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Potential Flow

Intuition

Suppose a flow has no "spin" anywhere — every little fluid blob translates and stretches but never rotates. Then the whole velocity field collapses into a single scalar function \(\phi\), like a landscape whose slopes are the velocities. Add incompressibility and \(\phi\) obeys the friendliest PDE in physics — Laplace's equation — whose solutions superpose. Suddenly you can build flows like Lego: uniform stream + doublet = flow past a cylinder.

Formal definition

A flow is a potential flow if it is irrotational:

\[\boldsymbol{\omega} = \nabla\times\mathbf{v} = 0 \quad\Longleftrightarrow\quad \mathbf{v} = \nabla\phi\]

(\(\nabla\times\nabla\phi \equiv 0\) guarantees consistency). If also incompressible (continuity: \(\nabla\cdot\mathbf{v} = 0\)), then

\[\boxed{\,\nabla^2\phi = 0\,}\]

— Laplace's equation, shared with electrostatics and gravity.

The stream function

In 2-D, define \(\psi\) by

\[u = \frac{\partial\psi}{\partial y}, \qquad v = -\frac{\partial\psi}{\partial x}\]

which satisfies continuity identically; irrotationality then gives \(\nabla^2\psi = 0\) too. Contours of \(\psi\) are streamlines, and the flow rate between two streamlines equals \(\Delta\psi\). \(\phi\)-lines and \(\psi\)-lines intersect at right angles.

Elementary solutions (the Lego bricks)

Flow Potential \(\phi\)
Uniform stream \(Ux + Vy\)
Source / sink (strength \(Q\)) \(\frac{Q}{2\pi}\ln r\)
Vortex (circulation \(\Gamma\)) \(-\frac{\Gamma}{2\pi}\theta\)
Doublet (strength \(K\)) \(\frac{K}{r}\cos\theta\)

Because Laplace's equation is linear, superposition works: uniform + source = flow past a blunt nose (Rankine half-body); uniform + doublet = flow past a circular cylinder; add a vortex = lifting cylinder (and, via conformal mapping, an airfoil).

With the velocity in hand, pressure follows from Bernoulli — which in irrotational flow holds with a single global constant.

Method of images

To satisfy the no-penetration condition at a solid wall, add fictitious mirror singularities. A source \(Q\) at \((a, 0)\) near a wall along the \(x\)-axis pairs with an image source at \((-a, 0)\):

\[\phi = \frac{Q}{2\pi}\ln\sqrt{(x-a)^2 + y^2} + \frac{Q}{2\pi}\ln\sqrt{(x+a)^2 + y^2}\]

By symmetry the normal velocity on the wall vanishes — boundary condition satisfied without solving anything. The same trick powers electrostatics and computational plasma boundary handling.

Limits of the theory

Potential flow ignores viscosity entirely, so it predicts zero drag for a cylinder in steady flow (d'Alembert's paradox) and misses boundary layers, separation and turbulence. It remains outstandingly useful outside boundary layers, for lift (with circulation), for waves, and as the skeleton that viscous corrections dress up.

Common mistakes

  • Assuming every incompressible flow has a \(\phi\). Only irrotational ones do. A shear flow \(u(y)\) has vorticity and no potential.
  • Expecting drag predictions. Potential flow gives pressure distributions and lift (with imposed circulation) but no viscous drag.
  • Confusing \(\phi\) and \(\psi\) roles: \(\phi\) exists in 3-D; the scalar \(\psi\) as defined here is a 2-D device.

Knowledge graph position

Prerequisites: Continuity, Euler's equation, Eulerian description. Leads to: Conformal mapping, interface instabilities.

Quiz

Q1 (computational). For the potential \(\phi = Ux + \frac{Q}{2\pi}\ln r\), find the stagnation point on the \(x\)-axis.

Answer

\(u = U + \frac{Q}{2\pi x} = 0 \Rightarrow x = -\frac{Q}{2\pi U}\) — upstream of the source: the incoming stream is exactly cancelled there (Rankine half-body nose).

Q2 (conceptual). Why can solutions of potential flow be superposed, while solutions of the full Navier–Stokes equation cannot?

Answer

Laplace's equation is linear; sums of solutions are solutions. Navier–Stokes contains the quadratic convective term \((\mathbf{v}\cdot\nabla)\mathbf{v}\), so the sum of two solutions generates cross terms that are not accounted for.

Q3 (multiple choice). Streamlines and equipotential lines in 2-D potential flow:

  • (a) coincide (b) are parallel (c) are orthogonal (d) never intersect
Answer

(c). \(\nabla\phi \parallel \mathbf{v}\) and \(\nabla\psi \perp \mathbf{v}\), so the two families cross at right angles (a consequence of the Cauchy–Riemann equations).