The layer that stops growing

Blasius’ boundary layer grows as $\sqrt{x}$ and never stops growing. A metre along a wing it is a couple of millimetres thick, ten metres along a fuselage it is a centimetre, and nothing in the solution suggests it would ever settle.

Make the wall porous and suck, gently and uniformly, and it settles at once — into a profile with no $x$ in it anywhere.

Why it stops #

The mechanism is a competition, and it is the same competition as the oscillating wall’s with a different clock.

Viscosity spreads momentum away from the wall. In Blasius’ problem nothing opposes that spreading, so the layer thickens for as long as the fluid stays beside the plate. Uniform suction gives the spreading something to work against: fluid is being drawn towards the wall at speed $V$, carrying the free stream’s momentum inwards at exactly the rate viscosity is carrying the wall’s deficit outwards.

Balance the two and the layer has nothing left to do. The momentum equation for a parallel layer with a wall-normal velocity $-V$ is

$$-V\frac{\mathrm{d}u}{\mathrm{d}y} = \nu\frac{\mathrm{d}^2u}{\mathrm{d}y^2}$$

which is a linear ordinary differential equation with an exponential solution:

$$u(y) = U\left(1 - e^{-Vy/\nu}\right)$$

There is no $x$ because there is nothing left for $x$ to do. A layer that is not growing cannot know how far along the wall it is.

What the solver computed, and how it was checked #

Four things, and the fourth is the one that carries the essay’s argument.

The momentum equation, differenced off the profile. Five-point stencils on the returned $u(y)$, at forty heights: worst relative residual $7\times10^{-10}$. A profile with the decay length 20 per cent wrong is still a smooth exponential rising to the free stream, and its residual is a hundred million times worse.

The thicknesses, by quadrature. $\delta^* = \int(1 - u/U)\,\mathrm{d}y$ integrates to $\nu/V$ and $\theta = \int (u/U)(1-u/U)\,\mathrm{d}y$ to $\nu/2V$, each to a part in a million of the analytic value. The shape factor $H = \delta^*/\theta$ is therefore exactly $2$ — at every suction rate, every speed and every viscosity.

Blasius at the same station, for scale. At $U = 30$ m/s and $x = 2$ m the Blasius displacement thickness would be 1.72 mm and still climbing; the suction layer sits at 0.50 mm forever.

The friction, twice. Once from the differenced wall slope, once from the identity below. They agree to seven figures, and the identity is the interesting half.

The friction is the suction, exactly #

The momentum-integral equation for a boundary layer with fluid crossing the wall is

$$\frac{c_f}{2} = \frac{\mathrm{d}\theta}{\mathrm{d}x} + \frac{V}{U}$$

Every boundary layer obeys it. Here $\theta$ does not depend on $x$, so the first term is zero and

$$\boxed{c_f = \frac{2V}{U} = 2c_q}$$

with nothing left over. The drag on a uniformly sucked wall is precisely the momentum of the fluid it swallowed, and it does not matter what the fluid is, how fast it is going, or how viscous it is.

That is a bookkeeping identity rather than a mechanism, and it has a blunt consequence: suction cannot reduce skin friction below its own price. Sucking harder makes the layer thinner, which makes the velocity gradient at the wall steeper, which makes the friction higher — in exact proportion to the suction. There is no free lunch in this direction and the algebra says so in one line.

So what is it for #

The payoff is not against a laminar layer. It is against a turbulent one.

An unsucked laminar layer has a friction coefficient of $0.664/\sqrt{\mathrm{Re}_x}$, which at $\mathrm{Re}_x = 4\times10^6$ is $0.00033$ — six times less than the sucked layer costs. If the laminar flow could be kept, suction would be a bad bargain.

It cannot be kept. Somewhere around $\mathrm{Re}_x$ of a few hundred thousand to a few million a laminar layer transitions, and the turbulent friction that follows is $0.0592/\mathrm{Re}_x^{1/5}$ — 0.0028 at the same station, and the ratio only worsens with Reynolds number. Against that, 0.0020 for the sucked layer is a real saving, and the reason suction wins is not that it removes slow fluid but that it changes which regime the wall is in.

There are two reasons it does so.

A thin layer is a stable layer. The stability of a boundary layer is governed by a Reynolds number built on its own displacement thickness, and holding $\delta^*$ at $\nu/V$ holds that number fixed however far along the wall the flow goes. An unsucked layer’s stability Reynolds number climbs as $\sqrt{x}$ and eventually passes any threshold; a sucked one never moves.

