Where the heat of a drag is made

Tow a body through still fluid at a steady speed and the power required is the drag times the speed. Nothing is accumulating — the flow is steady, the body’s kinetic energy is not changing, the fluid’s is not either — so that power has to be going somewhere, and the only place it can go is heat.

The equality is exact and it is also nearly useless as stated, because it is an equality over all the fluid there is. The moment a region is drawn on the picture, it stops holding, and the amount by which it stops holding is the subject of this essay.

Three ways out of a box #

The previous essay established the balance for a fixed region V with surface S:

$$\int_V \Phi\,dV \;=\; \oint_S \mathbf{u}\cdot(\boldsymbol{\sigma}\cdot\mathbf{n})\,dS \;-\; \oint_S \rho\tfrac12|\mathbf{u}|^2\,(\mathbf{u}\cdot\mathbf{n})\,dS.$$

Read from the right, it says a joule can leave a region in three ways. It can be destroyed inside it, which is the left-hand side. It can be handed across the boundary by the stresses, which is the first surface integral. Or it can be carried out by fluid that is still moving, which is the second.

The last two are what make a body’s heat non-local, and they behave completely differently. The traction term is present even when nothing is flowing across the surface at all: a fluid can push and shear on a boundary it is not crossing. The kinetic-energy term needs flow through the surface, and is identically zero in a creeping flow — there is no inertia, so nothing is carrying anything.

The clean case, where the shortfall is a formula #

Stokes’ sphere is the case in which the arithmetic can be done all the way, because the field is in closed form and the geometry is a sphere.

Integrate Φ from the body’s surface out to a radius $R$ and the answer is $6\pi\mu a U^2 (1 - 3a/2R)$ — less than the drag power at every finite radius, converging to it only as $R \to \infty$, and converging as $1/R$, which is as slowly as anything converges.

Three halves of a body radius over $R$. At two radii, a quarter of the heat is being made further out than that; at ten, a seventh; at a hundred, one part in sixty-seven. A sphere a millimetre across settling in oil is warming a region a centimetre across, appreciably, and a region ten centimetres across, measurably.

The reason is the 1/r decay of a Stokeslet’s velocity field, which is the same fact that makes creeping flow past a cylinder have no solution at all and makes a suspension of particles a problem about long-range interaction rather than about neighbours. A creeping flow has no near field, in the sense that no finite neighbourhood contains most of what is happening.

What inertia does to it #

Add inertia and the picture changes in a direction that is easy to guess wrongly. The obvious expectation is that the disturbance becomes more local — a wake is a definite object with edges, where a Stokes flow’s disturbance goes on for ever.

The disturbance does become more local. The dissipation does not, and the reason is the third term. At high Reynolds number a great deal of the energy leaving the neighbourhood of the body leaves as motion: fluid that has been set going and has not yet been slowed down again. That fluid is going to dissipate what it carries, but it is going to do so a long way downstream.

The two curves at the ends of that figure are the whole argument. Drag is a force on a body and is measured where the body is. The heat that force stands for is not there.

The flat plate, where the split can be written down #

The cylinder is a grid solve and its numbers are indicative. A flat plate is a similarity solution and its numbers are exact, so the same statement can be made with a coefficient attached.

Tow a plate of length $x$ through still air. The power is the drag times the speed, and the drag is the momentum thickness times $\rho U^2$. The heat made inside the boundary layer is half the free-stream kinetic energy density times the energy thickness. The ratio of the two is $\delta_3/2\theta$, which for a Blasius layer is 0.786.

Twenty-one per cent, on the simplest body there is, at the lowest possible drag. The third thickness is what that fraction is made of, and it turns out to be almost independent of what the pressure gradient is doing — which is a stronger and stranger result than the number itself.

Why the wake is where it is, and not further #

The energy left in a wake does not stay there. It is carried downstream, it spreads, and the spreading is what destroys it: a wake widens, its velocity deficit falls, and the shear that is doing the destroying weakens. The rate at which it is destroyed therefore falls with distance, and the total is finite — as it must be, since the total is the drag power and that is known in advance.

What is not fixed in advance is the length over which the paying happens, and it is set by how fast the wake mixes. A jet keeps its momentum and collects its mass by exactly the same mechanism, and the same integral quantity argument applies: the momentum deficit is conserved down the wake and the energy deficit is not, because energy is the thing being destroyed.

How slowly the debt is actually paid #

“Set by how fast the wake mixes” is the right answer and it is not a number. The far wake is self-similar, so a number is available, and it is worse than the qualitative statement suggests.

Two quantities describe the wake at a station: the velocity deficit $u_s$ on the axis and the width $\delta$. One relation between them is exact — the momentum deficit is conserved, because momentum is conserved — which for an axisymmetric wake means $u_s\delta^2$ is a constant of the flow and does not change from one diameter downstream to a thousand. The second relation comes from the spreading, and for a turbulent wake it gives $\delta \propto x^{1/3}$, hence $u_s \propto x^{-2/3}$.

