Isentropic relations

First law of thermodynamics

First law of thermodynamics:

$de=\delta q-\delta w$

(Eq. 1)

$de=\delta q-\delta w$

(Eq. 1)

$de=\delta q-\delta w$

(Eq. 1)

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

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

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