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Commits (5)
• Chapter 7: Major re-work + proof-read · f119a320
Olivier authored
The chapter is still unsatisfying in my eyes. I would like to see
a more systematic method for quantifying pressure difference
in pipes (mixing height, expansion/contraction, local losses,
and wall friction losses).
I would also like to see a good systematic (even if approximate)
exploration of the dependency between the main parameters at hand.

* Delta p_friction is now uniformly refered to as Delta p_loss
* Ditched entrance effects, focus in on fully-developed flow and
on methodology
* Summary added at end to better picture relevance of chapter in
view of the entire course
• Exercises 7: major clean-up · d26ee240
Olivier authored
* New exercise exploring practical implications of having laminar
flow in pipes
* Expanded/strengthened oil pipeline problem
* Kugel fountain exercise de-listed, moved to back
(fun not really critical)
* One unconvincing theory exercise moved to archive
* Wind tunnel design problem also moved to archive (not so well
suited to individual coursework)
• Chapter 9: completed re-structuring · e9210df4
Olivier authored
* Flow parameters (Re, Ma etc) come first, then force coefficients
* New brief section on building models
* Moved flow-parameters-as-force-ratios section to appendix
* Overall re-write & strenthening
• Minor fixes (broken links, numbering) · 379ed284
Olivier authored
 ... ... @@ -42,7 +42,7 @@ \begin{enumerate} \shift{1} \item What is the pressure force exerted on the right side of the plate?\\ \textit{[Hint: we will explore the required expression in \chapterfour as eq.~\ref{eq_pressure_force_scalar} p.\pageref{eq_pressure_force_scalar}]} \textit{[Hint: we will explore the required expression in \chapterfour as eq.~\ref{eq_pressure_force_twod_integration} p.\pageref{eq_pressure_force_twod_integration}]} \end{enumerate} ... ... @@ -63,7 +63,7 @@ \begin{enumerate} \item What is the shear force applying on the plate? \item What would be the shear force if the shear was not uniform, but instead was a function of $x$ expressed (in \si{pascals}) as $\tau_{zx} = \num{1,65} - \num{0,01} \times x^2$?\\ \textit{[Hint: we will explore the required expression in \chapterfive as eq.~\ref{eq_shear_force_twod_integration_general} p.\pageref{eq_shear_force_twod_integration_general}]} \textit{[Hint: we will explore the required expression in \chapterfive as eq.~\ref{eq_shear_force_twod_integration} p.\pageref{eq_shear_force_twod_integration}]} \end{enumerate} ... ... @@ -169,10 +169,10 @@ \tab If you adopt $\ma = \num{0,6}$ as an upper limit, you will obtain $V_\max = \SI{709}{\kilo\metre\per\hour}$ (eqs.~\ref{eq_def_ma} \& \ref{eq_speed_sound_perfect_gas} p.\pageref{eq_speed_sound_perfect_gas}). Note that propellers, fan blades etc. will meet compressiblity effects far sooner. \item [\ref{exo_pressure_induced_force}]% \tab 1) $F_\text{left} = \SI{400}{\kilo\newton}$ (eq.~\ref{eq_first_def_pressure} p.\pageref{eq_first_def_pressure}); \tab 2) $F_\text{right} = \SI{480}{\kilo\newton}$ (eq.~\ref{eq_pressure_force_scalar} p.\pageref{eq_pressure_force_scalar}). \tab 2) $F_\text{right} = \SI{480}{\kilo\newton}$ (eq.~\ref{eq_pressure_force_twod_integration} p.\pageref{eq_pressure_force_twod_integration}). \item [\ref{exo_shear_induced_force}]% \tab 1) $F_1 = \SI{14,85}{\newton}$ (eq.~\ref{eq_first_def_shear} p.\pageref{eq_first_def_shear}); \tab 2) $F_2 = \SI{14,58}{\newton}$ (eq.~\ref{eq_shear_force_twod_integration_general} p.\pageref{eq_shear_force_twod_integration_general}). \tab 2) $F_2 = \SI{14,58}{\newton}$ (eq.~\ref{eq_shear_force_twod_integration} p.\pageref{eq_shear_force_twod_integration}). \item [\ref{exo_speed_sound_newton}]% \tab \SI{26,7}{\degreeCelsius} \& \SI{5,6}{\degreeCelsius}. \item [\ref{exo_power_lost_to_drag}]% ... ...
