The entropy equation

From the second law of thermodynamics

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

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

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

The continuity equation on differential non-conservation form

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

and thus

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

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

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

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

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

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

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

Adiabatic flow:

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

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