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Daniel D'Orazio edited untitled.tex
over 9 years ago
Commit id: 286e4c5286bb36afb415892845e75d879341d073
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\end{equation}
where it is understoond that the gravitational potential $\Phi_G$ and the coefficient of kinematic viscosity $\nu$ are time independent. Now use the idenity $\left(\mathbf{v} \cdot \nabla \right)\mathbf{v} = \frac{1}{2} \nabla \left( \mathbf{v} \cdot \mathbf{v}\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right)$, neglect viscosity for now, and integrate the momentum equation along a streamline from a reference point to the point of evaluation
\begin{equation}
\int{\mathbf{ds}\cdot \left[\nabla \left( \frac{1}{2}v^2\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right) + \frac{1}{\rho}\nabla P + \nabla \Phi_G \right]}
=0 = 0.
\end{equation}
since Since $\mathbf{ds}$ is the line element of a streamline, it is in the same direction as $\mathbf{v}$,so you get
\begin{equation}
\frac{1}{2}v^2 - \Phi_G +\int{\frac{dP}{\rho}} = \rm{cst}
\end{equation}