Fluid Flow & Aerodynamics
why things fly, and why the usual explanation is wrong
Concept

Exact solution — where it appears

A solution of the equations of motion written in closed form, of which the Navier–Stokes equations have only a handful. Each is worth knowing precisely because it is exact: it can be used to check a solver, and its assumptions state where it applies.

Named by 17 essays across 7 fields — each of them below, with the objects they name alongside it.

The laminar line does not end; the flow leaves it. Friction factor against Reynolds number in a pipe. The laminar law f = 64/Re is exact and is drawn continuing past the transitional Reynolds number, faintly, because it remains a solution there — the flow simply stops taking it. The turbulent branch is Blasius' correlation and begins where experiments find transition, not where any calculation puts it.

The solutions stop being chosen

Hagen and Poiseuille's pipe profile is an exact solution of the Navier–Stokes equations at every Reynolds number, and it is linearly stable at every Reynolds number. Something else happens at 2300 anyway, and it is not that the solution stopped being one.

turbulence · Transition A wave that dies within one wavelength — 100 Hz in air. The velocity profile above an oscillating wall, at eight phases of one cycle, with depth in units of δ = √(2ν/ω). The motion is a wave travelling into the fluid, and its amplitude falls by 1/e in the same distance it turns by one radian — so it is dead within about one wavelength, and the fluid three δ up hardly knows the wall is moving at all. This is one of the very few exact solutions the Navier–Stokes equations have. The dashed line is one fixed distance above the wall in millimetres: in these units it climbs as the square root of the frequency, which is the whole of how far the motion reaches.

The wall that shakes

Slide a wall back and forth in its own plane and the fluid above it does not follow — a wave travels upwards into the fluid and dies within one wavelength. The depth it reaches is √(2ν/ω), it contains no length from the geometry at all, and the whole thing is one of the very few exact solutions the Navier–Stokes equations have.

viscous · Exact layer A boundary layer with no x in it. The velocity profile over a porous wall with uniform suction: U(1 − e^{−Vy/ν}), exactly, at every station along the wall. The displacement thickness is ν/V, the momentum thickness is half of it, and the shape factor is two — all of them constants, none of them a function of distance. It is the cleanest demonstration there is that a boundary layer's thickness is a balance rather than an accumulation.

The layer that stops growing

Blasius' boundary layer thickens as the square root of distance and never stops. Suck fluid through the wall at a uniform rate and it stops immediately — the profile becomes a single exponential with no x anywhere in it, and the friction comes out exactly equal to the momentum of the fluid that was taken away.

viscous · Exact layer The current at the surface is 45° from the wind, and nothing sets that angle. The Ekman spiral drawn as a hodograph: each point is the velocity at one depth, and depth runs along the curve. At the surface the flow is at exactly 45 degrees to the wind that drives it — not approximately, exactly, and independently of the wind, the viscosity and the latitude. By one Ekman depth the flow has turned another radian and lost 1/e of its speed; by three it is a hundredth of the surface value and pointing back the way it came. The angle is a property of the equation having two terms in it, and nothing else.

The layer that stops at a depth

Every other boundary layer grows. This one does not — rotation supplies a frequency, the balance against diffusion supplies a length, and the transport that comes out contains the stress on the surface and not the viscosity underneath it.

viscous · Rotating The hodograph plane, where the unknown boundary is the known one. The same flow drawn in the plane of its own velocity, ζ = (u − iv)/U. The plate, whose shape is known in the physical plane, becomes a segment of the imaginary axis; the axis of symmetry becomes a segment of the real one; and the free streamline — whose shape nobody knows — becomes an arc of the unit circle, because the speed on it is exactly the free stream. The unknown and the known have changed places, which is why the problem can be solved at all.

Where the unknown boundary is the known one

A free surface is the hardest kind of boundary — its shape is part of the answer, so the region the problem is posed in is not known until the problem is solved. Draw the same flow in the plane of its own velocity and the shape becomes an arc of a circle, known in advance and exactly.

inviscid · Free-streamline A cylinder in a uniform shear, K = 0.4. A stream whose velocity increases with height, meeting a circular cylinder. The oncoming profile is drawn at the left. The flow carries uniform vorticity −K, so it is a solution of Euler's equations and not of Laplace's, the pattern is no longer symmetric top to bottom, and the body feels a lift towards the fast side with no circulation anywhere.

