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Commits (4)
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%%%%
\subsection{Magnitude of the pressure force}
\youtubethumb{ke4Rf9FrHnw}{when pressure-induced forces in static fluids matter: 24 hours of heavy tonnage transit through the \textit{Miraflores} locks in Panama. Can you quantify the force applying on a single lock door?}{Y:XyliboxFrance (\styl)}
What is the force which which a fluid pushes against a wall?
What is the force with which a fluid pushes against a wall?
When the pressure $p$ exerted is uniform and the wall is flat, the resulting force $F$ is easily calculated:
\begin{IEEEeqnarray}{rCcCl}
......@@ -111,7 +111,7 @@
\subsection{The direction of pressure}
An important concept is that in continuum mechanics, the flat surface is imaginary. More precisely, a fluid is able to exert pressure on not only on solid surfaces, but also upon and within itself. In this context, we need to rework eq.~\ref{eq_first_def_pressure_two} so that now pressure is defined as perpendicular force per area on an \emph{infinitesimally small} surface of fluid:
An important concept is that in continuum mechanics, the flat surface is imaginary. More precisely, a fluid is able to exert pressure not only on solid surfaces, but also upon and within itself. In this context, we need to rework eq.~\ref{eq_first_def_pressure_two} so that now pressure is defined as perpendicular force per area on an \emph{infinitesimally small} surface of fluid:
\begin{IEEEeqnarray}{rCl}
p &\equiv& \lim_{A \to 0} \frac{F_\perp}{A} \label{eq_def_pressure}
\end{IEEEeqnarray}
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When the shear $\tau$ exerted by the fluid is not uniform (for example, because more friction is occurring on some parts of the surface than on others), the situation is more complex: the force must be obtained by integration. The surface is split in infinitesimal portions of area $\diff S$, and the corresponding forces are summed up as:
\begin{IEEEeqnarray}{rCcCl}
F_\text{{shear, direction } i} & = & \int_S \diff F_{\text{shear, direction } i} & = & \int_S \tau_{\text{direction } i} \diff S \label{eq_shear_force_scalar}
F_{\text{shear, direction } i} & = & \int_S \diff F_{\text{shear, direction } i} & = & \int_S \tau_{\text{direction } i} \diff S \label{eq_shear_force_scalar}
\end{IEEEeqnarray}
\begin{equationterms}
\item for a flat surface,
\item where the $S$-integral denotes an integration over the entire surface.
\end{equationterms}
What is required to calculate the scalar $F$ in eq.~\ref{eq_shear_force_scalar} is an expression of~$\tau$ as a function of~$S$. In a simple laminar flow, this expression will often be relatively easy to find, as we see later on. Typically, in two dimensions $x$ and $y$ we re-write $\diff S$ as $\diff S = \diff x \diff y$ and we may then proceed with the calculation starting from
What is required to calculate the scalar $F$ in eq.~\ref{eq_shear_force_scalar} is an expression of~$\tau$ as a function of~$S$. In a simple laminar flow, this expression will often be relatively easy to find, as we see later on. Typically, in two dimensions $x$ and $y$ we re-write $\tdiff S$ as $\tdiff S = \diff x \diff y$ and we may then proceed with the calculation starting from
\begin{mdframed}
\begin{IEEEeqnarray}{rCcCl}
F_\text{{shear, direction } i} & = & \iint \tau_{\text{direction } i (x, y)} \diff x \diff y \label{eq_shear_force_twod_integration}
\end{IEEEeqnarray}
\end{mdframed}
%%%%
\subsection{Direction and position of the shear force}
The above equations work only for a flat surface, and in a chosen direction $i$. When we consider a two- or three-dimensional object immersed in a fluid with non-uniform shear, the integration must be carried out with vectors. We will not attempt this in this course, but the expression is worth writing out in order to understand how computational fluid dynamics (\cfd) software will proceed with the calculation.
......@@ -144,13 +145,13 @@
\end{IEEEeqnarray}
Given that $S_3 = S_6 = \diff x \diff y$, that $S_2 = S_5 = \diff x \diff z$ and that $S_1 = S_4 = \diff z \diff y$, this is re-written as:
\begin{IEEEeqnarray}{rCl}
\vec F_{\text{shear}\ x} &=& \diff x \diff y (\vec \tau_{zx\ 3} - \vec \tau_{zx\ 6})\nonumber\\
&& + \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}
\vec F_{\text{shear}\ x} &=& \diff x \diff y \ (\vec \tau_{zx\ 3} - \vec \tau_{zx\ 6})\nonumber\\
&& + \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:
\begin{IEEEeqnarray}{rCl}
\vec F_{\text{shear}\ x} &=& \diff x \diff y \left(\diff z \partialderivative{\vec \tau_{zx}}{z}\right) + \diff x \diff z \left(\diff y \partialderivative{\vec \tau_{yx}}{y}\right) + \diff z \diff y \left(\diff x \partialderivative{\vec \tau_{xx}}{x}\right)\nonumber\\
\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}
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[0] Config.pm:302> INFO - This is Biber 2.9
[1] Config.pm:305> INFO - Logfile is 'fluidmech.blg'
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[46] biber:313> INFO - === Fri May 10, 2019, 10:07:34
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