{equation}

therefore

\begin{equation}
 T = f(\rho, u, a)
\end{equation}

\item We can relate $\rho$ to $P_o$ using the isentropic relations

\item We can also relate $u_e$ to $P_o$ using the momentum equation

\item Therefore, we find thrust as a function of pressure:

\begin{equation}
T=f(P)
\end{equation}

\item Since we know that pressure is a function of time,

\begin{equation}
P=f(t)
\end{equation}

we can explicitly conclude with thrust as a function of time:

\begin{equation}
T=f(t)
\end{equation}
\end{enumerate}

This means that given $P_o$ and $V_o$, we can calculate $T$ as a
function of time.

For the static case, the rocket does not move. We therefore have no
rocket-related values to optimize such as height or momentum change of
the rocket.  However, we can try to optimize the total momentum change
of the propellant $(L_{prop})$. This can be found by integrating the
thrust over the length of time the rocket is thrusting. 
\begin{equation}
L_{prop}=\int_{0}^{t} T(t) dt
\end{equation}

For the dynamic case, the rocket is moving and the total momentum
change of the propellant is not so obvious to calculate. Nor can the
momentum of the rocket alone be easily considered, since at any time
during the thrusting there is some fuel inside the rocket.

In this case, it is much easier to optimize the final height $(h_f)$ of
the rocket. We can relate height to $P_o$ and $V_o$ by summing the
accelerations in the upward direction. We obtain an equation of the
form

\begin{equation}
\ddot{h}(t)=\frac{T(t)}{m(t)}-\frac{\frac{1}{2}\rho u_e A c_T}{m(t)}-g
\end{equation}

where $T(t)$ is the same thrust we derived for the static case,
divided by the mass of the rocket. The second term is the drag on the
rocket due to air resistance, and $g$ is gravity. If we now integrate
this equation twice with respect to time, noting that the initial
conditions (position and velocity of the rocket at $t=0$) are both
zero, we obtain the height.

We see that $L_prop$ and $h_f$ will be related to $V_o$ and $P_o$
in a manner suggested by Figure~\ref{charts1}.
\begin{figure}[h]
\vspace{3in}
\caption{Relationship between $h_f$ or $L_{prop}$, $P_o$ and $V_o$}
\label{charts1}
\end{figure}
Notice that by allowing $V_o$ to go to zero, we default to the
air-only case (see Figure~\ref{charts2}).
\begin{figure}[h]
\vspace{3in}
\caption{Relationship between $h_f$ or $L_{prop}$ and $P_o$ ($V_o=0$)}
\label{charts2}
\end{figure}
We can see immediately that the peak of each curve represents the
conditions for maximum $L_prop$ or $h_p$. 

\input apparatus.tex

\begin{figure}[p]
\vspace{3in}
\caption{Timetable for 16.622}
\end{figure}

\newpage

\input materials.tex

\end{document}
