Whether the droplet turns

A wing flies through cloud. The air divides at the leading edge and goes round; the droplets suspended in that air have their own opinion, and the difference between the two is the whole of icing, of dust sampling, of cyclone separation and of why a car windscreen collects rain but not fog.

The number that decides it is a ratio of two times: how long a droplet takes to be brought to the speed of the air around it, against how long the air has to make its turn.

One number, from two times #

A small particle in a flow is dragged towards the local fluid velocity. For a sphere small enough for Stokes’ drag law to hold,

$$m\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 3\pi\mu d\left(\mathbf{u} - \mathbf{v}\right) \quad\Longrightarrow\quad \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = \frac{\mathbf{u} - \mathbf{v}}{\tau}, \qquad \tau = \frac{\rho_p d^2}{18\mu}$$

$\tau$ is the relaxation time: the time constant with which a particle forgets its own velocity and adopts the fluid’s. It goes as the square of the diameter, so a droplet twice as wide takes four times as long to be persuaded.

The flow’s own time is $a/U$ — how long the fluid takes to negotiate a body of size $a$. The ratio is the Stokes number,

$$\mathrm{St} = \frac{\tau U}{a}$$

and it is precisely the kind of object the counting argument produces: a dimensionless group formed from the particle’s properties and the flow’s, with the whole of the answer depending on it and on nothing else.

At $\mathrm{St} \ll 1$ the particle relaxes long before the flow turns, so it goes wherever the fluid goes. At $\mathrm{St} \gg 1$ it has not noticed the body at all and flies straight through where the body is.

What the solver computed, and how it was checked #

The trajectories are integrated with RK4 through the exact potential flow round a cylinder, and a path that reaches the surface is landed on it by bisecting the final step — because a path drawn through the body would be the one artefact that makes every figure here a lie.

The two limits, asserted. A particle at $\mathrm{St} = 10^{-3}$ must stay on its streamline: the check measures the departure using the streamfunction, which the tracker never touches, and it comes out at $0.0014$ of $\psi$. A particle at $\mathrm{St} = 60$ must be ballistic: the efficiency comes out at $0.93$.

Monotonicity. The collection efficiency must rise with the Stokes number at every step of a sweep; a solver whose paths wandered would show up here first.

The efficiency itself, by bisection on the release offset: the largest height at which a released particle still strikes, divided by the body’s own half-width. At $\mathrm{St} = 1$ it is 39 per cent — so a cylinder in a stream of $\mathrm{St} = 1$ droplets sweeps out a column of them, and collects less than half of what passes through its own frontal area.

The threshold is a discriminant #

The interesting end of that curve is the left one, and it is not a numerical artefact.

Follow the stagnation streamline. Near the front of the body the flow along it is a pure deceleration: writing $\xi$ for the distance to the surface, $u = -A\xi$, with $A$ a strain rate that the solver measures off the velocity field rather than assuming. For a cylinder in ideal flow it comes out at $2.000000\,U/a$.

The particle equation on that line is then linear:

$$\mathrm{St}\,\ddot\xi + \dot\xi + A\xi = 0$$

whose roots are $\left(-1 \pm \sqrt{1 - 4A\,\mathrm{St}}\right)/2\mathrm{St}$. While $4A\,\mathrm{St} < 1$ the roots are real and the approach is a sum of decaying exponentials, which tends to the wall and never reaches it. When $4A\,\mathrm{St} > 1$ the roots are complex, the approach becomes a decaying oscillation, and an oscillation crosses zero — which means the particle arrives.

$$\mathrm{St}_{\text{crit}} = \frac{1}{4A} = \frac{1}{8}$$

for a cylinder, exactly, with the eighth coming from the strain rate and nothing else. A different body has a different $A$ and therefore a different threshold, computed the same way.

The numerical search does not reproduce 0.125, and the disagreement is the honest part. On the stagnation streamline the search returns 0.1265, and it must land above the closed form: at the threshold itself the oscillation is arbitrarily slow, so a particle that theory says will strike takes arbitrarily long to do it, and any finite integration reports a threshold that is too high. Released a hundredth of a radius off the axis, the threshold rises further, to 0.19 — because an off-axis particle is swept aside before the oscillation can bring it in.

Three numbers, then: an exact threshold for the stagnation streamline, a slightly higher one for a finite integration on it, and a higher one still for anything released off it. All three are computed and all three are reported.

