...
 
Commits (4)
  • Olivier's avatar
    5e43b11d
  • Olivier's avatar
    Exercises 3: proof-read · 5cf84547
    Olivier authored
    * refreshed formula sheet
    * labeled non-examinable exercises
    * fixed page layout somewhat
    5cf84547
  • Olivier's avatar
    Exercises 4: Major cleanup, -5+1 problems · cf5a5a8e
    Olivier authored
    * Major cleanup. The problems that are not clearly about calculating
      a pressure force on a wall are gone.
    * Added one problem with non-static pressure distribution and
      non-flat surface (taken down from problem sheet 11)
    cf5a5a8e
  • Olivier's avatar
    Chapter 4: cleaner structure, integrates better with main plan · 4fad4984
    Olivier authored
    * Calculation of forces on wall come first. They are actually the
      most useful equations in the chapter
    * The derivation of the "grad p = rho g" equation is now done in
      3D, in line with the other chapters. I never completely understood
      the method I used to use (taken from classic textbooks) and suspect
      it just is worth nothing
      Part of the new derivation is taken back from chapter 6.
    * Static fluids are now clearly presented as a special case. This is
      now also reflected in the problem sheet.
    4fad4984
\renewcommand{\lastedityear}{2018}
\renewcommand{\lasteditmonth}{06}
\renewcommand{\lasteditday}{18}
\renewcommand{\lastedityear}{2019}
\renewcommand{\lasteditmonth}{03}
\renewcommand{\lasteditday}{30}
\renewcommand{\numberofthischapter}{11}
\renewcommand{\titleofthischapter}{Large- and small-scale flows}
\atstartofexercises
\fluidmechexercisestitle
......@@ -54,23 +53,23 @@
%%%%%%
\subsubsection{Hangar roof}
\subsubsection{Idealized flow over a hangar roof}
\wherefrom{based on White \smallcite{white2008} P8.54}
\label{exo_hangar_roof}
Certain flows in which both compressibility and viscosity effects are negligible can be described using the potential flow assumption (the hypothesis that the flow is everywhere irrotational). If we compute the two-dimensional laminar steady fluid flow around a cylinder profile, we obtain the velocities in polar coordinates as:
\begin{IEEEeqnarray}{rCl}
v_r &=& \frac{1}{r} \partialderivative{\psi}{\theta} = U_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\ztag{\ref{eq_ur_cylinder_nolift}}\\
v_\theta &=& - \partialderivative{\psi}{r} = - U_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\ztag{\ref{eq_utheta_cylinder_nolift}}
v_r &=& V_\infty \cos \theta \left(1 - \frac{R^2}{r^2}\right)\ztag{\ref{eq_ur_cylinder_nolift}}\\
v_\theta &=& - V_\infty \sin \theta \left(1 + \frac{R^2}{r^2}\right)\ztag{\ref{eq_utheta_cylinder_nolift}}
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab the origin ($r = 0$) is at the center of the cylinder profile;
\item \tab $\theta$ \tab\tab\tab is measured relative to the free-stream velocity vector;
\item \tab $U_\infty$ \tab is the incoming free-stream velocity;
\item \tab $V_\infty$ \tab is the incoming free-stream velocity;
\item and \tab $R$ \tab\tab\tab is the (fixed) cylinder radius.
\end{equationterms}
Based on this model, in this exercise, we study the flow over a hangar roof.
In this exercise, we study the air flow over a hangar roof with this model. We use the equations above to describe the air velocity everywhere, pretending the as the wind blows about a large semi-cylindrical solid structure — an idealized description of an otherwise complex flow.
\begin{figure}
\begin{center}
......@@ -84,10 +83,15 @@
\begin{enumerate}
\item If the pressure inside the hangar is the same as the pressure of the faraway atmosphere, and if the wind closely follows the hangar roof geometry (without any flow separation), what is the total lift force on the hangar?\\
(hint: we accept that $\int \sin^3 x \diff x = \frac{1}{3} \cos^3 x - \cos x + k$).
