Isentropic relations

First law of thermodynamics

First law of thermodynamics:

$de=\delta q-\delta w$

(Eq. 1.6)

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

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

(Eq. 1.7)

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

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

(Eq. 1.8)

Eliminate $de$ in (Eq. 1.7) using (Eq. 1.8)

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

(Eq. 1.9)

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

(Eq. 1.10)

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

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

(Eq. 1.11)

Integrating (Eq. 1.11) gives

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

(Eq. 1.12)

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

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

(Eq. 1.13)

An alternative form of (Eq. 1.13) is obtained by using $de=C_{v}dT$ in (Eq. 1.7), which gives

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

(Eq. 1.14)

Again, for a calorically perfect gas, we get

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

(Eq. 1.15)

Isentropic Relations

Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus (Eq. 1.13) reduces to

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

(Eq. 1.16)

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

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

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

(Eq. 1.17)

In the same way, (Eq. 1.15) gives

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

(Eq. 1.18)

Eqn. (Eq. 1.17) and Eqn. (Eq. 1.18) constitutes the isentropic relations

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

(Eq. 1.19)