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The thread: What is conserved — page 42

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384 essays carry this thread — page 42 of 43.

Mark's criterion lets the wall gas through above Mach 6.5; the gas inside the layer never gets through. The stagnation pressure of the incident shock's boundary-layer gas in the reflected shock's frame, over the reservoir pressure p₅, against the incident shock's Mach number in air: for the gas at the wall, which is Mark's criterion, and for the lowest anywhere in the layer. Below one the gas cannot pass into the reservoir and the shock bifurcates. The wall gas is refused from Mach 1.32 to 6.5; the layer's lowest from 1.32 on, levelling near 0.69. At the tailored helium condition, Mach 3.41, both give 0.51; at hydrogen's, Mach 6, the wall gas gives 0.89 and the layer 0.64.

The gas a reflected shock refuses is inside the layer

A reflected shock in a shock tube bifurcates when the gas in the wall's boundary layer cannot be pushed into the reservoir behind it: its stagnation pressure, in the shock's frame, is below the reservoir's. Mark's criterion asks that of the gas at the wall, and in air it says the bifurcation stops above an incident Mach number of 6.5. But the gas that fails at high Mach numbers is not at the wall. It is inside the layer, heated by friction until it meets the shock too slowly for its sound speed, and on that measure a reflected shock in air bifurcates at every Mach number above 1.3.

$Half the forced summit is gone in a fifth of an eddy time. The third moment of the velocity differences as a fraction of four-fifths εr, against separation in Kolmogorov lengths of the forced flow, at the instant the forcing is switched off and 0.1, 0.2, 0.5, 1 and 4 large-eddy times later. Forced, the curve peaks at 0.94 near 67 η. A tenth of an eddy time later the summit is 0.899, at a fifth 0.838, at a half 0.759, and after that it hardly moves: 0.742 at one and 0.72 at four, where the flow is simply decaying. The large separations fall first and farthest.$ Transition and turbulence

The summit forgets the forcing before the dissipation does

A flow forced at its largest scales carries the four-fifths law closer to exact than a decaying one. Switch the forcing off and the third moment has to pass from one value to the other. It does not wait for the cascade to drain: half the forced summit is gone in a fifth of an eddy turnover, while the dissipation has hardly moved, and the largest separations overshoot the decaying value before they settle on it.

Below a quarter the cloud settles on the cell walls; above it, it crosses them. Paths of particles released with the air's velocity inside one cell of the vortex lattice, at Stokes numbers of 0.15 and 0.4, over 25 time units; the cell's walls, the separatrices joining the saddles, are drawn. At 0.15 the particles are flung outwards by the rotation and approach the walls ever more closely without crossing, so the cloud is compressed onto lines. At 0.4 they reach a wall with enough speed to overshoot it into the next cell, where they meet particles coming the other way: the cloud folds.

Flows and fields

A lattice of vortices folds a cloud at the saddles' quarter

A cloud of heavy particles folds in still air above a Stokes number of one, and in converging air above a quarter. Air that turns as well as converges might have settled between them. In a lattice of vortices it settles exactly on the quarter: the vortices fling the particles to the cell walls, but only the saddles where the walls meet can make them cross. Below the quarter nothing folds, and the cloud is gathered onto the walls without limit instead.

Helium wastes least for most stiffnesses; xenon reaches furthest. The film's loss tangent — damping over stiffness — against its stiffness, each gas traced from a tenth of an atmosphere up to the pressure of its greatest stiffness. Raising the pressure buys stiffness and costs loss for every gas. Up to about a hundred newtons per unit relative amplitude helium's film loses least at any stiffness — 0.29 at 50 N against air's 0.4 and xenon's 0.46. Beyond, helium is near its peak of 127 N and the heavy monatomic gases take over: xenon reaches 149 N, and at 120 N loses 0.73 against helium's 0.78.

Viscosity

The gas sets a squeeze film's loss, not its stiffness

A squeeze-film levitator wastes part of every cycle as heat that crosses the gap, and the gas and its pressure decide how much. Fill one levitator with six gases at one atmosphere and its stiffness hardly changes while its loss nearly doubles from helium to xenon. Raise the pressure and every gas stiffens towards a peak near ten atmospheres. Helium wastes least at any stiffness up to a hundred newtons; above that, only the heavy monatomic gases get there.

With a small plenum, stall feedback closes the hysteresis loop. The stall amplitude against the throttle as it is closed slowly past the peak and opened again, at B = 0.2, with no stall feedback and with feedback at twice the critical gain; closing solid, opening dashed. Without feedback the cell appears suddenly near γ₀ = 0.6094 and vanishes only at 0.6428, and the loop between the two legs has an area of 0.078. With feedback the cell grows gradually as the throttle closes and shrinks along the same curve as it opens: the area is 0.0014.

