Governing equations on differential form: Difference between revisions

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==== Total Enthalpy Formulation ====
==== Total Enthalpy Formulation ====


\[h_o=h+\frac{1}{2}\mathbf{v}\mathbf{v}\Rightarrow\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\frac{D\mathbf{v}}{Dt}\]\\
<math display="block">
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 (Eqn. \ref{eq:governing:mom:non})


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


\noindent which gives\\
which gives


\[\frac{Dh_o}{Dt}=\frac{Dh}{Dt}+\mathbf{v}\cdot\mathbf{f} -\frac{1}{\rho}\mathbf{v}\cdot\nabla p \]\\
<math display="block">
\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 Eqn. \ref{eq:governing:energy:non:c} gives


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


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


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


\noindent and thus\\
and thus


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


\noindent which gives\\
which gives


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


\[\frac{Dh_o}{Dt}=\frac{1}{\rho}\frac{\partial p}{\partial t}\]\\
<math display="block">
\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


\[\frac{Dh_o}{Dt}=0\]\\
<math display="block">
\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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