Commit f71c4b97 authored by Olivier's avatar Olivier

Slides for chapter 3

parent be496308
\documentclass[17pt]{beamer}
\usepackage{fluidmechslides} % from https://framagit.org/olivier/sensible-styles
% Syntax for single-image slides:
% (the first argument (number) being the maximum fraction of the
% slide width that the image is allowed to have.)
% \figureframe{1}{filename}{Title}{Attribution}
% Do I want the print version? (no \pause, larger preamble)
\printversion
\begin{document}
\figureframe{0.9}{example_non-uniform_integral_1}{}{\wcfile{Integral analysis with a nonuniform flow.svg}{Figure} \cczero \oc}
\begin{frame}{Balance of momentum}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net
& = &\ \timederivative{} \iiint_\cv \rho {\color{nicegreen}\vec V} \diff \vol \nonumber\\
&& + \iint_\cs \rho {\color{nicegreen}\vec V} \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\label{eq_rtt_linearmom}
\end{IEEEeqnarray*}
\end{mdframed}
% \vspace{-1cm}
{\footnotesize the vector sum of forces on the fluid\\
= the rate of change of momentum within the control volume\\
+ the net flow of momentum through the control surface\\\par}
\end{frame}
\figureframe{1}{barge1}{}{\wcfile{Barges from the St Louis Arch.jpg}{Photo} \ccby by \flickrname{55363387@N00}{Tyler Craft}}
\figureframe{1}{barge2}{}{\wcfile{Barges from the St Louis Arch.jpg}{Photo} \ccby by \flickrname{55363387@N00}{Tyler Craft}}
\figureframe{1}{barge3}{}{\wcfile{Barges from the St Louis Arch.jpg}{Photo} \ccby by \flickrname{55363387@N00}{Tyler Craft}}
\figureframe{1}{barge4}{}{\wcfile{Barges from the St Louis Arch.jpg}{Photo} \ccby by \flickrname{55363387@N00}{Tyler Craft}}
\begin{frame}
\begin{centering}
$\sim$\\
A worked-out example\\
for the momentum equation\\
with non-uniform flow\\
$\sim$
\end{centering}
\end{frame}
\figureframe{0.9}{profile_integral_analysis}{}{\wcfile{Velocity profiles integral analysis.svg}{Figure} \cczero \oc}
\figureframe{0.9}{example_non-uniform_integral_1}{}{\wcfile{Integral analysis with a nonuniform flow.svg}{Figure} \cczero \oc}
\figureframe{0.9}{example_non-uniform_integral_2}{}{\wcfile{Integral analysis with a nonuniform flow.svg}{Figure} \cczero \oc}
\figureframe{0.9}{example_non-uniform_integral_3}{}{\wcfile{Integral analysis with a nonuniform flow.svg}{Figure} \cczero \oc}
\begin{frame}{}
\small
Remove unsteady terms, split inlet/outlet
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & \iint_{\cs\ \inn} \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A + \iint_{\cs\ \out} \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & + \iint_{\cs\ \inn} \rho \vec V \ (V_\perp) \diff A + \iint_{\cs\ \out} \rho \vec V \ (V_\perp) \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Hoop \#1: sign of $V_\perp$\pause