The two limits, drawn #

The extremes are worth seeing because they are what the middle is between, and because each is a different picture of the same equation.

At small Stokes number the particle paths and the streamlines are the same curves. That is not an approximation of the tracking; it is the tracking, run at a relaxation time so short that the slip velocity has nowhere to accumulate. The check on it is the one arrangement this site trusts most: the departure is measured with the streamfunction, a quantity the solver computes and the particle tracker never touches.

At large Stokes number the paths straighten into lines and the flow field might as well not be there. The collection efficiency then approaches one, and the body catches everything in its own frontal area — the case a reader’s intuition assumes and the flow only reaches when the particles are heavy or the body is small.

What a droplet has to be, in micrometres #

The threshold becomes a size once $\tau$ is written out.

Three consequences follow directly, and they are the reason the calculation is done in earnest.

Icing. A wing in cloud collects droplets above a few micrometres and misses the rest, so the ice that forms is fed by a fraction of the water in the air — and the fraction depends on the leading edge radius. It is also why the ice forms where it does: the collection is heaviest near the stagnation streamline, exactly where the flow is brought to rest, and thins away towards the shoulders. A thin, sharp wing has a small $a$, therefore a large Stokes number for the same droplet, therefore a higher collection efficiency: sharp wings ice up worse than blunt ones, which is the opposite of most intuitions about size.

Sampling. A probe drawn through a dusty flow measures the concentration of what it collects, and below its own threshold it collects nothing. Isokinetic sampling — matching the probe’s intake speed to the stream — exists precisely to keep the local Stokes number small so that the particles follow the air into the instrument.

Rain on a windscreen. A car at 30 m/s has a windscreen whose effective radius of curvature is of order a metre, so the flow time is about a thirtieth of a second. Rain drops of a millimetre have a relaxation time of seconds and are collected almost ballistically; fog droplets of ten micrometres have a relaxation time of a third of a millisecond and are swept round with the air. That is why a car in fog stays dry and a car in drizzle does not, and it is the same threshold as the wing’s, read at a different size.

Separation. A cyclone works by making the Stokes number large: spin the flow tightly, so $a$ is small and the turning is sharp, and the heavy particles fail to follow. Its cut-off size is this threshold read backwards, and a cyclone that must catch finer particles must be made smaller, which is why industrial units are banks of many small cyclones rather than one large one.

The same ratio, in three other places #

A Stokes number is a relaxation time against a flow time, and the shape of that comparison recurs all over this collection with different things being relaxed.

A parcel of fluid catching up with a pressure change. The convective time against the acoustic time is the Mach number, and the incompressible limit is the statement that the pressure relaxes instantly — which is an assumption with the same structure as St → 0.

Momentum diffusing across a layer. The diffusion time against the convection time is the Reynolds number, and the boundary layer is the region where the first has not yet won.

A vortex settling into a strain field. The viscous time against the stretching time sets the core radius of a stretched vortex, which is the same balance again with vorticity as the thing being relaxed.

What makes the particle case unusually clean is that the two times are separately visible — one belongs to the droplet, the other to the flow, and neither is a property of the fluid alone. It is the reason the Stokes number can be varied in an experiment by changing the particles and leaving the flow alone, which is a luxury the Reynolds number does not offer.

What the picture cannot show #

The flow is inviscid, so every number here is a lower bound. A real cylinder has a boundary layer in which the fluid slows to zero at the wall, so the streamlines pass closer to the surface than the potential ones do, and particles that just miss in this model would strike in reality. Collection efficiency computed on a potential flow underestimates what a real body collects, and the correction grows as the Reynolds number falls.

The body is a cylinder. An aerofoil’s leading edge is not, and the practical calculation is done on the real geometry — but the structure of the answer is the same, with the strain rate at the stagnation point standing in for the eighth. It is the same reason the same shape behaves differently at different Reynolds numbers: what matters is a rate, not a picture.

Stokes drag is a small-particle law. It holds while the particle’s own Reynolds number $\rho|u - v|d/\mu$ is below about one; a 100 µm drop in a 50 m/s stream is far past that, and its drag is nearer the bluff-body curve than the linear law. Nothing in this model corrects for it, and the essay’s numbers should be read as the small-drop end of the problem.

Nothing here bounces, splashes or evaporates. A droplet that arrives may shatter, run back along the surface, or freeze; a dust grain may rebound. The trajectory calculation ends at the surface and what happens next is a different subject entirely.

