Thermodynamic processes

Specific Heat Relations

For thermally perfect and calorically perfect gases

$$\begin{array}{cc}& C_p=\frac{dh}{dT}\\ & C_v=\frac{de}{dT}\end{array}$$

From the definition of enthalpy and the equation of state $p=\rho RT$

$$h=e+\frac{p}{\rho }=e+RT$$

Differentiate Eqn. \ref{eq:enthalpy} with respect to temperature gives

$$\frac{dh}{dT}=\frac{de}{dT}+\frac{d(RT)}{dT}$$

Inserting the specific heats gives

$$C_p=C_v+R$$

Dividing Eqn. \ref{eq:specificheat:b} by $C_v$ gives

$$\frac{C_p}{C_v}=1+\frac{R}{C_v}$$

Introducing the ratio of specific heats defined as

$$\gamma =\frac{C_p}{C_v}$$

Now, inserting Eqn. \ref{eq:gamma} in Eqn. \ref{eq:specificheat:c} gives

$$C_v=\frac{R}{\gamma -1}$$

In the same way, dividing Eqn. \ref{eq:specificheat:b} with $C_p$ gives

$$1=\frac{C_v}{C_p}+\frac{R}{C_p}=\frac{1}{\gamma }+\frac{R}{C_p}$$

and thus

$$C_p=\frac{\gamma R}{\gamma -1}$$

Isentropic Relations

First law of thermodynamics:

$$de=\delta q-\delta w$$

For a reversible process: $\delta w=pd(1/\rho )$ and $\delta q=Tds$

$$de=Tds-pd\left(\frac{1}{\rho }\right)$$

Enthalpy is defined as: $h=e+p/\rho$ and thus

$$dh=de+pd\left(\frac{1}{\rho }\right)+\left(\frac{1}{\rho }\right)dp$$

Eliminate $de$ in Eqn. \ref{eq:firstlaw:b} using Eqn. \ref{eq:dh}

$$Tds=dh-\cancel{pd\left(\frac{1}{\rho }\right)}-\left(\frac{1}{\rho }\right)dp+\cancel{pd\left(\frac{1}{\rho }\right)}$$

$$ds=\frac{dh}{T}-\frac{dp}{\rho T}$$

Using $dh=C_{p}T$ and the equation of state $p=\rho RT$, we get

$$ds=C_p\frac{dT}{T}-R\frac{dp}{p}$$

Integrating Eqn. \ref{eq:ds} gives

$$s_2-s_1=\underset{1}{\overset{2}{\int }}{C}_p\frac{dT}{T}-R\ln \left(\frac{p_2}{p_1}\right)$$

For a calorically perfect gas, $C_p$ is constant (not a function of temperature) and can be moved out from the integral and thus

$$s_2-s_1=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$$

An alternative form of Eqn. \ref{eq:ds:c} is obtained by using $de=C_{v}dT$ Eqn. \ref{eq:firstlaw:b}, which gives

$$s_2-s_1=\underset{1}{\overset{2}{\int }}{C}_v\frac{dT}{T}-R\ln \left(\frac{{\rho }_2}{{\rho }_1}\right)$$

Again, for a calorically perfect gas, we get

$$s_2-s_1=C_v\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{{\rho }_2}{{\rho }_1}\right)$$

Isentropic Relations

Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus Eqn. \ref{eq:ds:c} reduces to

$$\frac{C_p}{R}\ln \left(\frac{T_2}{T_1}\right)=\ln \left(\frac{p_2}{p_1}\right)$$

$$\frac{C_p}{R}=\frac{\gamma }{\gamma -1}$$

$$\frac{\gamma }{\gamma -1}\ln \left(\frac{T_2}{T_1}\right)=\ln \left(\frac{p_2}{p_1}\right)\Rightarrow$$

$$\frac{p_2}{p_1}={\left(\frac{T_2}{T_1}\right)}^{\gamma /(\gamma -1)}$$

In the same way, Eqn. \ref{eq:ds:e} gives

$$\frac{{\rho }_2}{{\rho }_1}={\left(\frac{T_2}{T_1}\right)}^{1/(\gamma -1)}$$

Eqn. \ref{eq:isentropic:a} and Eqn. \ref{eq:isentropic:b} constitutes the isentropic relations

$$\frac{p_2}{p_1}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma /(\gamma -1)}$$

Flow Processes

$$ds=C_v\frac{dT}{T}+R\frac{d\nu }{\nu }$$

$$d\nu =\frac{\nu }{R}ds-C_v\frac{\nu }{RT}dT=\frac{\nu }{R}ds-\frac{C_v}{p}dT$$

for an isentropic process ($ds=0$), $d\nu <0$ for positive values of $dT$.

$$ds=C_p\frac{dT}{T}-R\frac{dp}{p}$$

$$dp=-\frac{p}{R}ds+C_p\frac{p}{RT}dT=-\frac{p}{R}ds+C_p\rho dT$$

for an isentropic process ($ds=0$), $dp>0$ for positive values of $dT$.

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

$$0=\frac{\nu }{R}(ds-C_v\frac{dT}{T})\Rightarrow \frac{dT}{ds}=\frac{T}{C_v}$$

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 ($dp=0$) leads to the following relation

$$0=\frac{p}{R}(C_p\frac{dT}{T}-ds)\Rightarrow \frac{dT}{ds}=\frac{T}{C_p}$$

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 $C_p>C_v$.