Fluid Flow & Aerodynamics
why things fly, and why the usual explanation is wrong
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The thread: The exact theory is wrong — page 12

Page 12 of 22, continuing through the 193 essays this motif runs through.

193 essays carry this thread — page 12 of 22.

The shock radius against time, from an equation that was given no exponent. The thin-shell energy balance integrated forward from a small initial radius, on logarithmic axes. The two-fifths power is not put in: the ordinary differential equation is Ṙ = √(E/AρR³), and the straight line is what it does. The fitted slope is 0.39983 and the fitted prefactor is 0.90721 against a closed form of 0.90702 — the small residuals being the integration's memory of where it started, which the similarity solution has no equivalent of. Compressible flow

A radius that gives the energy away

Four quantities, three dimensions, one group. The radius of a strong blast must be a constant times (Et²/ρ)¹⁄⁵, and nothing about the device, the chemistry or the initial size can appear. The exponent is free and the constant is not.

 Two decay laws from two invariants, and nothing in the equations to choose. The energy of a decaying turbulence against time, integrated from dK/dt = −A K^(3/2)/l with the large scales conserving u² l³ in one case and u² l⁵ in the other. The exponents come out at 1.1997 and 1.4282 against the closed forms 6/5 and 10/7. Which invariant holds is decided by the shape of the spectrum at the very largest scales, at the moment the stirring stops. Transition and turbulence

What decay never forgets

Stir a box of fluid and stop. The turbulence decays, at a rate with no viscosity in it — so the rate cannot come from the fluid. It comes from an invariant of the very largest scales, fixed at the moment the stirring stops, and never revisited.

 The wall the outer flow is really solving for. A flat plate, the edge of its boundary layer, and the line the outer flow behaves as though the plate were on. The displacement thickness is the mass deficit divided by ρU — checked here against the profile's own integral rather than quoted — and moving the wall out by that much reproduces exactly the flow rate the viscous layer lets past. It is a third of the visible thickness of the layer and it is the only part of the layer the outer problem knows about. Ideal flow

The body the outer flow actually sees

A boundary layer lets less fluid past than an inviscid one would. The outer flow can be given exactly the same reduced flow rate by leaving the fluid inviscid and moving the wall out — so the potential flow that matters is not the flow past the body, but the flow past the body plus a thickness the boundary layer computes.

 Four bodies of identical length and volume, and their wave drags. Each body has the same length and the same volume; only the distribution of area along it differs. The Sears–Haack body — the spindle whose area goes as the three-halves power of x(L−x) — has the least wave drag of the four, and every other shape pays between thirty-seven per cent and a hundred and seventy-five per cent more for carrying the same volume the same distance. Nothing about the cross-sections' shape enters: only the area distribution does. Compressible flow

The least drag a volume can have

A body's supersonic wave drag depends on nothing about it except how its cross-sectional area is distributed along its length. Minimising that for a given volume gives one shape — and the answer goes as the volume squared over the fourth power of the length.

 Model spectra at four Reynolds numbers, compensated. The spectrum multiplied by k^(5/3) and divided by eps^(2/3), so that a true inertial range is a horizontal line at the Kolmogorov constant. What a finite Reynolds number has instead is a single maximum: it reaches 1.4996 at the highest and 1.49 at the lowest, and the band over which it is flat to one per cent goes from a third of a decade to two. Transition and turbulence

The range a real Reynolds number does not have

Kolmogorov's minus five thirds is a statement about a band of scales that has forgotten the forcing and does not feel the viscosity. Both conditions are about separation, and separation is exactly what a finite Reynolds number does not have much of.

 Two-thirds to one, and nothing outside it. The drag on a spherical drop as a fraction of Stokes' drag on a rigid sphere of the same size, against the ratio of the viscosities. It runs from exactly two-thirds for a clean gas bubble to exactly one for a rigid particle and takes no value outside that range, however light or heavy the drop is — the formula has no density in it anywhere. A drop of water in air and a drop of mercury in water sit at opposite ends of it. Viscosity

The surface that moves with the flow

A clean gas bubble feels two-thirds of the drag a rigid sphere of the same size would, and the formula has no density in it anywhere. What buys the third is that the bubble's surface is free to move — and real bubbles in ordinary water do not get it, for a reason that is a millionth of a per cent of the water by mass.

 The velocity through a shock at Mach two, as a function of position. The Becker profile, integrated outwards from its own inflection point. It runs from the upstream velocity to the downstream one — the two states the jump conditions give, which appear here as the equilibria of a first-order differential equation — and its steepest gradient matches the closed form to one part in ten thousand. The horizontal axis is in units of the thickness, which for this shock is 139 nanometres. Compressible flow

The discontinuity that has a thickness

The jump conditions do not contain the viscosity, which is why they are exact. The thickness is entirely viscosity — 289 nanometres at Mach 1.5, 35 at Mach 5, against a mean free path of 64. At Mach five the continuum equations have produced a structure thinner than the distance between collisions.

 Why a flat plate has no drag, drawn as a triangle. A flat plate at incidence with the three forces that must balance. Pressure can only act along the plate's normal, so the pressure force is the arrow perpendicular to the plate. Kutta–Joukowski says the resultant is perpendicular to the free stream. The difference between the two directions is the suction force at the leading edge, which acts forwards along the plate and is exactly L sin α. Without it the plate would have a drag of L sin α, and an inviscid fluid does not permit one. Circulation and lift

A finite force from an infinite speed

Pressure on a flat plate can only act along the plate's normal. Kutta–Joukowski says the force is perpendicular to the free stream. Those two directions differ by the angle of attack, and the discrepancy is made up at a single point where the velocity is infinite and the area is zero.

 Four corners, and what the flow does in each. The local flow in corners of four different interior angles, drawn from the exact local solution ψ = r^(π/α) sin(πθ/α). The exponent of the speed is π/α − 1 and depends on nothing else: at a right angle the corner is stagnant, at a flat wall nothing happens, and at any angle greater than a straight line the speed has no bound at the corner. The last panel, at 360 degrees, is the flow round the edge of a plate. Ideal flow

Nothing turns a sharp corner

Near a corner the flow is fixed by the angle and by nothing else — not by the size of the corner, not by the flow far away, not by the fluid. The exponent is π/α − 1, and every sharp edge in aerodynamics is the one case where it comes out at minus one half.

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