Why Draining Water Swirls (Free Vortex)

Equivalent forms

v_1 r_1 = v_2 r_2

\Gamma = 2\pi r v\ (\text{circulation})

\Delta h = \frac{v^2}{2g}\ (\text{surface dip})

An ice skater's pirouette written for water — angular momentum, conserved.

Symbol	Name	SI unit	Dimension	Range
$v_1$	Outer Speed Tangential speed at the outer radius	m/s	L·T⁻¹	0 – 2
$r_1$	Outer Radius Radius where the outer speed is measured	m	L	0.05 – 2
$r_2$	Inner Radius Radius near the drain	m	L	0.005 – 1
$v_2$	Inner Speedoutput Tangential speed near the drain	m/s	L·T⁻¹	0 – 20

Unit systems

Symbol	SI	CGS	Imperial
$v_1$	m/s	cm/s	ft/s
$r_1$	m	cm	ft
$r_2$	m	cm	ft
$v_2$	m/s	cm/s	ft/s

Where it holds

Ideal free (irrotational) vortex from inviscid angular-momentum conservation, valid outside a small viscous core. The true centre is a Rankine vortex where v rises linearly, capping the 1/r blow-up.

Dimensional analysis

$[v r] = (L\cdot T^{-1})(L) = L^{2}\cdot T^{-1}$ = const $\Rightarrow v = (L^{2}\cdot T^{-1})/L = L\cdot T^{-1} \checkmark$ (velocity)

Discovery

Hermann von Helmholtz · 1858

Helmholtz's vortex theorems established that, absent friction, circulation is conserved — giving the free (irrotational) vortex v ∝ 1/r. The popular belief that the hemisphere fixes the swirl direction was debunked by careful Ascher Shapiro experiments (1962); Coriolis matters only for huge, undisturbed tanks.

Try this

Pull the plug and the water spins faster as it nears the hole — conserved angular momentum, not the Coriolis force.

Water circulates with speed 0.1 m/s at radius 0.5 m from a drain. Using conservation of angular momentum (v·r = constant), how fast does it swirl at radius 0.05 m?

Research status: well studied

Real-world applications

Tornado and hurricane core dynamics
Cyclone separators and vortex tubes
Spillway and intake vortex suppression
Bathtub and sink drainage

Common misconceptions

The Coriolis effect fixes which way your sink drains — in a normal basin leftover motion dominates entirely
The water speeds up because it 'falls' faster — the speed-up is angular-momentum conservation, independent of the fall
The centre speed is truly infinite — viscosity forms a finite solid-body core

Experimental verification

Ascher Shapiro's carefully isolated, large, still tanks did show a weak Northern-Hemisphere counterclockwise preference from Coriolis, but everyday basins are dominated by residual swirl; particle tracking confirms

v \propto 1/r

outside the core.

Derivation

With negligible friction, the angular momentum of a fluid ring

L = m v r

is conserved as it spirals inward, so

v r =

constant, i.e.

v = \Gamma /(2\pi r)

where

\Gamma = 2\pi r v

is the (conserved) circulation.

Applying Bernoulli to the speeding flow shows the free surface must

dip by \Delta h = v^{2}/2g

, producing the funnel.

Limiting cases

r_2 \to 0⟶ v_2 \to \inftyThe ideal free vortex predicts unbounded speed at the centre; in reality viscosity forms a small solid-body core (a Rankine vortex).

r_2 = r_1

⟶

v_2 = v_1

No change in radius means no change in swirl speed.

Tiny initial swirl⟶ Direction is set by leftover motionAny residual circulation, not the hemisphere, sets the spin direction in a normal-sized basin.

What if…

What if the basin is on the equator?

Coriolis vanishes there, so any swirl comes purely from how the water was set moving — underscoring that residual motion, not latitude, rules a sink.

What if there were no viscosity at all?

The 1/r law would hold all the way in and the centre speed would diverge; real water caps it with a viscous Rankine core.

1

Swirl speed near the drain

Given ·

v 1:: 0.1
r 1:: 0.5
r 2:: 0.05

Find · v_2

Steps

Conserve angular momentum: $v_1\,\mathrm{r}_1 = v_2\,\mathrm{r}_2$
$v_2 = v_1\,\mathrm{r}_1 / r_2 = (0.1)(0.5)/(0.05)$
$v_2 = 1.0\,\mathrm{m/s}$ — hence the fast, narrow funnel at the plughole

Result ·

v_2 = v_1\,\mathrm{r}_1 / r_2 = 0.1\times 0.5/0.05 = 1.0\,\mathrm{m/s}

— a tenfold speed-up as radius shrinks tenfold.

Bernoulli's Equation→Kutta-Joukowski Lift Theorem→Continuity Equation→