Fluids at work

A throttle can hold the peak only in a small plenum

A compressor makes its most pressure at the peak of its characteristic, and at the peak it either surges or stalls. A throttle moved by feedback can fight both. Driven by the plenum's pressure it damps surge, with a gain that grows as the square of Greitzer's B; driven by the flow, it cannot. Driven by the stall cell's amplitude it turns the sudden jump into deep stall into a gradual one and closes the hysteresis loop — but only while the plenum is small. Above B of about a third the same feedback turns the jump into a cycle.

Four of six decayed states sit on the sinh branch; the other two lie above it. The streamfunction's flatness against the enstrophy-to-energy ratio Z/E: the sinh and tanh relaxed dipoles as curves, from the linear dipole at Z/E = 1 outwards, and the six decayed states at t = 600 as points. The tanh branch falls below 9/4 and stays near Z/E of one; the sinh branch rises. The two runs begun as two-level patches lie on the sinh branch at their own Z/E, 0.0049 to 0.055 from it, and so do the two begun on a k⁻³ spectrum, 0.04 to 0.072. The two random-phase runs lie above it, by 0.3 to 0.31: on the sinh side of 9/4, but not sinh dipoles.

Transition and turbulence

Decaying flows end on the sinh side

Three theories predict the state a decaying two-dimensional flow relaxes to, and the streamfunction's flatness tells them apart: above 9/4 for the point-vortex theory's sinh, below for the two-level theory's tanh. Run the decay from three kinds of start and every run ends above 9/4 — including the one begun as two-level patches, the case the tanh theory was built for. Four of six land on the sinh branch itself; the other two are sinh-like without being sinh.

The current along the shore runs fastest three Ekman depths down. The depth-averaged Eulerian current along the shore, positive to the left looking shoreward, against the local depth in Ekman depths, for a swell arriving 20° off the shore-normal at the shelf edge and for one arriving head-on. Both make a jet to the right, fastest at 2.9 Ekman depths for the oblique swell (-1.42 mm/s at 37.6 m) and 2.6 for the head-on one (-0.948 mm/s), and a weak current the other way in water shallower than one Ekman depth. Rotation turns the transport the shelf returns, so even a swell with no alongshore component drives one.

Flows and fields

A sloping shelf puts the swell's current three Ekman depths down

A swell's Stokes transport arrives at a coast and the sea must send it back or turn it aside. On a flat shelf one depth decided which. On a real shelf, deepening from the beach to its edge, every depth is present at once, and the sea does both: it sets up a few centimetres against the beach, dips a tenth of a millimetre where the depth passes two Ekman depths, and runs a current along the shore that is fastest about three Ekman depths down — wherever the friction puts it.

A lagged re-timing passes no error down the V. The variance of each bird's phase error, in radians squared, against its place in one arm of a V, for fore-and-aft wander of half a span correlated over four beats. Holding a fixed phase, every bird's error is its offset from the bird ahead, 3.16. Re-timing with a one-beat lag, every bird's error is 0.632 — the first follower's and the thirtieth's alike, to the last digit. Re-timing as smoothly but through two half-beat lags, the error grows from 0.819 at the first follower to 1.05 at the tenth and 1.08 at the thirtieth, and is still growing slowly there.

Circulation and lift

A wandering flock passes no error down the V

In a flapping V every bird re-times its beat to the wake of the bird ahead, whose own beat is imperfectly timed, so the errors ought to pile up along the arm. With the simplest way of re-timing they do not, at all: the thirtieth bird is off its phase by exactly as much as the first. A one-beat lag and its complement add to one at every frequency, and that identity telescopes the whole arm. Re-time more smoothly and the errors do accumulate — by about a third, and then they stop.

The water climbs the body, and the wetted width outruns the drawing. The wetted half-width against penetration for a wedge of 10° deadrise and a circular cylinder of unit radius: where the body crosses the undisturbed level, von Kármán's width, and where the risen water meets it, Wagner's. For the wedge Wagner's width is π/2 = 1.571 times the geometric one at every depth; for the circle it is √2 = 1.414 times, at small penetration.

Ideal flow

Water climbs a falling wedge, and the load comes from the climbing

A hull that strikes the sea sets the water under it moving, and the force is the rate at which it does so. The obvious estimate measures the wetted width where the hull crosses the undisturbed surface. But the water does not wait: pushed aside, it rises up the hull and wets it sooner, a factor π/2 wider for a wedge and √2 for a round bottom. The force carries that factor twice, and the peak pressure, where a thin jet leaves the hull, carries it squared on a cotangent that grows without bound as the bottom flattens.

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