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Vortex and drain
Circulation sets the swirl; the vortex model sets the core size.
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The science
A vortex over a drain
Water leaving a drain keeps whatever swirl it had. Angular momentum is conserved as fluid moves inward, so the swirl speed rises towards the axis. In a Rankine vortex the core turns like a solid body, vθ = Γ r / (2π rc²) inside rc, and outside it follows the free vortex vθ = Γ / (2π r), where Γ is the circulation. The Lamb-Oseen (Burgers) vortex smooths the corner with (1 − e−r²/rc²) and, for a draining tank, balances inward strain a against viscous spreading to give rc = 2√(ν/a).
The funnel
The free surface is where pressure is atmospheric, so the radial pressure balance dp/dr = ρ vθ²/r gives dz/dr = vθ² / (g r). For a Rankine vortex the depth at the axis is exactly Γ² / (4π² g rc²) = vc²/g. If this exceeds the water height, an air core reaches the drain. The cross-section is the numerical solution of that equation.
Draining time
Torricelli's law gives outflow speed √(2 g h), so dh/dt = −k √h with k = Cd (Ahole/Atank) √(2g) and the closed form t = 2 (√h0 − √h1) / k.
Honest simplifications: a steady axisymmetric velocity field (no waves or turbulence), a height-field surface, a constant discharge coefficient, and tracers that follow the velocity field exactly. The bathtub vortex direction is not set by the Coriolis force at tank scale; the circulation here is whatever you give it.
What to try
- Raise the circulation: the funnel depth grows with the square of Γ and an air core appears when it reaches the drain.
- Switch between Rankine, Lamb-Oseen and Burgers drain and compare the core size on the speed graph.
- Open the drain and watch the water-height curve follow the Torricelli square-root curve.
- Try Moon gravity: the funnel is six times deeper for the same swirl.