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & + \iint_{\cs\ \inn} \rho \vec V \ (V_\perp) \diff A + \iint_{\cs\ \out} \rho \vec V \ (V_\perp) \diff A\\\pause
& = & - \iint_{\cs\ \inn} \rho \vec V \ |V_{\perp\ \inn}| \diff A + \iint_{\cs\ \out} \rho \vec V \ |V_{\perp \out}| \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
All vectors in $x$-direction: drop vectors
Hoop \#2: sign of $V$\pause
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & - \iint_{\cs\ \inn} \rho \vec V \ |V_{\perp\ \inn}| \diff A + \iint_{\cs\ \out} \rho \vec V \ |V_{\perp \out}| \diff A\\\pause
F_\net &=& - \iint_{\cs\ \inn} \rho ({\color{niceblue}V_\inn}) \ |V_{\perp\ \inn}| \diff A\\
&& + \iint_{\cs\ \out} \rho ({\color{niceblue}V_\out}) \ |V_{\perp \out}| \diff A
\end{IEEEeqnarray*}\pause
\footnotesize($V$ positive or negative depending on direction). Here,\small
\begin{IEEEeqnarray*}{cCl}
F_\net &=& - \iint_{\cs\ \inn} \rho |V_\inn| \ |V_\inn| \diff A\\
&& + \iint_{\cs\ \out} \rho |V_\out| \ |V_\out| \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Time to integrate:
\begin{IEEEeqnarray*}{cClCl}
F_\net &=& - \iint_{\cs\ \inn} \rho |V_\inn| \ |V_\inn| \diff A &+& \iint_{\cs\ \out} \rho |V_\out| \ |V_\out| \diff A\\\pause
&=& - \iint_{\cs\ \inn} \rho V_\inn^2 \diff y \diff z &+& \iint_{\cs\ \out} \rho V_\out^2 \diff y \diff z\\\pause
&=& - \Delta z \int_{\cs\ \inn} \rho V_\inn^2 \diff y &+& \Delta z \int_{\cs\ \out} \rho V_\out^2 \diff y\\\pause
&=& - \Delta z \ \rho \int_{y_{1 \inn}}^{y_{2 \inn}} V_\inn^2 \diff y &+& \Delta z \ \rho \int_{y_{1 \out}}^{y_{2 \out}} V_\out^2 \diff y\pause
\end{IEEEeqnarray*}
\footnotesize caution: $y_n$ integration limits may differ between inlet and outlet
\end{frame}
\begin{frame}{}
\small
Integrating in our case:
\begin{IEEEeqnarray*}{cClCl}
F_\net &=& - \Delta z \ \rho \int_{y_{1 \inn}}^{y_{2 \inn}} V_\inn^2 \diff y &+& \Delta z \ \rho \int_{y_{1 \out}}^{y_{2 \out}} V_\out^2 \diff y\\\pause
&=& - \Delta z \ \Delta y_\inn \ \rho V_\inn^2 &+& \Delta z \ \rho \int_{y_{1 \out}}^{y_{2 \out}} V_\out^2 \diff y\\\pause
\end{IEEEeqnarray*}
Numbers!
\begin{IEEEeqnarray*}{cClCl}
&=& - \num{1} \times \num{2} \times \num{1,225} \times \num{20}^2\\
&&+ \num{1} \times \num{1,225} \times \int_{0}^{2} (19 + y)^2 \diff y\\\pause
F_\net &=& \SI{+885,3}{\newton}
\end{IEEEeqnarray*}
(wow!!!)
\end{frame}
\figureframe{0.9}{example_non-uniform_integral_1}{}{\wcfile{Integral analysis with a nonuniform flow.svg}{Figure} \cczero \oc}
\end{document}
\documentclass[17pt]{beamer}
\usepackage{fluidmechslides} % from https://framagit.org/olivier/sensible-styles
% Syntax for single-image slides:
% (the first argument (number) being the maximum fraction of the
% slide width that the image is allowed to have.)
