Equation of the funnel's surface of a forced whirlpool in a cylinder I was wondering what function describes the form of a whirlpool funnel in a liquid (say, water)  rotated by some sort of a paddle sunken at depth in a cylindrical barrel.
To simplify the problem, assume:


*

*water in a cylindrical vessel of large height and a $R$ radius 

*a cylindrical rod of radius $r$ ($\ll R$), along to the axis of the vessel, rotating with an angular velocity $\omega$ - suppose it driven by a motor.

*all in the gravitation field with the typical $g$ acceleration
I expect that after a certain amount of time, the entire volume of water will be involved more or less in a stable/stationary rotation and the surface of the water will show a typical shape of a whirlpool funnel (intersected by the driving central rod).
Q: I'm interested in the equation of the generating curve of the whirlpool funnel once the stationary conditions are reached (cylindrical symmetry, right?). I'm not interested in the time required to reach stationary conditions.
Ignore secondary effect, like driving rod wetting meniscus, the potential heat that viscosity would create and the effect of temperature variations on water viscosity

Extra references I could find:


*

*Batchelor vortex - infinite boundary conditions, with axial flow.

*Rankine vortex - swirling forced flow (yes! yes!!) but unbound radially (ahhhh)

Practical note (not included in the approximation model above): aeration of water tanks. How fast should I spin a stirrer at a certain depth to have the whirlpool funnel barely reaching the stirrer and "inject" bubbles of air into the mass of water. Alternatively, given a rotation speed, at what depth  should I place the stirrer.
 A: What you're asking about is Couette flow or more specifically Taylor–Couette flow. Plenty is known about this and, surprisingly, contrary to your assumption

I expect that after a certain amount of time, the entire volume of water will be involved more or less in a stable/stationary rotation and the surface of the water will show a typical shape of a whirlpool funnel

The resulting flow is often unstable and can exhibit cool and strange properties (e.g. Taylor vortices aka "fluid donuts").
If we assume our cylinders are infinitely long, we can exactly solve for the pressure in the system and use the criteria of the equilibrium surface being the surface of constant pressure.
We'll use the Euler equations for incompressible flow, which ignores fluid viscosity, together with the expression for $v_\theta(r)$ from the Couette flow article above
$$
v_\theta(r)=\frac{r^2 \left(r_1^2 \omega _1-r_2^2 \omega _2\right)+r_1^2 r_2^2 \left(\omega _2-\omega _1\right)}{r \left(r_1^2-r_2^2\right)}\quad\text{and}\quad(\vec{v}\cdot\nabla)\vec{v}=\vec{g}-\frac{1}{\rho}\nabla p
$$
Where $\omega_1$ is the angular velocity of the rod, $\omega_2$ is the angular velocity of the container (if any), $r_1$ is the radius of the rod, $r_2$ is the radius of the container, $\rho$ is the density of the fluid and $\vec{g}=-g\hat{z}$ is gravity.
Rearranging and computing the gradient gives
$$
\nabla p=\frac{\rho  \left(r_1^4 r_2^4 \left(\omega _1-\omega _2\right){}^2-r^4 \left(r_1^2 \omega _1-r_2^2 \omega _2\right){}^2\right)}{r^3 \left(r_1^2-r_2^2\right){}^2}\hat{r}-\rho g\hat{z}
$$
Which can be solved to give
$$
p(r,z)=C-g \rho  z-\frac{\rho  \left(r^4 \left(r_1^2 \omega _1-r_2^2 \omega _2\right){}^2+r_1^4 r_2^4 \left(\omega _1-\omega _2\right){}^2\right)}{2 r^2 \left(r_1^2-r_2^2\right){}^2}
$$
Finally the surface shape can be found from the free surface condition which tells us that the surface is at constant pressure. Call $z=0$ the height of the surface at the outer wall to find
$$
p(r,z_s)=p(r_2,0)\to z_s(r)=\frac{\left(r_2^2-r^2\right) \left(r^2 \left(r_1^2 \omega _1-r_2^2 \omega _2\right){}^2-r_1^4 r_2^2 \left(\omega _1-\omega _2\right){}^2\right)}{2 g r^2 \left(r_1^2-r_2^2\right){}^2}
$$
For your situation we have 
$$
\omega_2=0\to z_s(r)=\frac{r_1^4 \left(r_2^2-r^2\right){}^2 \omega _1^2}{2 g r^2 \left(r_1^2-r_2^2\right){}^2}
$$
And we see that $\omega_1$ and $g$ don't affect the shape. Taking $r_2/r_1=10$ the situation looks something like

This is likely as good as you're going to do if you want an analytic equation. To get more accurate we should really drop the assumption of infinitely long cylinders since finite size effects are obviously important at the surface! The derivation for $v_{\theta}(r)$ in that case is given here (paywall?) and the result for free here. It's an infinite sum of Bessel functions so it's a bit annoying to work with but should be doable by simple truncation. To get even more accurate, you could drop the inviscid and even incompressible assumptions (important if you want to consider aeration) and "upgrade" from Euler to Navier-Stokes.
If you recall at the beginning I mentioned that these things can exhibit strange and cool properties but then the result we found was neither strange nor particularly cool. This is because real fluids have viscosity and the interplay between instabilities and viscous damping creates the interesting meta-stable states of Taylor–Couette flow. However, those calculations are much more involved.
That being said I think this is a fairly good approximation if you want something qualitative and cool that it has a closed form solution. I also honestly have no idea how to approach the question of aeration besides fully simulating the microscopic hydrodynamics. At that point you're fully turbulent and you've really no luck of any nice formulae.