\item At which position on the roof could we drill a hole to negate the aerodynamic lift force?
\item Propose two reasons why the aerodynamic force measured in practice on the hangar roof may be lower than calculated with this model.
\item Starting from eqs.~\ref{eq_ur_cylinder_nolift} and~\ref{eq_utheta_cylinder_nolift}, show that the pressure $p_s$ on the surface on the roof is distributed 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}
\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})\\
(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.
\end{enumerate}
......@@ -252,8 +256,8 @@
\item [\ref{exo_water_drop}]%
\tab Same as previous exercise: $U_1 = \SI{4,578e-2}{\metre\per\second}$ and $U_2 = \SI{0,183}{\metre\per\second}$, with Reynolds numbers of \num{0,113} and \num{0,906} respectively (thus creeping flow hypothesis valid).
\item [\ref{exo_hangar_roof}]%
\tab 1) express roof pressure as a function of $\theta$ using eq.~\ref{eq_pressure_surface_cylinder} p.\pageref{eq_pressure_surface_cylinder} on eq.~\ref{eq_utheta_cylinder_nolift}, then integrate the vertical component of force due to pressure: $F_\text{L roof} = \SI{1,575}{\mega\newton}$.
\tab 2) $\left.\theta\right|_{F=0} = \SI{54,7}{\degree}$
\tab 1) Integrate the vertical component of force due to pressure: $F_\text{L roof} = \SI{1,575}{\mega\newton}$.
%\tab 2) $\left.\theta\right|_{F=0} = \SI{54,7}{\degree}$
\item [\ref{exo_cabling_wright_flyer}]%
\tab A simple reading gives $F_\D = \SI{6,9}{\newton}$, $\dot W = \SI{76}{\watt}$.
\item [\ref{exo_ping_pong_ball}]
......
......@@ -56,6 +56,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of mass}
\label{ch_balance_mass_int_simple}
How much mass flow is coming in and out of the control volume? We answer this question by writing a mass balance equation. It compares the rate of change of the mass of the system (which by definition is zero, see eq.~\ref{eq_massconservation} p.\pageref{eq_massconservation}), to the rate of change of mass in the control volume:
\begin{IEEEeqnarray}{CCCCCCC}
......@@ -129,6 +130,7 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of momentum}
\label{ch_balance_momentum_simple}
What force is applied to the fluid for it to travel through the control volume? We answer this question by writing a momentum balance equation. It compares the rate of change of the momentum of the system (which by definition is the net force applying to it, see eq.~\ref{eq_secondlaw} p.\pageref{eq_secondlaw}), to the rate of change of momentum in the control volume:
\begin{IEEEeqnarray}{CCCCCCC}
......
This diff is collapsed.
\renewcommand{\lastedityear}{2019}
\renewcommand{\lasteditmonth}{03}
\renewcommand{\lasteditday}{26}
\renewcommand{\lasteditday}{30}
\atstartofexercises
\fluidmechexercisestitle
......@@ -10,35 +10,30 @@
\mecafluexboxen
\begin{boiboiboite}
Reynolds Transport Theorem:
\begin{IEEEeqnarray}{rCl}
\timederivative{B_\sys} & = & \timederivative{} \iiint_\cv \rho b \diff \vol + \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt}}
\end{IEEEeqnarray}
Mass conservation:
\begin{IEEEeqnarray}{rCCCCCl}
\timederivative{m_\sys} & = & 0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}}
\end{IEEEeqnarray}
Change in linear momentum:
\begin{IEEEeqnarray}{rCCCCCl}
\timederivative{(m \vec V_{sys})} & = & \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}}
\end{IEEEeqnarray}
Change in angular momentum:
\begin{equation}
\timederivative{(\vec r_{\X m} \wedge m \vec V)_\sys} = \vec M_{\net, \X} = \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol + \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \tag{\ref{eq_rtt_angularmom}}
\end{equation}
Mass balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
0 & = & \timederivative{} \iiint_\cv \rho \diff \vol & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_mass}}
\end{IEEEeqnarray}
Momentum balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \ztag{\ref{eq_rtt_linearmom}}
\end{IEEEeqnarray}
Angular momentum balance through an arbitrary volume:
\begin{IEEEeqnarray}{rCCCl}
\vec M_{\net, \X} &=& \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol &+& \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \ztag{\ref{eq_rtt_angularmom}}
\end{IEEEeqnarray}
\end{boiboiboite}
\clearpage%handmade
%%%%
\subsubsection{Pipe bend}
\wherefrom{\cczero \oc}
\label{exo_pipe_bend}
A pipe with diameter~\SI{30}{\milli\meter} has a bend with angle $\theta = \SI{130}{\degree}$, as shown in \cref{fig_pipe_bend}. Water enters and leaves the pipe with the same speed $V_1 = V_2 = \SI{1,5}{\metre\per\second}$. The velocity distribution at both inlet and outlet is uniform.