The profile itself is fuller. The suction profile has no inflection point and a steeper gradient at the wall than Blasius’, and both make it harder to destabilise — a connection this site’s turbulence field makes precisely in Rayleigh’s inflection criterion. The critical Reynolds number for the asymptotic suction profile is quoted in the literature at about 47,000 against Blasius’ 520 — a borrowed pair of numbers, from stability calculations this site does not perform, and drawn here only as a statement of scale.

How hard the sucking has to be #

The suction coefficient $c_q = V/U$ is a strikingly small number, and the arithmetic is worth doing because it is the reason anybody ever tried this.

At $U = 30$ m/s a suction coefficient of $10^{-3}$ means drawing air through the surface at 30 millimetres a second — a drift, not a wind. It holds the displacement thickness at $\nu/V = 0.5$ mm. Halving it to $5\times10^{-4}$ doubles the layer to 1 mm and halves the friction penalty to $c_f = 10^{-3}$; doubling it to $2\times10^{-3}$ gives a quarter-millimetre layer at $c_f = 4\times10^{-3}$, which is already worse than leaving the flow turbulent.

So there is an optimum, and it is not subtle: the friction cost rises linearly with the suction while the benefit — staying laminar — is a threshold that is either met or not. The right suction rate is the smallest one that keeps the layer stable, and every millimetre-per-second beyond it is wasted drag.

Blowing, where there is no answer at all #

The solver refuses a negative $V$, and the refusal is not a guard against a typing error.

Reverse the sign — blow through the wall rather than sucking — and the exponential in the solution becomes a growing one. There is no steady layer to compute: the wall is now feeding low-momentum fluid into the flow, the layer thickens without limit downstream, and the profile that would satisfy the equation runs off to infinity. The model does not merely become inaccurate, it stops having a solution of the assumed form.

That is a fair description of what blowing does in practice, too. It thickens the layer, reduces the wall shear towards zero, and makes separation more likely rather than less — which is why blowing is used where those are the objectives: film cooling of a turbine blade, where a blanket of cool slow air next to the metal is the whole point, and transpiration cooling of a re-entry surface.

The one case where blowing helps aerodynamically is quite different and is not this model: blowing a high-speed jet tangentially along a surface adds momentum to the layer rather than mass to it, and that can delay separation very effectively. The distinction is whether what comes through the wall is slower or faster than the flow it joins, and the exact solution above covers neither case — it covers the one where fluid is quietly removed.

What the suction costs that the identity does not show #

The friction identity accounts for the momentum of the swallowed air and not for the work of swallowing it.

The sucked fluid has to be drawn through the surface against a pressure difference and then either dumped or returned to the stream, and the pump doing it consumes power. That power is a real drag penalty, it does not appear anywhere in $c_f = 2c_q$, and it is what has kept laminar flow control out of service aircraft rather than any doubt about the aerodynamics.

The honest accounting is therefore a system accounting: friction saved, pump power spent, weight of the ducting carried, and the reliability of a surface with millions of holes in it. Nothing in this essay’s solution has anything to say about any of those, and it is worth being explicit that the exact result covers the easy half of the problem.

The same identity, without the suction #

The relation $c_f/2 = \mathrm{d}\theta/\mathrm{d}x + V/U$ is worth keeping after the suction is switched off, because with $V = 0$ it says something about every boundary layer on this site.

$$\frac{c_f}{2} = \frac{\mathrm{d}\theta}{\mathrm{d}x}$$

The friction on a plate is the rate at which its momentum thickness grows, and nothing else. That is why the momentum thickness rather than the height of the layer is the quantity worth measuring: it is the drag, expressed as a length. A wake survey far behind a body measures the same thing — the momentum deficit is the drag — and the two statements are the same integral evaluated in two places.

Read that way, the suction result is almost obvious in hindsight. Holding $\theta$ constant means the friction can only be whatever crosses the wall, because there is nowhere else for the momentum to go. The exact solution is a demonstration that a layer can be held constant, and the identity does the rest.

What the picture cannot show #

Real suction is not uniform. It is applied through discrete slots or through a perforated skin, and near each hole the flow is not parallel — which is exactly the assumption that made this solvable. The asymptotic layer is what the flow settles to a long way downstream of the start of the suction, and “a long way” is itself a length the solution cannot supply.

The approach is missing. A real wall does not start with the asymptotic profile; it starts with whatever was there and approaches this solution exponentially. The distance that takes is a few hundred $\nu/V$, which is small but not zero, and the figures here draw only the settled state.

Nothing here is turbulent. The whole solution is laminar by construction. If the layer does trip — because of surface contamination, a bug strike, a step in the skin — the suction rate needed to recover it is far higher, and the exponential profile is not what a sucked turbulent layer looks like.

