Governing equations on differential form: Difference between revisions

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[[Category:Compressible flow]]
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[[Category:Governing equations]]
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__TOC__
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=== Conservation of Mass ===
--><noinclude><!--
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Apply Gauss's divergence theorem on the surface integral in Eqn. \ref{eq:governing:cont:int} gives
-->
=== The Differential Equations on Conservation Form ===


<math display="block">
==== Conservation of Mass ====
 
The continuity equation on integral form reads
 
{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>}}
 
Apply Gauss's divergence theorem on the surface integral gives
 
{{NumEqn|<math>
\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})dV
</math>
</math>}}


Also, if <math>\Omega</math> is a fixed control volume
Also, if <math>\Omega</math> is a fixed control volume


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV
\frac{d}{dt}\iiint_{\Omega} \rho dV=\iiint_{\Omega} \frac{\partial \rho}{\partial t} dV
</math>
</math>}}


The continuity equation can now be written as a single volume integral.
The continuity equation can now be written as a single volume integral.


<math display="block">
{{NumEqn|<math>
\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0
\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]dV=0
</math>
</math>}}


<math>\Omega</math> is an arbitrary control volume and thus
<math>\Omega</math> is an arbitrary control volume and thus


<math display="block">
{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>
</math>|label=eq-cont-pde}}


which is the continuity equation on partial differential form.
which is the continuity equation on partial differential form.


=== Conservation of Momentum ===
==== Conservation of Momentum ====
 
The momentum equation on integral form reads
 
{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
</math>}}


As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})dV
\iint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})dV
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV
\iint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pdV
</math>
</math>}}


Also, if <math>\Omega</math> is a fixed control volume
Also, if <math>\Omega</math> is a fixed control volume


<math display="block">
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega}  \frac{\partial}{\partial t}(\rho \mathbf{v}) dV
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV=\iiint_{\Omega}  \frac{\partial}{\partial t}(\rho \mathbf{v}) dV
</math>
</math>}}


The momentum equation can now be written as one single volume integral
The momentum equation can now be written as one single volume integral


<math display="block">
{{NumEqn|<math>
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]dV=0
\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]dV=0
</math>
</math>}}


<math>\Omega</math> is an arbitrary control volume and thus
<math>\Omega</math> is an arbitrary control volume and thus


<math display="block">
{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}  
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}  
</math>
</math>|label=eq-mom-pde}}


which is the momentum equation on partial differential form
which is the momentum equation on partial differential form


=== Conservation of Energy ===
==== 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\\
The energy equation on integral form reads


\[\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})d\mathscr{V}\]\\
{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>}}


\noindent Fixed control volume \\
Gauss's divergence theorem applied to the surface integral term in the energy equation gives


\[\frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) d\mathscr{V}\]\\
{{NumEqn|<math>
\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})dV
</math>}}


\noindent The energy equation can now be written as\\
Fixed control volume


\[\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]d\mathscr{V}=0\]\\
{{NumEqn|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) dV
</math>}}


\noindent $\Omega$ is an arbitrary control volume and thus\\
The energy equation can now be written as


\begin{equation}
{{NumEqn|<math>
\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]dV=0
</math>}}
 
<math>\Omega</math> is an arbitrary control volume and thus
 
{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
\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}
</math>|label=eq-energy-pde}}
\end{equation}\\


\noindent which is the energy equation on partial differential form\\
which is the energy equation on partial differential form


\subsection{Summary}
==== Summary ====


\noindent The governing equations for compressible inviscid flow on partial differential form:\\
The governing equations for compressible inviscid flow on partial differential form:


\[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0\]
<div style="border: solid 1px;">
{{OpenInfoBox|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|description=Continuity:}}


\[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}\]
{{OpenInfoBox|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|description=Momentum:}}


\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]
{{OpenInfoBox|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|description=Energy:}}
</div>


\section{The Differential Equations on Non-Conservation Form}
=== The Differential Equations on Non-Conservation Form ===


\subsection{The Substantial Derivative}
==== The Substantial Derivative ====


\noindent The substantial derivative operator is defined as\\
The substantial derivative operator is defined as


\begin{equation}
{{NumEqn|<math>
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla
\label{eq:substantial:derivative}
</math>|label=eq-cont-pde-non-cons}}
\end{equation}\\


\noindent where the first term of the right hand side is the local derivative and the second term is the convective derivative.\\
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}
==== Conservation of Mass ====


\noindent If we apply the substantial derivative operator to density we get\\
If we apply the substantial derivative operator to density we get


\[\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho\]\\
{{NumEqn|<math>
\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho
</math>}}


