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.
