Transport with nothing transported

Taylor’s dispersion requires two ingredients and the essay that computes it is careful about both: a velocity that varies across the tube, and a molecular diffusion that carries tracer between the fast part and the slow part. A mean velocity is not one of them. It appears in the answer because the calculation is done in a steady Poiseuille flow, and a steady flow with shear in it has a mean.

Take the mean away and keep the shear. Oscillate the fluid in the tube, backwards and forwards, so that over a cycle nothing goes anywhere at all. The mechanism is untouched: fluid near the axis still moves further than fluid near the wall on each stroke, tracer still diffuses between them, and tracer that crosses from a fast streamline to a slow one during the forward stroke does not come all the way back on the return.

A mechanism with no net motion in it #

The result of that arrangement is a genuinely strange object: an axial transport with no axial flow. A slug of tracer released at one place spreads symmetrically in both directions, at a rate hundreds of times the molecular one, while the fluid it is in returns to its starting position every tenth of a second.

Nothing is being carried. Every fluid particle ends the cycle within a stroke-length of where it started; the tracer moves because it changes streamlines while the streamlines are moving relative to one another, and a change of streamline is not undone by the flow reversing.

It is worth watching a single molecule to see why the reversal does not undo it. Start it on the axis at the beginning of a forward stroke. It is carried forward faster than its neighbours near the wall. During that stroke it wanders sideways, by diffusion, and finds itself nearer the wall. The stroke reverses; the fluid near the wall moves back less far than the fluid on the axis, so the molecule returns a shorter distance than it went. It ends the cycle displaced, and nothing about its own motion was irreversible — the diffusion was as likely to take it the other way, and a molecule that happened to wander towards the axis ends up displaced backwards by the same argument.

That is the point at which this stops being transport and becomes dispersion. The mean displacement over the population is zero, exactly, because the flow is symmetric and the wandering is unbiased. What is not zero is the variance, and a variance growing linearly in time is a diffusion with a coefficient. A cloud of tracer spreads; its centre does not move.

That is the same class of result as three others computed elsewhere, and the family resemblance is exact. Steady streaming is a mean flow left behind by an oscillation with no mean; the drift in a wave that has none is a mean displacement left behind by an orbit that nearly closes; a swimming sheet moves because the mean of a product of two first-order quantities is not zero. All of them are second-order means of first-order oscillations, all of them are invisible to any calculation that stops at first order, and the transport here is quadratic in the stroke for exactly that reason — checked to machine precision rather than argued.

The calculation, and the limit that checks it #

A plane channel of half-width $h$, an oscillating pressure gradient, and a tracer with an imposed mean axial gradient. The velocity is the Stokes profile,

$$\hat u(y) = A\left[1 - \frac{\cosh(\lambda y/h)}{\cosh\lambda}\right], \qquad \lambda = \alpha\,e^{i\pi/4},\ \ \alpha = h\sqrt{\omega/\nu},$$

and the tracer perturbation satisfies the same operator with the diffusivity in place of the viscosity, which changes $\alpha$ to $\alpha\sqrt{\mathrm{Sc}}$ and nothing else. Both are elementary, and the extra diffusivity is the cycle-averaged, cross-section-averaged product of the velocity deficit with the concentration perturbation.

The check is the slow limit. As the frequency goes to zero the profile is Poiseuille’s at every instant, so the answer must be Taylor’s coefficient with $u^2$ replaced by its mean over the cycle — that is, $W^2/2$ for a sinusoid of amplitude $W$. It is, to four parts in $10^9$, which is the quadrature’s own precision.

That limit is worth more than a check, because it settles a question the steady calculation leaves open. Taylor’s coefficient is quadratic in the velocity, so a flow whose velocity varies in time does not disperse at the rate its mean velocity would give — it disperses at the rate its mean square velocity gives, which is larger. A pipe carrying a steady flow of one metre a second and one carrying a square wave between zero and two disperse differently by a factor of two, and nothing in the steady analysis says so.

The general form of that statement is the one worth carrying: a quantity quadratic in a fluctuating input responds to the variance of the input as well as to its mean, and a flow whose velocity varies by a factor of two about its mean disperses about twenty per cent more than a steady one at the same average. The zero-mean case is the extreme of that, in which the entire transport is variance and none of it is mean.

