Skip to content

The Rankine Vortex

Source lecture(s): PC316 Ch. 8

Intuition

A real vortex cannot have \(v_\theta = \Gamma/2\pi r\) all the way to the axis — that would be infinite velocity at \(r = 0\). The Rankine vortex is the simplest repair: a solid-body rotating core stitched onto an irrotational exterior. Crude, but it captures the essential structure of a tornado, a bathtub drain and a wingtip core.

The model

\[v_\theta(r) = \begin{cases} \dfrac{\Gamma}{2\pi a^2}\,r, & r \le a \quad\text{(core: } \omega = \Gamma/\pi a^2\text{)}\\[8pt] \dfrac{\Gamma}{2\pi r}, & r > a \quad\text{(exterior: } \omega = 0\text{)} \end{cases}\]

All the vorticity is confined to \(r \le a\); outside, the flow is irrotational even though the streamlines are circles. The peak velocity is at the core edge:

\[v_{\max} = \frac{\Gamma}{2\pi a}\]

The pressure field

Radial force balance is centripetal: \(dp/dr = \rho v_\theta^2/r\). Integrating inward from \(p_\infty\), the exterior contributes \(\tfrac12\rho v_{\max}^2\) and the core contributes another \(\tfrac12\rho v_{\max}^2\) — equal halves, which is the memorable part:

\[\boxed{\;p_\infty - p(0) = \rho\,v_{\max}^2\;}\]

The vortex centre is a pressure deficit, and the deficit is twice what Bernoulli alone would suggest from the exterior. On a free surface that deficit carves the familiar funnel of depth

\[\Delta h = \frac{v_{\max}^2}{g}\]

Worked example: a tornado

A tornado has a core radius \(a = 50\) m and peak wind \(v_{\max} = 90\) m/s. Find the circulation and the central pressure drop.

\[\Gamma = 2\pi a v_{\max} = 2\pi(50)(90) = 2.83\times10^{4}\ \text{m}^2/\text{s}\]
\[\Delta p = \rho v_{\max}^2 = 1.2\times 90^2 = 9.7\times10^{3}\ \text{Pa} \approx 97\ \text{mbar}\]

About 10% of atmospheric pressure — consistent with the ~100 mbar deficits measured in strong tornadoes by instrumented probes. Note that this pressure drop is not what damages buildings (the old advice to open windows was based on that misconception); wind loading and debris do the damage.

Why the corner is fictional

The vorticity is discontinuous at \(r = a\), which no real fluid tolerates: the viscous term \(\nu\nabla^2\omega\) smooths it immediately. The exact viscous solution is the Lamb–Oseen vortex,

\[v_\theta = \frac{\Gamma}{2\pi r}\left(1 - e^{-r^2/4\nu t}\right)\]

whose core grows as \(a \sim \sqrt{4\nu t}\). Rankine is the \(t\to0\) idealisation with the corner left in; it is used because the algebra is trivial and the pressure result is unchanged in substance.

Common mistakes

  • Using Bernoulli across the core. Bernoulli's constant differs between streamlines in rotational flow. The exterior is irrotational so Bernoulli applies there; inside the core you must integrate the radial momentum equation. Skipping this gives half the correct pressure drop.
  • Expecting \(v \to \infty\) to be physical. The point vortex is a mathematical convenience; the core is what makes it a fluid.
  • Confusing core radius with vortex size. The velocity field extends far beyond \(a\); only the vorticity is confined.

Knowledge graph position

Prerequisites: vorticity, potential flow, pressure. Leads to: point vortices, tornado and wingtip-core modelling.

Quiz

Q1 (conceptual). Why is the pressure drop \(\rho v_{\max}^2\) rather than \(\tfrac12\rho v_{\max}^2\)?

Answer

The exterior (irrotational, Bernoulli applies) contributes \(\tfrac12\rho v_{\max}^2\) between infinity and \(r = a\). The core is rotational, so Bernoulli does not carry across it; integrating \(dp/dr = \rho v_\theta^2/r\) with \(v_\theta \propto r\) contributes a second \(\tfrac12\rho v_{\max}^2\). The halves sum to \(\rho v_{\max}^2\).

Q2 (computational). A bathtub vortex has \(v_{\max} = 0.3\) m/s. How deep is the surface dimple?

Answer

\(\Delta h = v_{\max}^2/g = 0.09/9.81 = 9.2\) mm — about a centimetre, which matches what you see over a draining plughole.

Q3 (MCQ). Outside the core of a Rankine vortex, the flow is:

  • (a) rotational, since the streamlines are circles
  • (b) irrotational, despite circular streamlines
  • (c) at rest
  • (d) turbulent
Answer

(b). With \(v_\theta \propto 1/r\) the vorticity vanishes: the shear exactly cancels the orbital rotation of a fluid element. Circular streamlines never by themselves imply vorticity.