 ### Maintenance (examinable tags, numbering, typography)

parent e6902816
 ... ... @@ -39,7 +39,7 @@ \label{fig_pressure_distribution_plate} \end{figure} \begin{enumerate} \item What is the pressure force (i.e. the force resulting from the pressure) exerted on the left side of the plate? \item What is the pressure force (\ie the force resulting from the pressure) exerted on the left side of the plate? \end{enumerate} On the right side of the plate, the water exerts a pressure which is not uniform: it increases with depth. The relation, expressed in \si{pascals}, is: ... ...
 ... ... @@ -182,7 +182,7 @@ %%%%%%%%%%%%%%%%% \subsection{Blasius’ solution} \we{Heinrich Blasius} undertook a PhD thesis under the guidance of Prandtl, in which he focused on the characterization of laminar boundary layers. As part of his work, he showed that the geometry of the velocity profile (i.e. the velocity distribution) within such a layer is \emph{always the same}, and that regardless of the flow velocity or the position, $u$ can be simply expressed as a function of non-dimensionalized distance away from the wall termed~$\eta$: \we{Heinrich Blasius} undertook a PhD thesis under the guidance of Prandtl, in which he focused on the characterization of laminar boundary layers. As part of his work, he showed that the geometry of the velocity profile (\ie the velocity distribution) within such a layer is \emph{always the same}, and that regardless of the flow velocity or the position, $u$ can be simply expressed as a function of non-dimensionalized distance away from the wall termed~$\eta$: \begin{IEEEeqnarray}{rCl} \eta &\equiv& y \ \sqrt{\frac{\rho U}{\mu x}} \end{IEEEeqnarray} ... ... @@ -198,10 +198,10 @@ Based on this work, it can be shown that for a laminar boundary layer flowing along a smooth wall, the four parameters about which we are interested are solely function of the distance-based Reynolds number $\rex$: \begin{IEEEeqnarray}{rCl} \frac{\delta}{x} &=& \frac{\num{4,91}}{\sqrt{\rex}} \IEEEyessubnumber\label{eq_delta_lam}\\\nonumber\\ \frac{\delta^*}{x} &=& \frac{\num{1,72}}{\sqrt{\rex}} \IEEEyessubnumber\label{eq_deltastar_lam}\\\nonumber\\ \frac{\delta^{**}}{x} &=& \frac{\num{0,664}}{\sqrt{\rex}} \IEEEyessubnumber\label{eq_deltastarstar_lam}\\\nonumber\\ c_{f_{(x)}} &=& \frac{\num{0,664}}{\sqrt{\rex}} \IEEEyessubnumber\label{eq_cf_lam} \frac{\delta}{x} &=& \frac{\num{4,91}}{\sqrt{\rex}} \label{eq_delta_lam}\\\nonumber\\ \frac{\delta^*}{x} &=& \frac{\num{1,72}}{\sqrt{\rex}} \label{eq_deltastar_lam}\\\nonumber\\ \frac{\delta^{**}}{x} &=& \frac{\num{0,664}}{\sqrt{\rex}} \label{eq_deltastarstar_lam}\\\nonumber\\ c_{f_{(x)}} &=& \frac{\num{0,664}}{\sqrt{\rex}} \label{eq_cf_lam} \end{IEEEeqnarray} \begin{comment} ... ... @@ -256,7 +256,7 @@ \section{Boundary layer transition} \label{ch_bl_transition} After it has traveled a certain length along the wall, the boundary layer becomes very unstable and it transits rapidly from a laminar to a turbulent regime (\cref{fig_bl_transition}). We have already described the characteristics of turbulence in broadly in \chapterseven and more extensively in \chapternine; they apply to turbulence within the boundary layer. It is worth reminding ourselves that the boundary layer may be turbulent in a globally laminar flow (e.g.\ around an aircraft in flight, the boundary layer is turbulent, but the main flow is laminar). Here, we refer to the regime of the boundary layer only, not the outer flow. After it has traveled a certain length along the wall, the boundary layer becomes very unstable and it transits rapidly from a laminar to a turbulent regime (\cref{fig_bl_transition}). We have already described the characteristics of turbulence in broadly in \chapterseven and more extensively in \chapternine; they apply to turbulence within the boundary layer. It is worth reminding ourselves that the boundary layer may be turbulent in a globally laminar flow (\eg around an aircraft in flight, the boundary layer is turbulent, but the main flow is laminar). Here, we refer to the regime of the boundary layer only, not the outer flow. \begin{figure}[ht!] \begin{center} \includegraphics[width=\textwidth]{boundary_layer_transition.png} ... ... @@ -270,7 +270,7 @@ \re_{x\ \text{transition}} &\approx& \num{5e5} \end{IEEEeqnarray} Transition can be generated earlier if the surface roughness is increased, or if obstacles (e.g.\ turbulators, vortex generators, trip wires) are positioned within the boundary layer. Conversely, a very smooth surface and a very steady, uniform incoming flow will result in delayed transition. Transition can be generated earlier if the surface roughness is increased, or if obstacles (\eg turbulators, vortex generators, trip wires) are positioned within the boundary layer. Conversely, a very smooth surface and a very steady, uniform incoming flow will result in delayed transition. ... ... @@ -304,10 +304,10 @@ In the same way that we have worked with the laminar boundary layer profiles, we can derive models for our characteristics of interest from this velocity profile: \begin{IEEEeqnarray}{rCl} \frac{\delta}{x} &\approx& \frac{\num{0,16}}{\rex^{\frac{1}{7}}} \IEEEyessubnumber\label{eq_delta_turb}\\\nonumber\\ \frac{\delta^*}{x} &\approx& \frac{\num{0,02}}{\rex^{\frac{1}{7}}} \IEEEyessubnumber\label{eq_deltastar_turb}\\\nonumber\\ \frac{\delta^{**}}{x} &\approx& \frac{\num{0,016}}{\rex^{\frac{1}{7}}} \IEEEyessubnumber\label{eq_deltastarstar_turb}\\\nonumber\\ c_{f_{(x)}} &\approx& \frac{\num{0,027}}{\rex^{\frac{1}{7}}} \IEEEyessubnumber\label{eq_cf_turb} \frac{\delta}{x} &\approx& \frac{\num{0,16}}{\rex^{\frac{1}{7}}} \label{eq_delta_turb}\\\nonumber\\ \frac{\delta^*}{x} &\approx& \frac{\num{0,02}}{\rex^{\frac{1}{7}}} \label{eq_deltastar_turb}\\\nonumber\\ \frac{\delta^{**}}{x} &\approx& \frac{\num{0,016}}{\rex^{\frac{1}{7}}} \label{eq_deltastarstar_turb}\\\nonumber\\ c_{f_{(x)}} &\approx& \frac{\num{0,027}}{\rex^{\frac{1}{7}}} \label{eq_cf_turb} \end{IEEEeqnarray} ... ... @@ -353,7 +353,7 @@ We shall remember two crucial points regarding the separation of boundary layers: \begin{enumerate} \item \textbf{Separation occurs in the presence of a positive pressure gradient}, which is sometimes named \vocab{adverse pressure gradient}.\\ Separation points along a wall (e.g. a car bodywork, an aircraft wing, rooftops, mountains) are always situated in regions where pressure increases (positive $\inlinederivative{p}{x}$ or negative $\inlinederivative{U}{x}$). If pressure remains constant, or if it decreases, then the boundary layer cannot separate. Separation points along a wall (\eg a car bodywork, an aircraft wing, rooftops, mountains) are always situated in regions where pressure increases (positive $\inlinederivative{p}{x}$ or negative $\inlinederivative{U}{x}$). If pressure remains constant, or if it decreases, then the boundary layer cannot separate. \item \textbf{Laminar boundary layers are much more sensitive to separation} than turbulent boundary layers (\cref{fig_reynolds_separation_airfoil}).\\ A widely-used technique to reduce or delay the occurrence of separation is to make boundary layers turbulent, using low-height artificial obstacles positioned in the flow. By doing so, we increase shear-based friction (which increases with turbulence) as a trade-off for better resistance to stall. \end{enumerate} ... ...
