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

Every essay — page 33

Page 33 of 39, 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

Transition and turbulence

Where the laminar solutions stop being the ones the flow takes, and what can honestly be said about what follows — which is less than the textbooks imply and more than nothing.

The roughness function, and the asymptote in which the viscosity has gone. The whole effect of a rough wall on a turbulent boundary layer is one number: the downward shift of the logarithmic profile. It vanishes on a smooth wall, rises through a transitional band, and becomes (1/kappa)ln(k+) + B − 8.5 — at which point substituting it back leaves u+ = (1/kappa)ln(y/k) + 8.5, with the fluid's own length gone from the answer entirely.

A second length at the wall

The logarithm in a turbulent wall profile exists because a region of the flow is not allowed to know about any length except the distance to the wall. Roughen the surface and there is one it does know about, which belongs neither to the fluid nor to the flow — and the slope does not change at all.

8 figures 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.

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.

8 figures 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.

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.

8 figures Four sets of scaling exponents, all of them exact at the third moment. zeta_p against p for K41, the beta-model, the log-normal model and She–Leveque. Every one of them passes through zeta_3 = 1 exactly, because the four-fifths law is a consequence of the equations and a model that missed it would be wrong about the one thing that is known. What they disagree about is every other moment.

The exponents that stop being thirds

Kolmogorov's 1941 theory says every moment of the velocity difference scales with the same exponent, p over three, so the distribution keeps its shape at every scale. It does not. The exponents fall below the line, by more the higher the moment, and what the departure measures is a dimension.

9 figures The mean flux, flat across the inertial shells and equal to the dissipation. The time-averaged transfer out of the first n shells, computed as the rate at which the nonlinearity changes their energy rather than from a remembered formula. It is constant to a tenth across the middle of the ladder and equal to the dissipation, which is the cascade — and it is an average.

A flux that runs both ways

The cascade is a statement about a mean. Kolmogorov's four-fifths law fixes an average and the constant flux through the inertial range is an average, and neither says anything about what the transfer is doing at any instant — which turns out to be running backwards a substantial part of the time.

9 figures The condensate, and the limit of no friction which does not exist. In a steady state the friction must remove everything the forcing puts in, so the energy is eps/(2 alpha) and the coherent velocity is sqrt(eps/alpha) — exactly a minus one half power, checked to 10⁻¹². As the friction is weakened the condensate grows without bound: the limit alpha to zero is not a flow with a weak condensate, it is a flow with no steady state at all.

Where the inverse cascade stops

Two-dimensional turbulence sends its energy upward in scale, and the upward direction has an end: the box. Without something to remove the energy before it arrives, it accumulates there in a pair of vortices filling the domain, and the limit of no friction has no steady state at all.

8 figures A normal stress the closure makes negative. The Boussinesq closure's first normal stress in a plane strain, against the strain measured in units of the turbulence's own time scale. It crosses zero at S k/eps = 1/(3 C_mu) = 3.704 — eleven per cent above the value the constant was calibrated at — and goes on falling. A variance below zero is not a small error; it is a statement that cannot be true.

The constant that makes a variance negative

Every engineering turbulence calculation in the world rests on one number, C-mu equals 0.09. It is not a property of turbulence. It is the assertion that a particular ratio is ten thirds, which is true in one flow — and eleven per cent above that flow the same closure reports a mean square below zero.

8 figures Every unstable mode of every profile, inside one circle. The complex phase speeds of the unstable modes, scaled so that each profile's own semicircle is the unit one. Howard's theorem says every one of them must lie inside — the centre and the radius are the mean and half-range of the velocity profile and nothing else — and every one of them does, with the closest approach at 0.915 of the radius.

Every unstable wave is inside one circle

Before solving anything, you know where the answer is. Howard's theorem says the complex phase speed of any growing disturbance in a shear flow lies inside a circle fixed by the fastest and slowest parts of the profile — and by nothing else about it at all.

9 figures Instability up to a quarter, and none past it. The fastest growth rate of a stratified shear layer against its Richardson number, on a profile whose gradient Richardson number is the same at every height. It falls smoothly towards zero and reaches it at a quarter: at Ri = 0.2499 the fastest mode still grows at 0.00106, and at 0.26 the solver finds no unstable mode at all.

Sufficient, and not necessary

A stratified shear layer whose Richardson number exceeds a quarter everywhere cannot go unstable. That is a theorem with an exact number in it. What it does not say — and what it is constantly read as saying — is that a layer below a quarter will.

9 figures The two dissipations, side by side. The strain form on the left and the enstrophy form on the right, for the same field, on the same scale. They have their maxima in different places — 0.46 apart on a box of side 2 pi — and neither is a smoothed version of the other. One says the dissipation is in the strained regions and the other says it is in the rotating ones, which is nearly a complete disagreement about what a turbulent flow is doing.

Equal on average, and nothing else

The rate at which a fluid turns motion into heat can be written two ways, and every textbook says the two are equivalent. Their averages are equal to fourteen decimal places. Point by point they are uncorrelated, and their maxima are in different places.

9 figures The scalar spectrum, with its two ranges. A model scalar spectrum at a Schmidt number of two thousand — dye in water. Below the Kolmogorov wavenumber it is Obukhov and Corrsin's five-thirds, inherited from the velocity; above it there is no turbulence left and the spectrum is Batchelor's minus one, which contains no velocity spectrum at all.

The scalar has its own cascade

Below the Kolmogorov scale there is no turbulence left, and a dye stirred into the flow goes on cascading anyway — on a spectrum whose exponent is minus one and whose amplitude contains no velocity spectrum at all. Resolving it costs the three-halves power of the Schmidt number, which for dye in water is a factor of ninety thousand.

10 figures Four wall models, through the buffer layer. Van Driest's damped mixing length, Reichardt's fit, Spalding's implicit law, and a control with no buffer layer at all — the viscous sublayer joined straight to the logarithm where they cross, at y+ = 11.6. The three fitted models agree with each other to a per cent and a half; the control is nineteen per cent above them at y+ = 10.

Three buffer layers, one friction

Four wall models are put through a pipe. The one with no buffer layer at all is nineteen per cent wrong where the turbulence production peaks and two and a half per cent wrong in the friction; the three respectable ones are within two per cent of each other in the buffer layer and spread by seven in the friction. The answer is not where it was expected.

10 figures