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Daniel D'Orazio edited untitled.tex
over 9 years ago
Commit id: e4d41b3c2f6a5a360e24c556333e9e371d468edb
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\begin{equation}
dG = SdT + \frac{dP}{\rho}
\end{equation}
Now we assume to special cases, adiabatic and isothermal flows. First assume an adiabatic flow, $dQ =0$. Then
\begin{equation}
\int{\frac{dP}{\rho}} = \frac{\gamma }{\gamma -1} RT
\end{equation}
which we could have also seen from the expression for the enthalpy when $dQ = TdS =0$.
For the isothermal case use the Gibbs free energy with $dT=0$ to find
\begin{equation}
\int{\frac{dP}{\rho}} = G = RT \ln{\frac{P}{P_0}}.
\end{equation}