 ... ... @@ -44,7 +44,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \includegraphics[width=\textwidth]{viscosities_horizontal} \vspace{-1cm} \end{center} \supercaption{Viscosity of various fluids at a pressure of \SI{1}{\bar} (in practice viscosity is almost independent of pressure).}{Figure repeated from \cref{fig_viscosities_various_fluids} p.\pageref{fig_viscosities_various_fluids}; \copyright\xspace White 2008~\cite{white2008}} \supercaption{Viscosity of various fluids at a pressure of \SI{1}{\bar} (in practice viscosity is almost independent of pressure).}{Figure repeated from \cref{fig_viscosities_various_fluids} p.~\pageref{fig_viscosities_various_fluids}; \copyright\xspace White 2008~\cite{white2008}} \label{fig_viscosities_various_fluids_three} \end{figure} ... ... @@ -75,8 +75,8 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \item and at a point further downstream where the boundary layer is turbulent. \end{itemize} \item Draw a few streamlines, indicate the boundary layer thickness~$\delta$, and the displacement thickness~$\delta^*$. \item Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime can be triggered. \item Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime could instead be delayed. \item Explain shortly (\eg in 30 words or less) how the transition to turbulent regime can be triggered. \item Explain shortly (\eg in 30 words or less) how the transition to turbulent regime could instead be delayed. \end{enumerate} ... ... @@ -119,7 +119,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \begin{enumerate} \item If the flow over the wings can be treated as if they were flat plates, what is the power necessary to compensate the shear exerted by the airflow on the wings during flight? \item Which other forms of drag would also be found on the aircraft? (give a brief answer, e.g. in 30 words or less) \item Which other forms of drag would also be found on the aircraft? (give a brief answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -136,7 +136,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \item What is approximately the maximum boundary layer thickness around the fuselage? \item What is approximately the average shear applying on the fuselage skin? \item Estimate the power dissipated to friction on the cylindrical part of the fuselage. \item In practice, in which circumstances could flow separation occur on the fuselage skin? (give a brief answer, e.g. in 30 words or less) \item In practice, in which circumstances could flow separation occur on the fuselage skin? (give a brief answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -148,7 +148,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo Air at \SI{1}{\bar} and \SI{20}{\degreeCelsius} flows along a smooth surface, and decelerates slowly, with a constant rate of $\SI{-0,25}{\metre\per\second\per\metre}$. According to the Pohlhausen model (eq.~\ref{eq_modele_Pohlhausen} p.\pageref{eq_modele_Pohlhausen}), at which distance downstream will separation occur? Is the boundary layer still laminar then? According to the Pohlhausen model (eq.~\ref{eq_modele_Pohlhausen} p.~\pageref{eq_modele_Pohlhausen}), at which distance downstream will separation occur? Is the boundary layer still laminar then? How could one generate such a deceleration in practice? \end{comment} ... ... @@ -200,7 +200,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \partialderivative{u}{x} + \partialderivative{v}{y} & = & 0 \ztag{\ref{eq_ns_bl_lam_deux}} \end{IEEEeqnarray} Identify these two equations, list the conditions in which they apply, and explain shortly (e.g. in 30 words or less) why a boundary layer cannot separate when a favorable pressure gradient is applied along the~wall. Identify these two equations, list the conditions in which they apply, and explain shortly (\eg in 30 words or less) why a boundary layer cannot separate when a favorable pressure gradient is applied along the~wall. \begin{comment} %%%% ... ... @@ -230,11 +230,11 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \NumTabs{2} \begin{description} \item [\ref{exo_water_air_flow}]% \tab 1) At trailing edge $\rex = \num{5348}$ thus the layer is laminar everywhere. $\delta$ will grow from~0 to~\SI{2,01}{\centi\metre} (eq.~\ref{eq_delta_lam} p.\pageref{eq_delta_lam}); \tab 1) At trailing edge $\rex = \num{5348}$ thus the layer is laminar everywhere. $\delta$ will grow from~0 to~\SI{2,01}{\centi\metre} (eq.~\ref{eq_delta_lam} p.~\pageref{eq_delta_lam}); \tab 2) For water: $\delta_\text{trailing edge} = \SI{4,91}{\milli\metre}$. \item [\ref{exo_bl_sketches}]% \tab 1) See \cref{fig_bl_transition} p.\pageref{fig_bl_transition}. At the leading-edge the velocity is uniform. Note that the $y$-direction is greatly exaggerated, and that the outer velocity~$U$ is identical for both regimes; \tab 2) See \cref{fig_bl_deltastar} p.\pageref{fig_bl_deltastar}. Note that streamlines penetrate the boundary layer; \tab 1) See \cref{fig_bl_transition} p.~\pageref{fig_bl_transition}. At the leading-edge the velocity is uniform. Note that the $y$-direction is greatly exaggerated, and that the outer velocity~$U$ is identical for both regimes; \tab 2) See \cref{fig_bl_deltastar} p.~\pageref{fig_bl_deltastar}. Note that streamlines penetrate the boundary layer; \tab 3) and 4) See \S\ref{ch_bl_transition} p.~\pageref{ch_bl_transition}. \item [\ref{exo_bl_shear}]% \tab $x_\text{transition, air} = \SI{4,898}{\metre}$ and $x_\text{transition, water} = \SI{0,4}{\metre}$. In a laminar boundary layer, inserting equation~\ref{eq_cf_lam} into equation~\ref{eq_def_shear_coeff} into equation~\ref{eq_force_tau} yields\\ ... ... @@ -247,7 +247,7 @@ Solutions to the turbulent boundary layer along a smooth surface yield the follo \tab Using the expressions developed in exercise \ref{exo_bl_shear}, $\dot W_\text{friction} \approx \SI{255}{\watt}$. \item [\ref{exo_bl_friction_fuselage}]% \tab 1) $x_\text{transition} = \SI{7,47}{\centi\metre}$ (the laminar part is negligible). With the equations developed in exercise 7.3, we get $F = \SI{24,979}{\kilo\newton}$ and $\dot W = \SI{6,09}{\mega\watt}$. Quite a jump from the Wright Flyer I! \tab\tab 2) When the longitudinal pressure gradient is zero, the boundary layer cannot separate. Thus separation from the fuselage skin can only happen if the fuselage is flown at an angle relative to the flight direction (e.g. during a low-speed maneuver). \tab\tab 2) When the longitudinal pressure gradient is zero, the boundary layer cannot separate. Thus separation from the fuselage skin can only happen if the fuselage is flown at an angle relative to the flight direction (\eg during a low-speed maneuver). % \item [\ref{exo_bl_separation_model}]% % \tab Once the puzzle pieces are put together, this is an algebra exercise: $\left(\frac{x}{L}\right)_\text{separation} = \num{0,1231}$. Bryan beats Ernst! \end{description} ... ...
 ... ... @@ -53,7 +53,7 @@ What can be done in the other two branches of fluid mechanics? \begin{itemize} \item Large-scale flows are difficult to investigate experimentally. As we have seen in \chaptereight, scaling down a flow (e.g. so it may fit inside a laboratory) while maintaining constant $\re$ requires increasing velocity by a corresponding factor. \item Large-scale flows are difficult to investigate experimentally. As we have seen in \chaptereight, scaling down a flow (\eg so it may fit inside a laboratory) while maintaining constant $\re$ requires increasing velocity by a corresponding factor. \item Large-scale flows are also difficult to investigate numerically. At high $\re$, the occurrence of turbulence makes for either an exponential increase in computing power (we saw in \chapternineshort that direct numerical simulation computing power increases with $\re^{\num{3,5}}$), or for increased reliance on hard-to-calibrate turbulence models (in Reynolds-averaged simulations). \end{itemize} ... ... @@ -367,7 +367,7 @@ \gradient{p} &=& \mu \laplacian{\vec V} \label{eq_stokes} \end{IEEEeqnarray} In this type of flow, the pressure field is entirely dictated by the Laplacian of velocity, and the fluid density has no importance. Micro-organisms, for which the representative length $L$ is very small, spend their lives in such flows (\cref{fig_microorganisms}). At the human scale, we can visualize the effects of these flows by moving an object slowly in highly-viscous fluids (e.g.\ a spoon in honey), or by swimming in a pool filled with plastic balls. The inertial effects are almost inexistent, drag is extremely important, and the object geometry has comparatively small influence. In this type of flow, the pressure field is entirely dictated by the Laplacian of velocity, and the fluid density has no importance. Micro-organisms, for which the representative length $L$ is very small, spend their lives in such flows (\cref{fig_microorganisms}). At the human scale, we can visualize the effects of these flows by moving an object slowly in highly-viscous fluids (\eg a spoon in honey), or by swimming in a pool filled with plastic balls. The inertial effects are almost inexistent, drag is extremely important, and the object geometry has comparatively small influence. \begin{figure} \begin{center} ... ...