\noindent From before we have the continuity equation on differential form as\\
From before we have the continuity equation on differential form as


\[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0\]\\
{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>}}


\noindent which can be rewritten as\\
which can be rewritten as


\[\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0\]\\
{{NumEqn|<math>
\frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0
</math>}}


\noindent and thus\\
and thus


\begin{equation}
{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
\label{eq:governing:cont:non}
</math>|label=eq-pde-noncons-cont}}
\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.\\
{{EquationNote|label=eq-pde-noncons-cont|nopar=1}} 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}
==== Conservation of Momentum ====


\noindent We start from the momentum equation on differential form derived above\\
We start from the momentum equation on differential form derived above


\[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}\]\\
{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>}}


\noindent Expanding the first and the second terms gives\\
Expanding the first and the second terms gives


\[\rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f}\]\\
{{NumEqn|<math>
\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}
</math>}}


\noindent Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.\\
Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.


\[\rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f}\]\\
{{NumEqn|<math>
\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}
</math>}}


\noindent which gives us the non-conservation form of the momentum equation\\
which gives us the non-conservation form of the momentum equation


\begin{equation}
{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
\label{eq:governing:mom:non}
</math>|label=eq-mom-pde-non-cons}}
\end{equation}\\


\subsection{Conservation of Energy}
==== 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\\
The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form {{EquationNote|label=eq-energy-pde}}, repeated here for convenience


\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
{{NumEqn|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|nonumber=1}}


\noindent Total enthalpy, $h_o$, is replaced with total energy, $e_o$\\
Total enthalpy, <math>h_o</math>, is replaced with total energy, <math>e_o</math>


\[h_o=e_o+\frac{p}{\rho}\]\\
{{NumEqn|<math>
h_o=e_o+\frac{p}{\rho}
</math>}}


\noindent which gives\\
which gives


\[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
{{NumEqn|<math>
\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
</math>}}


\noindent Expanding the two first terms as\\
Expanding the two first terms as


\[\rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
{{NumEqn|<math>
\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})=</math><br><br><math>= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>}}


\noindent Collecting terms, we can identify the substantial derivative operator applied on total energy, $De_o/Dt$ and the continuity equation\\
Collecting terms, we can identify the substantial derivative operator applied on total energy, <math>De_o/Dt</math> and the continuity equation


\[\rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}}  + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
{{NumEqn|<math>
\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
</math>}}


\noindent and thus we end up with the energy equation on non-conservation differential form\\
and thus we end up with the energy equation on non-conservation differential form


\begin{equation}
{{NumEqn|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
\label{eq:governing:energy:non}
</math>|label=eq-energy-pde-non-cons}}
\end{equation}\\
 
==== Summary ====


%\section*{The Governing Equations on Differential Non-Conservation Form}
<div style="border: solid 1px;">
%
{{OpenInfoBox|<math>
%\vspace*{1cm}
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
%
</math>|description=Continuity:}}
%\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}
{{OpenInfoBox|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|description=Momentum:}}


\subsection{Internal Energy Formulation}
{{OpenInfoBox|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|description=Energy:}}
</div>


\noindent Total internal energy is defined as\\
=== Alternative Forms of the Energy Equation ===


\[e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\]\\
==== Internal Energy Formulation ====


\noindent Inserted in Eqn. \ref{eq:governing:energy:non}, this gives
Total internal energy is defined as


\[\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
{{NumEqn|<math>
e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}
</math>}}


\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}).\\
Inserted in {{EquationNote|label=eq-energy-pde-non-cons|nopar=1}}, this gives


\[\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho\]\\
{{NumEqn|<math>
\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
</math>}}


\noindent Now, expand the term $\nabla\cdot(p\mathbf{v})$ gives\\
Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form {{EquationNote|label=eq-mom-pde-non-cons}}.


\[\rho\frac{De}{Dt} \cancel{-\mathbf{v}\cdot\nabla p} + \cancel{\mathbf{v}\cdot\nabla p} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\Rightarrow \rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho\]\\
{{NumEqn|<math>
\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
</math>}}


\noindent Divide by $\rho$\\
Now, expand the term <math>\nabla\cdot(p\mathbf{v})</math> gives


\begin{equation}
{{NumEqn|<math>
\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</math><br><br><math>\Rightarrow\rho\frac{De}{Dt} + p(\nabla\cdot\mathbf{v}) = \dot{q}\rho
</math>}}
 
