Commit f71c4b97 by Olivier

### Slides for chapter 3

parent be496308
3/3e2.tex 0 → 100644
 \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}
3/3v1.tex 0 → 100644
 \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}
3/3v2.tex 0 → 100644
 \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}
3/3v3.tex 0 → 100644
 \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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