Thermodynamic Relations

Specific Heat Relations¶

For thermally perfect and calorically perfect gases

\begin{aligned} &C_p=\frac{dh}{dT}\\ &C_v=\frac{de}{dT} \end{aligned}

(2.8)

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

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

(2.9)

Differentiate Eqn. (2.9) with respect to temperature gives

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

(2.10)

Inserting the specific heats gives

C_p=C_v+R

(2.11)

Dividing Eqn. (2.11) by $C_v$ gives

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

(2.12)

Introducing the ratio of specific heats defined as

\gamma=\frac{C_p}{C_v}

(2.13)

Now, inserting Eqn. (2.13) in Eqn. (2.12) gives

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

(2.14)

In the same way, dividing Eqn. (2.11) with $C_p$ gives

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

(2.15)

and thus

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

(2.16)

Isentropic Relations¶

First law of thermodynamics:

de=\delta q - \delta w

(2.17)

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

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

(2.18)

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

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

(2.19)

Eliminate $de$ in Eqn. (2.18) using Eqn. (2.19)

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

(2.20)

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

(2.21)

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}

(2.22)

Integrating Eqn. (2.22) gives

s_2-s_1=\int_1^2 C_p\frac{dT}{T}-R\ln\left(\frac{p_2}{p_1}\right)

(2.23)

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)

(2.24)

An alternative form of Eqn. (2.24) is obtained by using $de=C_v dT$ Eqn. (2.18), which gives

s_2-s_1=\int_1^2 C_v\frac{dT}{T}-R\ln\left(\frac{\rho_2}{\rho_1}\right)

(2.25)

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)

(2.26)

Isentropic Relations:

Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus Eqn. (2.24) reduces to

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

(2.27)

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

(2.28)

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

(2.29)

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

(2.30)

In the same way, Eqn. (2.26) gives

\frac{\rho_2}{\rho_1}=\left(\frac{T_2}{T_1}\right)^{1/(\gamma-1)}

(2.31)

Eqn. (2.30) and Eqn. (2.31) 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)}

(2.32)

2.3 Thermodynamic Relations

Specific Heat Relations¶

Isentropic Relations¶