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The thread: The exact theory is wrong — page 37

Page 37 of 38, continuing through the 335 essays this motif runs through.

335 essays carry this thread — page 37 of 38.

A travelling bubble's B-factor grows as the square root of its age, and the bubble turns heat-limited when it reaches B*. The vapour a growing bubble has made over the liquid it has cooled, against its age, in water at 80 °C under a tension 10 kPa below the vapour pressure, while inertia limits its growth; the dashed line is B, at which the cooling would use up the whole tension. B is 4.3 at a microsecond and 136 at a millisecond, and meets B = 31.6 at 53.8 µs, when its radius is 0.141 mm. Fluids at work

A bubble is throttled by its age, and a sheet by its length

A cavity cools the liquid it boils, and the cooling lowers the vapour pressure it needed. Whether that matters was a question about one number, the B-factor — vapour made over liquid cooled — and no real cavity's was known. Computed, the two kinds a pump has are nothing alike. A travelling bubble cools a thin shell that thins further as it stretches, so its B grows as the square root of its age and reaches the critical value at Brennen's thermal time: a millisecond in water at 60 °C, four microseconds at the boiling point. An attached sheet keeps renewing the liquid it cools, so its B is a few, fixed by its length and speed. In warm water the bubbles are suppressed forty degrees before the sheet.

On a NACA section thickness multiplies the lift by more than on a Joukowski section, by nearly one plus its thickness. The lift of a NACA four-digit section by panels over thin-aerofoil theory's, against its thickness: the 2412-type at 4°, and the symmetric section's lift slope over 2π; beside the Joukowski law 1 + 0.77 t/c. At 12 per cent the cambered section's factor is 1.115 and the symmetric slope's 1.101, against Joukowski's 1.092; at 18 per cent, 1.174 and 1.152. Ideal flow

A NACA section multiplies its lift by nearly one plus its thickness, and a nose correction cannot supply it

On Joukowski's exact sections, thickness — which lifts nothing alone — multiplies the lift of camber and incidence by 1 + 0.77 t/c. Solved by panels, the NACA four-digit sections that real wings use do more: their lift slope grows as about 1 + 0.84 t/c, and a cambered 2412 lifts 11.5 per cent more than thin-aerofoil theory says. The more of the thickness sits near the nose, the more it multiplies. Added as thin-aerofoil theory adds them, the parts lose a quarter of the suction peak and put the centre of pressure two hundredths of the chord too far aft. Riegels's nose factor, the classical repair, makes the nose speed finite and moves the peak, and adds no lift at all: the missing lift is not a nose effect.

A stagnation point hangs above the skin friction's saddle while the saddle is a separation, and is gone once it is not. Streamlines in the plane y = 0 of a wall flow with no axis of symmetry, at μ = 0.4: the wall's saddle of skin friction at x = -0.632, its node at 0.632. An interior stagnation point sits at height 0.15 above x = −0.5, and the saddle is a separation: the flow leaves the wall there. Flows and fields

A stagnation point off the axis can land only on a saddle or a node, and turns it inside out

In a swirling flow onto a plate every stagnation point lived on the axis, so one sinking to the wall always met the wall's single critical point. Without the symmetry it might seem free to land anywhere — on an ordinary stretch of skin-friction line, leaving a pair of new points behind. It cannot. Divided by the height, the velocity is continuous down to the wall, where it becomes the skin friction, so a point can only reach the wall where the skin friction already vanishes. It lands on a saddle or a node, and the count does the rest: the point it lands on turns from separation to attachment. An oil-flow picture of the wall, which shows the same pattern before and after, cannot tell which it is looking at.

The second gradient separates the layer soonest after a short release, latest after a dozen thicknesses, and sooner again after a long one. The distance past the second gradient's start at which the layer separates, in its thickness there, against the release's length in the thickness at the first release: Green's lagged layer, the same layer restarted at the second gradient with no memory, and the shape-only layer. Lagged: 2.91 after no release, 5.07 after 12 thicknesses, 3.95 after 60; with no memory, 3.16, 4.81 and 3.91. Transition and turbulence

A layer's memory of one gradient helps it through the next, unless the next comes too soon

A turbulent layer released from an adverse gradient keeps the gradient's eddies for a dozen thicknesses, entraining more than its shape asks. Decelerate it again before they fade — behind the next compressor stage, or past a flap — and the question is whether that memory helps or hurts. Both, depending on when. A second gradient that arrives within about three thicknesses of the release meets a layer whose entrainment is still catching up, and it separates eight per cent sooner than a layer of the same shape with no history. One that arrives later meets eddies still working from the first gradient, and the layer holds on up to five per cent longer. The crossing sits at three thicknesses for every pair of gradients tried.

