Governing equations on differential form: Difference between revisions

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==== Internal Energy Formulation ====
==== Internal Energy Formulation ====


\noindent Total internal energy is defined as\\
Total internal energy is defined as


\[e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}\]\\
<math display="block">
e_o=e+\frac{1}{2}\mathbf{v}\cdot\mathbf{v}
</math>


\noindent Inserted in Eqn. \ref{eq:governing:energy:non}, this gives
Inserted in Eqn. \ref{eq:governing:energy:non}, this gives


\[\rho\frac{De}{Dt} + \rho\mathbf{v}\cdot\frac{D \mathbf{v}}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho\]\\
<math display="block">
\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, let's replace the substantial derivative $D\mathbf{v}/Dt$ using the momentum equation on non-conservation form (Eqn. \ref{eq:governing:mom:non}).\\
Now, let's replace the substantial derivative <math>D\mathbf{v}/Dt</math> using the momentum equation on non-conservation form (Eqn. \ref{eq:governing:mom:non}).


\[\rho\frac{De}{Dt} -\mathbf{v}\cdot\nabla p + \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \nabla\cdot(p\mathbf{v}) = \cancel{\rho\mathbf{f}\cdot\mathbf{v}} + \dot{q}\rho\]\\
<math display="block">
\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 Now, expand the term $\nabla\cdot(p\mathbf{v})$ gives\\
Now, expand the term <math>\nabla\cdot(p\mathbf{v})</math> gives


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


\noindent Divide by $\rho$\\
Divide by <math>\rho</math>


\begin{equation}
<math display="block">
\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>
\end{equation}\\


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


\[\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0\Rightarrow \nabla\cdot\mathbf{v} = -\frac{1}{\rho}\frac{D\rho}{Dt}\]
<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 Insert in Eqn. \ref{eq:governing:energy:non:b}\\
Insert in Eqn. \ref{eq:governing:energy:non:b}


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


==== Enthalpy Formulation ====
==== Enthalpy Formulation ====
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