For a liquid whose velocity potential changes from ϕto ϕ^' owing to an impulsive pressure p^* an

Question

For a liquid whose velocity potential changes from $\phi$ to $\phi^{\prime}$ owing to an impulsive pressure $p^$ and impulsive forces $\mathrm{F}^=-\nabla V^$, show that $$ p^+\rho V^*+\rho\left(\phi^{\prime}-\phi\right)=\text { const. } $$ Apply the equations of impulsive action to show that, if liquid be contained within a closed surface, the circulation and the vorticity cannot be altered by any impulse applied to the boundary.

   For a liquid whose velocity potential changes from $\phi$ to $\phi^{\prime}$ owing to an impulsive pressure $p^*$ and impulsive forces $\mathrm{F}^*=-\nabla V^*$, show that

$$
p^*+\rho V^*+\rho\left(\phi^{\prime}-\phi\right)=\text { const. }
$$


Apply the equations of impulsive action to show that, if liquid be contained within a closed surface, the circulation and the vorticity cannot be altered by any impulse applied to the boundary.

Advanced Vector Analysis with Application to Mathematical Physics

C.E. Weatherburn 1st Edition

Chapter 4, Problem 14

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Step 1

Step 1: Let's start by recalling the Euler equation for inviscid fluid flow: $$\frac{\partial \vec{v}}{\partial t} + (\vec{v} \cdot \nabla)\vec{v} = -\frac{1}{\rho}\nabla p + \vec{F}$$ where $\vec{v}$ is the velocity, $p$ is the pressure, $\rho$ is the density, and Show more…

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For a liquid whose velocity potential changes from $\phi$ to $\phi^{\prime}$ owing to an impulsive pressure $p^*$ and impulsive forces $\mathrm{F}^*=-\nabla V^*$, show that $$ p^*+\rho V^*+\rho\left(\phi^{\prime}-\phi\right)=\text { const. } $$ Apply the equations of impulsive action to show that, if liquid be contained within a closed surface, the circulation and the vorticity cannot be altered by any impulse applied to the boundary.

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