The collection

Every essay — page 40

Page 40 of 40, continuing through the fields in the same order.

Flows and fields Ideal flow Circulation and lift Viscosity Regimes and numbers Compressible flow Transition and turbulence Fluids at work What is taught wrongly Series Concepts Regimes Refutations Search

Viscosity

The thin layer next to a surface that ideal flow ignores, and which supplies drag, separation and the wake.

What the fluid at one height is listening to. The weight the fluid two millimetres above a moving wall gives to the wall's velocity a given delay earlier, in water. It peaks at two thirds of a second and has a tail that falls as the delay to the power minus three halves — so the fluid is responding to a broad stretch of the wall's past rather than to a moment of it.

The wall the fluid is listening to

Water two millimetres above a moving wall is responding to what the wall did two thirds of a second ago — most likely. Half of its response is older than four and a half seconds, a tenth is older than two minutes, and the average age of what it is responding to does not exist at all.

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The flow stops and the stress does not. Shear stress against time for a fluid sheared at a constant rate for two seconds and then left alone. Nothing is moving after the second mark and the fluid is still stressed: 37 per cent of the peak one relaxation time later, and 0.7 per cent after five.

The fluid that has not finished its last deformation

Shear a polymer solution for two seconds and stop. Nothing is moving and the fluid is still stressed — 37 per cent of the peak one relaxation time later, and still measurably stressed after five. Two histories imposing exactly the same total strain leave it in states differing by a factor of 3.7.

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Two external flows that agree where it matters and nowhere else. The velocity just outside the boundary layer, for two pressure distributions, against distance along the surface. They cross at the half-way station with the same value and the same gradient, and they have nothing else in common: one accelerates steadily and the other does most of its accelerating at once.

A layer that is an integral of everything upstream

Two surfaces are given external velocity distributions that agree exactly at one station — the same speed and the same gradient. The boundary layers there differ by 38 per cent in momentum thickness, and the two surfaces separate five per cent of their length apart.

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How long a fluid takes to forget it was not rotating. The fraction of solid-body rotation a container's interior has reached, against time, by the two available routes. The Ekman layers on the end walls pump fluid radially and carry angular momentum inwards in a hundred seconds; diffusion alone would need ten thousand.

How long a fluid takes to forget it was not rotating

Spin a container of water and the fluid inside reaches solid-body rotation in a hundred seconds rather than the three hours diffusion would need. The shortcut is the thin layers on the end walls, and the advantage they give is exactly the reciprocal of the square root of the Ekman number.

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How fast a duct forgets each part of its inlet. The decay rate along the duct for each mode of a disturbance to the developed profile. It goes as the square of the mode number, so the third mode is forgotten nine times faster than the first and the tenth a hundred times faster.

A duct that forgets everything but one number

Whatever is fed into a pipe, what survives a little way down it is one shape. The disturbance's higher modes decay as the square of their mode number, so the sixth is gone in thirteen centimetres where the first survives four and a half metres — and the entrance length is that one mode's decay rate and an arbitrary threshold.

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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.

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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.

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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.

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Where a film stops being a damper and starts being a spring. The stiffness and the damping of a circular gas film against its squeeze number, both in units of ambient pressure times disc area. At the left the stiffness is nothing and the damping is Stefan's incompressible law; at the right the damping has gone and the stiffness has reached π, which is an isothermal gas trapped in the gap with no way out. They cross at σ = 6.23, and neither limit was put in by hand — both fall out of a Bessel function of a complex argument.

A damper that turns into a spring

A film of oil squeezed between two plates resists motion and stores nothing, and that is a property of the oil rather than of the film. Fill the same gap with air and the same equation gives a film that stores and resists nothing — above a squeeze number of six, with nothing changed but the frequency and the gap.

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Below a Stokes number of ten a ball does not come back. The restitution of an impact through a liquid film, as a fraction of the same impact's dry restitution, against Stokes number. Nothing rebounds below the threshold and the recovery above it is a hyperbola: the restitution, as a fraction of its dry value, is one minus the critical Stokes number over the Stokes number, with that critical value 10.18 computed from the film and the dry restitution alone. The measured curve, drawn beside it, is the same expression with ten in place of that number — and ten is what four decades of viscosity and four materials all give.

A ball that bounces in water and not in oil

A squeeze film cannot be closed with any finite energy, so nothing should ever touch anything. A sphere dropped into a tank nevertheless rebounds, and whether it does is decided by a number near ten that four materials and four decades of viscosity all agree on.

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Three strokes, and only the flat one is a theorem. Three cycles drawn in the swimmer's shape space: a square, a circle of the same width, and an out-and-back along the diagonal. The first two enclose area and carry the swimmer forward; the third encloses none and carries it exactly nowhere, which is the scallop theorem with no symmetry argument in it. What the third lacks is not a broken symmetry but an interior.

A stroke is worth the area it encloses

The usual account of swimming without inertia is a symmetry argument about reciprocal strokes, which says what cannot work and nothing about what does. Draw the stroke in the space of the swimmer's own shapes and the displacement is a line integral — so it is an area, it does not depend on how fast the stroke is played, and the scallop theorem is Stokes' theorem.

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The only term that turns spin into thrust, and what it is made of. The coupling term of the propulsion matrix against the drag anisotropy, with the geometry held fixed. It is exactly proportional to the difference of the two drag coefficients, so it is zero when they are equal — not small, zero — and no helix of any pitch turned at any rate would move. A real filament sits at 1.66, which is a third of the way from useless to the unreachable limit.

Two drags, or nothing swims

A bacterium turns a corkscrew and goes forward, and the reason is not the corkscrew. It is that a thin filament dragged broadside resists more than the same filament dragged end-on. Make the two resistances equal and the thrust is not small but exactly zero, for every pitch and every rate.

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