 ... ... @@ -207,19 +207,19 @@ We can finally rewrite this as: \begin{mdframed} \begin{IEEEeqnarray}{rCl} \gradient{p} &=& \rho \vec g \label{gradp} \gradient{p} &=& \rho \vec g \label{eq_gradp} \end{IEEEeqnarray} \end{mdframed} This is a very useful equation, which states that in a static fluid, the only parameter affecting pressure is gravity. More precisely, the fluid density times the gravity vector is equal to the change in space of the pressure. We will see in \chaptersix that equation~\ref{gradp} is the specific case for a much larger general and powerful equation, the \vocab{Navier-Stokes equation}. But more on that later! We will see in \chaptersix that equation~\ref{eq_gradp} is the specific case for a much larger general and powerful equation, the \vocab{Navier-Stokes equation}. But more on that later! \subsection{Pressure and depth} \label{ch_pressure_and_depth} It is now easy to quantify pressure everywhere inside a static fluid. Very often in studies of static fluids, the $z$-axis is oriented vertically, positive downwards. With this convention, there is no need for a vector equation to quantify pressure, and equation~\ref{gradp} becomes: Very often in studies of static fluids, the $z$-axis is oriented vertically, positive downwards. With this convention, there is no need for a vector equation to quantify pressure, and equation~\ref{eq_gradp} becomes: \begin{IEEEeqnarray}{rCl} \derivative{p}{z} & = & \rho g \label{eq_verticalgradientp} \end{IEEEeqnarray} ... ...
 ... ... @@ -137,7 +137,7 @@ The hinge stands \SI{1,5}{\metre} below the water surface. The window has a leng \label{exo_burj_khalifa} \wherefrom{non-examinable} The integration we carried out in \S\ref{ch_atmospheric_pressure} p.\pageref{ch_atmospheric_pressure} to model the pressure distribution in the atmosphere was based on the hypothesis that the temperature was uniform and constant ($T = T_\cst$). In practice, this may not always be the case. The integration we carried out in with equation~\ref{eq_atmtemp} p.\pageref{eq_atmtemp} to model the pressure distribution in the atmosphere was based on the hypothesis that the temperature was uniform and constant ($T = T_\cst$). In practice, this may not always be the case. \begin{enumerate} \item If the atmospheric temperature decreases with altitude at a constant rate (e.g. of~\SI{-7}{\kelvin\per\kilo\metre}), how can the pressure distribution be expressed analytically? ... ...
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 ... ... @@ -261,6 +261,89 @@ Thus, we can see that if we follow a particle along its path, in a steady, incompressible, frictionless flow with no heat or work transfer, its change in kinetic energy is due only to the result of gravity and pressure, in accordance with the Navier-Stokes equation. \clearpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Flow parameters as force ratios} \label{appendix_flow_parameters} \coveredin{Massey \cite{massey1983}} Instead of the mathematical approach covered in \S\ref{ch_non_dim_ns} p.\pageref{ch_non_dim_ns}, the concept of \vocab{flow parameter} can be approached by \emph{comparing forces} in fluid flows. Fundamentally, understanding the movement of fluids requires applying Newton’s second law of motion: the sum of forces which act upon a fluid particle is equal to its mass times its acceleration. We have done this in an aggregated manner with integral analysis (in \chapterthreeshort, eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}), and then in a precise and all-encompassing way with differential analysis (in \chaptersixshort, eq.~\ref{eq_navierstokes} p.\pageref{eq_navierstokes}). With the latter method, we obtain complex mathematics suitable for numerical implementation, but it remains difficult to obtain rapidly a quantitative measure for what is happening in any given flow. In order to obtain this, an engineer or scientist can use force ratios. This involves comparing the magnitude of a type of force (pressure, viscous, gravity) either with another type of force, or with the mass-times-acceleration which a fluid particle is subjected to as it travels. We are not interested in the absolute value of the resulting ratios, but rather, in having a measure of the parameters that influence them, and being able to compare them across experiments. \subsection[Acceleration vs. viscous forces: the Reynolds number]{Acceleration vs. viscous forces:\\ the Reynolds number} The net sum of forces acting on a particle is equal to its mass times its acceleration. If a representative length for the particle is $L$, the particle mass grows proportionally to the product of its density $\rho$ and its volume $L^3$. Meanwhile, its acceleration relates how much its velocity $V$ will change over a time interval~$\Delta t$: it may be expressed as a ratio $\Delta V/ \Delta t$. In turn, the time interval~$\Delta t$ may be expressed as the representative length $L$ divided by the velocity $V$, so that the acceleration may be represented as proportional to the ratio $V \Delta V/L$. Thus we obtain: \begin{IEEEeqnarray*}{rCl} |\text{net force}| = |\text{mass} \times \text{acceleration}| &\sim& \rho L^3 \frac{V \Delta V}{L}\\ |\vec{F}_\net| &\sim& \rho L^2 V \Delta V \end{IEEEeqnarray*} We now observe the viscous force acting on a particle: it is proportional to the the shear effort and a representative acting surface $L^2$. The shear can be modeled as proportional to the viscosity $\mu$ and the rate of strain, which will grow proportionally to $\Delta V/L$. We thus obtain a crude measure for the magnitude of the shear force: \begin{IEEEeqnarray*}{rCl} |\text{viscous force}| &\sim& \mu \frac{\Delta V}{L} L^2\\ |\vec{F}_\text{viscous}| &\sim& \mu \Delta V L \end{IEEEeqnarray*} The magnitude of the viscous force can now be compared to the net force: \begin{IEEEeqnarray}{rCl} \frac{|\text{net force}|}{|\text{viscous force}|} &\sim& \frac{\rho L^2 V \Delta V}{\mu \Delta V L} = \frac{\rho V L}{\mu} = \re \label{eq_re_forces} \end{IEEEeqnarray} and we recognize the ratio as the Reynolds number (\ref{eq_def_re} p.