Commit cf5a5a8e authored by Olivier's avatar Olivier

Exercises 4: Major cleanup, -5+1 problems

* 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)
parent 5cf84547
\renewcommand{\titleofthischapter}{Large- and small-scale flows}
......@@ -54,23 +53,23 @@
\subsubsection{Hangar roof}
\subsubsection{Idealized flow over a hangar roof}
\wherefrom{based on White \smallcite{white2008} P8.54}
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:
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}}
\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.
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.
......@@ -84,10 +83,15 @@
\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:
p_s &=& p_\infty + \frac{1}{2} \rho \left(V_\infty^2 - 4 V_\infty^2 \sin^2 \theta\right)
\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.
......@@ -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}]
This diff is collapsed.
\subsubsection{Water lock}
%\wherefrom{Exam \textsc{ws}2015/16}%homemade
A system of lock doors is set up to allow boats to travel up the side of a hill (figure~\ref{fig_locks_perpendicular}).
\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}
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.
\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)
\subsubsection{Buoyancy force on a tin can}
A student contemplates a tin can of height~\SI{10}{\centi\metre} and diameter~\SI{7}{\centi\metre}.
\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?
Instead of uniform density, we now wish to calculate the atmospheric buoyancy under the hypothesis of uniform temperature (room temperature~\SI{20}{\degreeCelsius}).
\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:
\frac{p_2}{p_1} & = & \exp \left[\frac{g \Delta z}{RT_\text{cst.}} \right] \ztag{\ref{eq_atmtemp}}
\item What is the buoyancy generated by the room atmosphere?
\subsubsection{Atmospheric buoyancy force}
\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 ($\rho_\text{cruise} = \SI{0,4}{\kilogram\per\metre\cubed}, T_\text{cruise} = \SI{-40}{\degreeCelsius}$).
\supercaption{Airbus A380-800}{\wcfile{A380-800v1.0.png}{Drawing} \ccbysa by Julien Scavini}
\begin{tabularx}{10cm}{r|c} % the @{something} kills the inter-column space and replaces it with "something"
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} \\
\caption{Characteristics of the Airbus A380-800.}
\subsubsection{Lock with diagonally-mounted doors}
\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?
\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}
\subsubsection{Reservoir door}
\wherefrom{Munson \& al. \smallcite{munsonetal2013} 2.87}
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.
\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}
\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)
\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, $F_\text{cable} = \SI{81,13}{\kilo\newton}$.
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