The third thickness

The boundary layer has no edge. It approaches the free stream and never arrives, so any thickness quoted for it is a convention — and the two conventions in general use are integrals of the velocity deficit against two different weights.

The displacement thickness weights the deficit by how much fluid is missing, and says how far the outer flow is pushed aside. The momentum thickness weights it by how much momentum is missing, and is the drag. Both are in the essay that established them, and between them they carry nearly all of boundary-layer practice.

There is a third, and it answers a different question.

The definition, and what it is for #

$$\delta_3 = \int_0^\infty \frac{u}{U}\left(1 - \frac{u^2}{U^2}\right)dy.$$

It is the thickness of free stream whose kinetic energy flux equals the kinetic energy flux the layer is missing. Where the momentum thickness converts to a force, the energy thickness converts to a power: $\tfrac12\rho U^3\delta_3$ is the rate at which the layer has destroyed mechanical energy up to that station.

That last statement is the von Kármán energy integral equation, and it is the reason this thickness exists. It is the exact analogue of the momentum integral, derived by multiplying the boundary-layer momentum equation by $u$ instead of by one and integrating across.

An identity that checks the solve #

This collection likes a number that two unrelated calculations both produce, and the energy thickness brings one that nothing in the code arranges.

The dissipation integral of a similarity profile, $\int f''(\eta)^2 d\eta$, is a squared second derivative integrated across the layer. The energy thickness is a first derivative integrated against a cubic. There is no reason for those to be related — and for a flat plate,

$$\int f''^2\,d\eta = \frac{\delta_3}{4},$$

which holds here to a part in ten million on this site’s own Blasius solve. It holds because the energy integral equation is a consequence of the momentum equation the profile satisfies. A profile that is not a solution parts them.

That claim is worth testing rather than asserting, and there is a natural candidate: the Pohlhausen quartic, which satisfies the boundary conditions at the wall and at the edge and is not a solution of anything. It gets the energy shape factor right to a tenth of a per cent — 1.5712 against 1.5726 — and misses the energy identity by five and a half per cent. A profile can be a very good approximation to the shape and a poor one to the energetics, and only the second test notices.

The constant that was nearly wrong twice #

The identity above carries a factor of a quarter, and getting that quarter right took two attempts, both of which are worth recording because both are the kind of error that makes a correct solver look broken.

The first version applied the flat-plate quarter to the whole Falkner–Skan family, where the outer flow accelerates as $x^m$. The energy integral equation acquires a term in the pressure gradient there, and the constant becomes $(5m+1)/4$ — so at a stagnation point the check failed by a factor of six.

The second version fixed the exponent and used the wrong normalisation. This site’s Blasius solver is written $f''' + \tfrac12 ff'' = 0$ and its Falkner–Skan solver $f''' + ff'' + \beta(1-f'^2) = 0$, which differ by a factor $\sqrt2$ in the similarity variable — a fact the solver’s own docstring records — and the constant is $\delta_3/4$ in one and $\delta_3(5m+1)/2(m+1)$ in the other. At the same physical layer, a quarter and a half.

A constant that depends on a change of variable is exactly the kind that gets carried out of one book into another problem. Both forms are now checked, across the whole family.

The number: 78.6 per cent #

Now the result the third thickness exists for.

Tow a flat plate of length $x$ through still fluid. The power required is the drag times the speed, and the drag is $\rho U^2\theta$. The heat made inside the layer is $\tfrac12\rho U^3\delta_3$. So

$$\frac{\text{dissipated}}{\text{delivered}} = \frac{\delta_3}{2\theta} = \frac{H_{32}}{2},$$

and for a Blasius layer $H_{32} = 1.5726$, so the share is 0.7863.

Twenty-one per cent, on the simplest body there is, at the lowest possible drag, in a laminar flow with no separation and no pressure gradient. That is the best case for locality, and a fifth of the bill is somewhere else.

What the share does across the family #

The obvious next question is how much that fraction moves, and the answer is the one genuinely surprising result in this essay: almost not at all.

Run the Falkner–Skan family from a strongly favourable gradient to separation. The ordinary shape factor $H_{12}$ — the one every integral boundary-layer method is built on, the one that tells a designer separation is coming — goes from 2.22 to 3.99, nearly doubling. Over the same range $H_{32}$ goes from 1.626 to 1.515, a change of seven per cent.

So the split between heat and wake is very nearly a constant of the boundary layer, whatever the layer is doing. A layer about to separate is in most respects a completely different object from one in a favourable gradient — different profile, different stability, different wall slope — and it banks its energy in almost the same proportions.

