Governing equations on differential form: Difference between revisions

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The continuity equation on integral form reads
The continuity equation on integral form reads


{{NumEqn|<math>
{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
\frac{d}{dt}\iiint_{\Omega} \rho dV+\iint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0
</math>|nonumber=1|infobox=1|numw=2em}}
</math>}}


Apply Gauss's divergence theorem on the surface integral gives
Apply Gauss's divergence theorem on the surface integral gives
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The momentum equation on integral form reads
The momentum equation on integral form reads


{{NumEqn|<math>
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\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} dV+\iint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}dV
</math>|nonumber=1|infobox=1|numw=2em}}
</math>}}


As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
As for the continuity equation, the surface integral terms are rewritten as volume integrals using Gauss's divergence theorem.
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The energy equation on integral form reads
The energy equation on integral form reads


{{NumEqn|<math>
{{InfoBox|<math>
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
\frac{d}{dt}\iiint_{\Omega}\rho e_o dV+\iint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=</math><br><br><math>\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}dV+\iiint_{\Omega} \dot{q}\rho dV
</math>|nonumber=1|infobox=1|numw=2em}}
</math>}}


Gauss's divergence theorem applied to the surface integral term in the energy equation gives
Gauss's divergence theorem applied to the surface integral term in the energy equation gives
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<div style="border: solid 1px;">
{{NumEqn|<math>
{{OpenInfoBox|<math>
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0
</math>|infobox=1|description=Continuity:|nonumber=1|noborder=1}}
</math>|description=Continuity:}}


{{NumEqn|<math>
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\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
\frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f}
</math>|infobox=1|description=Momentum:|nonumber=1|noborder=1}}
</math>|description=Momentum:}}


{{NumEqn|<math>
{{OpenInfoBox|<math>
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
\frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|infobox=1|description=Energy:|nonumber=1|noborder=1}}
</math>|description=Energy:}}
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</div>


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<div style="border: solid 1px;">
{{NumEqn|<math>
{{OpenInfoBox|<math>
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
\frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0
</math>|infobox=1|description=Continuity:|nonumber=1|noborder=1}}
</math>|description=Continuity:}}


{{NumEqn|<math>
{{OpenInfoBox|<math>
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
\frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f}
</math>|infobox=1|description=Momentum:|nonumber=1|noborder=1}}
</math>|description=Momentum:}}


{{NumEqn|<math>
{{OpenInfoBox|<math>
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
\rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho
</math>|infobox=1|description=Energy:|nonumber=1|noborder=1}}
</math>|description=Energy:}}
</div>
</div>