Vortex Potential Energy, VPE

Consider a cone inverted in a fluid,


The force on an elemental area $\triangle A$, at depth $x$ on the side of such a cone is given by

${ F }_{ s }=\rho g.x\triangle A$,

where $g$ is gravity, $\rho$ is density of the fluid.

$\triangle A$ is an elemental ring of radius $\triangle y$, height $\triangle x$,

$\triangle A=2\pi \triangle y. \triangle x$

$\therefore { F }_{ s }=2\pi \rho g. x.\triangle y  \triangle x$

The work done against such a force in moving the fluid out to the perimeter $y$, at a given depth $x$ is,

$\triangle W_r = F_s . \triangle y =  2\pi \rho g. x. \triangle y^2  \triangle x$

$W_r = 2\pi \rho g. x. \triangle x \int_{0}^{y} (dy)^2$

$W_r = 2\pi \rho g. x. \triangle x |_{0}^{y} \frac {y^2}{2}$

$W_r = \pi \rho g. x.y^2 \triangle x$

And so, total work done in establishing such a cone of height, $h$ in the fliud is,

$W = \int_0^h dW_r = \pi \rho g. \int_0^h x.y^2 dx$, since $y=\frac r h x$

$W = \pi \rho g. \frac {r^2}{ h^2} \int _{0}^{h} x^3 dx$

$W =  \frac{1}{4} \pi \rho g r^2h^2$

If we model a vortex as a inverted cone, than this expression is the Potential Energy of the vortex.