The reason is an identity worth having. Since $1 - (u/U)^2 = (1 - u/U)(1 + u/U)$, the energy thickness is the momentum thickness’s integrand weighted by one more factor, so

$$H_{32} = 1 + \left\langle \frac{u}{U} \right\rangle,$$

where the average is taken over the layer with the local momentum deficit $u(U-u)$ as its weight. So $H_{32}$ is one plus a mean velocity, and it therefore lies strictly between 1 and 2 whatever the profile does. A linear profile gives exactly $3/2$; Blasius gives 1.573; a profile at the point of separation gives 1.515. The whole family lives in the narrow band those bounds leave.

The dissipation coefficient, and a number worth carrying #

There is a second quantity the energy equation produces and it is the one an engineer actually uses.

Define the dissipation coefficient $c_D = D/\rho U^3$, with $D$ the dissipation per unit area of plate — the exact analogue of the skin-friction coefficient $c_f = \tau_w/\tfrac12\rho U^2$. For a Blasius layer

$$c_D = \frac{0.2611}{\sqrt{\mathrm{Re}_x}}, \qquad c_f = \frac{0.664}{\sqrt{\mathrm{Re}_x}},$$

and the ratio of the two is 0.393, which is $H_{32}/4$.

Those two coefficients are worth putting side by side because they are so often conflated. $c_f$ is a stress at the wall: a local quantity, measurable with a floating element, and the thing a drag is integrated from. $c_D$ is an integral across the layer: not measurable at any point, not a wall quantity at all, and the thing a heat budget is integrated from.

They differ by a factor of two and a half on a flat plate, and the factor is not universal. In a strongly accelerated layer the wall stress rises and the dissipation coefficient rises less; near separation the wall stress goes to zero and the dissipation coefficient emphatically does not, because a separating profile is still being sheared vigorously away from the wall. At the point of separation the skin friction is zero and the dissipation is not, which is as clean a demonstration as there is that the two are different quantities.

What the turbulent layer does with the same accounting #

The identity above makes the turbulent case answerable without a closure, which is worth doing because the answer runs against the direction most readers expect.

Take the standard one-seventh power profile, $u/U = (y/\delta)^{1/7}$. Its three integrals are elementary:

$$\frac{\delta^*}{\delta} = \frac18, \qquad \frac{\theta}{\delta} = \frac{7}{72} = 0.0972, \qquad \frac{\delta_3}{\delta} = \frac78 - \frac{7}{10} = 0.175,$$

so $H_{12} = 1.29$ and $H_{32} = 1.80$. The energy shape factor has risen from Blasius’ 1.573 towards its ceiling of two, and the heat share $H_{32}/2$ is therefore 90 per cent against the laminar 78.6.

A turbulent boundary layer leaves ten per cent of the towing power in its wake where a laminar one leaves twenty-one. It costs several times as much in total — that is the whole of the price of going turbulent — and it is markedly more local about spending it.

The identity says why in one line. $H_{32} - 1$ is the mean speed of the fluid whose momentum is missing, weighted by how much is missing. A turbulent profile is full: it reaches nearly free-stream speed within a small fraction of its thickness, so the momentum deficit lives in fluid that is itself moving fast, and $\langle u/U\rangle$ comes out at 0.80 rather than 0.57. Fast-moving deficit means a large energy thickness for a given momentum thickness, and the energy thickness is precisely the part that has already been destroyed.

Which is a slightly uncomfortable pairing for anyone drawing conclusions about efficiency. The layer that is cheaper to carry is the one that hands more of its bill to the wake, and the layer that is expensive settles up on the spot. Neither fact is available from the drag, because the drag is the momentum thickness and says nothing about where in the profile the deficit sits.

Two cautions on the number. The one-seventh profile is a fit rather than a solution — it has infinite wall slope and does not satisfy anything at the edge — so 1.80 is an indication and not a computed result, which is why this collection quotes the turbulent value with a range around it rather than to four figures. And the whole calculation still assumes the profile is a function of $y/\delta$ alone, which a turbulent layer only approximately is.

There is a design reading of all this, and it is the one worth carrying off the page. A wing kept laminar over its forward chord saves a great deal of drag, and a part of what it saves it does not save at all — it defers, into a wake that will dissipate somewhere behind the aircraft. The saving is real, because the deferred part is still smaller in absolute terms than the turbulent layer’s much larger bill. But the two layers are not being compared on the same footing when only their skin friction is quoted, and the quantity that puts them on one is the third thickness.

Why integral methods use the third thickness #

The practical reason the energy thickness exists is that it makes a second equation available, and two equations allow a two-parameter profile family.

Momentum-integral methods carry one equation and one unknown — the momentum thickness — and have to guess the shape factor from a correlation. Energy-integral methods carry both equations, close them with a dissipation coefficient, and let the shape factor be a computed quantity rather than a fitted one. That is what every modern integral boundary-layer code does, and it is why they predict separation as well as they do.