Inviscid does not mean irrotational

Dropping viscosity gives Euler's equations. Assuming nothing is spinning gives Laplace's — one scalar, linear, unique. The second step is a separate hypothesis about the flow's history, and a flow that fails it is still an inviscid flow with exact solutions of its own.

inviscid · Euler rotational Hill's spherical vortex. A sphere of rotating fluid travelling steadily through fluid at rest, drawn in the frame that moves with it. Outside the sphere the flow is the ordinary potential flow past a sphere; inside, the vorticity is proportional to the distance from the axis and the fluid recirculates. The two solutions match in value and in slope across the surface, and there is no body anywhere — the boundary is a streamline and nothing else.

The one rotational solution anybody can write down

A sphere of spinning fluid travelling steadily through fluid at rest, with no body anywhere in it — the boundary is a streamline and nothing else. It is exact, it is two lines long, and the reason it is the famous one turns out to be the reason it is the only one a real fluid can settle into.

inviscid · Euler rotational One streamline, sectioned, in two steady flows. Every time a single streamline crosses the plane z ≡ 0 going upwards, a point is plotted. On the left the flow is integrable and the points lie on a curve, however long the trajectory is run. On the right one coefficient of the same exact solution has been changed and the same single streamline scatters over a sixth of the plane. Both flows are steady, both are incompressible to machine precision, and both are exact solutions of the Euler equations.

Steady, three-dimensional, and mixing anyway

A steady flow that solves the Euler equations exactly, with its vorticity equal to its velocity to six parts in ten thousand million — and one of its streamlines wanders through a sixth of the box while another, started nearby, lies on a curve for ever.

kinematics · Advection Two exact solutions of the same problem. Two steady Euler flows in the same square cell with the same boundary condition, differing only in the function relating vorticity to streamfunction. On the left the vorticity is proportional to the streamfunction, which is the textbook cellular flow; on the right it is uniform. Both satisfy the equations exactly. Nothing in the ideal theory prefers either, and at the same peak streamfunction their kinetic energies differ by thirty-one per cent.

The vorticity nothing decides

A streamline that comes from upstream carries its vorticity with it. A closed one comes from nowhere, so nothing determines what it carries — the ambiguity is not one number per body but a whole function. What closes it is a limit, and setting the viscosity to zero gives a different answer from letting it go to zero.

inviscid · Euler rotational A compression piston, and where its characteristics first cross. Sixty C+ characteristics from an accelerating piston, drawn in the distance-time plane. Each is a straight line, because the invariant makes the state along it constant; later ones are faster, because the gas ahead of them has been compressed; so they converge, and the first crossing is the shock. The envelope formula gives 2.7529 and the first actual crossing is at 2.7510.

Two numbers that do not change

One-dimensional unsteady gas flow carries two quantities that are exactly constant along two families of curves. That single fact turns a pair of coupled partial differential equations into a family of straight lines, and gives an exact speed at which a gas outruns its own expansion.

compressible · Characteristics The marginal Taylor number against the axial wavenumber. The smallest Taylor number at which a disturbance of a given axial wavenumber is neutral. Its minimum is 1707.757 at a wavenumber of 3.1158, which are the critical Rayleigh number and critical wavenumber of a layer of fluid heated between two rigid walls — the same numbers, because in the narrow-gap limit the two problems are the same sixth-order eigenvalue problem. This curve is computed by that essay's own solver.

A transition that needs a second number

Fluid between rotating cylinders goes unstable at a Taylor number of 1707.762 — which is the same number, to every digit, as a layer of fluid heated between two rigid walls. It is not an analogy. And two things the number cannot carry decide whether the transition happens at all and what it looks like when it does.

regimes · Taylor One shape, at every station and for every jet. The axial velocity of Schlichting's round jet at four distances, each scaled on its own centreline speed and its own half-width. The four curves are one curve: the shape function (1 + eta²/4)⁻² has no parameter in it at all, and every station of every jet — laminar or modelled, weak or strong — collapses onto it exactly.