Kicked at ninety per cent of its peak flow, the compressor is held only when both modes have an actuator. The flow through the compressor, as a fraction of the peak flow, after a stall cell of amplitude 0.05 is started at 90 per cent of the peak flow, at B = 2, under four controllers: a tip injector of gain 0.7 and capacity 2 per cent of the peak flow with the throttle fixed; the injector with a pressure-driven throttle; the pressure-and-flow throttle with no injector; and both. injector, throttle fixed: surges or runs away; injector, pressure throttle: surges or runs away; throttle pair, no injector: stalls; injector and throttle pair: recovers. Fluids at work

A compressor needs one actuator for each way it can fail

A throttle alone cannot hold a compressor left of its peak in a large plenum: it has one degree of freedom and the machine has two ways to fail, and the throttle's own surge gain runs into a ceiling that the plenum's size does not move. Give the stall cell its own actuator — air injected at the rotor tips against the cell — and give the throttle a second signal, the flow, and a machine at B = 2 is held at ninety per cent of its peak flow, recovering from a kicked cell nearly three times the injector's capacity. Let the injection lag the cell by a radian and the line still holds, but almost no disturbance is survived.

Inboard the spar is sized by its load, outboard by a stiffness floor its load cannot lower. The spar's weight per unit span along the semi-span, as a share of the payload's weight per unit of the reference span, at the best wing when the stiffness floor is 0.4: the weight the bending moment needs, and the floor. The floor governs the outer 53.1 per cent of the semi-span; the best loading has a third-harmonic ratio of -0.0993 and the spar is 32.1 per cent of the lift. Circulation and lift

A stiffness floor keeps the span and gives the tips their load back

A spar sized by strength alone thins to nothing at the tips, and charged as lift it stops the best span where it weighs a quarter of everything the wing carries, with the tips a little unloaded. A real outer wing is sized by stiffness, and stiffness sets a floor under the spar's weight that no reduction of the tip load can lower. Put that floor in and the tips get their load back — the best loading slides from a sixth of the way to the bell towards the ellipse — the quarter does not survive, rising to two fifths as the floor spreads over the outer two thirds of the span, and the span itself barely moves. A weight per metre cannot stop a span.

At its design Mach number the biplane's channel lifts nothing: all its lift is on the two outer faces. The lift coefficient of the pair at 2° incidence against Mach number, per unit of one element's chord, beside what the two outer faces give alone, 4α/β, and two isolated flat plates, 8α/β. At the design Mach number, 1.944, the pair's lift is 0.08378 and the outer faces' 0.08378: the channel's net lift is zero to rounding. At Mach 1.7 the channel adds 13.9 per cent, at 2.2 21.1. Compressible flow

Busemann's biplane lifts only on its outer faces, and its channel turns the lift into a quarter-chord load

Tilt Busemann's biplane to make it lift and the same cancellation that removed its wave drag removes the lift of its channel: at the design Mach number the waves the incidence sends across the gap meet and cancel just as the thickness's did, and every bit of lift is carried by the two outer faces — half what two separate plates would give. The channel does not go idle. It carries a couple, lift on the front half and download on the rear, which moves the centre of pressure from mid-chord, where every supersonic plate has it, to the quarter chord. The pair pays for its lift two and a half times as steeply as flat plates do, and still beats a diamond fourfold.

Along a block's wall the pressure stagnates at each foot and plunges at each lip; between two lips close together it never recovers. The pressure coefficient along the wall unrolled, against the distance from the front foot in block heights: up the front face, along the top, down the back, for a block as long as it is tall and one six times as long; the lips' suction is infinite in ideal flow and is drawn to −6. On the cube-like block the pressure at the middle of its top is -1.89; on the long one, -0.365. Ideal flow

A block's front face is pulled upstream by its own lip

A forward-facing step's face is pushed at its foot and pulled at its lip by exactly equal amounts. Give the step a back — make it a block on the floor — and the balance goes. The two lips, a short top apart, each sit in the other's fast flow, and the suction at the front lip grows while the push at the foot hardly changes: a block as long as it is tall has its front face pulled upstream by three tenths of the dynamic pressure times its height, a block a quarter as long by two thirds. The back is pulled downstream by exactly as much, so the ideal block still has no drag; but every force that ideal flow assigns is now on the wrong face for separation to undo.

An oval that sheds its fluid is braked less, so the pair sinks further and stops later, until it sheds too fast to stop at all. The depth of a trailing vortex pair below its starting level, in spacings, against time in spacings over its initial speed, in air of N = 0.22: Saffman's oval, which keeps its fluid, and one detraining with α = 0.05. Saffman's stops at depth 6.75 after 10.6; this one at 6.93 after 12.4. What is taught wrongly

A wake that sheds its fluid sinks further, and past a point does not stop

In stable air the pair of trailing vortices behind a wing carries a column of lighter air down with it, and buoyancy pushes the column back until the pair has given up its momentum and stops. That picture keeps every parcel of the column. A real pair sheds the light air as it sinks, and shedding does two things at once: it takes away part of the momentum the buoyancy had to return, and it takes away the buoyancy that was returning it. The second wins. A pair that sheds a little stops later and deeper; one that sheds faster than a threshold loses its column before buoyancy can stop it. The shed air keeps a quarter of the momentum swinging up and down behind the aeroplane.

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