 ... ... @@ -29,7 +29,7 @@ \begin{enumerate} \item What is the terminal velocity of the particle? \item Will this terminal velocity increase or decrease as the particle progresses towards the ground? (briefly justify your answer, e.g.\ in 30 words or less) \item Will this terminal velocity increase or decrease as the particle progresses towards the ground? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -89,10 +89,10 @@ p_s &=& p_\infty + \frac{1}{2} \rho \left(V_\infty^2 - 4 V_\infty^2 \sin^2 \theta\right) \end{IEEEeqnarray} \item The pressure inside the hangar is set to $p_\infty$. What is the total lift force on the hangar?\\ (see also problem~\ref{exo_pressure_force_cylinder} p.\pageref{exo_pressure_force_cylinder})\\ (see also problem~\ref{exo_pressure_force_cylinder} p.~\pageref{exo_pressure_force_cylinder})\\ (a couple of hints to help with the algebra: $\int \sin x \diff x = -\cos x + k$ and $\int \sin^3 x \diff x = \frac{1}{3} \cos^3 x - \cos x + k$). \item At which position on the roof is the $p_s = p_\infty$? \item Describe briefly (e.g.\ in 30 words or less) two reasons why the results above would not correspond to reality. \item Describe briefly (\eg in 30 words or less) two reasons why the results above would not correspond to reality. \end{enumerate} ... ... @@ -113,7 +113,7 @@ \includegraphics[width=\textwidth]{images/cylinder_sphere_drag_munson_1} \includegraphics[width=0.6\textwidth]{images/cylinder_sphere_drag_munson_2} \end{center} \supercaption{Experimental measurements of the drag coefficient applying to a cylinder and to a sphere as a function of the diameter-based Reynolds number $\reD$, shown together with schematic depictions of the flow around the cylinder. By convention, the \vocab{drag coefficient} $C_\D \equiv C_{F\ \D} \equiv \frac{F_\D}{\frac{1}{2} \rho S U_\infty^2}$ (eq.~\ref{eq_def_force_coefficient} p.\pageref{eq_def_force_coefficient}) compares the drag force~$F_\D$ with the frontal area~$S$.}{Both figures \copyright~from Munson \& al.\cite{munsonetal2013}} \supercaption{Experimental measurements of the drag coefficient applying to a cylinder and to a sphere as a function of the diameter-based Reynolds number $\reD$, shown together with schematic depictions of the flow around the cylinder. By convention, the \vocab{drag coefficient} $C_\D \equiv C_{F\ \D} \equiv \frac{F_\D}{\frac{1}{2} \rho S U_\infty^2}$ (eq.~\ref{eq_def_force_coefficient} p.~\pageref{eq_def_force_coefficient}) compares the drag force~$F_\D$ with the frontal area~$S$.}{Both figures \copyright~from Munson \& al.\cite{munsonetal2013}} \label{fig_sphere_cylinder_drag} \end{figure} ... ... @@ -142,7 +142,7 @@ %%%%%% \clearpage \subsubsection{Flow field of a tornado} \wherefrom{Non-examinable. From Çengel \& al. \smallcite{cengelcimbala2010} E9-5, E9-14 \& E10-3} \wherefrom{Çengel \& al. \smallcite{cengelcimbala2010} E9-5, E9-14 \& E10-3} \label{exo_tornado} In this problem, we attempt to model a very large-scale flow: that of a tornado (\cref{fig_tornado}). We begin by pretending the tornado is one perfectly straight, stationary structure. We divide the flow into two regions: a core cylinder that rotates almost like a solid body, and an outer region where flow spins in an irrotational matter. This model is called the \we{Rankine vortex} (displayed in \cref{fig_rankine_vortex}) and is used widely as a simple, first approximation to model flows as large as a hurricane and as small as turbulence-induced vortices. ... ... @@ -171,7 +171,7 @@ \begin{enumerate} \item The mass balance equation for incompressible flow (eq.~\ref{eq_continuity_der_inc} p.\pageref{eq_continuity_der_inc}) is developed in cylindrical coordinates as follows: \item The mass balance equation for incompressible flow (eq.~\ref{eq_continuity_der_inc} p.~\pageref{eq_continuity_der_inc}) is developed in cylindrical coordinates as follows: \begin{IEEEeqnarray}{rCl} \frac{1}{r} \partialderivative{r v_r}{r} + \frac{1}{r} \partialderivative{v_\theta}{\theta} + \partialderivative{v_z}{z} = 0 \end{IEEEeqnarray} ... ... @@ -185,7 +185,7 @@ \begin{enumerate} \shift{1} \item The momentum balance equation for incompressible flow (eq.~\ref{eq_navierstokes} p.\pageref{eq_navierstokes}) is developed in cylindrical coordinates are as follows: \item The momentum balance equation for incompressible flow (eq.~\ref{eq_navierstokes} p.~\pageref{eq_navierstokes}) is developed in cylindrical coordinates are as follows: \begin{IEEEeqnarray}{lr} \rho \left[ \partialtimederivative{v_r} + v_r \partialderivative{v_r}{r} + \frac{v_\theta}{r} \partialderivative{v_r}{\theta} - \frac{v_\theta^2}{r} + v_z \partialderivative{v_r}{z} \right] &\nonumber\\ \ \ \ \ \ = \ \rho g_r - \partialderivative{p}{r} + \mu \left[ \frac{1}{r} \partialderivative{}{r} \left(r \partialderivative{v_r}{r} \right) - \frac{v_r}{r^2} + \frac{1}{r^2} \secondpartialderivative{v_r}{\theta} -\frac{2}{r^2} \partialderivative{v_\theta}{\theta} + \secondpartialderivative{v_r}{z} \right] \nonumber \\ ... ... @@ -211,8 +211,8 @@ \shift{2} \item What is the radial velocity $v_\theta$ distribution? \item What is the pressure field within the rotational core of the tornado?\\ (hint: you may start directly from an energy balance equation, eq.~\ref{eq_sfee} p.\pageref{eq_sfee}, without having to use the Navier-Stokes equations above). \item Make a simple, qualitative sketch (i.e.\ without numerical data) of the pressure as a function of radius throughout the entire tornado flow field. (hint: you may start directly from an energy balance equation, eq.~\ref{eq_sfee} p.~\pageref{eq_sfee}, without having to use the Navier-Stokes equations above). \item Make a simple, qualitative sketch (\ie without numerical data) of the pressure as a function of radius throughout the entire tornado flow field. \end{enumerate} It is finally time to calibrate and exploit our model. We estimate the tornado diameter to be \SI{50}{\metre} and the maximum wind velocity to be \SI{180}{\kilo\metre\per\hour}. ... ... @@ -230,7 +230,7 @@ \subsubsection{Lift on a symmetrical object} \wherefrom{non-examinable} Briefly explain (e.g. with answers 30 words or less) how lift can be generated on a sphere or a cylinder, Briefly explain (\eg with answers 30 words or less) how lift can be generated on a sphere or a cylinder, \begin{itemize} \item with differential control boundary layer control; \item with the effect of rotation. ... ... @@ -240,7 +240,7 @@ %%%%%% \subsubsection{Air flow over a wing profile} \wherefrom{Non-examinable. From Munson \& al. \smallcite{munsonetal2013} 9.109} \wherefrom{From Munson \& al. \smallcite{munsonetal2013} 9.109} The characteristics of a thin, flat-bottomed airfoil are examined by a group of students in a wind tunnel. The first investigations focus on the boundary layer, and the research group evaluate the boundary layer thickness and make sure that it is fully attached. ... ...
This diff is collapsed.