% \figureframe{1}{filename}{Title}{Attribution}
% Do I want the print version? (no \pause, larger preamble)
\printversion
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Balance of momentum}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net \pause
& = &\ \timederivative{} \iiint_\cv \rho {\color{nicegreen}\vec V} \diff \vol \nonumber\\\pause
&& + \iint_\cs \rho {\color{nicegreen}\vec V} \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\label{eq_rtt_linearmom}
\end{IEEEeqnarray*}
\end{mdframed}
% \vspace{-1cm}
{\footnotesize the vector sum of forces on the fluid\\
= the rate of change of momentum within the control volume\\
+ the net flow of momentum through the control surface\\\par}
\end{frame}
\figureframe{0.8}{simple_cv_fnet0}{}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
\figureframe{0.8}{simple_cv_fnet}{}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
\begin{comment}
\begin{frame}
If $\left\{\rho \vec V (\vec V_\rel \cdot \vec n)\right\}$ is uniform at each inlet/outlet: \pause
\begin{IEEEeqnarray*}{rClcl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol &+& \sum_\out \left\{ (\rho |V_\perp| A) \vec V\right\} \nonumber\\
&&& -& \sum_\inn \left\{ (\rho |V_\perp| A) \vec V\right\} \nonumber\\\label{eq_rtt_linearmom_simple}\\\pause
& = & \timederivative{} \left(m \vec V \right)_\cv &+& \sum_\out \left\{ |\dot m| \vec V\right\}\nonumber\\
&&& - &\sum_\inn \left\{ |\dot m| \vec V\right\}\label{eq_rtt_linearmom_simple_two}
\end{IEEEeqnarray*}
\end{frame}
\end{comment}
\begin{frame}{One inlet (1), one outlet (2)}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{}\iiint_\cv \rho \vec V \diff \vol \nonumber\\
&&+ \iint_\out \rho_2 |V_{\perp 2}| \vec V_2 \diff A_2\nonumber\\
&& - \iint_\out \rho_1 |V_{\perp 1}| \vec V_1 \diff A_1 \nonumber\\\label{eq_fnet_twovectors_unsteady}
\end{IEEEeqnarray*}
What could create a force $\vec F_\net$?
\end{frame}
\begin{frame}{}
The first term, $\timederivative{}\iiint_\cv \rho \vec V \diff \vol$:\pause
Momentum inside the control volume may change if:\pause
\begin{itemize}
\item the distribution of velocities is changing;\pause
\item the distribution of the mass following those velocities is changing.
\end{itemize}
\end{frame}
\pictureframe{1}{Slosh_in_Pool_20150719}{}{\wcfile{Slosh_in_Pool_20150719}{Photo} \ccbysa by \wcu{Ka23 13}}
\begin{frame}{One inlet (1), one outlet (2)}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{}\iiint_\cv \rho \vec V \diff \vol \nonumber\\
&&+ \iint_\out \rho_2 |V_{\perp 2}| \vec V_2 \diff A_2\nonumber\\
&& - \iint_\out \rho_1 |V_{\perp 1}| \vec V_1 \diff A_1 \nonumber\\\label{eq_fnet_twovectors_unsteady}
\end{IEEEeqnarray*}
What could create a force $\vec F_\net$?
\end{frame}
\begin{frame}{}
The second term, $|\dot m|_2 \vec V_2 - |\dot m|_1 \vec V_1$:\pause
Possible reasons why not $\vec 0$:\pause
\begin{itemize}
\item mass flow $\dot m$ is different;\pause
\item $\vec V_1$ and $\vec V_2$ have different lengths;\pause
\item $\vec V_1$ and $\vec V_2$ have different directions;\pause
\item $\vec V_1$ and $\vec V_2$ have different \emph{distributions}.
\end{itemize}
\end{frame}
\pictureframe{1}{130828-N-ZZ999-013}{V2 longer than V1}{\wcfile{Sailors practice firefighting in Timor-Leste. (9623221318).jpg}{Photo} by Jon Marzullo, U.S. Navy (\pd)}
