Thermodynamic processes
\section{Thermodynamics}
\subsection{Specific Heat Relations}
\noindent For thermally perfect and calorically perfect gases\\
\begin{equation} \begin{aligned} &C_p=\frac{dh}{dT}\\ &C_v=\frac{de}{dT} \end{aligned} \label{eq:specificheat} \end{equation}\\
\noindent From the definition of enthalpy and the equation of state $p=\rho RT$\\
\begin{equation} h=e+\frac{p}{\rho}=e+RT \label{eq:enthalpy} \end{equation}\\
\noindent Differentiate Eqn. \ref{eq:enthalpy} with respect to temperature gives\\
\begin{equation} \frac{dh}{dT}=\frac{de}{dT}+\frac{d(RT)}{dT} \label{eq:enthalpy:b} \end{equation}\\
\noindent Inserting the specific heats gives\\
\begin{equation} C_p=C_v+R \label{eq:specificheat:b} \end{equation}\\
\noindent Dividing Eqn. \ref{eq:specificheat:b} by $C_v$ gives\\
\begin{equation} \frac{C_p}{C_v}=1+\frac{R}{C_v} \label{eq:specificheat:c} \end{equation}\\
\noindent Introducing the ratio of specific heats defined as\\
\begin{equation} \gamma=\frac{C_p}{C_v} \label{eq:gamma} \end{equation}\\
\noindent Now, inserting Eqn. \ref{eq:gamma} in Eqn. \ref{eq:specificheat:c} gives\\
\begin{equation} C_v=\frac{R}{\gamma-1} \label{eq:specificheat:d} \end{equation}\\
\noindent In the same way, dividing Eqn. \ref{eq:specificheat:b} with $C_p$ gives\\
\begin{equation} 1=\frac{C_v}{C_p}+\frac{R}{C_p}=\frac{1}{\gamma}+\frac{R}{C_p} \label{eq:specificheat:e} \end{equation}\\
\noindent and thus\\
\begin{equation} C_p=\frac{\gamma R}{\gamma-1} \label{eq:specificheat:f} \end{equation}\\
\subsection{Isentropic Relations}
First law of thermodynamics:\\
\begin{equation} de=\delta q - \delta w \label{eq:firstlaw} \end{equation}\\
\noindent For a reversible process: $\delta w=pd(1/\rho)$ and $\delta q=Tds$\\
\begin{equation} de=Tds-pd\left(\frac{1}{\rho}\right) \label{eq:firstlaw:b} \end{equation}\\
\noindent Enthalpy is defined as: $h=e+p/\rho$ and thus\\
\begin{equation} dh=de+pd\left(\frac{1}{\rho}\right)+\left(\frac{1}{\rho}\right)dp \label{eq:dh} \end{equation}\\
\noindent Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh}\\
\[Tds=dh-\cancel{pd\left(\frac{1}{\rho}\right)}-\left(\frac{1}{\rho}\right)dp+\cancel{pd\left(\frac{1}{\rho}\right)}\]\\
\[ds=\frac{dh}{T}-\frac{dp}{\rho T}\]\\
\noindent Using $dh=C_p T$ and the equation of state $p=\rho RT$, we get\\
\begin{equation} ds=C_p\frac{dT}{T}-R\frac{dp}{p} \label{eq:ds} \end{equation}\\
\noindent Integrating Eqn. \ref{eq:ds} gives\\
\begin{equation} s_2-s_1=\int_1^2 C_p\frac{dT}{T}-R\ln\left(\frac{p_2}{p_1}\right) \label{eq:ds:b} \end{equation}\\
\noindent For a calorically perfect gas, $C_p$ is constant (not a function of temperature) and can be moved out from the integral and thus\\
\begin{equation} s_2-s_1=C_p\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{p_2}{p_1}\right) \label{eq:ds:c} \end{equation}\\
\noindent An alternative form of Eqn. \ref{eq:ds:c} is obtained by using $de=C_v dT$ Eqn. \ref{eq:firstlaw:b}, which gives\\
\begin{equation} s_2-s_1=\int_1^2 C_v\frac{dT}{T}-R\ln\left(\frac{\rho_2}{\rho_1}\right) \label{eq:ds:d} \end{equation}\\
\noindent Again, for a calorically perfect gas, we get\\
