The entropy equation: Difference between revisions

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Created page with "Category:Compressible flow Category:Governing equations __TOC__ \section{The Entropy Equation} \noindent From the second law of thermodynamics\\ \begin{equation} \frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho}\right) \label{eq:second:law} \end{equation}\\ \noindent From the energy equation on differential non-conservation form internal energy formulation\\ \[\frac{De}{Dt} = \dot{q} - \frac{p}{\rho}(\nabla\cdot\mathbf{v})\]\\ \noindent The co..."
 
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\section{The Entropy Equation}
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\noindent From the second law of thermodynamics\\
-->
From the second law of thermodynamics


\begin{equation}
{{NumEqn|<math>
\frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho}\right)
\frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho}\right)
\label{eq:second:law}
</math>}}
\end{equation}\\


\noindent From the energy equation on differential non-conservation form internal energy formulation\\
From the energy equation on differential non-conservation form internal energy formulation


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


\noindent The continuity equation on differential non-conservation form
The continuity equation on differential non-conservation form


\[\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 \Rightarrow \nabla\cdot\mathbf{v}=-\frac{1}{\rho}\frac{D\rho}{Dt}\]
{{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 and thus\\
and thus


\[\frac{De}{Dt} = \dot{q} +\frac{p}{\rho^2}\frac{D\rho}{Dt}\]\\
{{NumEqn|<math>
\frac{De}{Dt} = \dot{q} +\frac{p}{\rho^2}\frac{D\rho}{Dt}
</math>}}


\[\frac{D\rho}{Dt}=-\frac{1}{\nu^2}\frac{D\nu}{Dt}\]\\
{{NumEqn|<math>
\frac{D\rho}{Dt}=-\frac{1}{\nu^2}\frac{D\nu}{Dt}
</math>}}


\[\rho\frac{De}{Dt} = \rho\dot{q} -\frac{p}{\rho\nu^2}\frac{D\nu}{Dt} = \rho\dot{q} -\rho p\frac{D\nu}{Dt} \]\\
{{NumEqn|<math>
\rho\frac{De}{Dt} = \rho\dot{q} -\frac{p}{\rho\nu^2}\frac{D\nu}{Dt} = \rho\dot{q} -\rho p\frac{D\nu}{Dt}
</math>}}


\[\rho\left[\frac{De}{Dt}+p\frac{D\nu}{Dt}-\dot{q}\right]=0\Rightarrow\frac{De}{Dt}=\dot{q}-p\frac{D\nu}{Dt}\]\\
{{NumEqn|<math>
\rho\left[\frac{De}{Dt}+p\frac{D\nu}{Dt}-\dot{q}\right]=0\Rightarrow\frac{De}{Dt}=\dot{q}-p\frac{D\nu}{Dt}
</math>}}


\noindent Insert $De/Dt$ in Eqn. \ref{eq:second:law} \\
Insert <math>De/Dt</math> in Eqn. \ref{eq:second:law}


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


\begin{equation}
{{NumEqn|<math>
T\frac{Ds}{Dt}=-\dot{q}
T\frac{Ds}{Dt}=-\dot{q}
\label{eq:second:law:b}
</math>}}
\end{equation}\\


\noindent Adiabatic flow:\\
Adiabatic flow:


\begin{equation}
{{NumEqn|<math>
T\frac{Ds}{Dt}=0
T\frac{Ds}{Dt}=0
\label{eq:second:law:b}
</math>}}
\end{equation}\\


\noindent In an adiabatic, steady-state, inviscid flow, entropy is constant along a streamline.
In an adiabatic, steady-state, inviscid flow, entropy is constant along a streamline.
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