Vortex Kinetic Energy, VKE

Consider now the centripetal force ${F}_{c}(x)$ at depth $x ≤ h$,

${F}_{c}(x) = Fc - (h-x)\rho .g.\triangle A$

where $\rho$ is the density of the fluid and $g$ is gravity and $\triangle A$, an elemental area at depth $(h-x)$ and radius $y$.  As depth increases, ${F}{c}(x)$ decreases and the opening of the vortex closes up, this models the vortex as a inverted cone.  We know that $F_c(0) = 0$, therefore,

$F_c = h. \rho .g.\triangle A$

${F}_{c}(x) = \rho .g.\triangle A.x = \triangle m.y.w^2$,

$w^2 = \rho .g.\triangle A.\frac { x }{\triangle m.y}$

where $w$ is the angular velocity of an elemental volume at depth $(h-x)$ just on the slant surface of the cone at a radius of $y$ and $\triangle m$ an elemental mass of fluid in circular motion.

$\triangle A = y.\triangle \theta .\triangle x$,   $\triangle m = y\triangle \theta . \triangle x. \triangle y. \rho$

where $\theta$ is the angle at the axis of rotation.  If we consider a elemental ring of radius $y$ and thickness $\triangle y$ and integrate over $d\theta$ for $2\pi$ then,

$\triangle A = 2\pi.y.\triangle x$,  $\triangle m = 2\pi.y.\triangle x .\triangle y. \rho$

$w^2 = g.\frac { x}{y.\triangle y}$

The angular kinetic energy of such a ring is,

$\triangle VKE_y = \frac { 1 }{ 2 }I(y)w^{ 2 }$

where $I(y)$ is the moment of Inertia of a thin ring of mass $m$, radius y spinning at its center axis.

$I(y)= m.y^2$,  where $m = 2\pi.y.\triangle y . \triangle x.\rho$

$I(y)=2\pi.\rho.y^3.\triangle y . \triangle x$

So,

$VKE_y = \pi.\rho.g.{ x }.y^{ 2 }. \triangle x$

For total VKE over all depth $x$, we consider,

$\triangle VKE =  VKE_y =  \pi.\rho .g.{ x }. y^2 .\triangle x$

$VKE =  \pi.\rho .g.\int^{h}_{0}{ x }y^2 dx$, since $y=\frac{r}{h}x$

$VKE = \pi. \rho .g.\int^{h}_{0}{ x }{(\frac{r}{h}x)^2} dx$,

$VKE = \pi. \rho .g.\frac{r^2}{h^2}|^{h}_{0}\frac {x^4}{4}$

$VKE = \frac{1}{4} \pi \rho g{r^2}{h^2}$

This is the vortex kinetic enegry.