Flight and Staging Lecture
Författare:
Brandt
Last Updated:
för 9 år sedan
Licens:
Creative Commons CC BY 4.0
Sammanfattning:
Staging and Flight lecture
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
Staging and Flight lecture
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb,textcomp, esvect,esint}
\usepackage{graphicx,mathtools}
\usepackage{subcaption}
\usepackage[onehalfspacing]{setspace}
\usepackage{mathrsfs,comment}
% place figures and tables where it seems like they should be
\usepackage{float}
\floatplacement{figure}{H}
\floatplacement{table}{H}
% automatically center figures/tables
\makeatletter
\g@addto@macro\@floatboxreset{\centering}
\makeatother
\usepackage{tocvsec2}
\usepackage[bookmarksdepth=subsection]{hyperref}
\usepackage{bookmark}
\usepackage[margin=1in]{geometry}
\usepackage{authblk}
\usepackage{titling}
\setlength{\droptitle}{-1in}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% NEW COMMANDS
\newcommand{\bvec}[1]{\ensuremath{\mathbf{#1}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%The Class the Homework is for!
\pretitle{\begin{center}\fontsize{14pt}{1em}\bfseries\selectfont AA 462 Rocket Propulsion \\ \vspace{14pt}}
\posttitle{\end{center}}
\preauthor{\begin{center}\fontsize{12pt}{1em}\selectfont}
\postauthor{\end{center}}
\predate{\begin{center}\fontsize{12pt}{1em}\selectfont}
\postdate{\end{center}}
% section heading formatting
\renewcommand{\thesection}{\Roman{section}.}
\usepackage{sectsty}
\sectionfont{\centering\fontsize{12pt}{1em}\selectfont}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% The Homework Set
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Flight and Staging Lecture}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Your names go here (one should be underlined)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\author{Brandt Monson}
\affil{William E. Boeing Department of Aeronautics \& Astronautics\\
University of Washington, Seattle, WA 98195-2400}
\begin{document}
\maketitle
\noindent
The material used for this lecture can be found in \textit{Rocket Propulsion Elements} written by George Sutton, Oscar Biblarz, as well as, \textit{Orbital Mechanics for Engineering Students} written by Howard Curtis. \\
\noindent
Rocket science is hard. It's almost darn near impossible. To help reduce some of this complexity engineers make assumptions about the system to reduce equations into solvable forms. A common assumption made for rocket propulsion is that the rocket has a constant mass flow as seen in the equation below.
\begin{equation} \label{const_mass}
\dot m = const. = \frac{m_p}{t_p}
\end{equation}
\noindent
Where $m_p$ is the total propellant left (i.e. usable propellant) and $t_p$ is the total burn time. Lets use this assumption to examine the velocity change of a rocket while it's in space. Since the rocket at question is assumed to be in space, we will make two more assumptions; the space environment is zero-gravity and there is no drag. Recall from elementary physics, thrust can be expressed as
\begin{equation} \label{thrust}
F = m \cdot \frac{du}{dt}
\end{equation}
\noindent
The mass is a function of time and is shown in Eq. \ref{mass_func1}. Here $m_0$ is the starting initial mass, the mass of the propellant plus the mass of the rocket final rocket structure.
\begin{equation} \label{mass_func1}
m = m_0 - \dot m t
\end{equation}
\noindent
A substitution can be made for $\dot m$ and Eq. \ref{const_mass}. The result is Eq. \ref{mass_func}.
\begin{equation} \label{mass_func}
m = m_0 - \frac{m_p}{t_p} t
\end{equation}
\noindent
From here we can define two new terms the propellant mass fraction and mass ratio, $\zeta$ and $MR$ respectively. The mass fraction, $\zeta$, is defined as the remaining propellant mass divided by the initial mass. By factoring out $m_0$ from Eq. \ref{mass_func}, we can represent the mass as a function of the mass fraction.
\begin{equation} \label{mass_frac}
m = m_0\left( 1 - \zeta \frac{t}{t_p} \right)
\end{equation}
\noindent
This can be done for the mass ratio shown in Eq. \ref{mass_ratio}. Note that the mass ratio is the final mass to the initial mass.