There is one quantity in the slow limit that a reader should be careful with. The Womersley number carries the viscous diffusion across the channel and the Schmidt number converts it to the tracer’s; for a gas the two are within thirty per cent, so both relaxations happen at about the same frequency, and for a liquid the tracer is four decades slower. A liquid in an oscillating tube is therefore already past its own slow limit at frequencies where the velocity profile is still quasi-steady, which is a regime this calculation contains and no intuition does.

Why the fast limit is a cube #

The naive estimate gives the wrong power, and following the correction is the most informative part of the calculation.

At high frequency the shear is confined to a Stokes layer of thickness $h\sqrt2/\alpha$ and the core moves as a plug. Only the tracer inside that layer is sheared at all, so one expects the transport to fall in proportion to the layer’s share of the channel, which is $\alpha^{-1}$. It falls as $\alpha^{-3}$.

The two missing powers are a phase. Inside a thin sheared layer, the concentration perturbation satisfies $-i\omega\hat c \approx \hat u'$ — the diffusion term is negligible across so thin a layer — so $\hat c$ is $i\hat u'/\omega$, which is exactly a quarter of a cycle out of phase with the velocity. A quarter-cycle phase difference contributes nothing at all to a cycle-averaged product. The transport therefore comes entirely from the small in-phase correction that cross-stream diffusion supplies, and that correction is smaller by $D/(\omega\delta^2) \sim \alpha^{-2}$.

The same structure turns up in every oscillatory transport and it is worth naming once. A flux is a cycle average of a product of two oscillating quantities, and a cycle average of two sinusoids in quadrature is zero. So whenever the leading-order response is a quadrature response — which it is whenever the forcing is fast enough that the responding quantity has not had time to relax — the leading order contributes nothing and the answer is a correction. That is why fast oscillatory transports fall away far more steeply than a share-of-the-fluid argument predicts, and it is the same reason a fast squeeze film stores rather than dissipates and why the dissipation there peaks in the middle of the range rather than at the end of it.

That the exponent does not move when the Schmidt number is changed by a factor of ten is the evidence that this reading is right. If the decay came from the tracer’s own diffusion distance, the Schmidt number would enter the power; it does not, and the Schmidt number appears only in the prefactor.

The layer that does the work, and how little of it there is #

The cube has a second consequence, which is about where in the channel the transport is being done.

At a Womersley number of twenty the Stokes layer is a fourteenth of the half-width. Ninety-three per cent of the fluid is moving as a plug, is not sheared, and contributes nothing whatever. All of the axial transport is taking place in a ring near the wall that holds seven per cent of the tracer, and the rest of the channel is along for the ride.

That is an inefficient way to move anything, and it is the reason the enhancement collapses so fast. It is also the reason the mechanism is so sensitive to anything that changes the wall region: a layer of mucus, a partial obstruction, a change in the airway’s compliance that alters the local velocity amplitude. A transport mechanism that lives in seven per cent of the fluid is a transport mechanism whose rate is decided by the state of the wall, and that is not how a diffusive process usually behaves.

The contrast with the slow limit is complete. There the whole cross-section participates, the transport is a property of the bulk, and the wall matters only through the no-slip condition that generates the shear in the first place. Between those two descriptions the mechanism migrates from the middle of the channel to its edge, and the migration is what the cube is measuring.

What a ventilator holds fixed #

Everything above holds the velocity amplitude constant while the frequency is changed, and no machine does that. A ventilator holds the tidal volume constant — the stroke, the distance the gas actually moves — and the velocity amplitude is then the stroke times the frequency.

Clinical accounts of high-frequency ventilation quote carbon dioxide elimination as going with the square of the tidal volume and a power of frequency between a half and one. The square is exactly this — it is the second-order mean, and it is the least negotiable part of the whole calculation. The range on the frequency exponent is this curve seen through the band of settings the machines are used at, plus the fact that a real airway carries the load by several mechanisms at once.

One airway, one machine #

Four hundred times molecular diffusion is the number that makes the technique possible. Conventional ventilation works by bulk flow: a breath larger than the dead space pushes fresh gas past it into the alveoli. High-frequency oscillation uses tidal volumes smaller than the dead space — one to three millilitres per kilogram against a dead space of two — so no fresh gas reaches the alveoli by bulk flow at all, and the exchange happens entirely by mechanisms of this kind.