\pictureframe{1}{RRobertson_Sumburgh_Rescue_IMG_6242_(23395830674).jpg}{}{\wcfile{RRobertson_Sumburgh_Rescue_IMG_6242_(23395830674).jpg}{Photo} \ccbysa \flickrname{16633132@N04}{Ronnie Robertson}}
\pictureframe{1}{Arabsat-6A_Mission_(40628435833)}{}{\wcfile{File:Arabsat-6A Mission (40628435833).jpg}{Photo} \cczero by SpaceX}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{centering}
$\sim$\\
A worked-out example\\
for the momentum equation\\
$\sim$
\end{centering}
\end{frame}
\oldpictureframe{}{blast_deflector1}{1}{\wcfile{US Navy 100329-N-4236E-356 Lt. Scott Ryan inspects the jet blast deflectors.jpg}{Photo} by Chad R. Erdmann, U.S. Navy (\pd)}
\oldpictureframe{}{blast_deflector2}{1}{\wcfile{US Navy 030914-N-2009B-028 F-A 18C Hornets from the Argonauts of Strike Fighter Squadron One Four Seven (VFA-147) are positioned for launch.jpg}{Photo} by Bridgette Beaudion, U.S. Navy (\pd)}
\figureframe{0.7}{example_blast_deflector_1}{}{\wcfile{Flat exhaust deflector simplified flow.svg}{Figure} \cczero}
\figureframe{0.7}{example_blast_deflector_2}{}{\wcfile{Flat exhaust deflector simplified flow.svg}{Figure} \cczero}
\figureframe{0.6}{example_blast_deflector_3}{}{\wcfile{Flat exhaust deflector simplified flow.svg}{Figure} \cczero}
\begin{frame}{}
\small
Remove unsteady terms, split inlet/outlet
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & \iint_{\cs\ \inn} \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A + \iint_{\cs\ \out} \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & + \iint_{\cs\ \inn} \rho \vec V \ (V_\perp) \diff A + \iint_{\cs\ \out} \rho \vec V \ (V_\perp) \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Hoop \#1: sign of $V_\perp$
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & + \iint_{\cs\ \inn} \rho \vec V \ (V_\perp) \diff A + \iint_{\cs\ \out} \rho \vec V \ (V_\perp) \diff A\\\pause
& = & - \iint_{\cs\ \inn} \rho \vec V \ |V_{\perp\ \inn}| \diff A + \iint_{\cs\ \out} \rho \vec V \ |V_{\perp \out}| \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Velocities are uniform: drop integrals
\begin{IEEEeqnarray*}{rCccc}
\vec F_\net & = & - \iint_{\cs\ \inn} \rho \vec V \ |V_{\perp\ \inn}| \diff A &+& \iint_{\cs\ \out} \rho \vec V \ |V_{\perp \out}| \diff A\\\pause
&=& - \rho \vec V_\inn \ |V_{\perp\ \inn}| A_\inn &+& \rho \vec V_\out \ |V_{\perp \out}| A_\out\\\pause
&=& - |\dot m_\inn| \vec V_\inn &+& |\dot m_\out| \vec V_\out\\\pause
&=& |\dot m| \left(\vec V_\out - \vec V_\inn \right)\\\pause
&=& |\dot m| \left(\vec V_2 - \vec V_1 \right)
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Split into components
\begin{IEEEeqnarray*}{rCl}
\vec F_\net &=& |\dot m| \left(\vec V_2 - \vec V_1 \right)\\\pause
\end{IEEEeqnarray*}
\begin{IEEEeqnarray*}{cClCl}
\left\{ \begin{array}{rcl}
F_{\net x} & = & |\dot m| \left(V_{2x} - V_{1x} \right)\\
F_{\net y} & = & |\dot m| \left(V_{2y} - V_{1y} \right)\\
\end{array}\right.
\end{IEEEeqnarray*}
Hoop \#2: sign of components $V_x$ and $V_y$\pause
\begin{IEEEeqnarray*}{cCccc}
F_{\net\ x} &=& |\dot m| \left[\left({\color{niceblue}V_{2 x}}\right) - \left({\color{niceblue}V_{1 x}}\right) \right]\\
F_{\net\ y} &=& |\dot m| \left[\left({\color{niceblue}V_{2 y}}\right) - \left({\color{niceblue}V_{1 y}}\right) \right]\\
\end{IEEEeqnarray*}
\footnotesize positive or negative according to coordinate system
\end{frame}
\begin{frame}{}
\small
Numbers!