\begin{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[width=7.5cm]{pipe_bend}
\vspace{-0.5cm}
......@@ -53,9 +48,6 @@ Change in angular momentum:
\item What would be the new force if all of the speeds were doubled?
\end{enumerate}
%%%%
\subsubsection{Exhaust gas deflector}
\label{exo_exhaust_gas_deflector}
......@@ -86,8 +78,7 @@ Change in angular momentum:
\label{exo_pelton_turbine}
A water turbine is modeled as the following system: a water jet exiting a stationary nozzle hits a blade which is mounted on a rotor (\cref{fig_water_turbine}). In the ideal case, viscous effects can be neglected, and the water jet is deflected entirely with a~\SI{180}{\degree} angle.
\begin{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[width=12cm]{water_turbine_blade}
\end{center}
......@@ -258,6 +249,7 @@ Change in angular momentum:
%%%%
\subsubsection{Moment on gas deflector}
\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.
\begin{figure}[ht!]
......@@ -270,10 +262,11 @@ Change in angular momentum:
What is the moment generated by the gas flow about the axis of the wheel labeled “\textbf{A}”?
\clearpage%handmade
%%%%
\subsubsection{Helicopter tail moment}
\label{exo_helicopter_tail_moment}
\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.
......@@ -297,7 +290,7 @@ Change in angular momentum:
\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 chapter~8 (\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}).}
%%%%
......@@ -407,7 +400,7 @@ Change in angular momentum:
\tab $V_\text{center} = \num{1,2245} U$; $F_\net = \SI{+393}{\newton}$ (positive!)
\item [\ref{exo_thrust_reverser}]%
\tab $\dot m_\text{cold} = \SI{297}{\kilogram\per\second}$, $\dot m_\text{hot} = \SI{59,4}{\kilogram\per\second}$;
\tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$, $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$;
\tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$,\\ $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$;
\tab $F_\text{cold flow, reverse, bench \& runway} = \SI{-24,35}{\kilo\newton}$ and $F_\text{hot flow, normal, bench \& runway} = \SI{+8,316}{\kilo\newton}$.
\tab Adding the net pressure force due to the (lossless) flow acceleration upstream of the inlet, we obtain, on the bench: $F_\text{engine bench, normal} = \SI{-93,26}{\kilo\newton}$, $F_\text{engine bench, reverse} = \SI{+5,344}{\kilo\newton}$;
and on the runway: $F_\text{engine runway, normal} = \SI{-92,5}{\kilo\newton}$ and $F_\text{engine bench, reverse} = \SI{+6,094}{\kilo\newton}$.
......
This diff is collapsed.
This diff is collapsed.
