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 ... ... @@ -13,6 +13,7 @@ \renewcommand{\theequation}{A/\arabic{equation}} \setcounter{equation}{0} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Notation} ... ... @@ -41,6 +42,7 @@ \clearpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Field operators} \label{appendix_field_operators} ... ... @@ -172,7 +174,57 @@ \end{IEEEeqnarray} \clearpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Derivations of the Bernoulli equation} \label{appendix_bernoulli} \subsection{The Bernoulli equation from the energy equation} \label{appendix_benoulli_energy} This is covered in section~\ref{ch_bernoulli} p.\pageref{ch_bernoulli}. \subsection{The Bernoulli equation from the integral momentum equation} We begin with the integral linear momentum equation (eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}): \begin{IEEEeqnarray*}{rCl} \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \end{IEEEeqnarray*} When considering a fixed, infinitely short control volume along a known streamline $s$ of the flow, this equation becomes: \begin{IEEEeqnarray*}{rCl} \diff \vec F_\text{pressure} + \diff \vec F_\text{shear} + \diff \vec F_\text{gravity} & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \rho V A \diff \vec V \end{IEEEeqnarray*} \begin{equationterms} \item along a streamline, where the velocity $\vec V$ is aligned (by definition) with the streamline. \end{equationterms} Now, adding the restrictions of steady flow ($\diff/\diff{t}=0$) and no friction ($\diff \vec F_\text{shear} = \vec 0$), we already obtain: \begin{IEEEeqnarray*}{rCl} \diff \vec F_\text{pressure} + \diff \vec F_\text{gravity} & = & \rho V A \diff \vec V \end{IEEEeqnarray*} The projection of the net force due to gravity $\diff \vec F_\text{gravity}$ on the streamline segment $\diff s$ has norm $\diff \vec F_\text{gravity} \cdot \diff \vec{s} = -g \rho A \diff z$, while the net force due to pressure is aligned with the streamline and has norm $\diff F_{\text{pressure}, s} = - A \diff p$. Along this streamline, we thus have the following scalar equation, which we integrate from points~1 to~2: \begin{IEEEeqnarray*}{rCl} - A \diff p - \rho g A \diff z & = & \rho V A \diff V \nonumber\\ -\frac{1}{\rho} \diff p - g \diff z & = & V \diff V \nonumber\\ -\int_1^2 \frac{1}{\rho} \diff p - \int_1^2 g \diff z & = & \int_1^2 V \diff V \end{IEEEeqnarray*} The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density $\rho$ is constant. In this way, we arrive to equation.~\ref{eq_bernoulli} p.\pageref{eq_bernoulli} again: \begin{IEEEeqnarray*}{rCl} \frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2\\ \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2 \end{IEEEeqnarray*} \clearpage %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{List of references} \mecafluboxen ... ...
 \documentclass[12pt,a4paper,twoside]{book} \usepackage{fluidmechnotes} % from https://git.framasoft.org/u/olivier/sensible-styles \usepackage{a/texttemplates-fluidmech} %\usepackage{showframe} % Debug mode to work on layout \input{a/steer} % May contain \includeonly command genrated with bash script ... ...