\begin{IEEEeqnarray*}{rCl}
F_{\net\ x} &=& |\dot m| \left[\left({\color{niceblue}V_{2 x}}\right) - \left({\color{niceblue}V_{1 x}}\right) \right]\\
F_{\net\ y} &=& |\dot m| \left[\left({\color{niceblue}V_{2 y}}\right) - \left({\color{niceblue}V_{1 y}}\right) \right]\\
&\\\pause
F_{\net\ x} &=& \num{10} \times \left[\left(\num{-20} \cos \SI{40}{\degree}\right) - \left(\num{-20}\right) \right]\\
F_{\net\ y} &=& \num{10} \times \left[\left(\num{+20} \sin \SI{40}{\degree}\right) - \left(\num{0}\right) \right]\\
&\\\pause
\vec F_\net &=& \left(\begin{array}{c}
\num{+46,8}\\
\num{+128,6}\end{array}\right) ~~(\si{\newton})
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{Balance of momentum}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net
& = &\ \timederivative{} \iiint_\cv \rho {\color{nicegreen}\vec V} \diff \vol \nonumber\\
&& + \iint_\cs \rho {\color{nicegreen}\vec V} \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\label{eq_rtt_linearmom}
\end{IEEEeqnarray*}
\end{mdframed}
% \vspace{-1cm}
{\footnotesize the vector sum of forces on the fluid\\
= the rate of change of momentum within the control volume\\
+ the net flow of momentum through the control surface\\\par}
\end{frame}
\end{document}
\documentclass[17pt]{beamer}
\usepackage{fluidmechslides} % from https://framagit.org/olivier/sensible-styles
% Syntax for single-image slides:
% (the first argument (number) being the maximum fraction of the
% slide width that the image is allowed to have.)
% \figureframe{1}{filename}{Title}{Attribution}
% Do I want the print version? (no \pause, larger preamble)
\printversion
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Balance of momentum}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net \pause
& = &\ \timederivative{} \iiint_\cv \rho {\color{nicegreen}\vec V} \diff \vol \nonumber\\\pause
&& + \iint_\cs \rho {\color{nicegreen}\vec V} \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\label{eq_rtt_linearmom}
\end{IEEEeqnarray*}
\end{mdframed}
% \vspace{-1cm}
{\footnotesize the vector sum of forces on the fluid\\
= the rate of change of momentum within the control volume\\
+ the net flow of momentum through the control surface\\\par}
\end{frame}
\begin{frame}{What is the force on Olivier’s hand?}
\begin{enumerate}
\item \SI{40}{\newton}
\item \SI{20}{\newton}
\item \SI{2}{\newton}
\item \SI{0,2}{\newton}
\end{enumerate}
\end{frame}
\figureframe{0.7}{example_water_pistol_1}{}{}
\figureframe{0.7}{example_water_pistol_2}{}{}
\figureframe{0.7}{example_water_pistol_3}{}{}
\begin{frame}{}
\small
Remove unsteady terms, focus on only outlet, drop integrals (uniform velocity)
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & \iint_{\cs\ \out} \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & + \iint_{\cs\ \out} \rho \vec V \ (V_\perp) \diff A\\\pause
& = & + \iint_{\cs\ \out} \rho \vec V \ |V_{\perp\ \inn}| \diff A\\\pause
& = & + |\dot m| \vec V_\out
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Split into components
\begin{IEEEeqnarray*}{cClCl}
\vec F_\net &=& |\dot m| \vec V_\out\\\pause
F_{\net\ x} &=& |\dot m| \left({\color{niceblue}V_{\out x}}\right)
\end{IEEEeqnarray*}\pause
Numbers!
\begin{IEEEeqnarray*}{rCl}\pause
F_\net &=& \num{0,1} \times \left(\num{-2}\right)\\\pause
&=& \SI{-0,2}{\newton}
\end{IEEEeqnarray*}\pause
ha
\begin{IEEEeqnarray*}{rCl}\pause
F_\text{hand} = -F_\net &=& \SI{+0,2}{\newton}
\end{IEEEeqnarray*}
\end{frame}
\figureframe{0.7}{example_water_pistol_4}{What is the net force now?}{}
\begin{frame}{What is the force now?}
\begin{enumerate}
\item \SI{+0,4}{\newton}
\item \SI{0}{\newton}
\item \SI{-0,2}{\newton}
\item \SI{-0,1}{\newton}
\end{enumerate}
\end{frame}
\begin{frame}{}
\small
Now, one inlet + one outlet
\begin{IEEEeqnarray*}{cCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \\\pause
& = & + |\dot m| \left(\vec V_\out - \vec V_\inn\right)
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Split into components
\begin{IEEEeqnarray*}{cClCl}
\vec F_\net &=& |\dot m| \left(\vec V_\out - \vec V_\inn\right) \\\pause
F_{\net\ x} &=& |\dot m| \left[\left({\color{niceblue}V_{\out x}}\right) - \left({\color{niceblue}V_{\inn x}}\right)\right]
\end{IEEEeqnarray*}\pause
Numbers!
