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Since I am not going ot upgrade yet, I will make these notes public. public.\\  The steady-state Euler equation reads  \begin{equation}  \left(\mathbf{v} \cdot \nabla \right)\mathbf{v} + \frac{1}{\rho}\nabla P + \nabla \Phi_G -\nu\left[\nabla^2\mathbf{v} + \frac{1}{3}\left(\nabla \cdot \mathbf{v} \right) \right] = 0  \end{equation}  where it is understoond that the  gravitational potential $\Phi_G$ and the coeficcient coefficient  of kinematic viscosity $\nu$ are time independent. Now use the idenity $\left(\mathbf{v} \cdot \nabla \right)\mathbf{v} = \frac{1}{2} \nabla \left( \mathbf{v} \cdot \mathbf{v}\right) - \mathbf{v} \times \left( \nabla \times \mathbf{v}\right)$ \mathbf{v}\right)$,  neglect viscosity for now, and integrate the momentum equation along a streamline from a refernce 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  \end{equation}  since $\mathbf{ds}$ is the line element of a streamline, it is in teh 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}