And there is no gravity. Adding it introduces a second dimensionless group — the settling velocity against the flow speed — and the trajectories stop being scale-free.

Where the model stops #

The whole calculation is one-way: the flow moves the particles and the particles do not move the flow. That is right at low concentration and wrong in a dense spray, where the momentum the particles take out of the air is enough to change it, and where the calculation has to be done both ways at once.

The model is also steady. A particle in an unsteady flow — behind a rotor, in a gust, in turbulence — sees a spectrum of flow timescales rather than one, so a single Stokes number is not enough. The turbulent version of exactly this question, how a particle samples an eddy field, is a subject of its own with the same number at the centre of it.

And the threshold argument holds only where the flow near the stagnation point is a pure strain, which requires a smooth body. A sharp edge has no such expansion, and its collection is governed by the geometry rather than by an eigenvalue.

One more limit deserves naming because it is the one most likely to be forgotten. The particle is treated as a point with a drag law attached, so it has no size in the flow field: it does not displace fluid, it is not affected by the gradient of the flow across its own diameter, and it feels no force from the pressure field it sits in. Each of those adds a term to the particle equation, and each becomes significant as the particle approaches the scale of the body — at which point the object being tracked is a second body in the flow rather than something carried by it.

The number that is missing, and the case where it reverses everything #

Every trajectory above was integrated from one equation with one parameter in it, and the equation quietly assumes something about what the particle is made of. The relaxation time $\tau = \rho_p d^2/18\mu$ contains the particle’s density and not the fluid’s, which is legitimate only when the first is enormously larger than the second — a water drop in air, at a ratio near a thousand. A second dimensionless group has been set to zero, and outside that case it is not small.

Two terms come back when it is not. Added mass: accelerating a particle also accelerates the fluid that has to move out of its way, so the inertia to be overcome is not $\rho_p$ but $\rho_p + \tfrac12\rho_f$ times the volume. And the pressure-gradient term: the particle occupies a region where the surrounding fluid is itself accelerating, so it feels the force the displaced fluid would have felt, which is buoyancy generalised to an accelerating frame.

Both scale with $\rho_f$, so both vanish for a drop in air and both dominate for a bubble in water.

And in that limit the whole picture of this essay inverts. A heavy particle lags the flow, cannot make the turn, and is flung outwards — which is every trajectory drawn above, and is what makes a cyclone work. A light particle is pushed by the fluid’s own acceleration towards wherever the pressure is lowest, and in a rotating flow the pressure is lowest at the centre. So a bubble does not fail to follow a vortex: it is drawn into its core and stays there.

That is visible without apparatus. Bubbles in a stirred glass collect on the axis; the trailing vortex behind a propeller blade is made visible by the vapour that forms in its core, which is the same migration with the pressure low enough to boil the water; and a bubble column in a swirling flow concentrates rather than disperses.

The consequence for a mixture is that the two populations separate by density in the same flow, with the crossing at neutral buoyancy. Heavy particles gather where the strain is strong and the vorticity weak; light ones gather where the vorticity is strong. In a turbulent flow, where both regions are everywhere and constantly rearranging, that produces a spatial clustering far more pronounced than a uniform suspension would have — strongest, as the argument here would suggest, when the relaxation time matches the smallest eddies’ own turnover time.

So the Stokes number is the right single parameter for the problem this essay solves and only for that problem. What decides whether it is enough is the density ratio, and the honest statement of the threshold above is that it belongs to a body in a stream of things much heavier than the stream.

Who found it, and when #

Taylor set out the trajectory problem in 1940 for the practical case of an aircraft in cloud. Langmuir and Blodgett’s 1946 report is the classic treatment: they computed collection efficiencies for cylinders and spheres numerically, by hand, on the potential flow field, and their tables were what icing certification used for decades. The threshold behaviour — that below a critical inertia parameter nothing is collected at all — is theirs.

The stagnation-point argument that turns the threshold into the discriminant of a quadratic is a later tidying-up of the same result, and it is the version worth carrying, because it says what the threshold depends on: one number, the rate at which the flow decelerates on the way in.

Where the ladder goes next #

Two essays in this field have now been about numbers that decide what a flow does. What has not been questioned is whether the flow field they act in is being described correctly at all — and the oldest theory of how air pushes on a surface, which was wrong enough to make flight look impossible, turns out to be exactly right in a regime its author could not have imagined.