Commit cd03c378 authored by Olivier's avatar Olivier

Chapters 1-4: added comics

I had a pretty awful week so I rewarded myself with a couple of hours working
on something fun, adding links to some of my favorite comics to the script.
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Like all matter, fluids are made of discrete, solid molecules. However, in fluid mechanics, we work at the macroscopic scale: at that scale, matter can be treated like a \vocab{continuum}, in which all physical properties of interest can be continuously differentiated.
\youtubethumb{37K7GxYnebk}{why not to calculate the movement of molecules}{\oc (\ccby)}
\youtubethumb{37K7GxYnebk}{why not to calculate the movement of molecules}{\oc (\ccby)}{}
There are about \num{2e22} molecules in the air within an “empty” 1-\si{liter} bottle at ambient temperature and pressure. Even when the air within the bottle is completely still, these molecules are constantly colliding with each other and with the bottle walls; on average, their speed is equal to the speed of sound: approximately \SI[per-mode = symbol]{1000}{\kilo\metre\per\hour}.
Despite the complexity of individual molecule movements, even the most turbulent flows can be accurately described and solved by considering the velocities of \emph{groups} of several millions of molecules collectively, which we name \vocab{fluid particles}.\wupen{Fluid parcel} By doing so, we never find out the velocity of individual molecules: instead, those are averaged in space and time and result in much simpler and smoother trajectories, which are those we can observe with macroscopic instruments such as video cameras and pressure probes.
\xkcdthumb{2283}{how (not) to picture big numbers}{0.5}{}
Our selection of an appropriate fluid particle size (in effect defining the lower boundary of the \vocab{macroscopic scale}), is illustrated in \cref{fig_particle}. We choose to reduce our volume of study to the smallest possible size before the effect of individual molecules becomes meaningful.
\begin{figure}
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Adopting this point of view, which is named the \vocab{continuum abstraction}, is not a trivial decision, because the physical laws which determine the behavior of molecules are very different from those which determine the behavior of elements of fluid. For example, in fluid mechanics we never consider any inter-element attraction or repulsion forces; while new forces “appear” due to pressure or shear effects that do not exist at a molecular level.
\xkcdthumb{2283}{how (not) to picture big numbers}{0.5}
A direct benefit of the continuum abstraction is that the mathematical complexity of our problems is greatly simplified. Finding the solution for the bottle of “still air” mentioned above, for example, requires only a single equation (eq.~\ref{eq_gradp} p.~\pageref{eq_gradp}) instead of a system of \num{2e22} equations with \num{2e22} unknowns (all leading up to $\vec{V}_{\text{average}, x, y, z, t} = \vec{0}$!).
Another consequence is that we cannot treat a fluid as if it were a mere set of marbles with no interaction which would hit objects as they move by. Instead we must think of a fluid –even a low-density fluid such as atmospheric air– as an infinitely-flexible medium able to fill in almost instantly all of the space made available to it.
......@@ -57,6 +57,7 @@
\subsection{Theory, numerics, and experiment}
\smbcthumb{2010-08-29}{cooperation between theoretical and experimental scientists}{0.5}{}
Today, fluid dynamicists are typically specializing in any one of three sub-disciplines:
\begin{description}
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\clearpage %homemade
\subsection{Viscosity}
\youtubethumb{5YUFt-V_kdk}{how to make sense of viscosity}{\oc (\ccby)}
\youtubethumb{5YUFt-V_kdk}{how to make sense of viscosity}{\oc (\ccby)}{}
We have said above that a fluid element can deform continuously under pressure and shear efforts: it will never “snap” or break apart. However this deformation is not “for free”: it will require force and energy inputs which are not reversible (they are not reversed if the motion is reversed). Resistance to straining in a fluid is measured with a property named \vocab{viscosity}.\wupen{Viscosity}
......@@ -336,7 +337,7 @@
\label{ch_basic_flow_quantities}
\youtubethumb{H5l--WlCZQ8}{how to deal with the $\perp$ symbol when calculating mass flow}{\oc (\ccby)}
\youtubethumb{H5l--WlCZQ8}{how to deal with the $\perp$ symbol when calculating mass flow}{\oc (\ccby)}{}
A few fluid-flow related quantities can be quantified easily and are worth listing here.
\begin{description}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{One-dimensional flow problems}
\agthumb{237}{how many dimensions are enough for you?}{0.5}{}
The method we develop here is called \vocab{integral analysis}, because it involves calculating the overall (integral) effect of the fluid flowing through a considered volume. In this chapter, we consider one-dimensional flows (at least in a loose definition); we will consider more advanced cases in \chapterthree.
For now, we are interested in flows where four conditions are met:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of momentum}
\agthumb{338}{Newton’s laws of motion almost didn’t happen}{0.7}{}
What force is applied to the fluid for it to travel through the control volume? We answer this question by writing out a mass balance equation in the template provided by the Reynolds transport theorem (eq.~\ref{eq_rtt}).
