\documentclass{book}
\usepackage{amsmath,amsthm}
\newtheorem{example}{Example}[chapter]
\begin{document}
\chapter{Overview of Hydromechanics}
\label{ch02:chap02}
This chapter provides a broad overview of the basics of hydromechanics (also referred to as \textit{fluid mechanics}) that are needed background for subsequent chapters. The purpose of this chapter is to highlight fundamental concepts and equations rather than to substitute for textbooks of fluid mechanics. The overview covers water properties, hydrostatic pressure and force, conservations of mass, energy and momentum in flowing water, and dimensional analysis and similitude.
\begin{example}\label{ch02:exa2.1}
For the hydraulic structures submerged in water $(\gamma = 62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})$ shown in Figure~\ref{ch02:fig2.2}, determine and sketch the hydrostatic pressures on the submerged surfaces.
\begin{figure}[!h]
\centerline{\fbox{\vbox to 30pt{\hbox to 30pt{}}}}
\caption{\textbf{The submerged structures for Example~\ref{ch02:exa2.1}.} (Not drawn to scale.)}
\label{ch02:fig2.2}
\end{figure}
$\gamma = 62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3}$. Use Eq.~(\ref{ch02:eqn2.6}) to compute pressures at different water depths.
\underline{For the structure in Figure~\ref{ch02:fig2.2}}a:
At $\mathrm{h} = 0\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (0\,\mathrm{ft}) = 0\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
At $\mathrm{h} = 15\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (15\,\mathrm{ft}) = 936\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
At $\mathrm{h} = 15\,\mathrm{ft} + 75\,\mathrm{ft} = 90\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (90\,\mathrm{ft}) = 5616\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
The pressure distributions between 0 and 15 ft as well as between 15 and 90 ft are linear. The results are shown in Figure~\ref{ch02:fig2.3}a.
\underline{For the structure in Figure~\ref{ch02:fig2.2}}b:
At $\mathrm{h} = 0\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (0\,\mathrm{ft}) = 0\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
At $\mathrm{h} = 80\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (80\,\mathrm{ft}) = 4992\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
At the midpoint of the submerged arch, $\mathrm{h} = 80\,\mathrm{ft} + (20\,\mathrm{ft})\ast \sin(\pi/4) = 94.14\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (94.14\,\mathrm{ft}) = 5874.3\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
At $\mathrm{h} = 80\,\mathrm{ft} + 20\,\mathrm{ft} = 100\,\mathrm{ft}$, $\mathrm{p} = (62.4\,\mathrm{lbf}\,\mathrm{ft}^{-3})\ast (100\,\mathrm{ft}) = 6240\,\mathrm{lbf}\,\mathrm{ft}^{-2}$.
\begin{figure}[!h]
\centerline{\fbox{\vbox to 30pt{\hbox to 30pt{}}}}
\caption{\textbf{The computed pressures for Example~\ref{ch02:exa2.1}.}}
\label{ch02:fig2.3}
\end{figure}
The pressure distributions between 0 and 80 ft as well as between 80 and 100 ft are linear. However, the directions of the pressures between 80 and 100 ft vary from point to point and always point to the center of the circle. The results are shown in Figure~\ref{ch02:fig2.3}b.
\end{example}
This chapter provides a broad overview of the basics of hydromechanics (also referred to as \textit{fluid mechanics}) that are needed background for subsequent chapters. The purpose of this chapter is to highlight fundamental concepts and equations rather than to substitute for textbooks of fluid mechanics. The overview covers water properties, hydrostatic pressure and force, conservations of mass, energy and momentum in flowing water, and dimensional analysis and similitude.
\end{document}
