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

Every essay — page 38

Page 38 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

Viscosity

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

The only candidate the far field allows, and the wall it slips past. The general Stokes solution has four constants; the condition at infinity kills two of them and fixes a third, leaving one to satisfy two conditions at the wall. Setting the stream function to zero there uses it up, and the tangential velocity that remains is exactly twice the free stream — for every radius, every speed, and every fluid.

The flow with no solution

Creeping flow past a sphere has a solution and everybody knows it. Creeping flow past a cylinder has none — not a difficult one, not one needing a clever method. The equations, the no-slip condition and the uniform stream at infinity are inconsistent, and the residual is exactly twice the free stream.

8 figures The wall shear, marched to the station where it stops. Howarth's linearly retarded outer flow, marched with an implicit finite-difference scheme from a Blasius profile. The wall shear falls, its slope steepens, and at x = 0.11983 it reaches zero — against Howarth's 0.1198, which is a quarter of a per cent. There is nothing downstream of it: the solution does not continue.

The singularity a layer makes for itself

March Prandtl's equations into an adverse pressure gradient and the wall shear reaches zero with an infinite slope at a finite station, and the solution cannot be continued past it. The singularity is real, it is not a numerical difficulty, and it belongs to the boundary condition rather than to the equations.

8 figures The core that does not move, and where its edge is. Four velocity profiles at the same pressure gradient and four yield stresses. The shear stress in a pipe is G r/2 whatever the fluid is — that is a force balance and not a constitutive law — so the fluid is unyielded exactly inside r = 2 tau_y/G, and the plug radius is known before anything is solved.

The core that does not move

Toothpaste in a tube has a region in the middle that is not being sheared at all, and its edge is at a radius you can write down before solving anything. Four things about that core are exact, and one of them is that the fluid does not flow below a threshold — not slowly, at all.

9 figures A pure number with no fluid in it. The load a plane slider bearing carries, divided by everything dimensional in it. What is left is a function of the convergence ratio alone — no viscosity, no speed, no size, no material — so the tilt that carries the most load is decided before anything about the machine is known. It is the root of one transcendental equation, at 2.1887048.

The gap that carries the most

A bearing's most consequential dimension is decided by a number with no oil in it, no speed and no size — the root of one transcendental equation, at 2.1887048. And the maximum it sits on is flat enough that the value printed in every handbook is not it.

8 figures One shape, at every station and for every jet. The axial velocity of Schlichting's round jet at four distances, each scaled on its own centreline speed and its own half-width. The four curves are one curve: the shape function (1 + eta²/4)⁻² has no parameter in it at all, and every station of every jet — laminar or modelled, weak or strong — collapses onto it exactly.

Exactly similar, and one number short

The round jet has an exact solution of the Navier–Stokes equations, and a turbulent jet is modelled by the same formula with the viscosity replaced by a fitted constant. The shape is identical. Nothing a measurement of the profile can do will tell the two apart.

8 figures Four guesses at a boundary-layer profile. A straight line, a parabola, Pohlhausen's cubic and a quarter sine, each rising from zero at the wall to the free stream at the edge. Two of them also satisfy the conditions the true profile satisfies — no curvature at the wall, no slope at the edge — and two do not, which is what sorts them.

Four profiles, one drag

The momentum integral is exact and asks nothing about the shape of the velocity profile. Four guesses at that shape span twenty-three per cent in the drag they give — and the two that satisfy the conditions the true profile satisfies are within three, which says the freedom belongs to the family rather than to the constraint.

9 figures Four fluids under the same stress, and the four profiles that result. Each fluid carries the same linear stress and answers it differently: a Newtonian fluid with a parabola, a shear-thinning one with a blunter profile, a shear-thickening one with a sharper one, and a Bingham plastic with a plug in the middle where the stress is below its yield point.

The stress a pipe knows

A capillary viscometer measures a pressure drop and a flow rate and reports a viscosity. The first half of that inference is a force balance and is exact for any fluid there is; the second needs the slope of a whole flow curve, which is the experiment the instrument was bought to avoid.

10 figures A particle released in still fluid, with the history and without. The velocity of a hundred-micron particle released at one metre a second in air, computed with the unsteady Stokes force and with the quasi-steady drag alone. After five particle time constants the one that remembers is still moving nearly four times as fast.

The drag that integrates a whole history

A particle released in still air does not stop exponentially. The unsteady drag on it carries a term that is an integral over everything the particle has already done, weighted by the inverse square root of how long ago — so there is no time constant, and five particle time constants later it is still moving four times faster than the quasi-steady answer allows.

8 figures 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.

8 figures 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.

8 figures 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.

8 figures 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.

8 figures