 ... ... @@ -46,7 +46,7 @@ Angular momentum balance through an arbitrary volume: \begin{enumerate} \item What is the mass flow traveling through the pipe? \item What is the force exerted by the pipe bend on the water? \item Represent the force vector qualitatively (i.e.\ without numerical data). \item Represent the force vector qualitatively (\ie without numerical data). \item What would be the new force if all of the speeds were doubled? \end{enumerate} ... ... @@ -69,7 +69,7 @@ Angular momentum balance through an arbitrary volume: \begin{enumerate} \item What is the force exerted on the ground by the deflection of the exhaust gases? \item Describe qualitatively (i.e.\ without numerical data) a modification to the deflector that would reduce the horizontal component of force. \item Describe qualitatively (\ie without numerical data) a modification to the deflector that would reduce the horizontal component of force. \item What would the force be if the deflector traveled rearwards (positive $x$-direction) with a velocity of~\SI{10}{\metre\per\second}? \end{enumerate} ... ... @@ -109,7 +109,7 @@ Angular momentum balance through an arbitrary volume: \shift{5} \item Show that the maximum power that can be transmitted to the generator occurs for $V_\text{blade} = \frac{1}{3} V_\text{water}$. \item What is the maximum power that can be transmitted to the generator? \item How would the above result change if viscous effects were taken into account? (briefly justify your answer, e.g. in 30 words or less) \item How would the above result change if viscous effects were taken into account? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -168,7 +168,7 @@ Angular momentum balance through an arbitrary volume: \begin{itemize} \item What is the drag force applying on the cylinder? \item How would this value change if the flow in the cylinder wake was turbulent, and the function $u_{2(y)}$ above only modeled \emph{time-averaged values} of the horizontal velocity? (briefly justify your answer, e.g. in 30 words or less) \item How would this value change if the flow in the cylinder wake was turbulent, and the function $u_{2(y)}$ above only modeled \emph{time-averaged values} of the horizontal velocity? (briefly justify your answer, \eg in 30 words or less) \end{itemize} ... ... @@ -199,7 +199,7 @@ Angular momentum balance through an arbitrary volume: \begin{enumerate} \item What is the drag force applying on the plate? \item What is the power required to compensate the drag? \item Under which form is the kinetic energy lost by the flow carried away? Can this new form of energy be measured? (briefly justify your answer, e.g. in 30 words or less) \item Under which form is the kinetic energy lost by the flow carried away? Can this new form of energy be measured? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} %%%% ... ... @@ -253,12 +253,12 @@ Angular momentum balance through an arbitrary volume: \label{exo_moment_gas_deflector} \wherefrom{non-examniable} We revisit the exhaust gas deflector of exercise \ref{exo_exhaust_gas_deflector} p.\pageref{exo_exhaust_gas_deflector}. \Cref{fig_deflector_sideview} below shows the deflector viewed from the side. The midpoint of the inlet is \SI{2}{\metre} above and \SI{5}{\metre} behind the wheel labeled~“\textbf{A}”, while the midpoint of the outlet is \SI{3,5}{\metre} above and \SI{1,5}{\metre} behind it. We revisit the exhaust gas deflector of exercise \ref{exo_exhaust_gas_deflector} p.~\pageref{exo_exhaust_gas_deflector}. \Cref{fig_deflector_sideview} below shows the deflector viewed from the side. The midpoint of the inlet is \SI{2}{\metre} above and \SI{5}{\metre} behind the wheel labeled~“\textbf{A}”, while the midpoint of the outlet is \SI{3,5}{\metre} above and \SI{1,5}{\metre} behind it. \begin{figure}[ht!] \begin{center} \includegraphics[width=\textwidth]{deflector_sideview} \end{center} \supercaption{Side view of the mobile exhaust gas deflector which was shown in \cref{fig_deflector} p.\pageref{fig_deflector}}{\wcfile{Exhaust gas deflector 2.svg}{Figure} \cczero \oc} \supercaption{Side view of the mobile exhaust gas deflector which was shown in \cref{fig_deflector} p.~\pageref{fig_deflector}}{\wcfile{Exhaust gas deflector 2.svg}{Figure} \cczero \oc} \label{fig_deflector_sideview} \end{figure} ... ... @@ -268,7 +268,7 @@ Angular momentum balance through an arbitrary volume: %%%% \subsubsection{Helicopter tail moment} \label{exo_helicopter_tail_moment} \wherefrom{non-examinable} %\wherefrom{non-examinable} In a helicopter, the role of the tail is to counter exactly the moment exerted by the main rotor about the main rotor axis. This is usually done using a tail rotor which is rotating around a horizontal axis. ... ... @@ -292,7 +292,7 @@ Angular momentum balance through an arbitrary volume: \item Propose and quantify a modification to the tail geometry or operating conditions that would allow the tail to produce no thrust (that is to say, zero force in the $x$-axis), while still generating the same moment. \end{enumerate} \textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in \chaptereleven (\S\ref{ch_circulating_cylinder} p.\pageref{ch_circulating_cylinder}).} \textit{Remark: this system is commercialized by MD Helicopters as the \wed{NOTAR}{\textsc{notar}}. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this effect is explored in \chaptereleven (\S\ref{ch_circulating_cylinder} p.~\pageref{ch_circulating_cylinder}).} %%%% ... ... @@ -394,7 +394,7 @@ Angular momentum balance through an arbitrary volume: \item [\ref{exo_drag_wing_profile}]% \tab $F_{\net x} \approx \rho L \Sigma_y \left[\left(u_2^2 - U u_2\right) \delta y\right] = \SI{-64,8}{\newton}$. \item [\ref{exo_moment_gas_deflector}]% \tab Re-use $\dot m = \SI{67,76}{\kilogram\per\second}$, $V_1 = \SI{166,7}{\metre\per\second}$, $V_2 = \SI{33,94}{\metre\per\second}$ from ex.\ref{exo_exhaust_gas_deflector}. From trigonometry, find $R_{2 \perp V_2} = \SI{1,717}{\metre}$. Then, plug in eq.~\ref{eq_rtt_angularmom_simple} p.\pageref{eq_rtt_angularmom_simple}: $M_\net = \SI{+18,64}{\kilo\newton\metre}$ in $z$-direction. \tab Re-use $\dot m = \SI{67,76}{\kilogram\per\second}$, $V_1 = \SI{166,7}{\metre\per\second}$, $V_2 = \SI{33,94}{\metre\per\second}$ from ex.\ref{exo_exhaust_gas_deflector}. From trigonometry, find $R_{2 \perp V_2} = \SI{1,717}{\metre}$. Then, plug in eq.~\ref{eq_rtt_angularmom_simple} p.~\pageref{eq_rtt_angularmom_simple}: $M_\net = \SI{+18,64}{\kilo\newton\metre}$ in $z$-direction. \item [\ref{exo_helicopter_tail_moment}]% \tab 1) Work eq.~\ref{eq_rtt_angularmom} down to scalar equation (in $y$-direction), solve for $\theta$: $\theta = \SI{123,1}{\degree}$. \tab 2) There are multiple solutions which allow both moment and force equations to be solved at the same time. $r_\C$ can be shortened, the flow in C can be split into forward and rearward components, or tilted downwards etc. Reductions in $\dot m_\B$ or $V_\C$ are also possible, but quantifying them requires solving both equations at once. ... ...