\begin{IEEEeqnarray*}{rCl}\pause
F_\net &=& \num{0,1} \times \left[\left(\num{2}\right) - \left(\num{-2}\right)\right]\\\pause
&=& \SI{+0,4}{\newton}
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{Balance of momentum}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec F_\net
& = &\ \timederivative{} \iiint_\cv \rho {\color{nicegreen}\vec V} \diff \vol \nonumber\\
&& + \iint_\cs \rho {\color{nicegreen}\vec V} \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\label{eq_rtt_linearmom}
\end{IEEEeqnarray*}
\end{mdframed}
% \vspace{-1cm}
{\footnotesize the vector sum of forces on the fluid\\
= the rate of change of momentum within the control volume\\
+ the net flow of momentum through the control surface\\\par}
\end{frame}
\end{document}
\documentclass[17pt]{beamer}
\usepackage{fluidmechslides} % from https://framagit.org/olivier/sensible-styles
% Syntax for single-image slides:
% (the first argument (number) being the maximum fraction of the
% slide width that the image is allowed to have.)
% \figureframe{1}{filename}{Title}{Attribution}
% Do I want the print version? (no \pause, larger preamble)
\printversion
\begin{document}
\begin{frame}{Angular momentum \& RTT}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec M_{\net, \X} &=& \pause \timederivative{} \iiint_\cv {\color{nicegreen}\vec r_{\X m} \wedge } \rho {\color{nicegreen} \vec V} \diff \vol \nonumber\\\pause
&& + \iint_\cs {\color{nicegreen}\vec r_{\X m} \wedge } \rho (\vec V_\rel \cdot \vec n) {\color{nicegreen}\vec V } \diff A
\end{IEEEeqnarray*}
\end{mdframed}
{\footnotesize \setlength{\parskip}{0.2\parskip} The vector sum of moments on the fluid\\
= the rate of change of angular momentum in the \textsc{cv}\\
+ the net flow of angular momentum through the \textsc{cs}\par}
\end{frame}
\figureframe{0.8}{simple_cv_mnet0.png}{}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
\figureframe{0.8}{simple_cv_mnet.png}{}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
\begin{frame}
If $ \vec r_{\X m} \wedge \rho (\vec V_\rel \cdot \vec n) \vec V$ is uniform across each inlet/outlet: \pause
\begin{IEEEeqnarray*}{rCllll}
\vec M_\net & = & \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol \nonumber\\
&&+ \sum_\out \left\{ \vec r_{\X m} \wedge |\dot m| \vec V\right\} - \sum_\inn \left\{ \vec r_{\X m} \wedge |\dot m| \vec V\right\} \nonumber\\&&&&&\ztag{3/13}
\end{IEEEeqnarray*}
\end{frame}
\pictureframe{1}{Nelson_A3000_Accelerator}{}{\wcfile{Nelson_A3000_Accelerator.png}{Photo} \ccbysa by \wcu{Timhall}}
\pictureframe{1}{Impact_Sprinkler_Mechanism_2}{}{\wcfile{Impact Sprinkler Mechanism 2.jpg}{Photo} \ccby by \wcun{JJ_Harrison}{JJ Harrison}}
\oldfigureframe{}{f1352}{1}{\wcfile{Jet engine F135(STOVL variant)'s thrust vectoring nozzle N.PNG}{Drawing} \ccbysa U:Tosaka}
\oldpictureframe{}{f1351}{1}{\wcfile{USMC-100317-C-0000W-002.jpg}{Photo} by Andy Wolfe, US Army (\pd)}
