 ### Chapter 5: minor typography fixes

parent deb649f0
 \renewcommand{\lastedityear}{2019} \renewcommand{\lasteditmonth}{03} \renewcommand{\lasteditday}{31} \renewcommand{\lasteditmonth}{05} \renewcommand{\lasteditday}{10} \renewcommand{\numberofthischapter}{5} \renewcommand{\titleofthischapter}{\namechapterfive} ... ... @@ -23,7 +23,7 @@ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Shear forces on walls} %%%% %%%% \subsection{Magnitude of the shear force} What is the force which which a fluid shears (i.e.\ “rubs”) against a wall? ... ... @@ -35,20 +35,21 @@ 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. ... ... @@ -80,7 +81,7 @@ 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: \begin{IEEEeqnarray}{rCl} \vec \tau &\equiv& \lim_{A \to 0} \frac{\vec F_\parallel}{A} \label{eq_def_shear} \end{IEEEeqnarray} \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}: ... ... @@ -90,7 +91,7 @@ \tau_y\\ \tau_z \end{array}\right)_{(x, y, z, t)} \end{IEEEeqnarray} \end{IEEEeqnarray} \subsection{Shear on an infinitesimal volume} ... ... @@ -107,13 +108,13 @@ In order to express the efforts on any given face, we express a component of shear with two subscripts, the first indicating the direction normal to the surface of interest, and the second indicating the direction of the effort. For example, $\vec \tau_{xy}$ represents the shear in the $y$-direction on a surface perpendicular to the $x$-direction. On this face, the shear vector would be: \begin{IEEEeqnarray}{rCl} \vec \tau_{xj} &=& \vec \tau_{xx} + \vec \tau_{xy} + \vec \tau_{xz} \label{eq_shear_x}\\ &=& \tau_{xx} \vec i + \tau_{xy} \vec j + \tau_{xz} \vec k \end{IEEEeqnarray} \begin{equationterms} &=& \tau_{xx} \vec i + \tau_{xy} \vec j + \tau_{xz} \vec k \end{IEEEeqnarray} \begin{equationterms} \item where the subscript $xj$ indicates all of the directions ($j = x, y, z$) on a face perpendicular to the $x$-direction. \end{equationterms} In eq.~\ref{eq_shear_x}, the reader may be surprised to see the term $\tau_{xx}$ appear — a shear effort perpendicular to the surface of interest. This is because the faces of the infinitesimal cube studied here (shown in \cref{fig_tau_cube}) are not solid. They are permeable, and the local velocity on each one may (in fact, must, if there is to be any flow) include a component of velocity through the face of the cube. Thus, there is no reason for the shear effort, which is three-dimensional, to be aligned along each flat surface. As the fluid travels across any face, it can be sheared (which results in strain) in any arbitrary direction, regardless of the local pressure — and thus shear can and most often does have a component ($\tau_{ii}$) perpendicular to an arbitrary surface inside a fluid. \end{equationterms} In eq.~\ref{eq_shear_x}, the reader may be surprised to see the term $\tau_{xx}$ appear — a shear effort perpendicular to the surface of interest. This is because the faces of the infinitesimal cube studied here (shown in \cref{fig_tau_cube}) are not solid. They are permeable, and the local velocity on each one may (in fact, must, if there is to be any flow) include a component of velocity through the face of the cube. Thus, there is no reason for the shear effort, which is three-dimensional, to be aligned along each flat surface. As the fluid travels across any face, it can be sheared (which results in strain) in any arbitrary direction, regardless of the local pressure — and thus shear can and most often does have a component ($\tau_{ii}$) perpendicular to an arbitrary surface inside a fluid. Now, the net shear effect on the cube will have \emph{eighteen} components: one tree-dimensional vector for each of the six faces. Each of those components may take a different value. The net shear could perhaps be represented as en entity —a \vocab{tensor}— containing six vectors $\vec \tau_1, \vec \tau_2, \vec \tau_3 … \vec \tau_6$. By convention, however, shear is notated using only three vector components: one for each pair of faces. Shear efforts on a volume are thus represented with a \vocab{tensor field} $\vec \tau_{ij}$: \begin{IEEEeqnarray}{rCl} ... ... @@ -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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