We now state that the placeholder variable $B$ is momentum $m \vec V$. It follows that $\inlinetimederivative{B}$ becomes $\inlinetimederivative{m \vec V_\sys)}$, which by definition is the net force $\vec F_\net$ applying to the system (see eq.~\ref{eq_secondlaw} p.~\pageref{eq_secondlaw}). Also, $b \equiv B/m = m \vec V/m = \vec V$ and now the Reynolds transport theorem becomes:\dontbreakpage
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\end{IEEEeqnarray}
\end{mdframed}
\youtubethumb{iDCpqoJJSI4}{Making sense of the 3D linear momentum balance equation}{\oc (\ccby)}
\youtubethumb{iDCpqoJJSI4}{Making sense of the 3D linear momentum balance equation}{\oc (\ccby)}{}
Sometimes, the control volume has well-defined inlets and outlets through which the flow $\rho \vec V (\vec V_\rel \cdot \vec n)$ is uniform: this is illustrated again in figure~\ref{fig_cv_linearmom_simple}. In that case equation~\ref{eq_rtt_linearmom} reduces to forms that we have already identified in the previous chapter (see \S\ref{ch_balance_momentum_simple} p.~\pageref{ch_balance_momentum_simple}):
\begin{IEEEeqnarray}{rCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \sum_\out \left\{ (\rho |V_\perp| A) \vec V\right\} - \sum_\inn \left\{ (\rho |V_\perp| A) \vec V\right\} \label{eq_rtt_linearmom_simple}\\
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\label{fig_cv_linearmom_simple}
\end{figure}
\youtubethumb{S6JKwzK37_8}{as a person walks, the deflection of the air passing around their body can be used to sustain the flight of a paper airplane (a \textit{walkalong glider}\wupen{Walkalong glider}). Can you figure out the momentum flow entering and leaving a control volume surrounding the glider, and the resulting net force?}{Y:sciencetoymaker (\styl)}
\youtubethumb{S6JKwzK37_8}{as a person walks, the deflection of the air passing around their body can be used to sustain the flight of a paper airplane (a \textit{walkalong glider}\wupen{Walkalong glider}). Can you figure out the momentum flow entering and leaving a control volume surrounding the glider, and the resulting net force?}{Y:sciencetoymaker (\styl)}{}
To make clear a few things, let us focus on the simple case where a considered volume has only one inlet (point~1) and one outlet (point~2). From equation~\ref{eq_rtt_linearmom}, the net force $\vec F_\net$ applying on the fluid is:
\begin{IEEEeqnarray}{rCl}
\vec F_\net & = & \timederivative{}\iiint_\cv \rho \vec V \diff \vol + \iint \rho_2 |V_{\perp 2}| \vec V_2 \diff A_2 - \iint \rho_1 |V_{\perp 1}| \vec V_1 \diff A_1 \nonumber\\\label{eq_fnet_twovectors_unsteady}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of angular momentum}
\youtubethumb{4cvGGxTsQx0}{rocket landing gone wrong. Can you compute the moment exerted by the top thruster around the base of the rocket as it (unsuccessfully) attempts to compensate for the collapsed landing leg?}{Y:SciNews (\styl)}
\youtubethumb{4cvGGxTsQx0}{rocket landing gone wrong. Can you compute the moment exerted by the top thruster around the base of the rocket as it (unsuccessfully) attempts to compensate for the collapsed landing leg?}{Y:SciNews (\styl)}{}
What moment (“twisting effort”) is applied to the fluid for it to travel through the control volume? We answer this question by writing an angular momentum balance (see eq.~\ref{eq_secondlawmom} p.~\pageref{eq_secondlawmom}) in the template provided by the Reynolds transport theorem (eq.~\ref{eq_rtt}).
We position ourselves at a point $X$, about which we measure all moments. All positions are measured with a position vector $\vec r_{\X m}$. We now state that the placeholder variable $B$ is angular momentum $\vec r_{\X m} \wedge m \vec V$. It follows that $\inlinetimederivative{B}$ becomes $\inlinetimederivative{\vec r_{\X m} \wedge m \vec V_\sys}$, which by definition is the net moment $\vec M_\net$ applying to the system (see again eq.~\ref{eq_secondlawmom} p.~\pageref{eq_secondlawmom}). Also, $b \equiv B/m = \vec r_{\X m} \wedge \vec V$ and now the Reynolds transport theorem becomes:\dontbreakpage
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\youtubethumb{VR8LGr6PuRY}{Making sense of the angular momentum balance equation}{\oc (\ccby)}
\youtubethumb{VR8LGr6PuRY}{Making sense of the angular momentum balance equation}{\oc (\ccby)}{}
Sometimes, the control volume has well-defined inlets and outlets through which the flow $\vec r_{\X m} \wedge \rho \vec V (\vec V_\rel \cdot \vec n)$ is uniform: this is illustrated in figure~\ref{fig_cv_angularmomentum_simple}. In that case equation~\ref{eq_rtt_angularmom} reduces to a more readable form:
\begin{IEEEeqnarray}{rCCCCCl}
\vec M_{\net, \X} & = & \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol &+& \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\\\label{eq_rtt_angularmom_simple}
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%%%%
\subsection{Magnitude of the pressure force}
\youtubethumb{ke4Rf9FrHnw}{when pressure-induced forces in static fluids matter: 24 hours of heavy tonnage transit through the \textit{Miraflores} locks in Panama. Can you quantify the force applying on a single lock door?}{Y:XyliboxFrance (\styl)}
\youtubethumb{ke4Rf9FrHnw}{when pressure-induced forces in static fluids matter: 24 hours of heavy tonnage transit through the \textit{Miraflores} locks in Panama. Can you quantify the force applying on a single lock door?}{Y:XyliboxFrance (\styl)}{}
What is the force with which a fluid pushes against a wall?