Exactly similar, and one number short

The round jet has an exact solution of the Navier–Stokes equations, and a turbulent jet is modelled by the same formula with the viscosity replaced by a fitted constant. The shape is identical. Nothing a measurement of the profile can do will tell the two apart.

viscous · Free shear Fourteen pure numbers, and where each came from. Every one of these is dimensionless, exact and quoted as a fact about fluids. None of them comes from dimensional analysis, which says which numbers an answer may depend on and never what any of them is. They come in four kinds — algebra, an integral, a root and an optimum — and the last two are not equally knowable.

Where a pure number comes from

Sixty-four for a round pipe, sixteen twenty-sevenths for a wind turbine, 0.332 for Blasius. Counting dimensions produces none of them — it produces the list of arguments, and the function has to be solved. Which of the four ways it was solved decides how many digits are worth printing.

regimes · Dimensional 121 m after the cavity closes, against 69 m from the closure. The head at a valve shut instantly on water flowing at 0.36 m/s through 600 m of 100 mm pipe, a = 1200 m/s, with a steady head of 25 m. The closure raises it to 69.1 m, the Joukowsky head; the reflection returns at one round trip, 1.00 s, and takes the head down to the vapour head, −10.1 m, where a cavity opens (shaded). It closes 2.146 round trips after the closure, and the first pulse after it reaches 121.3 m — 52.3 m above the Joukowsky head — for 146 ms. The step line is the exact solution between events; the thin line is a 240-reach grid solver that was told nothing about it and agrees with its first pulse to better than a millimetre.

Twice the margin, on top of the hammer

Shut a valve on a line whose pressure is low and the returning wave boils the water beside it. When that cavity closes, the head at the valve can pass the Joukowsky rise — by up to twice the margin that let the water boil, in a sawtooth that jumps each time one more round trip fits into the cavity's life, and for a time that is shortest exactly when the pulse is tallest.

applied · Water hammer Three profiles that do not depend on the radius. The radial, azimuthal and axial velocities of the flow above a rotating disc, as functions of one similarity variable. The radial one is a jet: fluid thrown outward by the swirl it has picked up, peaking at 0.181 of the local disc speed a fifth of the way through the layer. The azimuthal one falls from the disc's own speed to nothing. And the axial one is the surprise — it does not vanish far from the disc but tends to a constant, so the disc draws fluid down onto itself at 0.8845 times the square root of the viscosity times the rotation rate, at every radius and for ever.

The solution that keeps its nonlinear term

Every exact solution before this one has been exact because the nonlinear term vanished. A rotating disc's does not vanish — at the wall it is the whole of the balance — and the reduction is exact anyway, because the radius divides out of all three momentum equations at once.

viscous · Exact layer The profile a diverging channel flattens into, and then cannot hold. Five purely outward profiles in a wedge of 0.2 radians, at rising flux, each normalised to its own centreline value. As the flux rises the profile flattens in the middle and steepens at the walls — and then it stops. The last one has zero slope at the wall, which is separation, and beyond it no purely outward profile of this form exists at all. Nothing was added to the equation to make that happen: the wall shear is the square root of a cubic and the cubic runs out.

One channel, one flux, two flows

Flow between two plane walls meeting at a line has an exact solution. Past a threshold that turns out to be a ratio of gamma functions, it has two — the same wedge carrying the same flux, once outward everywhere and once with the fluid running backwards along both walls, and nothing in the equations chooses.

viscous · Exact layer Six that are symmetries and five that look like them. Each transformation applied to an exact solution, with the Navier-Stokes residual recomputed from the transformed field by finite differences — nothing differentiated by hand. The six symmetries leave the residual at the differencing floor, a few parts in 10^8. The five near-misses leave between 0.048 and 4.3, which is six to nine orders of magnitude larger. The gap is what makes this a test rather than an illustration: a transformation that is nearly a symmetry does not exist here, and every one of the five is something a reader might reasonably believe.

Why the list is this long

Every textbook list of exact solutions of the Navier–Stokes equations is about a dozen long, and the usual explanation is that the equations are hard. It is not the reason. A similarity reduction is a solution invariant under a subgroup of the equations' own symmetries, so the catalogue of possible reductions is the catalogue of subgroups — and that is a finite, countable object.

viscous · Exact layer

Named alongside it

The objects these essays reach for when they reach for this one.

Model limitBoundary conditionLaminar flowMeasurementNavier–Stokes equationsSimilarity solutionBoundary layerIrrotationalStreamfunctionTransportVorticityNonlinearity

All concepts