Governing equations on differential form
Conservation of Mass
Apply Gauss's divergence theorem on the surface integral in Eqn. \ref{eq:governing:cont:int} gives
Also, if is a fixed control volume
The continuity equation can now be written as a single volume integral.
is an arbitrary control volume and thus
which is the continuity equation on partial differential form.
Conservation of Momentum
As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
Also, if is a fixed control volume
The momentum equation can now be written as one single volume integral
is an arbitrary control volume and thus
which is the momentum equation on partial differential form
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.\\