Divide by <math>\rho</math>
 
{{NumEqn|<math>
\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
\frac{De}{Dt} + \frac{p}{\rho}(\nabla\cdot\mathbf{v}) = \dot{q}
\label{eq:governing:energy:non:b}
</math>|label=eq-energy-pde-non-cons-b}}
\end{equation}\\


\noindent Conservation of mass gives\\
Conservation of mass gives


\[\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}\]
{{NumEqn|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}
</math>}}


\noindent Insert in Eqn. \ref{eq:governing:energy:non:b}\\
Insert in {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}


\[\frac{De}{Dt} - \frac{p}{\rho^2}\frac{D\rho}{Dt} = \dot{q}\Rightarrow \frac{De}{Dt} + p\frac{D}{Dt} \left(\frac{1}{\rho}\right)= \dot{q}\]\\
{{NumEqn|<math>
\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}
</math>}}


\begin{equation}
{{NumEqn|<math>
\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}
\frac{De}{Dt} + p\frac{D\nu}{Dt} = \dot{q}
\label{eq:governing:energy:non:b}
</math>}}
\end{equation}\\


\noindent Compare with the first law of thermodynamics: $de=\delta q-\delta w$\\
Compare with the first law of thermodynamics: <math>de=\delta q-\delta w</math>


%\newpage
==== Enthalpy Formulation ====


\subsection{Enthalpy Formulation}
{{NumEqn|<math>
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)
</math>}}


\vspace*{1cm}
with <math>De/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-b|nopar=1}}


\[h=e+\frac{p}{\rho}\Rightarrow \frac{Dh}{Dt}=\frac{De}{Dt}+\frac{1}{\rho}\frac{Dp}{Dt}+p\frac{D}{Dt}\left(\frac{1}{\rho}\right)\]\\
{{NumEqn|<math>
\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)}
</math>}}


\noindent with $De/Dt$ from Eqn. \ref{eq:governing:energy:non:b}\\
{{NumEqn|<math>
 
\[\frac{Dh}{Dt}=\dot{q} - \cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)} +\frac{1}{\rho}\frac{Dp}{Dt}+\cancel{p\frac{D}{Dt}\left(\frac{1}{\rho}\right)}\]\\
 
\begin{equation}
\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
\frac{Dh}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}
\label{eq:governing:energy:non:c}
</math>|label=eq-energy-pde-non-cons-c}}
\end{equation}\\
 
\subsection{Total Enthalpy Formulation}


\vspace*{1cm}
==== Total Enthalpy Formulation ====


\[h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}\]\\
{{NumEqn|<math>
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}
</math>}}


\noindent From the momentum equation (Eqn. \ref{eq:governing:mom:non})\\
From the momentum equation {{EquationNote|label=eq-mom-pde-non-cons}}


\[\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p\]\\
{{NumEqn|<math>
\frac{D\mathbf{v}}{Dt}=\mathbf{f}-\frac{1}{\rho}\nabla p
</math>}}


\noindent which gives\\
which gives


\[\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p \]\\
{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p
</math>}}


\noindent Inserting $Dh/Dt$ from Eqn. \ref{eq:governing:energy:non:c} gives\\
Inserting <math>Dh/Dt</math> from {{EquationNote|label=eq-energy-pde-non-cons-c|nopar=1}} gives


\[\frac{Dh_o}{Dt}=\dot{q} + \frac{1}{\rho}\frac{Dp}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p = \frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}\]\\
{{NumEqn|<math>
\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 =</math><br><br><math>=\frac{1}{\rho}\left[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p\right] + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}


\noindent The substantial derivative operator applied to pressure\\
The substantial derivative operator applied to pressure


\[\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p\]\\
{{NumEqn|<math>
\frac{Dp}{Dt}=\frac{\partial p}{\partial t}+\mathbf{v}\cdot\nabla p
</math>}}


\noindent and thus\\
and thus


\[\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}\]\\
{{NumEqn|<math>
\frac{Dp}{Dt}-\mathbf{v}\cdot\nabla p=\frac{\partial p}{\partial t}
</math>}}


\noindent which gives\\
which gives


\[\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}\]\\
{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t} + \dot{q} + \mathbf{v}\cdot\mathbf{f}
</math>}}


\noindent If we assume adiabatic flow without body forces\\
If we assume adiabatic flow without body forces


\[\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}\]\\
{{NumEqn|<math>
\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}
</math>}}


\noindent If we further assume the flow to be steady state we get\\
If we further assume the flow to be steady state we get


\[\frac{Dh_o}{Dt}=0\]\\
{{NumEqn|<math>
\frac{Dh_o}{Dt}=0
</math>}}


\noindent This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.\\
This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant along a streamline.
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