Governing equations on differential form
\section{Governing Equations on Differential Form}
\subsection{Conservation of Mass}
\noindent Apply Gauss's divergence theorem on the surface integral in Eqn. \ref{eq:governing:cont:int} gives\\
\[\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})d\mathscr{V}\]\\
\noindent Also, if $\Omega$ is a fixed control volume\\
\[\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}=\iiint_{\Omega} \frac{\partial \rho}{\partial t} d\mathscr{V}\]\\
\noindent The continuity equation can now be written as a single volume integral.\\
\[\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]d\mathscr{V}=0\]\\
\noindent $\Omega$ is an arbitrary control volume and thus\\
\begin{equation} \frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0 \label{eq:governing:cont:pde} \end{equation}\\
\noindent which is the continuity equation on partial differential form.\\
\subsection*{Conservation of Momentum}
As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.\\
\[\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})d\mathscr{V}\]\\
\[\oiint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pd\mathscr{V}\]\\
Also, if $\Omega$ is a fixed control volume\\
\[\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}=\iiint_{\Omega} \frac{\partial}{\partial t}(\rho \mathbf{v}) d\mathscr{V}\]\\
\noindent The momentum equation can now be written as one single volume integral\\
\[\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]d\mathscr{V}=0\]\\
\noindent $\Omega$ is an arbitrary control volume and thus\\
\begin{equation} \frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f} \label{eq:governing:mom:pde} \end{equation}\\
\noindent which is the momentum equation on partial differential form
\subsection{Conservation of Energy}
\noindent Gauss's divergence theorem applied to the surface integral term in the energy equation (Eqn. \ref{eq:governing:energy:int}) gives\\
\[\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})d\mathscr{V}\]\\
\noindent Fixed control volume \\
\[\frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) d\mathscr{V}\]\\
\noindent The energy equation can now be written as\\
\[\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]d\mathscr{V}=0\]\\
\noindent $\Omega$ is an arbitrary control volume and thus\\
\begin{equation} \frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho \label{eq:governing:energy:pde} \end{equation}\\
\noindent which is the energy equation on partial differential form\\
\subsection{Summary}
\noindent The governing equations for compressible inviscid flow on partial differential form:\\
\[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0\]
\[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}\]
\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]
\section{The Differential Equations on Non-Conservation Form}
\subsection{The Substantial Derivative}
\noindent The substantial derivative operator is defined as\\
\begin{equation} \frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla \label{eq:substantial:derivative} \end{equation}\\
\noindent where the first term of the right hand side is the local derivative and the second term is the convective derivative.\\
\subsection{Conservation of Mass}
\noindent If we apply the substantial derivative operator to density we get\\
\[\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho\]\\
\noindent From before we have the continuity equation on differential form as\\
\[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0\]\\
\noindent which can be rewritten as\\
\[\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0\]\\
\noindent and thus\\
\begin{equation} \frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 \label{eq:governing:cont:non} \end{equation}\\
\noindent Eqn. \ref{eq:governing:cont:non} says that the mass of a fluid element with a fixed set of fluid particles is constant as the element moves in space.\\
\subsection{Conservation of Momentum}
\noindent We start from the momentum equation on differential form derived above\\
\[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}\]\\
\noindent Expanding the first and the second terms gives\\
\[\rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f}\]\\
\noindent Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.\\
\[\rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f}\]\\
\noindent which gives us the non-conservation form of the momentum equation\\
\begin{equation} \frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f} \label{eq:governing:mom:non} \end{equation}\\
\subsection{Conservation of Energy}
\noindent The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form (Eqn. \ref{eq:governing:energy:pde}), repeated here for convenience\\
\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
\noindent Total enthalpy, $h_o$, is replaced with total energy, $e_o$\\
\[h_o=e_o+\frac{p}{\rho}\]\\
\noindent which gives\\
\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
\noindent Expanding the two first terms as\\
\[\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
\noindent Collecting terms, we can identify the substantial derivative operator applied on total energy, $De_o/Dt$ and the continuity equation\\
\[\rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}} + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
\noindent and thus we end up with the energy equation on non-conservation differential form\\
\begin{equation} \rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho \label{eq:governing:energy:non} \end{equation}\\
%\section*{The Governing Equations on Differential Non-Conservation Form} % %\vspace*{1cm} % %\noindent Continuity: % %\begin{equation} %\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 %\label{eq:governing:cont:non} %\end{equation}\\ % %\noindent Momentum: % %\begin{equation} %\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f} %\label{eq:governing:mom:non} %\end{equation}\\ % %\noindent Energy: % %\begin{equation} %\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho %\label{eq:governing:energy:non} %\end{equation}\\
\section{Alternative Forms of the Energy Equation}
\subsection{Internal Energy Formulation}
\noindent Total internal energy is defined as\\
\[e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\]\\
\noindent Inserted in Eqn. \ref{eq:governing:energy:non}, this gives
\[\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
\noindent Now, let's replace the substantial derivative $D\mathbf{v}/Dt$ using the momentum equation on non-conservation form (Eqn. \ref{eq:governing:mom:non}).\\
\[\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho\]\\
\noindent Now, expand the term $\nabla\cdot(p\mathbf{v})$ gives\\
\[\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow \rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\]\\
\noindent Divide by $\rho$\\
\begin{equation} \frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q} \label{eq:governing:energy:non:b} \end{equation}\\
\noindent Conservation of mass gives\\
\[\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}\]
\noindent Insert in Eqn. \ref{eq:governing:energy:non:b}\\
\[\frac{De}{Dt} - \frac{p}{\rho^2}\frac{D\rho}{Dt} = \dot{q}\Rightarrow \frac{De}{Dt} + p\frac{D}{Dt} \left(\frac{1}{\rho}\right)= \dot{q}\]\\
\begin{equation} \frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q} \label{eq:governing:energy:non:b} \end{equation}\\
\noindent Compare with the first law of thermodynamics: $de=\delta q-\delta w$\\
%\newpage
\subsection{Enthalpy Formulation}
\vspace*{1cm}
\[h=e+\frac{p}{\rho}\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho}\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho}\right)\]\\
\noindent with $De/Dt$ from Eqn. \ref{eq:governing:energy:non:b}\\
\[\frac{Dh}{Dt}=\dot{q} - \cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)} +\frac{1}{\rho}\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)}\]\\
\begin{equation} \frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt} \label{eq:governing:energy:non:c} \end{equation}\\
\subsection{Total Enthalpy Formulation}
\vspace*{1cm}
\[h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}\]\\
\noindent From the momentum equation (Eqn. \ref{eq:governing:mom:non})\\
\[\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p\]\\
\noindent which gives\\
\[\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p \]\\
\noindent Inserting $Dh/Dt$ from Eqn. \ref{eq:governing:energy:non:c} gives\\
\[\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p = \frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}\]\\
\noindent The substantial derivative operator applied to pressure\\
\[\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p\]\\
\noindent and thus\\
\[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}\]\\
\noindent which gives\\
\[\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}\]\\
\noindent If we assume adiabatic flow without body forces\\
\[\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}\]\\
\noindent If we further assume the flow to be steady state we get\\
\[\frac{Dh_o}{Dt}=0\]\\
\noindent This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.\\