 ... ... @@ -44,7 +44,7 @@ \begin{center} \includegraphics[width=0.6\textwidth]{pressure_distribution_plate} \end{center} \supercaption{Pressure distribution on a flat plate. We already studied this situation in \chapteroneshort, problem~\ref{exo_pressure_induced_force} p.\pageref{exo_pressure_induced_force}.}{\wcfile{Pressure distribution on a flat plate.svg}{Figure} \cczero \oc} \supercaption{Pressure distribution on a flat plate. We already studied this situation in \chapteroneshort, problem~\ref{exo_pressure_induced_force} p.~\pageref{exo_pressure_induced_force}.}{\wcfile{Pressure distribution on a flat plate.svg}{Figure} \cczero \oc} \label{fig_pressure_distribution_plate_two} \end{figure} ... ... @@ -158,7 +158,7 @@ F_{\text{net, pressure}, z} & = & \diff \vol \frac{-\partial p}{\partial z} \end{IEEEeqnarray*} 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}): 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} \end{IEEEeqnarray} ... ... @@ -181,7 +181,7 @@ \subsection{Forces in static fluids} Fluid statics is the study of fluids at rest, i.e. whose velocity field~$\vec V$ is everywhere null and constant: Fluid statics is the study of fluids at rest, \ie whose velocity field~$\vec V$ is everywhere null and constant: \begin{IEEEeqnarray}{rCl} \left\{ \begin{array}{rcl} \vec V & = & \vec 0 \\ \partialtimederivative{\vec V} & = & \vec 0 ... ... @@ -224,7 +224,7 @@ \derivative{p}{z} & = & \rho g \label{eq_verticalgradientp} \end{IEEEeqnarray} The first consequence we draw from equation~\ref{eq_verticalgradientp} is that in a static fluid (e.g.\ in a glass of water, in a swimming pool, in a calm atmosphere), pressure depends solely on height. Within a static fluid, at a certain altitude, we will measure the same pressure regardless of the surroundings (\cref{fig_pressure_depth_statics}). The first consequence we draw from equation~\ref{eq_verticalgradientp} is that in a static fluid (\eg in a glass of water, in a swimming pool, in a calm atmosphere), pressure depends solely on height. Within a static fluid, at a certain altitude, we will measure the same pressure regardless of the surroundings (\cref{fig_pressure_depth_statics}). \begin{figure}[ht!] \begin{center} \includegraphics[width=10cm]{pressure_depth} ... ... @@ -273,7 +273,7 @@ \frac{p_2}{p_1} & = & \exp \left[\frac{g \Delta z}{R T_\cst} \right] \label{eq_atmtemp} \end{IEEEeqnarray} Of course, air temperature varies significantly within the atmosphere (at moderate altitudes the change with altitude is approximately~\SI{-6}{\kelvin\per\kilo\metre}). Adapting equation~\ref{eq_atmtemp} for a uniform temperature gradient (instead of uniform temperature) is the subject of problem \ref{exo_burj_khalifa} p.\pageref{exo_burj_khalifa}. Of course, air temperature varies significantly within the atmosphere (at moderate altitudes the change with altitude is approximately~\SI{-6}{\kelvin\per\kilo\metre}). Adapting equation~\ref{eq_atmtemp} for a uniform temperature gradient (instead of uniform temperature) is the subject of problem \ref{exo_burj_khalifa} p.~\pageref{exo_burj_khalifa}. In practice, the atmosphere also sees large lateral pressure gradients (which are strongly related to the wind) and its internal fluid mechanics are complex and fascinating. Equation~\ref{eq_atmtemp} is a useful and convenient model, but refinements must be made if precise results are to be obtained. ... ...
 ... ... @@ -77,7 +77,7 @@ The hinge stands \SI{1,5}{\metre} below the water surface. The window has a leng \begin{enumerate} \item Represent graphically the pressure of the water and atmosphere on each side of the window. \item What is the magnitude of the moment exerted by the pressure of the water about the axis of the window hinge? \item If the same door was positioned at the same depth, but the angle $\theta$ was decreased, would the moment be modified? (briefly justify your answer, e.g. in 30 words or less) \item If the same door was positioned at the same depth, but the angle $\theta$ was decreased, would the moment be modified? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -93,7 +93,7 @@ The hinge stands \SI{1,5}{\metre} below the water surface. The window has a leng \label{fig_potential_flow_cylinder_coordinates} \end{figure} It is possible to express the pressure distribution $p_s$ on the surface of a cylinder standing in a fluid flow with faraway velocity $V_\infty$ and pressure $p_\infty$ (see also problem \ref{exo_hangar_roof} p.\pageref{exo_hangar_roof}). The pressure distribution is expressed in cylindrical coordinates as: It is possible to express the pressure distribution $p_s$ on the surface of a cylinder standing in a fluid flow with faraway velocity $V_\infty$ and pressure $p_\infty$ (see also problem \ref{exo_hangar_roof} p.~\pageref{exo_hangar_roof}). The pressure distribution is expressed in cylindrical coordinates as: \begin{IEEEeqnarray}{rCl} p_s &=& p_\infty + \frac{1}{2} \rho \left(V_\infty^2 - 4 V_\infty^2 \sin^2 \theta\right) \end{IEEEeqnarray} ... ... @@ -137,12 +137,12 @@ The hinge stands \SI{1,5}{\metre} below the water surface. The window has a leng \subsubsection{Atmospheric pressure distribution} \label{exo_burj_khalifa} \wherefrom{non-examinable} %\wherefrom{non-examinable} 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. 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? \item If the atmospheric temperature decreases with altitude at a constant rate (\eg of~\SI{-7}{\kelvin\per\kilo\metre}), how can the pressure distribution be expressed analytically? \end{enumerate} A successful fluid dynamics lecturer purchases an apartment at the top of the \we{Burj Khalifa} tower (\SI{800}{\metre} above the ground). Inside the tower, the temperature is controlled everywhere at \SI{18,5}{\degreeCelsius}. Outside, the ground temperature is~\SI{30}{\degreeCelsius} and it decreases linearly with altitude (gradient: \SI{-7}{\kelvin\per\kilo\metre}). ... ...
 ... ... @@ -26,7 +26,7 @@ %%%% \subsection{Magnitude of the shear force} What is the force which which a fluid shears (i.e.\ “rubs”) against a wall? What is the force which which a fluid shears (\ie “rubs”) against a wall? When the shear $\tau$ exerted is uniform and the wall is flat, the resulting force $F$ in the direction $i$ is easily calculated: \begin{IEEEeqnarray}{rCcCl} ... ... @@ -62,7 +62,7 @@ Much like equation~\ref{eq_pressure_force_vector} in the previous chapter, eq.~\ref{eq_shear_force_vector} is not too hard to implement as a software algorithm to obtain numerically, for example, the force resulting from shear due to fluid flow around a body such as the body of a car. Its computation by hand, however, is far too tedious for us to even attempt. The position of the shear force is obtained with two moment vector equations, in a manner similar to that described in \S\ref{ch_position_pressure_force} p.\pageref{ch_position_pressure_force} with pressure. This is outside of the scope of this course. The position of the shear force is obtained with two moment vector equations, in a manner similar to that described in \S\ref{ch_position_pressure_force} p.~\pageref{ch_position_pressure_force} with pressure. This is outside of the scope of this course. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ... ... @@ -78,13 +78,13 @@ \subsection{The direction of shear} Already from the definition in eq.~\ref{eq_first_def_shear_two} we can appreciate that “parallel to a flat plate” can mean a multitude of different directions, and so that we need more than one dimension to represent shear. Furthermore, much in the same way as we did for pressure, we do away with the flat plate and accept that shear is a \vocab{field}, i.e. it is an effort applying not only upon solid objects but also upon and within fluids themselves. We replace eq.~\ref{eq_first_def_shear_two} with a more general definition: Already from the definition in eq.~\ref{eq_first_def_shear_two} we can appreciate that “parallel to a flat plate” can mean a multitude of different directions, and so that we need more than one dimension to represent shear. Furthermore, much in the same way as we did for pressure, we do away with the flat plate and accept that shear is a \vocab{field}, \ie it is an effort applying not only upon solid objects but also upon and within fluids themselves. We replace eq.