\figureframe{1}{F-35B_Joint_Strike_Fighter_(thrust_vectoring_nozzle_and_lift_fan)}{}{\wcfile{F-35B_Joint_Strike_Fighter_(thrust_vectoring_nozzle_and_lift_fan).PNG}{Figure} \ccby by \wcu{Tosaka}}
\pictureframe{1}{150618-F-EI321-275}{}{\wcfile{150618-F-EI321-275.JPG}{Photo} by Alex R. Lloyd, U.S. Air Force (\pd)}
%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{centering}
$\sim$\\
A worked-out example\\
for the angular momentum equation\\
$\sim$
\end{centering}
\end{frame}
\pictureframe{1}{Arabsat-6A_Mission_(40628438523)}{Falcon Heavy take-off on 2019-04-10}{\wcfile{Arabsat-6A_Mission_(40628438523)}{Photo} by \textsc{nasa} (\pd)}
\pictureframe{1}{Arabsat-6A_Mission_(40628435833)}{Falcon Heavy take-off on 2019-04-10}{\wcfile{Arabsat-6A_Mission_(40628435833)}{Photo} \cczero by SpaceX}
%\pictureframe{1}{Arabsat-6A_Mission_(47593269202)}{Falcon Heavy center stage landing on 2019-04-10}{\wcfile{Arabsat-6A_Mission_(47593269202)}{Photo} \cczero by SpaceX}
\pictureframe{1}{CRS-8_first_stage_landing_(25787998624)}{\textsc{crs}-8 landing on 2016-04-09}{\wcfile{CRS-8 first stage landing (25787998624).jpg}{Photo} \cczero by SpaceX}
\pictureframe{1}{CRS-8_(26239020092)}{\textsc{crs}-8 landing on 2016-04-09}{\wcfile{CRS-8 (26239020092).jpg}{Photo} \cczero by SpaceX}
\figureframe{0.9}{example_rocket_tipover_1}{}{\wcfile{Rocket tip-over.svg}{Figure} \cczero \oc}
\figureframe{0.9}{example_rocket_tipover_2}{}{\wcfile{Rocket tip-over.svg}{Figure} \cczero \oc}
\figureframe{0.9}{example_rocket_tipover_3}{}{\wcfile{Rocket tip-over.svg}{Figure} \cczero \oc}
\begin{frame}{}
\small
Remove unsteady terms, split inlet/outlet
\begin{IEEEeqnarray*}{cClllll}
\vec M_{\net, \X} &=& \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol \\
&& + \iint_\cs \vec r_{\X m} \wedge \rho (\vec V_\rel \cdot \vec n) \vec V \diff A\\\pause
&=& + \iint_\inn \vec r_{\X m} \wedge \rho (\vec V_\rel \cdot \vec n) \vec V \diff A \\
&& + \iint_\out \vec r_{\X m} \wedge \rho (\vec V_\rel \cdot \vec n) \vec V \diff A\\\pause
&=& + \iint_\inn \vec r_{\X m} \wedge \rho V_{\perp\ \inn} \vec V \diff A \\
&& + \iint_\out \vec r_{\X m} \wedge \rho V_{\perp\ \out} \vec V \diff A\\
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Hoop \#1: sign of $V_\perp$
\begin{IEEEeqnarray*}{cClllll}
\vec M_{\net, \X} &=& + \iint_\inn \vec r_{\X m} \wedge \rho V_{\perp\ \inn} \vec V \diff A \\
&& + \iint_\out \vec r_{\X m} \wedge \rho V_{\perp\ \out} \vec V \diff A\\\pause
&=& - \iint_\inn \vec r_{\X m} \wedge \rho |V_{\perp\ \inn}| \vec V \diff A \\
&& + \iint_\out \vec r_{\X m} \wedge \rho |V_{\perp\ \out}| \vec V \diff A
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Uniform velocity at inlets \& outlets
\begin{IEEEeqnarray*}{cClllll}
\vec M_{\net, \X} &=& - \iint_\inn \vec r_{\X m} \wedge \rho |V_{\perp\ \inn}| \vec V \diff A \\
&& + \iint_\out \vec r_{\X m} \wedge \rho |V_{\perp\ \out}| \vec V \diff A\\\pause