\pageref{eq_def_re}). We thus see that the Reynolds number can be interpreted as the inverse of the influence of viscosity. The larger $\re$ is, and the smaller the influence of the viscous forces will be on the trajectory of fluid particles. \subsection{Acceleration vs. gravity force: the Froude number} The weight of a fluid particle is equal to its mass, which grows with $\rho L^3$, multiplied by gravity $g$: \begin{IEEEeqnarray*}{rCl} |\text{weight force}| = |\vec{F}_\text{W}| &\sim& \rho L^3 g \end{IEEEeqnarray*} The magnitude of this force can now be compared to the net force: \begin{IEEEeqnarray}{rCl} \frac{|\text{net force}|}{|\text{weight force}|} &\sim& \frac{\rho L^2 V^2}{\rho L^3 g} = \frac{V^2}{L g} = \fr^2 \end{IEEEeqnarray} and here we recognize the square of the Froude number (\ref{eq_def_fr} p.\pageref{eq_def_fr}). We thus see that the Froude number can be interpreted as the inverse of the influence of weight on the flow. The larger $\fr$ is, and the smaller the influence of gravity will be on the trajectory of fluid particles. \subsection{Acceleration vs. elastic forces: the Mach number} \label{ch_mach_number_force_ratios} In some flows called \vocab{compressible flows} the fluid can perform work on itself, and and the fluid particles then store and retrieve energy in the form of changes in their own volume. In such cases, fluid particles are subject to an \vocab{elastic force} in addition to the other forces. We can model the pressure resulting from this force as proportional to the bulk modulus of elasticity $K$ of the fluid (formally defined as $K \equiv \rho \ \partial{p}/\partial{\rho}$); the elastic force can therefore be modeled as proportional to $K L^2$: \begin{IEEEeqnarray*}{rCl} |\text{elasticity force}| = |\vec{F}_\text{elastic}| &\sim& K L^2 \end{IEEEeqnarray*} The magnitude of this force can now be compared to the net force: \begin{IEEEeqnarray*}{rCl} \frac{|\text{net force}|}{|\text{elasticity force}|} &\sim& \frac{\rho L^2 V^2}{K L^2} = \frac{\rho V^2}{K} \end{IEEEeqnarray*} This ratio is known as the Cauchy number; it is not immediately useful because the value of $K$ in a given fluid varies considerably not only according to temperature, but also according to the type of compression undergone by the fluid: for example, it grows strongly during brutal compressions. During isentropic compressions and expansions (isentropic meaning that the process is fully reversible, i.e. without losses to friction, and adiabatic, i.e. without heat transfer), % %we will show in chapter~9 (with eq.~\ref{eq_speed_sound_elasticity_two} p.\pageref{eq_speed_sound_elasticity_two}) it can be shown that the bulk modulus of elasticity is proportional to the square of the speed of sound~$c$: \begin{IEEEeqnarray}{rCl} K|_\text{reversible} &=& c^2 \rho \label{eq_speed_sound_elasticity_one} \end{IEEEeqnarray} The Cauchy number calibrated for isentropic evolutions is then \begin{IEEEeqnarray}{rCl} \frac{|\text{net force}|}{|\text{elasticity force}|_\text{reversible}} &\sim& \frac{\rho V^2}{K} = \frac{V^2}{c^2} = \ma^2 \end{IEEEeqnarray} and here we recognize the square of the Mach number (\ref{eq_def_ma} p.\pageref{eq_def_ma}). We thus see that the Mach number can be interpreted as the influence of elasticity on the flow. The larger $\ma$ is, and the smaller the influence of elastic forces will be on the trajectory of fluid particles. \subsection{Other force ratios} The same method can be applied to reach the definitions for the Strouhal and Euler numbers given in \S\ref{ch_scaling_flows} p.\pageref{ch_scaling_flows}. Other numbers can also be used which relate forces that we have ignored in our study of fluid mechanics. For example, the relative importance of surface tension forces or of electromagnetic forces are quantified using similarly-constructed flow parameters.\\ In some applications featuring rotative motion, such as flows in centrifugal pumps or planetary-scale atmospheric weather, it may be convenient to apply Newton’s second law in a rotating reference frame. This results in the appearance of new reference-frame forces, such as the Coriolis or centrifugal forces; their influence can then be studied using additional flow parameters. In none of those cases can flow parameters give enough information to predict solutions. They do, however, provide quantitative data to indicate which forces are relevant in which places: this not only helps us understand the mechanisms at work, but also distinguish the negligible from the influential, a key characteristic of efficient scientific and engineering work. \clearpage ... ...
