The Entropy Equation - Shock Waves

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}(\nabla\cdot\mathbf{v})

The continuity equation on differential non-conservation form

\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 \Rightarrow \nabla\cdot\mathbf{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\left[\frac{De}{Dt}+p\frac{D\nu}{Dt}-\dot{q}\right]=0\Rightarrow\frac{De}{Dt}=\dot{q}-p\frac{D\nu}{Dt}

Insert $De/Dt$ in Eqn. (3.78)

\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.

3.6 The Entropy Equation