~\ref{eq_first_def_shear_two} with a more general definition: \begin{IEEEeqnarray}{rCl} \vec \tau &\equiv& \lim_{A \to 0} \frac{\vec F_\parallel}{A} \label{eq_def_shear} \end{IEEEeqnarray} \youtubetopthumb{LjWeYPEmCk8}{cloud movements in a time\--lapse video on an interesting day are evidence of a highly\--strained atmosphere: pilots and meteorologists refer to this as \vocab{wind shear}.}{Y:StormsFishingNMore (\styl)} Contrary to pressure, shear is not a scalar, i.e. it can (and often does) take different values in different directions. At a given \emph{point} in space we represent it as a vector $\vec \tau = \left(\tau_x, \tau_y, \tau_y\right)$, and in a fluid, there is a shear \vocab{vector field}: Contrary to pressure, shear is not a scalar, \ie it can (and often does) take different values in different directions. At a given \emph{point} in space we represent it as a vector $\vec \tau = \left(\tau_x, \tau_y, \tau_y\right)$, and in a fluid, there is a shear \vocab{vector field}: \begin{IEEEeqnarray}{rCl} \vec \tau_{(x, y, z, t)} &\equiv& \left(\begin{array}{c} \tau_x\\ ... ... @@ -149,13 +149,13 @@ && + \diff x \diff z \ (\vec \tau_{yx\ 2} - \vec \tau_{yx\ 5})\nonumber\\ && + \diff z \diff y \ (\vec \tau_{xx\ 1} - \vec \tau_{xx\ 4})\label{eq_fshear_xdir} \end{IEEEeqnarray} In the same way we did with pressure in \chapterfourshort (\S\ref{ch_pressure_and_depth} p.\pageref{ch_pressure_and_depth}), we express each pair of values as derivative with respect to space multiplied by an infinitesimal distance: In the same way we did with pressure in \chapterfourshort (\S\ref{ch_pressure_and_depth} p.~\pageref{ch_pressure_and_depth}), we express each pair of values as derivative with respect to space multiplied by an infinitesimal distance: \begin{IEEEeqnarray}{rCl} \vec F_{\text{shear}\ x} &=& \diff x \diff y \left(\tdiff z \partialderivative{\vec \tau_{zx}}{z}\right) + \diff x \diff z \left(\tdiff y \partialderivative{\vec \tau_{yx}}{y}\right) + \diff z \diff y \left(\tdiff x \partialderivative{\vec \tau_{xx}}{x}\right)\nonumber\\ &=& \diff \vol \left(\partialderivative{\vec \tau_{zx}}{z} + \partialderivative{\vec \tau_{yx}}{y} + \partialderivative{\vec \tau_{xx}}{x}\right)\label{eq_shear_force_x_tmp} \end{IEEEeqnarray} If we make use of the operator \vocab{divergent} (see also Appendix~\ref{appendix_field_operators} p.\pageref{appendix_field_operators}), written~$\divergent{}$~: If we make use of the operator \vocab{divergent} (see also Appendix~\ref{appendix_field_operators} p.~\pageref{appendix_field_operators}), written~$\divergent{}$~: \begin{IEEEeqnarray}{rCcCl} \divergent{} &\equiv& \partialderivative{}{x} \vec i \cdot \ + \ \partialderivative{}{y} \vec j \cdot \ + \ \partialderivative{}{z} \vec k \cdot \label{eq_def_divergent}\\ \divergent{\vec A} &\equiv& \partialderivative{A_x}{x} \ + \ \partialderivative{A_y}{y} \ + \ \partialderivative{A_z}{z}\\ ... ... @@ -235,7 +235,7 @@ ||\vec \tau_{ij}|| &=& \mu \partialderivative{V_j}{i} \label{eq_shear_velocity_gradient} \end{IEEEeqnarray} \begin{equationterms} \item in which the subscript $i$ is an arbitrary direction ($x$, $y$ or $z$) and $j$ is the direction following it in order (e.g.\ $j=z$ when $i=y$); \item in which the subscript $i$ is an arbitrary direction ($x$, $y$ or $z$) and $j$ is the direction following it in order (\eg $j=z$ when $i=y$); \item and where $\mu$ is the viscosity (or “dynamic viscosity”) (\si{\pascal\second}). \end{equationterms} \end{mdframed} ... ... @@ -259,7 +259,7 @@ Most fluids of interest in engineering fluid mechanics (air, water, exhaust gases, pure gases) can be safely modeled as Newtonian fluids. Their viscosity~$\mu$ varies slightly with pressure (a dependency which we ignore) and mildly with temperature (an effect we take into account by reading values in a diagram). The values of viscosity vary very strongly from one fluid to another: for example, honey is roughly ten thousand times more viscous than water, which is roughly a hundred times more viscous than ambient air. The viscosities of various fluids are quantified in \cref{fig_viscosities_various_fluids} p.\pageref{fig_viscosities_various_fluids}. The values of viscosity vary very strongly from one fluid to another: for example, honey is roughly ten thousand times more viscous than water, which is roughly a hundred times more viscous than ambient air. The viscosities of various fluids are quantified in \cref{fig_viscosities_various_fluids} p.~\pageref{fig_viscosities_various_fluids}. Oil-based paint, blood and jelly-based fluids are strongly non-Newtonian; they require more complex viscosity models (\cref{fig_viscosity_characteristics}). \begin{figure} ... ... @@ -277,7 +277,7 @@ In any ordinary fluid flow, the velocity field is complex, and it is difficult to express shear and its net effect on particles. Since this requires expressing three values at each point in (three-dimensional) space and time, this would require complex mathematics or large amounts of discrete data. In simple cases, however, it is possible to express and calculate shear relatively easily. This is especially true in simple, steady, laminar (smooth) flows —typically flows for which the Reynolds number (eq.\ref{eq_def_reynolds_number} p.\pageref{eq_def_reynolds_number}) is low. In simple cases, however, it is possible to express and calculate shear relatively easily. This is especially true in simple, steady, laminar (smooth) flows —typically flows for which the Reynolds number (eq.\ref{eq_def_reynolds_number} p.~\pageref{eq_def_reynolds_number}) is low. In those cases, we can \emph{guess} a reasonably realistic velocity distribution, and then derive an expression for the distribution of shear from it. ... ...
 ... ... @@ -69,7 +69,7 @@ Shear in the direction $j$, on a plane perpendicular to direction $i$: \begin{enumerate} \item Express the force $F_{yz}$ due to shear on the plate as a function of its velocity $U_\text{plate}$, the gap height~$H$, and the properties of the fluid. \item The plate speed is $U_\text{plate} = \SI{1}{\metre\per\second}$ and the gap height is $H = \SI{5}{\milli\meter}$. What is the shear force $F_{yz}$ when the fluid is air ($\mu_{\text{atm.}} = \SI{1,5e-5}{\newton\second\per\metre\squared}$), and when the fluid is honey ($\mu_{\text{honey}} = \SI{40}{\newton\second\per\metre\squared}$)? \item If a very long and thin plate with the same surface area was used instead of the A4-shaped plate, would the shear force be different? (briefly justify your answer, e.g. in 30 words or less) \item If a very long and thin plate with the same surface area was used instead of the A4-shaped plate, would the shear force be different? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} ... ... @@ -92,7 +92,7 @@ Shear in the direction $j$, on a plane perpendicular to direction $i$: \begin{enumerate} \item If the flow in between the cylinders corresponds to the simplest possible flow case (steady, uniform, fully-laminar), what is the viscosity of the fluid? \item Would a non-Newtonian fluid induce a higher moment? (briefly justify your answer, e.g. in 30 words or less) \item Would a non-Newtonian fluid induce a higher moment? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} \textit{[Note: in practice, when the inner cylinder is turned at high speed, the flow displays mesmerizing patterns called \wed{Taylor-Couette flow}{Taylor—Couette vortices}, the description of which is much more complex!]} ... ... @@ -145,7 +145,7 @@ Shear in the direction $j$, on a plane perpendicular to direction $i$: \item What is the moment imparted by one disk to the other? \item How would the moment change if the radius of each disk was doubled? \item What is the transmitted power and the clutch efficiency? \item Briefly (e.g. in 30 words or less) propose one reason why in practice the flow in between the two disks may be different from the simplest-case flow used in this exercise. \item Briefly (\eg in 30 words or less) propose one reason why in practice the flow in between the two disks may be different from the simplest-case flow used in this exercise. \end{enumerate} ... ...
This diff is collapsed.