 ... ... @@ -6,13 +6,6 @@ \input{a/steer} % May contain \includeonly command genrated with bash script %\includeonly{1/chap1} % to include only one document \newcommand{\mecafluboxtmp}{% \begin{boiboite} {\scriptsize \textcolor{vocabcolor}{You are reading a draft version. Completion expected mid-April 2019.}\par} \end{boiboite} } \begin{document} \include{0/titlepage} ... ...
 \subsubsection{Couette flow} \wherefrom{\cczero \oc} \label{exo_couette_flow} We consider laminar flow of a fluid between two parallel plates (named \vocab{Couette flow}), as shown in \cref{fig_twoplates_exo}. \begin{figure}[h!] \begin{center} \includegraphics[width=0.6\textwidth]{velocity_distribution_couette_flow.png} \end{center} \supercaption{Two-dimensional laminar flow between two plates, also called \vocab{Couette flow}.}{\wcfile{Couette flow flat plate laminar velocity distributions.svg}{Figure} \cczero \oc} \label{fig_twoplates_exo}\vspace{-0.5cm}%handmade \end{figure} \begin{enumerate} \item Starting from the Navier-Stokes equations for incompressible, two-dimensional flow, \begin{IEEEeqnarray}{cCc} \rho \left[ \partialtimederivative{u} + u \partialderivative{u}{x} + v \partialderivative{u}{y} \right] & = & \rho g_x - \partialderivative{p}{x} + \mu \left[ \secondpartialderivative{u}{x} + \secondpartialderivative{u}{y} \right] \\%\ztag{\ref{eq_ns_twodone}}\\ \rho \left[ \partialtimederivative{v} + u \partialderivative{v}{x} + v \partialderivative{v}{y} \right] & = & \rho g_y - \partialderivative{p}{y} + \mu \left[ \secondpartialderivative{v}{x} + \secondpartialderivative{v}{y} \right] \\%\ztag{\ref{eq_ns_twodtwo}} \end{IEEEeqnarray} show that the velocity profile in a horizontal, laminar, steady, fully-developed flow between two horizontal plates separated by a gap of height~$2H$ is: \begin{IEEEeqnarray}{rCl} u &=& \frac{1}{2 \mu} \left(\partialderivative{p}{x} \right) (y^2 - H^2)\ztag{\ref{eq_tmp5}} \end{IEEEeqnarray} \item Why would this equation fail to describe turbulent flow? (Briefly justify your answer, e.g.\ in 30 words or less) \end{enumerate} \subsubsection{Design of a wind tunnel} \wherefrom{non-examinable} \label{exo_wind_tunnel} %homemade Describe the main characteristics of a wind tunnel that could be installed and operated in the room you are standing in. In order to do this: \begin{itemize} \item Start by proposing key characteristics for the test section; \item From these dimensions, draw approximately an air circuit to feed the test section (while attempting to minimize the size of the fan, whose cost increases exponentially with diameter). \item Quantify the static and stagnation pressures along the air circuit, by estimating the losses generated by wall shear and in the bends (you may use data from \cref{fig_loss_coefficients_stator_vanes}); \item Quantify the minimum power required to generate your chosen test section flow characteristics. \end{itemize} \begin{figure} \begin{center} \includegraphics[width=0.5\textwidth]{wind_tunnel_guide_vanes_barlow1999.png}\\ Filter screen: $\eta = \num{0,05}$ \end{center} \supercaption{Loss coefficients $K_L$ (here noted $\eta$) generated by the use of various components within wind tunnel ducts.}{Figure \copyright\xspace Barlow, Rae \& Pope 1999~\cite{barlowraepope1999}} \label{fig_loss_coefficients_stator_vanes} \end{figure} %%%%%%%%%%%%%%%%%%%%%%% \item [\ref{exo_couette_flow}]% \tab The structure is given in the derivation of equation~\ref{eq_tmp5} p.~\pageref{eq_tmp5}, and more details about the math are given in the derivation of the (very similar) equation~\ref{eq_u_lam} p.~\pageref{eq_u_lam}.