Governing Equations on Differential Form - Shock Waves

Conservation of Mass¶

Apply Gauss’s divergence theorem on the surface integral in Eqn. (3.5) gives

\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=\iiint_{\Omega}\nabla\cdot(\rho\mathbf{v})d\mathscr{V}

Also, if $\Omega$ is a fixed control volume

\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}=\iiint_{\Omega} \frac{\partial \rho}{\partial t} d\mathscr{V}

The continuity equation can now be written as a single volume integral.

\iiint_{\Omega} \left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})\right]d\mathscr{V}=0

$\Omega$ is an arbitrary control volume and thus

\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0

which is the continuity equation on partial differential form.

Conservation of Momentum¶

As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss’s divergence theorem.

\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS=\iiint_{\Omega} \nabla\cdot(\rho \mathbf{v}\mathbf{v})d\mathscr{V}

\oiint_{\partial \Omega} p\mathbf{n}dS=\iiint_{\Omega} \nabla pd\mathscr{V}

Also, if $\Omega$ is a fixed control volume

\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}=\iiint_{\Omega} \frac{\partial}{\partial t}(\rho \mathbf{v}) d\mathscr{V}

The momentum equation can now be written as one single volume integral

\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p - \rho \mathbf{f}\right]d\mathscr{V}=0

$\Omega$ is an arbitrary control volume and thus

\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}

which is the momentum equation on partial differential form

Conservation of Energy¶

Gauss’s divergence theorem applied to the surface integral term in the energy equation (Eqn. (3.24)) gives

\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\nabla\cdot(\rho h_o\mathbf{v})d\mathscr{V}

Fixed control volume

\frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}=\iiint_{\Omega}\frac{\partial}{\partial t}(\rho e_o) d\mathscr{V}

The energy equation can now be written as

\iiint_{\Omega}\left[\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) - \rho\mathbf{f}\cdot\mathbf{v} - \dot{q}\rho \right]d\mathscr{V}=0

$\Omega$ is an arbitrary control volume and thus

\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho

which is the energy equation on partial differential form