Thermodynamic processes: Difference between revisions

Latest revision as of 13:30, 30 March 2026

d s = C_{v} \frac{d T}{T} + R \frac{d ν}{ν}

(Eq. 1.24)

d ν = \frac{ν}{R} d s - C_{v} \frac{ν}{R T} d T = \frac{ν}{R} d s - \frac{C_{v}}{p} d T

(Eq. 1.25)

for an isentropic process ( $d s = 0$ ), $d ν < 0$ for positive values of $d T$ .

d s = C_{p} \frac{d T}{T} - R \frac{d p}{p}

(Eq. 1.26)

d p = - \frac{p}{R} d s + C_{p} \frac{p}{R T} d T = - \frac{p}{R} d s + C_{p} ρ d T

(Eq. 1.27)

for an isentropic process ( $d s = 0$ ), $d p > 0$ for positive values of $d T$ .

Since $ν$ decreases with temperature and pressure increases with temperature for an isentropic process, we can see from (Eq. 1.25) that $d ν$ 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 $d v = 0$ which implies

0 = \frac{ν}{R} (d s - C_{v} \frac{d T}{T}) \Rightarrow \frac{d T}{d s} = \frac{T}{C_{v}}

(Eq. 1.28)

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.

In analogy, we can see that an isobar ( $d p = 0$ ) leads to the following relation

0 = \frac{p}{R} (C_{p} \frac{d T}{T} - d s) \Rightarrow \frac{d T}{d s} = \frac{T}{C_{p}}

(Eq. 1.29)

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 $C_{p} > C_{v}$ .

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 [[Category:Compressible flow]]
-[[Category:Thermodynamics]]
+[[Category:Thermodynamics]]<!--
+--><noinclude><!--
+-->[[Category:Compressible flow:Topic]]<!--
+--></noinclude><!--
-__TOC__
+--><nomobile><!--
+-->__TOC__<!--
+--></nomobile><!--
+--><noinclude><!--
+-->{{#vardefine:secno|1}}<!--
+-->{{#vardefine:eqno|23}}<!--
+--></noinclude><!--
+-->
+{{NumEqn|<math>
+ds=C_v\dfrac{dT}{T}+R\dfrac{d\nu}{\nu}
+</math>|label=eq_process_ds_a}}
-\section{Thermodynamics}
+{{NumEqn|<math>
-\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}
-=\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}
+</math>|label=eq_process_dnu}}
-\end{equation}
-\noindent for an isentropic process ($ds=0$), $d\nu < 0$ for positive values of $dT$.
+for an isentropic process (<math>ds=0</math>), <math>d\nu < 0</math> for positive values of <math>dT</math>.
-\[ds=C_p\dfrac{dT}{T} - R \dfrac{dp}{p}\]
+{{NumEqn|<math>
+ds=C_p\dfrac{dT}{T} - R \dfrac{dp}{p}
+</math>|label=eq_process_ds_b}}
-\[dp=-\dfrac{p}{R}ds+C_p\dfrac{p}{RT}dT=-\dfrac{p}{R}ds+C_p\rho dT\]
+{{NumEqn|<math>
+dp=-\dfrac{p}{R}ds+C_p\dfrac{p}{RT}dT=-\dfrac{p}{R}ds+C_p\rho dT
+</math>|label=eq_process_dp}}
-\noindent for an isentropic process ($ds=0$), $dp > 0$ for positive values of $dT$.
+for an isentropic process (<math>ds=0</math>), <math>dp > 0</math> for positive values of <math>dT</math>.
+<!--
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-\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
+Since <math>\nu</math> decreases with temperature and pressure increases with temperature for an isentropic process, we can see from {{EquationNote|label=eq_process_dnu}} that <math>d\nu</math> 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 <math>dv=0</math> which implies
-\[0=\dfrac{\nu}{R}\left(ds-C_v\dfrac{dT}{T}\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_v}\]
+{{NumEqn|<math>
+=\dfrac{\nu}{R}\left(ds-C_v\dfrac{dT}{T}\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_v}
+</math>}}
-\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.
+and thus we can see that the slope of an isochore in a <math>T-s</math>-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
+In analogy, we can see that an isobar (<math>dp=0</math>) leads to the following relation
-\[0=\dfrac{p}{R}\left(C_p\dfrac{dT}{T}-ds\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_p}\]
+{{NumEqn|<math>
+=\dfrac{p}{R}\left(C_p\dfrac{dT}{T}-ds\right) \Rightarrow \dfrac{dT}{ds}=\dfrac{T}{C_p}
+</math>}}
-\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 $C_p > C_v$.
+and consequently isobars will also have a positive slope that increases with temperature in a <math>T-s</math>-diagram. Moreover, isobars are less steep than ischores as <math>C_p > C_v</math>.
+<!--
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Thermodynamic processes: Difference between revisions

Latest revision as of 13:30, 30 March 2026

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