When the pressure $p$ exerted is uniform and the wall is flat, the resulting force $F$ is easily calculated:
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\end{IEEEeqnarray}
\youtubetopthumb{0d8gfsKllmU}{Two weird things about pressure in fluid dynamics}{\oc (\ccby)}
\youtubetopthumb{0d8gfsKllmU}{Two weird things about pressure in fluid dynamics}{\oc (\ccby)}{}
Equation~\ref{eq_def_pressure} may appear unsettling at first sight, because as $A$ tends to zero, $F_\perp$ also tends to zero; nevertheless, in any continuous medium, the ratio of these two terms tends to a single non-zero value: the local pressure.
This brings us to the second particularity of pressure in fluids: the pressure on either side of the infinitesimal flat surface is the same regardless of its orientation. In other words, \emph{pressure has no direction}: there is only one (scalar) value for pressure at any one point in space.
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F_{\text{net, pressure}, z} & = & \diff \vol \frac{-\partial p}{\partial z}
\end{IEEEeqnarray*}
\youtubethumb{wjRWNBOX3OM}{How many dimensions does it take to describe the effect of pressure?}{\oc (\ccby)}
\youtubethumb{wjRWNBOX3OM}{How many dimensions does it take to describe the effect of pressure?}{\oc (\ccby)}{}
This is tedious to write, but we recognize a pattern. And indeed, we introduce the concept of \vocab{gradient}, a mathematical operator, defined as so (see also Appendix~\ref{appendix_field_operators} p.~\pageref{appendix_field_operators}):
\begin{IEEEeqnarray}{rCl}
\gradient{} \equiv \vec i \partialderivative{}{x} + \vec j \partialderivative{}{y} + \vec k \partialderivative{}{z} \label{eq_def_gradient}
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Since the rate of pressure change depends on pressure, it also varies with altitude, and the calculation of pressure differences in the atmosphere is a little more complicated than for water.
\xkcdthumb{2153}{effects of high altitude}{0.7}{}
If we focus on a moderate height change, it may be reasonable to consider that temperature $T$, the gravitational acceleration $g$ and the gas constant~$R$ are uniform. In this (admittedly restrictive) case, equation~\ref{eq_dpdzatm} can be integrated as so:
\begin{IEEEeqnarray}{rCl}
\derivative{p}{z} & = & \frac{g}{R T_\cst} p \nonumber\\
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\subsection{Buoyancy}
\youtubethumb{ViuQKqUQ1U8}{playing around with an air pump and a vacuum chamber}{Y:Roobert33 (\styl)}
\youtubethumb{ViuQKqUQ1U8}{playing around with an air pump and a vacuum chamber}{Y:Roobert33 (\styl)}{}
Any solid body immersed within a fluid is subjected to pressure on its walls. When the pressure is not uniform (for example because the fluid is subjected to gravity, although this may not be the only cause), then the net force due to fluid pressure on the body walls will be non-zero.
When the fluid is purely static, this net pressure force is called \vocab{buoyancy}. Since in this case, the only cause for the pressure gradient is gravity, the net pressure force is oriented upwards. The buoyancy force is completely independent from (and may or may not compensate) the object’s weight.
\youtubethumb{PoDgQ6KsnOk}{What is so special about buoyancy?}{\oc (\ccby)}
\youtubethumb{PoDgQ6KsnOk}{What is so special about buoyancy?}{\oc (\ccby)}{}
Since it comes from equation~\ref{eq_gradp} that the variation of pressure within a fluid is caused solely by the fluid’s weight, we can see that the force exerted on an immersed body is equal to the weight of the fluid it replaces (that is to say, the weight of the fluid that would occupy its own volume were it not there). This relationship is sometimes named \vocab{Archimedes’ principle}. The force which results from the static pressure gradient applies to all immersed bodies: a submarine in an ocean, an object in a pressurized container, and of course, the reader of this document as presently immersed in the earth’s atmosphere.
\begin{figure}[ht!]
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Abstruse Goose is a comic by an anonymous artist, published at https://abstrusegoose.com/
It is published under a Creative Commons Attribution-Non-Commercial (CC-BY-NC) license, https://creativecommons.org/licenses/by-nc/3.0/us/
Saturday Morning Breakfast Cereal (SMBC) is a comic by Zach Weinersmith, published at https://www.smbc-comics.com/
It is fully copyrighted.
We the Robots is a comic by Chris Harding, published at http://www.wetherobots.com/
It is fully copyrighted.
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