Vortex and Exponential Drag

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.

$\because \triangle m = \triangle A. \triangle y. \rho$

 $w^2 = g.\frac { x }{y.\triangle y}$

Let the angular velocity profile (an assumption) beyond radius $y$ be

$w(r) = we^{-\frac{r-y}{y}}$, where $r ≥ y$

$[w(r)]^2 = g.\frac{(x)}{ y.\triangle y}.e^{-\frac{2r-2y}{y}}$, where $r ≥ y$

We consider the circular motion beyond $y$ as concentric elemental rings with decreasing speed.

The angular kinetic energy of one such ring is,

$\triangle VKE_r = \frac { 1 }{ 2 }I(r)w^{ 2 }$

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

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

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

So,

$\triangle VKE_r = \pi.\rho.g.\frac{x}{ y.\triangle y}.\triangle x.r^3e^{-\frac{2r-2y}{y}} \triangle r$

since $\frac{\triangle x}{\triangle y}=\frac{h}{r_c}$ and $\frac{x}{y}=\frac{h}{r_c}$,   where $r_c$ is the radius of the vortex.

$\triangle VKE_r = \pi.\rho.g.\frac{h^2}{r^2_c}.r^3e^{-\frac{2r-2y}{y}} \triangle r$  

Kinetic Energy of all rings at depth (h-x) beyond radius y is,

$VKE_r =\int^{\infty}_{y}\triangle VKE_r dr= \pi.\rho.g.\frac{h^2}{r^2_c}\int^{\infty}_{y} r^3e^{-\frac{2r-2y}{y}}dr$

since

$\int_y^\infty r^3 (e^{-\frac{2r-2y}{y}}) dr =|^{r \rightarrow \infty}_{r = y} -\frac { 1 }{ 8 } ye^{ { - }\frac { 2r-2y }{ y }}(4r^{ 3 }+6r^{ 2 }y+6ry^{ 2 }+3y^{ 3 }) = \frac {19}{8}y^4$

$VKE_r =\frac {19}{8}\pi.\rho.g.\frac{h^2}{r^2_c}.y^4=\frac {19}{8}\pi.\rho.g.\frac{r^2_c}{h^2}.x^4$,

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

$\triangle VKE = VKE_r = \frac {19}{8}\pi.\rho.g.\frac{r^2_c}{h^2}.x^4$

$VKE = \frac {19}{8}\pi.\rho.g.\frac{r^2_c}{h^2}\int^h_0 {x^4} dx$,

$VKE = \frac {19}{8}\pi.\rho.g.\frac{r^2_c}{h^2}|^h_0 \frac{x^5}{5}$

$VKE = \frac {19}{40}\pi\rho g r^2_c h^2.h$

This is the vortex kinetic enegry with exponential drag profile $w(r)=w.e^{-(\frac{r-y}{y})}$.  The problem with this derivation is that the drag profile is unproven and the unit for VKE in the last expression is not consistent.