\begin{equation} \label{mass_ratio}
m = m_0 \left[ 1 - (1 - MR) \frac{t}{t_p} \right]
\end{equation}
\noindent
The relationship between the mass ratio and the mass fraction is
$$
\zeta + MR = \frac{m_p}{m_0} + \frac{m_f}{m_0} = \frac{m_p + m_f}{m_0} = \frac{m_0}{m_0} = 1
$$
\begin{equation} \label{zetaMR}
\zeta + MR = 1
\end{equation}
\noindent
It was shown in the previous lecture that the characteristic exhaust velocity can be represented as the following
\begin{equation} \label{exhaust}
c = I_s \cdot g_0 = \frac{F}{\dot m}
\end{equation}
\noindent
Using Eq. \ref{thrust} and looking to solve for the velocity. yields the following
$$
du = \frac{F}{m} dt
$$
\noindent
Using equations \ref{mass_func1}, \ref{mass_func}, and \ref{exhaust} shown previously, we make several substitutions into the above equation
\begin{align*}
du &= \frac{F}{m} dt \\[6pt]
&= \frac{c\dot m}{m}dt \\[6pt]
&= \frac{c (m_p / t_p)}{m_0 - m_p \frac{t}{t_p}} dt
\end{align*}
\noindent
By factoring out $1/m_0$ from the numerator and denominator yields the equation in terms of the mass fraction
\begin{equation} \label{speed}
du = \frac{c\zeta /t_p}{1 - \zeta t/t_p} dt
\end{equation}
\noindent
Integrating the above with boundary conditions $u_0$ at $t = 0$ and $u = u_p$ at $t = t_p$ yields the following
$$
u_p - u_0 = -c\ln(1 - \zeta)
$$
\noindent
Typically $u_0$ is assumed to be zero and the velocity $u_p$ is usually designated as the change of velocity $\Delta v$. With this knowledge the above equation becomes the following
\begin{align}
\Delta v &= c \ln \frac{1}{1 - \zeta} \\[6pt]
&= c \ln \frac{1}{MR} \label{deltavMR} \\[6pt]
&= c \ln \frac{m_0}{m_f}
\end{align}
\noindent
Great! We have a simplified equation for the change in velocity for our spacecraft in space. What about when the spacecraft is on it's way to space and is being affected by gravity and atmospheric drag? These are great questions! Thanks for asking. Your welcome. The picture below displays the many forces acting on a spacecraft before it reaches the zero-gravity and zero drag we assumed at the beginning of the lecture.
\begin{figure}
\includegraphics[width= 0.8\textwidth]{flight}
\caption*{The forces on a vehicle in flight}
\end{figure}
\noindent
To help reduce the complexity of these equations we can make some assumptions about the direction of the thrust and that our rocket won't have lift. This reduced figure can be seen below
\begin{figure}
\includegraphics[width= 0.7\textwidth]{reducedFig}
\caption*{Vehicle in flight with assumptions}
\end{figure}
\noindent
Examining the sum of the forces in the direction of flight yields
$$
m \cdot \frac{du}{dt} = F - mgsin\theta - D
$$
\noindent
Dividing both sides by mass and making Eq. \ref{mass_func} substitution then it will be reduced to the following
$$
\frac{dV}{dt} = \frac{C \zeta /t_{p}}{1 - \zeta t/t_p} - g\cdot sin\theta - \frac{\frac{1}{2}C_D \rho V^2 A/m_0}{1 - \zeta t/t_p}
$$
\noindent
Integrating both sides for the same boundary conditions used earlier ($t=0$, $u=u_0$ and $t=t_p$, $u=u_p$).
\begin{equation} \label{speedoverall}
\Delta V = - \bar{C} \ln (1 - \zeta) - (\bar{g}\cdot sin\theta ) \cdot t_p - \frac{BC_DA}{m_0}
\end{equation}
\noindent
Where the average characteristic exhaust velocity and gravity were found. The B in the last term of the equation is the complicated integral below, that can be solved numerically or graphically.
\begin{equation} \label{Brandt}
B = \int\limits_0^{t_p} \frac{\frac{1}{2}\rho V^2}{1 - \zeta t /t_p} dt
\end{equation}
\noindent
Now that we have gone over the flight dynamics of a rocket lets look at the staging of rockets. Some definitions to start us off: where $m_f$ is the final weight of the spacecraft, $m_d$ is the dead weight, and $m_0$ is the starting mass of the rocket.\\
\noindent
Payload fraction: $\lambda$
$$
\lambda = \frac{m_{pay}}{m_0} = \frac{m_f - m_d}{m_0}
$$
\noindent
Dead weight fraction $\delta$
$$
\delta = \frac{m_d}{m_0} = \frac{m_f - m_{pay}}{m_0}
$$
\noindent
Relationships:
\begin{align*}
\frac{m_0}{m_0 - m_p} &= \frac{m_0}{m_d + m_{pay}} \\[6pt]
\frac{1}{MR} &= \frac{1}{\delta + \lambda}
\end{align*}
\noindent
Gravity free space for (N stages) would have us use Eq. \ref{deltavMR} and making the substitution from above. Then the change in velocity would be.