 ... ... @@ -133,15 +133,15 @@ Navier-Stokes equation for incompressible flow: \NumTabs{2} \begin{description} \item [\ref{exo_ns_revision}]% \tab 1) Continuity: eq.~\ref{eq_continuity_der_inc_dev} p.\pageref{eq_continuity_der_inc_dev}. Navier-Stokes: see eqs.~\ref{eq_ns_cartone}, \ref{eq_ns_carttwo} and~\ref{eq_ns_cartthree} p.~\pageref{eq_ns_cartone}; \tab 2) Read \S\ref{ch_continuity_der} p.~\pageref{ch_continuity_der} for continuity, and \S\ref{ch_navier-stokes} p.\pageref{ch_navier-stokes} for Navier-Stokes; \tab 1) Continuity: eq.~\ref{eq_continuity_der_inc_dev} p.~\pageref{eq_continuity_der_inc_dev}. Navier-Stokes: see eqs.~\ref{eq_ns_cartone}, \ref{eq_ns_carttwo} and~\ref{eq_ns_cartthree} p.~\pageref{eq_ns_cartone}; \tab 2) Read \S\ref{ch_continuity_der} p.~\pageref{ch_continuity_der} for continuity, and \S\ref{ch_navier-stokes} p.~\pageref{ch_navier-stokes} for Navier-Stokes; \tab 3) and 4) see \S\ref{ch_substantial_derivative} p.~\pageref{ch_substantial_derivative}. \item [\ref{exo_acceleration_field}]% \tab $\totaltimederivative{\vec V} = (\num{0,4} + \num{0,64}x)\vec i + (\num{-1,2} + \num{0,64}y)\vec j$. At the probe it takes the value $\num{1,68}\vec{i} + \num{0,08}\vec{j}$ (length \SI{1,682}{\metre\per\second\squared}). \item [\ref{exo_volumetric_dilatation_rate}]% \tab $\divergent{\vec V} = -x^2 + x - z$; thus at the probe it takes the value $\left(\divergent{\vec V}\right)_\text{probe} = \SI{-4}{\per\second}$. \item [\ref{exo_incompressiblity}]% \tab Apply equation~\ref{eq_continuity_der_inc_dev} p.\pageref{eq_continuity_der_inc_dev} to $\vec V$: the answer is yes. \tab Apply equation~\ref{eq_continuity_der_inc_dev} p.~\pageref{eq_continuity_der_inc_dev} to $\vec V$: the answer is yes. \item [\ref{exo_missing_components}]% \tab 1) Applying equation~\ref{eq_continuity_der_inc_dev}: $w_1 = \num{-3}xz - \frac{1}{2}z^2 + f_{(x,y,t)}$; \tab 2) idem, $v_2 = \num{-3}axy - bzy^2 + f_{(x,z,t)}$. ... ...
This diff is collapsed.
 ... ... @@ -20,7 +20,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for $\reD \lesssi \begin{IEEEeqnarray}{rCl} K_L &\equiv& \frac{|\Delta p_\text{loss}|}{\frac{1}{2} \rho V_\av^2} \ztag{\ref{eq_def_loss_coeff}} \end{IEEEeqnarray} Viscosities of various fluids are given in \cref{fig_viscosities_various_fluids}. Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in \cref{fig_moody} p.\pageref{fig_moody}. Viscosities of various fluids are given in \cref{fig_viscosities_various_fluids}. Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in \cref{fig_moody} p.~\pageref{fig_moody}. \end{boiboiboite} \begin{figure} ... ... @@ -45,7 +45,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for$\reD \lesssi \subsubsection{Revision questions} \wherefrom{non-examinable} The Moody diagram (\cref{fig_moody} p.\pageref{fig_moody}) is simple to use, yet it takes practice to understand it fully… here are three questions to guide your exploration. They can perhaps be answered as you work through the other examples. The Moody diagram (\cref{fig_moody} p.~\pageref{fig_moody}) is simple to use, yet it takes practice to understand it fully… here are three questions to guide your exploration. They can perhaps be answered as you work through the other examples. \begin{enumerate} \item Why is there no zero on the diagram? \item Why are the curves sloped downwards — should friction losses not instead \emph{increase} with increasing Reynolds number? ... ... @@ -84,7 +84,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for $\reD \lesssi The pump must be powerful enough to push \SI{1}{\metre\cubed\per\second} of water at \SI{20}{\degreeCelsius}. \Cref{fig_viscosities_various_fluids} p.\pageref{fig_viscosities_various_fluids} quantifies the viscosity of various fluids, and \cref{fig_moody} p.\pageref{fig_moody} quantifies losses in cylindrical pipes. \Cref{fig_viscosities_various_fluids} p.~\pageref{fig_viscosities_various_fluids} quantifies the viscosity of various fluids, and \cref{fig_moody} p.~\pageref{fig_moody} quantifies losses in cylindrical pipes. \begin{enumerate} \item Will the flow in the water pipe be turbulent? ... ... @@ -128,7 +128,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for$\reD \lesssi \shift{4} \item What would be the pressure loss if an \SI{8}{\centi\metre}-diameter plastic pipe was used? \item What would then be the pumping power required? \item What is one advantage of using a pipe with smaller diameter? (briefly justify your answer, e.g.\ in 30 words or less) \item What is one advantage of using a pipe with smaller diameter? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} \subsubsection{Major oil pipeline} ... ... @@ -144,7 +144,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for $\reD \lesssi \label{fig_trans_alaska} \end{figure} Your design must safely carry 700 thousand barrels of oil (\SI{110 000}{\metre\cubed}) per day along a length of \SI{1200}{\kilo\metre}. The crude oil has density \SI{900}{\kilogram\per\metre\cubed} and its viscosity is quantified in \cref{fig_viscosities_various_fluids} p.\pageref{fig_viscosities_various_fluids}. The average temperature of the oil during the transit is \SI{60}{\degreeCelsius}. Your design must safely carry 700 thousand barrels of oil (\SI{110 000}{\metre\cubed}) per day along a length of \SI{1200}{\kilo\metre}. The crude oil has density \SI{900}{\kilogram\per\metre\cubed} and its viscosity is quantified in \cref{fig_viscosities_various_fluids} p.~\pageref{fig_viscosities_various_fluids}. The average temperature of the oil during the transit is \SI{60}{\degreeCelsius}. The landscape is flat for most of the journey, with a \SI{200}{\kilo\metre}-wide mountain range in the middle that reaches \SI{1400}{\metre} altitude. ... ... @@ -159,11 +159,11 @@ In cylindrical pipe flow, we assume the flow is always laminar for$\reD \lesssi \item Propose a pumping station arrangement, and calculate the power required for each pump. \end{enumerate} Before you start building the pipeline, the operator would like to know how the system would perform at half-capacity (i.e.\ with half the volume flow). Before you start building the pipeline, the operator would like to know how the system would perform at half-capacity (\ie with half the volume flow). \begin{enumerate} \shift{5} \item If none of the other input data changes, what is the new pumping power? \item Propose one reason why in practice, the pumping power may be higher than you just calculated (briefly justify your answer, e.g.\ in 30 words or less). \item Propose one reason why in practice, the pumping power may be higher than you just calculated (briefly justify your answer, \eg in 30 words or less). \end{enumerate} \clearfloats ... ... @@ -217,7 +217,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for $\reD \lesssi \end{IEEEeqnarray} \item What is the volume flow between the sphere and the concrete foundation? \item What is the pumping power required for the fountain to work? \item How would this power change if the diameter of the concrete support (but not of the sphere) was increased? (briefly justify your answer, e.g. in 30 words or less) \item How would this power change if the diameter of the concrete support (but not of the sphere) was increased? (briefly justify your answer, \eg in 30 words or less) \end{enumerate} \begin{figure}[ht!] \begin{center} ... ... @@ -324,7 +324,7 @@ In cylindrical pipe flow, we assume the flow is always laminar for$\reD \lesssi \item [\ref{exo_kugel_fountain}]% \tab $\Delta p = \frac{F_{W \text{sphere}}}{A_\text{basin}} = \SI{71,51}{\kilo\pascal}$. Then $\dot \vol = \SI{0,494}{\liter\per\second}$ which enables us to obtain $\dot W_\text{pump} = \SI{35,3}{\watt}$. \item [\ref{exo_complex_ducted_flow}]% \tab Calculate the velocity distribution in the same way as for equation~\ref{eq_tmp5} p.\pageref{eq_tmp5}. Evaluate the pressure gradient $\partialderivative{p}{y}$ using mass conservation (total cross-section mass flow is zero). With the full velocity distribution, derivate $v$ with respect to $x$ to obtain $F_\tau = \rho S g h + \frac{2 \mu S}{h} v_\text{p}$. \tab Calculate the velocity distribution in the same way as for equation~\ref{eq_tmp5} p.~\pageref{eq_tmp5}. Evaluate the pressure gradient $\partialderivative{p}{y}$ using mass conservation (total cross-section mass flow is zero). With the full velocity distribution, derivate $v$ with respect to $x$ to obtain $F_\tau = \rho S g h + \frac{2 \mu S}{h} v_\text{p}$. \item [\ref{exo_dontexist}] \tab The author cannot remember which exercise you are referring to. \end{description} \atendofexercises