The dissipation coefficient they need is exactly the integral checked above, and its being a computed property of the profile rather than an empirical input is the whole advantage.

The wake, and when the debt is paid #

The twenty-one per cent left in the wake does become heat, and the question of where is the same question the previous essay in this ladder measured on a cylinder: it depends entirely on how fast the wake mixes.

For a laminar wake behind a plate, the velocity deficit decays as $x^{-1/2}$ and the dissipation as $x^{-3/2}$, so the integral converges and most of the debt is paid within a few plate lengths. For a turbulent wake it is faster in absolute terms and slower relative to the plate, because the plate is longer.

What does not happen in either case is that it disappears. The momentum deficit is conserved down a wake — that is what makes a wake survey a drag measurement — and the energy deficit is not, because energy is exactly the thing being destroyed. Measuring both at two stations and differencing gives the dissipation between them, which is one of the few direct experimental accesses to the quantity this whole phase is about.

What a wake survey actually measures #

The practical use of all three thicknesses at once is the wake survey, and it is worth setting out because it is where the distinction between them stops being a matter of definition.

Traverse a probe across the wake behind a body and record the velocity profile. From it:

• the momentum thickness gives the drag, exactly, at any station, because momentum is conserved and the deficit does not change downstream;
• the displacement thickness gives the blockage, which is what a tunnel correction needs;
• and the energy thickness gives what has not yet been dissipated, which is different at every station and falls with distance.

So a survey at two stations, differenced, measures the dissipation between them. That is one of the very few direct experimental accesses to the quantity this whole phase is about, and it requires no calorimetry, no thermometry and no assumption about where the heat went.

It also gives the cleanest possible demonstration that a drag and a dissipation are different things: the first column of that traverse is the same at every station and the third is not.

What the picture cannot show #

The layer is laminar and similar. Everything here is a similarity solution. A transitional or turbulent layer has a different profile shape, a different $H_{32}$ — around 1.8 rather than 1.57, as the power-law estimate above gives — and a correspondingly different split. The estimate is a fit rather than a solution, and computing the profile properly needs a closure this collection does not have.

Compressibility is absent. At high speed the layer’s kinetic energy is comparable with its thermal energy, the wall heats itself, and the energy integral equation acquires terms this essay does not carry.

And the wake is not resolved. The claim that the remaining 21 per cent becomes heat downstream is an inference from conservation and not a measurement on any solution computed here.

One consequence for a flying thing #

The 21 per cent has a consequence for anything that flies close behind something else, and it is the reason formation flight and drafting work at all.

The energy in a wake is, by construction, energy that has been paid for and not yet destroyed. A body that flies in it is flying in fluid that is already moving in a useful direction, and the fraction of the leader’s expenditure that is theoretically available to a follower is exactly the fraction that has not yet become heat.

For a plate that is 21 per cent at the trailing edge, falling with distance as the wake mixes. For a wing the far larger contribution is the induced part of the drag, whose energy sits in a trailing vortex pair and persists for kilometres — where a wing’s lift reaction actually is is the momentum half of the same statement, and the energy half is why a bird in formation saves what it saves.

The general principle is worth stating in this collection’s terms: a wake is a debt rather than a loss, and its recoverability is measured by the energy thickness rather than by the momentum thickness. A body whose wake had already dissipated would offer a follower nothing at all, and one whose wake is entirely kinetic offers a follower everything.

Who found it, and when #

Leibenson introduced the energy integral equation in 1935 and Wieghardt gave the form that is used now in 1948. The two-equation integral methods that depend on it are Head’s, Green’s and Drela’s, in that order, and the last of those — the basis of the aerofoil analysis code most small aircraft have been designed with — carries $H_{32}$ as a primary variable rather than as a derived one.

The surprising connection is with something a long way from aerodynamics. The three thicknesses are the first three moments of the velocity deficit against successive powers of the velocity, which makes them the beginning of an infinite family — and the shape factors are ratios of moments, which is exactly the structure of the cumulants of a probability distribution. A boundary-layer profile is being summarised the way a distribution is: the first moment says how much, the second how spread, the third how skewed. That is why $H_{12}$ is so informative about a profile’s shape and $H_{32}$ so uninformative: they are the analogues of a variance and a fourth moment, and the higher moments of a constrained distribution are always nearly determined by the lower ones.

Where the ladder goes next #

Below this rung is how thick is thin, which is where the first two thicknesses come from, and where the heat of a drag is made, which is the general form of the accounting this essay closes for one geometry.

Beside it is the price of a gradient, which supplies the dissipation function, and above it is the turbulent version, which this collection cannot compute honestly and does not.