&=& -\ \vec r_\text{X-inlet} \wedge \rho |V_{\perp\ \inn}| \vec V_\inn \ A_\inn \\
&& +\ \vec r_\text{X-outlet} \wedge \rho |V_{\perp\ \out}| \vec V_\out \ A_\out \\\pause
&=& -\ \rho |V_{\perp\ \inn}| \ A_\inn \ \vec r_\text{X-inlet} \wedge \vec V_\inn \\
&& +\ \rho |V_{\perp\ \out}| \ A_\out \ \vec r_\text{X-outlet} \wedge \vec V_\out \\\pause
&=& -\ \dot m_\inn \ \vec r_\text{X-inlet} \wedge \vec V_\inn \\
&& +\ \dot m_\out \ \vec r_\text{X-outlet} \wedge \vec V_\out\\
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
Vector cross-product: don’t panic
\begin{IEEEeqnarray*}{cClllll}
\vec M_{\net, \X} &=& -\ \dot m_\inn \ \vec r_\text{X-inlet} \wedge \vec V_\inn \ +\ \dot m_\out \ \vec r_\text{X-outlet} \wedge \vec V_\out\\\pause
&=& -\ \dot m_\inn \ \vec r_\text{X-inlet} \wedge \vec V_{\inn \perp r} \ +\ \dot m_\out \ \vec r_\text{X-outlet} \wedge \vec V_{\out \perp r}\\
\end{IEEEeqnarray*}
\end{frame}
\begin{frame}{}
\small
All right-hand-side vectors in same plane: we drop vectors
Hoop \#2: sign of $V_{\perp r}$
\begin{IEEEeqnarray*}{cClllll}
\vec M_{\net, \X} &=& -\ \dot m_\inn \ \vec r_\text{X-inlet} \wedge \vec V_{\inn \perp r} \ +\ \dot m_\out \ \vec r_\text{X-outlet} \wedge \vec V_{\out \perp r}\\\pause
M_{\net, \X} &=& -\ \dot m_\inn \ r_\text{X-inlet} \times \left({\color{niceblue}V_{\inn \perp r}}\right)\\
&& +\ \dot m_\out \ r_\text{X-outlet} \times \left({\color{niceblue}V_{\out \perp r}}\right)
\end{IEEEeqnarray*}
{\footnotesize($V_{\perp r}$ positive clockwise, negative anticlockwise)}\pause
\begin{IEEEeqnarray*}{cClllll}
M_{\net, \X} &=& 0 +\ \dot m_\out \ r_\text{X-outlet} \times \left(+\left|V_{\out \perp r}\right|\right)
\end{IEEEeqnarray*}\pause
\footnotesize{(length in $z$-direction)}
\end{frame}
\begin{frame}{}
\small
Numbers!
\begin{IEEEeqnarray*}{cClllll}
M_{\net, \X} &=& 0 +\ \dot m_\out \ r_\text{X-outlet} \times \left(+\left|V_{\out \perp r}\right|\right)\\\pause
&=& 0 +\ \num{2,5} \times \num{40} \times \left(+\left|\num{155} \cos \num{20}\right|\right)\\\pause
&=& \SI{+14,7}{\kilo\newton\metre}\\\pause
\vec M_{\net, \X} &=& \left(\begin{array}{c}
0\\
0\\
\num{+14,7e3}\end{array}\right)
\end{IEEEeqnarray*}\pause
Moment exerted on fluid leaving the rocket\\
(moment on rocket by fluid is opposite)
\end{frame}
\begin{frame}{Angular momentum \& RTT}
\begin{mdframed}
\begin{IEEEeqnarray*}{rCl}
\vec M_{\net, \X} &=& \timederivative{} \iiint_\cv {\color{nicegreen}\vec r_{\X m} \wedge } \rho {\color{nicegreen} \vec V} \diff \vol \nonumber\\
&& + \iint_\cs {\color{nicegreen}\vec r_{\X m} \wedge } \rho (\vec V_\rel \cdot \vec n) {\color{nicegreen}\vec V } \diff A
\end{IEEEeqnarray*}
\end{mdframed}
{\footnotesize \setlength{\parskip}{0.2\parskip} The vector sum of moments on the fluid\\
= the rate of change of angular momentum in the \textsc{cv}\\
+ the net flow of angular momentum through the \textsc{cs}\par}
\end{frame}
\end{document}
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