\subsubsection{Water lock}
%\wherefrom{Exam \textsc{ws}2015/16}%homemade
\label{exo_lock}
A system of lock doors is set up to allow boats to travel up the side of a hill (figure~\ref{fig_locks_perpendicular}).
\begin{figure}[h]
\begin{center}
%\vspace{-0.5cm}
\includegraphics[width=10cm]{locks1}\\
\vspace{0.5cm}
\includegraphics[width=9cm]{locks2}
%\vspace{-0.5cm}
\end{center}
\supercaption{Outline schematic of a water canal lock.}{\wcfile{Water lock door principle.svg}{Figure 1} \cczero \oc; \wcfile{Lock doors.svg}{figure 2} \ccbysa \oc}
\label{fig_locks_perpendicular}
\end{figure}
We are studying the force and moment exerted by the water on the lock door labeled \textbf{A}. At this instant, the water levels are as shown in figure~\ref{fig_locks_perpendicular}: \SI{2}{\metre} in the lower canal, \SI{5}{\metre} in the lock, and \SI{9}{\metre} in the upper canal. The door is \num{2,5}~\si{metres} wide.
\begin{enumerate}
\item Sketch the distribution of the pressure exerted by the water and the atmosphere on both sides of door A.
\item What is the net force exerted by the water on door A?
\item At which height is this force exerting?
%\item What is the net moment exerted by the water about the hinge of door A?
%\item If the water level was lowered by \SI{1}{\metre} everywhere, would the moment about the hinge be reduced? (briefly justify your answer)
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Buoyancy force on a tin can}
\label{exo_buoyancy_tin_can}
A student contemplates a tin can of height~\SI{10}{\centi\metre} and diameter~\SI{7}{\centi\metre}.
\begin{enumerate}
\item If one considers that the atmospheric density is uniform, what is the buoyancy force generated by the atmosphere on the can when it is positioned vertically?
\item What is the force generated when the can is immersed in water at a depth of~\SI{20}{\centi\metre}? At a depth of~\SI{10}{\metre}?
\item What is the buoyancy force generated when the can is immersed in the water in a horizontal position?
\end{enumerate}
Instead of uniform density, we now wish to calculate the atmospheric buoyancy under the hypothesis of uniform temperature (room temperature~\SI{20}{\degreeCelsius}).
\begin{enumerate}
\shift{3}
\item Starting from equation~\ref{eq_gradp}: $\gradient{p} = \rho \vec g$, show that when the temperature $T_\text{cst.}$ is assumed to be uniform, the atmospheric pressure~$p$ at two points~1 and~2 separated by a height difference $\Delta z$ is such that:
\begin{IEEEeqnarray}{rCl}
\frac{p_2}{p_1} & = & \exp \left[\frac{g \Delta z}{RT_\text{cst.}} \right] \ztag{\ref{eq_atmtemp}}
\end{IEEEeqnarray}
\item What is the buoyancy generated by the room atmosphere?
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Atmospheric buoyancy force}
\label{exo_buoyancy_airbus}
\wherefrom{non-examinable. \cczero \oc}
If the atmospheric density is considered uniform, estimate the buoyancy force exerted on an Airbus A380, both on the ground and in cruise flight ($\rho_\text{cruise} = \SI{0,4}{\kilogram\per\metre\cubed}, T_\text{cruise} = \SI{-40}{\degreeCelsius}$).
\begin{figure}[hb!]
\begin{center}
\includegraphics[width=\textwidth]{380}
\end{center}
\supercaption{Airbus A380-800}{\wcfile{A380-800v1.0.png}{Drawing} \ccbysa by Julien Scavini}
\vspace{-0.5cm}%handmade
\label{fig_athreeeighty}
\end{figure}
\begin{table}[hb!]
\begin{center}
\renewcommand{\tabcolsep}{0.8em}
\renewcommand{\arraystretch}{1.3}
\begin{tabularx}{10cm}{r|c} % the @{something} kills the inter-column space and replaces it with "something"
\hline
Length overall & {\SI{72,73}{\metre}} \\
Wingspan & {\SI{79,75}{\metre}} \\
Height & {\SI{24,45}{\metre}} \\
Wing area & {\SI{845}{\metre\squared}} \\
Aspect ratio & {\num{7,5}} \\
Wing sweep & {\SI{33,5}{\degree}} \\
Maximum take-off weight & \SI{560 000}{\kilogram} \\
Typical operating empty weight & \SI{276 800}{\kilogram} \\
\hline
\end{tabularx}
\caption{Characteristics of the Airbus A380-800.}
\label{tab_athreeeighty}
\end{center}
\end{table}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Lock with diagonally-mounted doors}
\label{exo_lock_diagonal}
\wherefrom{non-examinable. \cczero \oc}
The doors of the lock studied in exercise~\ref{exo_lock} p.\pageref{exo_lock} are replaced with diagonally-mounted doors at an angle $\alpha = \SI{20}{\degree}$, mounted such that no bending moment is sustained by the hinges (\cref{fig_lock_doors_diagonal}).
