The entropy equation: Difference between revisions

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Revision as of 06:36, 17 March 2026

From the second law of thermodynamics

$\frac{D e}{D t} = T \frac{D s}{D t} - p \frac{D}{D t} (\frac{1}{ρ})$

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

$\frac{D e}{D t} = \dot{q} - \frac{p}{ρ} (\nabla \cdot 𝐯)$

The continuity equation on differential non-conservation form

$\frac{D ρ}{D t} + ρ (\nabla \cdot 𝐯) = 0 \Rightarrow \nabla \cdot 𝐯 = - \frac{1}{ρ} \frac{D ρ}{D t}$

and thus

$\frac{D e}{D t} = \dot{q} + \frac{p}{ρ^{2}} \frac{D ρ}{D t}$

$\frac{D ρ}{D t} = - \frac{1}{ν^{2}} \frac{D ν}{D t}$

$ρ \frac{D e}{D t} = ρ \dot{q} - \frac{p}{ρ ν^{2}} \frac{D ν}{D t} = ρ \dot{q} - ρ p \frac{D ν}{D t}$

$ρ [\frac{D e}{D t} + p \frac{D ν}{D t} - \dot{q}] = 0 \Rightarrow \frac{D e}{D t} = \dot{q} - p \frac{D ν}{D t}$

Insert $D e / D t$ in Eqn. \ref{eq:second:law}

$\dot{q} - p \frac{D}{D t} (\frac{1}{ρ}) = T \frac{D s}{D t} - p \frac{D}{D t} (\frac{1}{ρ}) \Rightarrow$

$T \frac{D s}{D t} = - \dot{q}$

Adiabatic flow:

$T \frac{D s}{D t} = 0$

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

@@ Line 4: / Line 4: @@
 __TOC__
-\section{The Entropy Equation}
+From the second law of thermodynamics
-\noindent From the second law of thermodynamics\\
+<math display="block">
-\begin{equation}
 \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
-\[\frac{De}{Dt} = \dot{q} - \frac{p}{\rho}(\nabla\cdot\mathbf{v})\]\\
+<math display="block">
+\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
-\[\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 \Rightarrow \nabla\cdot\mathbf{v}=-\frac{1}{\rho}\frac{D\rho}{Dt}\]
+<math display="block">
+\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
-\[\frac{De}{Dt} = \dot{q} +\frac{p}{\rho^2}\frac{D\rho}{Dt}\]\\
+<math display="block">
+\frac{De}{Dt} = \dot{q} +\frac{p}{\rho^2}\frac{D\rho}{Dt}
+</math>
-\[\frac{D\rho}{Dt}=-\frac{1}{\nu^2}\frac{D\nu}{Dt}\]\\
+<math display="block">
+\frac{D\rho}{Dt}=-\frac{1}{\nu^2}\frac{D\nu}{Dt}
+</math>
-\[\rho\frac{De}{Dt} = \rho\dot{q} -\frac{p}{\rho\nu^2}\frac{D\nu}{Dt} = \rho\dot{q} -\rho p\frac{D\nu}{Dt} \]\\
+<math display="block">
+\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>
-\[\rho\left[\frac{De}{Dt}+p\frac{D\nu}{Dt}-\dot{q}\right]=0\Rightarrow\frac{De}{Dt}=\dot{q}-p\frac{D\nu}{Dt}\]\\
+<math display="block">
+\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}
-\[\dot{q}-p\frac{D}{Dt}\left(\frac{1}{\rho}\right)=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho}\right)\Rightarrow\]\\
+<math display="block">
+\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}
+<math display="block">
 T\frac{Ds}{Dt}=-\dot{q}
-\label{eq:second:law:b}
+</math>
-\end{equation}\\
-\noindent Adiabatic flow:\\
+Adiabatic flow:
-\begin{equation}
+<math display="block">
 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.

The entropy equation: Difference between revisions

Revision as of 06:36, 17 March 2026

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