$$
\Delta V _{tot} = \sum\limits_{i=1}^N \Delta V_i = \sum\limits _{i=1}^N C_i \ln \frac{1}{\delta _i + \lambda _i} = \sum\limits _{i=1}^N C_i \ln \frac{1}{1-\zeta _i}
$$
\noindent
Another helpful equation would be to write the overall payload fraction. This is demonstrated below.
$$
\lambda _0 = \frac{m_{pay}}{m_{01}} = \prod \limits _{i=1}^N \lambda _i = \frac{m_{pay}}{m_{0N}} \cdot \frac{m_{0N}}{M_{0N-1}} \ldots \frac{m_{03}}{m_{02}} \cdot \frac{m_{f1} - m_{d1}}{m_01}
$$
\noindent
The goal now is to maximize the change in velocity for a given set of $\lambda _0$, $C_i$, and $\delta _i$ with the restriction that
\begin{equation} \label{restrict}
\lambda _0 = \prod \limits_{i=1}^N \lambda _i
\end{equation}
\noindent
To do this, lets define the function $F$ with Lagrange multiplier $K$ on the payload restriction.
$$
F = \sum\limits_{i=1}^N C_i \ln \frac{1}{\delta _i + \lambda_i} + K \left[ ln\left(\prod\limits_{i=1}^N \lambda_i \right) - \ln \lambda _0 \right]
$$
\noindent
Note that natural log was taken to both sides of the restriction then one term was subtracted to the other side. This allows us to incorporate it into the F function. Taking the derivative with respect to $\lambda _i$ and setting it equal to zero.
$$
\frac{\partial F}{\partial \lambda _i} = \frac{-C_i}{\delta _i+ \lambda _i} + \frac{K}{\lambda _i} = 0
\qquad \longrightarrow \qquad
\lambda _i = \frac{\delta _i}{C_i/K -1}
$$
\noindent
Plugging this into Eq. \ref{restrict} yields the following
$$
\prod\limits _{i=1}^N \lambda _i = \lambda _0 = \frac{\prod\limits_{i=1}^N\delta_i}{\prod\limits_{i=1}^N \frac{C_i}{K} - 1}
$$
\noindent
The result is an $N^{th}$ order equation for K
\begin{equation} \label{Kequation}
\prod\limits_{i=1}^N \left( \frac{C_i}{K} - 1 \right) = \frac{1}{\lambda _0} \prod\limits_{i=1}^N\delta _i
\end{equation}
\noindent
Assuming constant C ($C_i = C$) Eq. \ref{Kequation} reduces to
$$
\left(\frac{C}{K} - 1 \right)^N = \frac{1}{\lambda _0} \prod\limits_{i=1}^N \delta_i = \prod\limits_{i=1}^N\left(\frac{\delta _i}{\lambda _i}\right)
$$
\noindent
Then $\delta_i/\lambda _i$ must be constant and can be expressed as
$$
\lambda _i = \lambda _0 ^{1/N} \cdot \frac{\delta _i}{\left( \prod\limits_{i=1}^N\delta _i \right) ^{1/N}}
$$
\noindent
Adding to the assumptions that the dead weight ratio is constant, $\delta _i = \delta$ then we have the following
\begin{equation} \label{equal}
\lambda _i = \lambda _0 ^{1/N}
\end{equation}
\noindent
From taking the derivative and plugging it back into the Lagrange equation $F$ we arrive to the relationship above. Which demonstrates for a maximum $\Delta V$ that all the stages need to have an equal payload ratio $\lambda$. Resulting in the final velocity equation
$$
\Delta V_{tot} = c \cdot \sum\limits_{i=1}^N \ln \frac{1}{\delta + \lambda _i} = c \cdot N \ln \frac{1}{\delta + \lambda _0^{1/N}}
$$
\end{document}
\begin{table}
\begin{tabular}{|c|c|}
\hline
Something! & Something! \\
\hline
\end{tabular}
\end{table}
\begin{table}
\centering
\makebox[0pt][c]{%
\begin{minipage}[c]{0.58\linewidth}
INPUT TEXT OR PICTURE
\end{minipage}%
\hspace{1cm}
\begin{minipage}[c]{0.35\linewidth}
\begin{figure}
\includegraphics[width= 0.8\textwidth]{}
\caption*{ }
\label{}
\end{figure}
\end{minipage}%
}
\end{table}
\begin{itemize}
\itemsep -4pt
\item[a.)]
\item[b.)]
\end{itemize}