What is the force exerted by door “A” on the other door?
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.8\textwidth]{waterlock2}
\end{center}
\supercaption{Lock doors mounted with an angle relative one to another. The angle relative to the case in exercise~\ref{exo_lock} is~$\alpha = \SI{20}{\degree}$. The width of the canal is still~\SI{5}{\metre}.}{\wcfile{Water lock door principle.svg}{Figure} \cczero \oc}
\label{fig_lock_doors_diagonal}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Reservoir door}
\wherefrom{Munson \& al. \smallcite{munsonetal2013} 2.87}
\label{exo_reservoir_door}
A water reservoir has a door of width~\SI{3}{\metre} which is held in place with a horizontal cable, as shown in \cref{fig_reservoir_door_cable}. The door has a mass of~\SI{200}{\kilogram} and the friction in the hinge is negligible.
\begin{figure}[h]
\begin{center}
\includegraphics[width=0.9\textwidth]{reservoir2}
\end{center}
\supercaption{A sealed, hinged door in a water reservoir. The width across the drawing (towards the reader) is~\SI{3}{\metre}}{\wcfile{Water reservoir door cable.svg}{Figure} \cczero \oc}
\label{fig_reservoir_door_cable}
\end{figure}
\begin{enumerate}
\item What is the force in the cable?
\item If the water height was decreased, how would this force be modified? (briefly justify your answer, e.g. in 30 words or less)
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item [\ref{exo_lock}]%
\tab 2) $F_\text{net, door A} = \rho g W \left[\frac{1}{2} l^2\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{+686,7}{\kilo\newton}$ (positive in right direction)
\tab 3) $M_\text{net, door A, about bottom axis} = \rho g W \left[\frac{1}{6} l^3\right]_{\SI{5}{\metre}}^{\SI{9}{\metre}} = \SI{2,4689}{\mega\newton\metre}$ (positive clockwise): $\vec F_\text{net, door A}$ exerts at $R_2 = \SI{3,595}{\metre}$ from the bottom.
\item [\ref{exo_buoyancy_tin_can}]%
\tab 1) $F_\text{vertical} = \SI{4,625}{\milli\newton}$ upwards;
\tab 2) $F_\text{vertical} = \SI{3,775}{\newton}$ in both cases;
\tab 3) There is no change;
\tab 4) See \S\ref{ch_atmospheric_pressure};
\tab 5) $F_\text{vertical} = \SI{4,487}{\milli\newton}$ upwards (and so in question 1 we were off by \SI{3}{\percent}).
\item [\ref{exo_buoyancy_airbus}]
\tab Assuming a volume of approx.~\SI{2600}{\metre\cubed}, we obtain approx.~\SI{30,9}{\kilo\newton} on the ground, \SI{10,4}{\kilo\newton} during cruise.
\item [\ref{exo_lock_diagonal}]
\tab The moment is brought to zero by an inter-door force $F_\text{sideways} = \SI{1,068}{\mega\newton}$ (perpendicular to the canal axis).
\item [\ref{exo_reservoir_door}] \tab $M_\text{water} = L \left(\derivative{p}{z}\right)_\text{water} \left[\frac{H}{2}r^2 - \frac{\sin{\theta}}{3}r^3 \right]^R_2 = \SI{0,4186}{\mega\newton\metre}$; so, $F_\text{cable} = \SI{81,13}{\kilo\newton}$.