The entropy equation

From the second law of thermodynamics

$\frac{De}{Dt}=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)$

(Eq. 2.76)

From the energy equation on differential non-conservation form internal energy formulation

$\frac{De}{Dt}=\dot{q}-\frac{p}{\rho }(\mathrm{\nabla }\cdot 𝐯)$

(Eq. 2.77)

The continuity equation on differential non-conservation form

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0\Rightarrow \mathrm{\nabla }\cdot 𝐯=-\frac{1}{\rho }\frac{D\rho }{Dt}$

(Eq. 2.78)

and thus

$\frac{De}{Dt}=\dot{q}+\frac{p}{{\rho }^2}\frac{D\rho }{Dt}$

(Eq. 2.79)

$\frac{D\rho }{Dt}=-\frac{1}{{\nu }^2}\frac{D\nu }{Dt}$

(Eq. 2.80)

$\rho \frac{De}{Dt}=\rho \dot{q}-\frac{p}{\rho {\nu }^2}\frac{D\nu }{Dt}=\rho \dot{q}-\rho p\frac{D\nu }{Dt}$

(Eq. 2.81)

$\rho [\frac{De}{Dt}+p\frac{D\nu }{Dt}-\dot{q}]=0\Rightarrow \frac{De}{Dt}=\dot{q}-p\frac{D\nu }{Dt}$

(Eq. 2.82)

Insert $De/Dt$ in Eqn. \ref{eq:second:law}

$\dot{q}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)=T\frac{Ds}{Dt}-p\frac{D}{Dt}\left(\frac{1}{\rho }\right)\Rightarrow$

(Eq. 2.83)

$T\frac{Ds}{Dt}=-\dot{q}$

(Eq. 2.84)

Adiabatic flow:

$T\frac{Ds}{Dt}=0$

(Eq. 2.85)

In an adiabatic, steady-state, inviscid flow, entropy is constant along a streamline.