The reason to do it is that the lung is then never inflated much. Ventilator-induced lung injury is caused by repeated large excursions of pressure and volume, and a machine that moves a few millilitres at ten hertz on top of a steady mean pressure avoids them; the alveoli sit at a constant volume and the gas exchange is done by stirring rather than by breathing.

The arithmetic of the exchange is worth doing once, because it says whether four hundred is enough. An effective diffusivity of about nine thousandths of a square metre a second — the same quantity Taylor computes for a steady tube, arrived at with no mean flow — acting over a dead space of roughly a quarter of a metre, across a carbon dioxide concentration difference of about five per cent by volume, through a tracheal cross-section of two and a half square centimetres, gives a carbon dioxide flux of order a tenth of a litre a minute at standard conditions. A resting adult produces about two hundred millilitres a minute.

So the mechanism computed here, taken alone and in a single straight channel, gets to within a factor of a few of what is required — which is the right answer for a calculation that has left out the branching, the turbulence, the profile asymmetry and every other airway generation. It would be a worse result if it came out exactly right, because then the other mechanisms would have nowhere to go.

What the picture cannot show #

The Reynolds number of the ventilator case is twelve thousand. The Stokes profile this calculation is built on is laminar, and an airway at that oscillation is not. Turbulence would mix the cross-section far faster than molecular diffusion does, which raises the transport — so the laminar answer is a floor rather than an estimate, and the mechanism is what survives rather than the number.

The airway is a plane channel here and a branching tree in fact. A tree adds a mechanism this cannot contain: gas entering a bifurcation on the forward stroke does not return through the same daughter on the back stroke, so there is a net exchange between branches — pendelluft — which has nothing to do with shear or diffusion and is thought to carry a substantial share of the load.

The velocity profile in a real airway is asymmetric between inspiration and expiration. A jet enters and a plug leaves, so the mean of the product over a cycle is not what a symmetric sinusoid gives, and that asymmetry alone produces a transport of the same order as everything computed here.

And the tracer is passive and the gas is one gas. Carbon dioxide is being carried against its own concentration gradient into a mixture whose composition is changing, and the diffusivity used is a binary one.

Who found it, and when #

Watson gave the closed-form solution for oscillatory dispersion in a tube in 1983, which is the calculation this essay does in a channel; Chatwin had treated the high-frequency limit in 1975, and Aris the oscillatory problem in 1960. High-frequency oscillatory ventilation reached clinical use in the 1980s, and the theoretical accounts and the clinical practice grew up alongside one another rather than one from the other.

The surprising connection is with a result about heat in a very different geometry. A wall whose temperature oscillates drives a thermal wave into the solid beside it that dies within a skin depth, and the mean temperature of the solid is unaffected — an oscillation carries no net heat, by symmetry. Put a fluid there instead and the symmetry breaks, because the fluid moves: the same oscillation that carries no net heat into a solid carries a great deal along a tube, and the difference is entirely that one medium has streamlines to change between. What makes the transport possible is not the oscillation but the shear, and a tube is the simplest object that has one without having a mean flow.

Still open: how much of a real airway’s transport this is #

The mechanism is unarguable and its share is not measured, and the gap between those two is where the subject has sat for forty years.

A real airway carries the tidal oscillation by at least four mechanisms: this one, the asymmetry between inspiratory and expiratory profiles, pendelluft between branches with different time constants, and direct bulk convection to the nearest generations of the tree. Each has been estimated in isolation, each estimate carries a factor of a few, and the sum is routinely larger than the measured elimination — which means at least one of them is being double-counted.

What would separate them is an experiment rather than a calculation, and it is an unusually clean one: the four mechanisms have different dependences on the gas. This one goes as the molecular diffusivity through the Schmidt number and through the prefactor; pendelluft does not involve diffusion at all; and the convective share scales with neither. Running the same oscillation on helium–oxygen and on sulphur hexafluoride–oxygen — a factor of about six in diffusivity and of four in density — and measuring the elimination would give two equations towards separating four mechanisms, which is not enough to solve but is enough to rule one of them out. That measurement appears to have been made for other purposes and never read this way.