\begin{equation} s_2-s_1=C_v\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{\rho_2}{\rho_1}\right) \label{eq:ds:e} \end{equation}\\
\section*{Isentropic Relations}
\noindent Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus Eqn. \ref{eq:ds:c} reduces to\\
\[\frac{C_p}{R}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)\]\\
\[\frac{C_p}{R}=\frac{\gamma}{\gamma-1}\]\\
\[\frac{\gamma}{\gamma-1}\ln\left(\frac{T_2}{T_1}\right)=\ln\left(\frac{p_2}{p_1}\right)\Rightarrow\]\\
\begin{equation} \frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)} \label{eq:isentropic:a} \end{equation}\\
\noindent In the same way, Eqn. \ref{eq:ds:e} gives\\
\begin{equation} \frac{\rho_2}{\rho_1}=\left(\frac{T_2}{T_1}\right)^{1/(\gamma-1)} \label{eq:isentropic:b} \end{equation}\\
\noindent Eqn. \ref{eq:isentropic:a} and Eqn. \ref{eq:isentropic:b} constitutes the isentropic relations\\
\begin{equation} \frac{p_2}{p_1}=\left(\frac{\rho_2}{\rho_1}\right)^{\gamma}=\left(\frac{T_2}{T_1}\right)^{\gamma/(\gamma-1)} \label{eq:isentropic:a} \end{equation}
\section{Flow Processes}
\[ds=C_v\dfrac{dT}{T}+R\dfrac{d\nu}{\nu}\]
\begin{equation} d\nu=\dfrac{\nu}{R}ds-C_v\dfrac{\nu}{RT}dT=\dfrac{\nu}{R}ds-\dfrac{C_v}{p}dT \label{eqn:process:dnu} \end{equation}
\noindent for an isentropic process ($ds=0$), $d\nu < 0$ for positive values of $dT$.
\[ds=C_p\dfrac{dT}{T} - R \dfrac{dp}{p}\]
\[dp=-\dfrac{p}{R}ds+C_p\dfrac{p}{RT}dT=-\dfrac{p}{R}ds+C_p\rho dT\]
\noindent for an isentropic process ($ds=0$), $dp > 0$ for positive values of $dT$.
\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter01/pdf/TS_process_trends.pdf} \caption{Isentropic process trends} \label{fig:isentropic:trends} \end{center} \end{figure}
\noindent Since $\nu$ decreases with temperature and pressure increases with temperature for an isentropic process, we can see from Eqn.~\ref{eqn:process:dnu} that $d\nu$ will be greater at lower temperatures and thus isochores (lines of constant specific volume) will be closely spaced at low temperatures and more sparse at higher temperatures. For an isochore $dv=0$ which implies
\[0=\dfrac{\nu}{R}\left(ds-C_v\dfrac{dT}{T}\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_v}\]
\noindent and thus we can see that the slope of an isochore in a $T-s$-diagram is positive and that the slope increases with temperature.
\noindent In analogy, we can see that an isobar ($dp=0$) leads to the following relation
\[0=\dfrac{p}{R}\left(C_p\dfrac{dT}{T}-ds\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_p}\]
\noindent and consequently isobars will also have a positive slope that increases with temperature in a $T-s$-diagram. Moreover, isobars are less steep than ischores as $C_p > C_v$.
\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter01/pdf/TS_process_iso.pdf} \caption{Isobar and isochor in a $T-s$-diagram} \label{fig:Ts:isobars:isochors} \end{center} \end{figure}
\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter01/pdf/TS_process_multi_iso.pdf} \caption{Isobar and isochor in a $T-s$-diagram} \label{fig:Ts:isobars:isochors:multi} \end{center} \end{figure}