 ... ... @@ -30,7 +30,7 @@ In many practical engineering situations which involve fluid flow, investigating the flow in the real application is impossible or impractical. For example, trying different designs of an airplane or submarine, each time attaching expensive measurement equipment to the device, is prohibitively expensive. Studying the flow in or around very small objects, or in inaccessible locations is also very hard: imagine for example having to study metal flow in a furnace, blood flow in a beating heart, or around a mosquito’s wings.\\ In those cases it is common practice to build scaled-up or scaled-down versions of the flow in a laboratory. This brings up the question: how should the model flow properties be adapted to represent the original flow? For example, if the model is half as small as the original, should the velocity be halved? Or perhaps doubled? The answer to this problem is as follows: two flows are \vocab{dynamically similar} (i.e. representative of one another) when their \vocab{flow parameters} are the same.\\ The answer to this problem is as follows: two flows are \vocab{dynamically similar} (\ie representative of one another) when their \vocab{flow parameters} are the same.\\ In order to understand what this means, we will need to look back at the Navier-Stokes equation (the momentum balance equation we derived in \chaptersixshort), re-writing it in a non-dimensional form. Onwards! \subsection{The non-dimensional Navier-Stokes equation} ... ... @@ -89,7 +89,7 @@ \end{IEEEeqnarray} \end{adjustwidth} In equation~\ref{eq_nsndtmp}, the terms in brackets each appear in front of non-di\-men\-sion\-al (unit) vectors. These bracketed terms all have the same dimension, i.e. \si{\kilogram\per\metre\squared\per\second\squared}. Multiplying each by $\frac{L}{\rho V^2}$ (of dimension \si{\metre\squared\second\squared\per\kilogram}), we obtain a purely non-dimensional equation: In equation~\ref{eq_nsndtmp}, the terms in brackets each appear in front of non-di\-men\-sion\-al (unit) vectors. These bracketed terms all have the same dimension, \ie \si{\kilogram\per\metre\squared\per\second\squared}. Multiplying each by $\frac{L}{\rho V^2}$ (of dimension \si{\metre\squared\second\squared\per\kilogram}), we obtain a purely non-dimensional equation: \begin{adjustwidth}{-0.1cm}{0cm}%handmade \begin{IEEEeqnarray}{rCl} \left[\frac{fL}{V}\right]\ \partialderivative{\vec V^*}{t^*} + \left[1\right]\ \left(\vec V^* \cdot \gradient^*\right) \vec V^* & = & \left[\frac{gL}{V^2}\right]\ \vec g^* - \left[\frac{p_0 - p_\infty}{\rho V^2}\right]\ \gradient^* p^* + \left[\frac{\mu}{\rho V L}\right]\ \gradient^{*2} \vec V^*\nonumber\\\label{eq_navierstokes_nondim_tmp} ... ... @@ -179,7 +179,7 @@ We now know that when we create a scaled-down or scaled-up version of one flow (as shown for example in \cref{fig_dynamic_similarity}), we attempt to keep all flow parameters identical. In practice, this is extremely difficult to do, as we will see while going through this chapter’s problem sheet, especially if ordinary fluids (air or water) are to be used. The practice in science and engineering is usually to focus on one or two key parameters, while ignoring others. If the budget allows for it, several models may be built, each focusing on one parameter (e.g.\ one model for compressibility effects, one model for viscous effects). In practice, this is extremely difficult to do, as we will see while going through this chapter’s problem sheet, especially if ordinary fluids (air or water) are to be used. The practice in science and engineering is usually to focus on one or two key parameters, while ignoring others. If the budget allows for it, several models may be built, each focusing on one parameter (\eg one model for compressibility effects, one model for viscous effects). \begin{figure}[ht!] \begin{center} \includegraphics[width=0.49\textwidth]{images/Lockheed_C-141_Model_in_TDT_-_GPN-2000-001741.jpg} ... ... @@ -193,7 +193,7 @@ In the design and construction of models, practical constraints must be balanced against the need to reproduce the dynamics of fluids accurately. They include: \begin{itemize} \item Cost of production. The volume of a model typically increases with its length \emph{cubed}, i.e.\ doubling its length multiplies its volume by a factor~8; \item Cost of production. The volume of a model typically increases with its length \emph{cubed}, \ie doubling its length multiplies its volume by a factor~8; \item Precision of manufacturing. Usually, the smaller the model, and the smaller the geometrical accuracy that can be reached; \item Ease of instrumenting. Carrying out measurements over extremely large or extremely small models may be challenging; \item Ease of optical access. Optical flow measurement devices are usually preferred because they do not obstruct the flow, but when the flow is internal to a model, they require transparent, flat walls to work correctly; ... ...
 ... ... @@ -38,7 +38,7 @@ \includegraphics[width=\textwidth]{viscosities_horizontal} \vspace{-1.1cm} \end{center} \supercaption{Viscosity of various fluids at a pressure of \SI{1}{\bar} (in practice viscosity is almost independent of pressure).}{Figure repeated from \cref{fig_viscosities_various_fluids} p.\pageref{fig_viscosities_various_fluids}; \copyright\xspace White 2008~\cite{white2008}} \supercaption{Viscosity of various fluids at a pressure of \SI{1}{\bar} (in practice viscosity is almost independent of pressure).}{Figure repeated from \cref{fig_viscosities_various_fluids} p.~\pageref{fig_viscosities_various_fluids}; \copyright\xspace White 2008~\cite{white2008}} \label{fig_viscosities_various_fluids_two} \end{figure} ... ... @@ -146,7 +146,7 @@ \shift{4} \item Your team considers modifying the air temperature to compensate for the limit in the air speed. If the temperature in the tunnel can be controlled between \SI{-10}{\degreeCelsius} and \SI{40}{\degreeCelsius}, but the pressure remains atmospheric, what is the maximum race-track speed that can be reproduced in the wind tunnel? \item In that case, by which factor should the model drag force measurements be multiplied in order to correspond to the real car? \item \textit{[non-examinable question]} The wind tunnel has a \SI{2}{\metre} by \SI{2}{\metre} square test cross-section. What is the power required to bring atmospheric air ($T_\atm = \SI{15}{\degreeCelsius}$, $V_\atm = \SI{0}{\metre\per\second}$) to the desired speed and temperature in the test section? \item The wind tunnel has a \SI{2}{\metre} by \SI{2}{\metre} square test cross-section. What is the power required to bring atmospheric air ($T_\atm = \SI{15}{\degreeCelsius}$, $V_\atm = \SI{0}{\metre\per\second}$) to the desired speed and temperature in the test section? \end{enumerate} {\small \textit{NB: this exercise is inspired by \href{https://www.youtube.com/watch?v=KC0E0wU6inU}{an informative and entertaining video by the Sauber F1 team about their wind tunnel testing}, which the reader is encouraged to watch at\\ \href{https://www.youtube.com/watch?v=KC0E0wU6inU}{https://www.youtube.com/watch?v=KC0E0wU6inU}.}} ... ... @@ -170,7 +170,7 @@ \tab 2) In water, the density cannot be reasonably controlled, and we need a velocity $V_3 = \SI{418}{\metre\per\second}$! \tab 3) In air, at $V_4 = \SI{80}{\metre\per\second}$, Mach number can be reproduced at $T_4 = \SI{-251}{\degreeCelsius}$ (although the Reynolds number is off). Wind tunnels used to investigate compressible flow around aircraft have very powerful coolers. \item [\ref{exo_scale_dragonfly}]% \tab 1) With e.g. a model of span $\SI{60}{\centi\metre}$, match the Reynolds number: $V_2 = \SI{0,67}{\metre\per\second}$; \tab 1) With \eg a model of span $\SI{60}{\centi\metre}$, match the Reynolds number: $V_2 = \SI{0,67}{\metre\per\second}$; \tab 2) Match the Strouhal number: $f_2 = \SI{0,56}{\hertz}$. Mach, Froude and Euler numbers will have no effect here; \tab 3) and 4) are left as a surprise for the student. \item [\ref{exo_formula_one}] ... ...
 ... ... @@ -90,7 +90,7 @@ Not all unsteady flows are turbulent. Well-known patterns such as a \wed{von Karman vortex street}{von Kármán vortex street} (figure~\ref{fig_not_turbulence}) or a series of \wed{Wave cloud}{wave clouds}, for example, are not turbulent. Those oscillations occur at a single recognizable frequency and scale, and once the phenomena has begun, their evolution is easily predictable.