What the suction is actually fighting on a swept wing #

The essay’s model is two-dimensional, and the wing the arithmetic above was applied to is not. That matters more than the usual caveat about geometry, because sweeping a wing introduces an instability the flat-plate problem has no term for, and it is the one that decides whether laminar flow is available at all.

On a swept wing the pressure gradient is not aligned with the local flow, so the streamlines inside the boundary layer curve relative to those outside it. The layer therefore carries a crossflow component — a velocity perpendicular to the external streamline — and that component must vanish both at the wall, by no slip, and at the edge, by definition. A profile that is zero at both ends and non-zero in between has an inflection point by construction, which by Rayleigh’s criterion means it is inviscidly unstable and therefore violently so.

The result is the crossflow instability: a row of co-rotating vortices lying nearly along the flow, growing rapidly, and taking the layer turbulent within a few per cent of chord. It dominates above about twenty degrees of sweep, it is why a transport wing cannot be kept laminar by shaping alone however carefully the pressure distribution is designed, and it is not in this essay’s equations anywhere — a two-dimensional layer has no crossflow to be unstable.

Suction is effective against it, and effective in a specific place. The crossflow is generated where the pressure gradient is strongest and the streamline curvature largest, which is the leading-edge region; further aft the flow has straightened and, with a favourable gradient, the layer can stay laminar unaided. So the modern arrangement applies suction only over the first fifteen or twenty per cent of chord and lets shaping do the rest — hybrid laminar flow control, which removes most of the plumbing the X-21 carried and is what has actually been flight-tested, on a 757’s wing and on an A320’s fin.

There is a third mechanism the flat-plate model cannot see either, and it defeated more than one early attempt. The attachment line — the streamline that divides at the leading edge and runs along the span — is a boundary layer in its own right, and if it is turbulent where it meets the fuselage the turbulence runs outboard along it and contaminates the whole wing before any of the careful design downstream has a chance. The remedy is a small bump near the root that forces the attachment-line layer to restart laminar, and it is a device whose entire purpose is to interrupt a mechanism that does not exist in two dimensions.

Which is the honest bound on this essay’s exact solution. It computes what suction costs and what it buys against a two-dimensional instability, and the instability that actually matters on a swept wing is one it has no variables for.

Where the model stops #

The solution is exact for an infinite flat plate with uniform suction, constant free-stream speed, and no pressure gradient. Every one of those is doing work.

A pressure gradient changes the profile completely; the suction analogue of the Falkner–Skan family exists and is no longer a single exponential. A curved wall introduces the centrifugal instabilities the flat case has none of. And suction strong enough to matter at high Reynolds number begins to violate the boundary-layer approximation itself, because the wall-normal velocity is no longer small compared with everything else.

What survives all of it is the identity. $c_f = 2c_q$ came from the momentum integral and the statement that $\theta$ is constant, so it holds for any layer held at a fixed thickness by any means — which is the useful thing to carry away.

The arithmetic of a wing, in passing #

It is worth putting the result on an aircraft, because the numbers decide whether anybody would bother.

A transport wing of 4 m chord at 240 m/s has $\mathrm{Re}_c = 6\times10^7$. Left alone, its boundary layer is turbulent over almost the whole chord at a mean friction coefficient around $0.0025$, and friction is roughly half the drag of such an aircraft. Held laminar by suction at $c_q = 5\times10^{-4}$, the wall friction is $0.001$ — sixty per cent less — and the wing’s total drag falls by something like a quarter once the pump work is subtracted.

A quarter of the drag of a long-haul aeroplane is an enormous prize, and it has been within reach on paper since the 1930s. What has kept it there is everything the exact solution says nothing about: the surface must stay clean to a few thousandths of a millimetre, the suction must be distributed over the whole wing, and the ducting and pumps must weigh less than the fuel they save. The aerodynamics has never been the difficulty.

Who found it, and when #

Prandtl mentioned suction as a way of controlling a boundary layer in the 1904 paper that introduced the boundary layer itself, and demonstrated it on a cylinder: sucking on one side kept the flow attached and killed the separation on that side.

The asymptotic suction profile as an exact solution belongs to the 1930s and is associated with Griffith and Meredith’s work in Britain, alongside Schlichting’s in Germany; the stability calculations that gave it its enormous critical Reynolds number followed in the 1940s. The largest practical test was the American X-21 programme of the early 1960s — two converted bombers with slotted, sucked wings, which achieved laminar flow over most of the wing and were defeated by insects, rain and the maintenance burden of keeping several million slots clear.

Where the ladder goes next #

Two exact solutions have now been built out of the same trick: kill the nonlinear term, and what remains can be integrated. Both were about a wall. The other place this site can be exact is about the numbers themselves — which quantities a flow can possibly depend on, and how many of them there are — and that turns out to be a question about the rank of a matrix.