Thermodynamics

Specific heat

For thermally perfect and calorically perfect gases

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

(Eq. 1.1)

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

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

(Eq. 1.2)

Differentiate (Eq. 1.2) with respect to temperature gives

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

(Eq. 1.3)

Inserting the specific heats gives

$C_p=C_v+R$

(Eq. 1.4)

Dividing (Eq. 1.4) by $C_v$ gives

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

(Eq. 1.5)

Introducing the ratio of specific heats defined as

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

(Eq. 1.6)

Now, inserting (Eq. 1.6) in Eqn. \ref{eq:specificheat:c} gives

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

(Eq. 1.7)

In the same way, dividing (Eq. 1.4) with $C_p$ gives

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

(Eq. 1.8)

and thus

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

(Eq. 1.9)

Isentropic relations

First law of thermodynamics

First law of thermodynamics:

$de=\delta q-\delta w$

(Eq. 1.10)

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

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

(Eq. 1.11)

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. 1.12)

Eliminate $de$ in (Eq. 1.11) using (Eq. 1.12)

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

(Eq. 1.13)

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

(Eq. 1.14)

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. 1.15)

Integrating (Eq. 1.15) 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. 1.16)

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. 1.17)

An alternative form of (Eq. 1.17) is obtained by using $de=C_{v}dT$ in (Eq. 1.11), 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. 1.18)

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. 1.19)

Isentropic Relations

Adiabatic and reversible processes, i.e., isentropic processes implies $ds=0$ and thus (Eq. 1.17) reduces to

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

(Eq. 1.20)

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

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

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

(Eq. 1.21)

In the same way, (Eq. 1.19) gives

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

(Eq. 1.22)

Eqn. (Eq. 1.21) and Eqn. (Eq. 1.22) 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. 1.23)

Thermodynamic processes

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

(Eq. 1.24)

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

(Eq. 1.25)

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

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

(Eq. 1.26)

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

(Eq. 1.27)

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 (Eq. 1.25) 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

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

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

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

(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 $C_p>C_v$.