Now form the quantity this essay is about. The kinetic energy still carried by the wake goes as $u_s^2\delta^2$, and putting the two exponents in,

$$\text{energy still in the wake} \;\propto\; x^{-2/3}.$$

That is the debt outstanding, and a two-thirds power is an appalling rate of repayment. Reducing the residual by a factor of ten takes thirty times the distance; reducing it by a hundred takes a thousand times. A wake that still holds a tenth of the towing power ten diameters back holds a hundredth at ten thousand. For a plane wake the arithmetic is worse still — $\delta \propto x^{1/2}$, $u_s \propto x^{-1/2}$, and the energy falls only as $x^{-1/2}$.

The contrast with the momentum is the whole point. One integral over the wake is exactly constant for ever and the other decays as a small fractional power, so a survey taken at any station whatever recovers the drag, and no survey at any practical station recovers the heat. That is the arithmetic behind the control-volume figure above, and it is why the two measurements were separated in the first place.

It also puts a number on the useful side of the same fact. A wake still carrying a substantial fraction of its energy is a wake worth flying in, and the slow decay is why the benefit persists so far: a bird in formation, a cyclist in a bunch and a lorry being drafted are all taking back energy that has not yet been destroyed, at separations of many body lengths, precisely because $x^{-2/3}$ has not got anywhere by then.

And it is why the figures above stop where they do rather than being drawn further out. Extending the frame by a factor of ten would capture, at Reynolds number 100, rather less than half of what is missing from it — so a larger picture would be a more expensive way of making the same admission.

The one case where the box does contain everything #

There is a geometry in which the accounting is local, and it is worth naming because it is the exception that shows what the others are missing: a confined flow.

Drive fluid along a pipe and the region between two cross-sections has no fluid entering or leaving it except through those two faces, and at both of them the profile is the same. So the kinetic energy carried in equals the kinetic energy carried out, the third term cancels exactly, and the pressure work at the two ends is the whole of the supply. Everything put in between two stations is destroyed between those two stations. A pump’s power and a pipe’s heating are the same number, at the same place, with no delay and no debt.

That is why the pipe is where this subject’s arithmetic was first done and why the intuition it builds is misleading everywhere else. A pipe has no outside. A body in a stream is nothing but outside, and the two facts that make its accounting non-local — a traction that acts across a surface nothing is crossing, and a flux of energy carried by fluid that has been set moving — are both identically absent from the case everybody learns first.

The turbulent case, and an honest limit #

At the Reynolds numbers most things actually operate at, the wake is turbulent, the shear layers roll up, and none of the maps above are a picture of what happens. The statement that survives is the integral one — the drag power equals the total dissipation, over all of the fluid and all of time — because it follows from the equations rather than from any solution of them.

What does not survive is any claim about where. A turbulent wake dissipates in a cascade, at scales this site’s grid cannot resolve and does not pretend to, and the essay on where the energy goes is the honest account of what that means: the rate is set by the large scales, and the place is set by the small ones.

What the numbers mean for something built #

A model in a tunnel is heating the tunnel, not the balance. Drag balances measure a force. The thermal load of a wind tunnel’s working section is the drag power of everything in it, and it is deposited over the whole circuit rather than at the model — which is why a closed-return tunnel needs a cooler and an open one does not.

A drag measurement and a calorimetry measurement disagree by design. Two ways of measuring the same drag — a balance under the model, and the temperature rise of the fluid leaving a duct — will not agree unless the duct is long enough for the wake to have finished paying. How long that is is not set by the body’s size; it is set by how fast the wake mixes, which is a Reynolds-number question and, above transition, a turbulence question.

A settling particle warms a region far larger than itself. In a dense suspension the regions overlap long before the particles do, which is the physical content of the observation that a suspension’s viscosity departs from Einstein’s dilute value at a few per cent by volume rather than at a few tens.

And a towed body’s wake is a debt, not a loss. The energy in it is recoverable in principle and is recovered in practice by anything that flies in formation or swims in a school. The reason a wake is worth exploiting at all is that its energy has not yet been destroyed.

What the picture cannot show #

The grid solve is steady and two-dimensional. Above about Reynolds number 47 a real cylinder sheds, and this site’s stepper does not; the Re = 100 map is a steady solution of the two-dimensional equations and is not a photograph of anything. What it is good for is the shape of the dissipation field and the comparison across Reynolds number, both of which are qualitative claims that a shed wake would strengthen rather than reverse.

The frame is finite and the integral is not. Every share quoted from the grid is a share of what is inside the frame, not of the total. That is stated on the figure rather than corrected, because correcting it would require knowing what is outside, which is the thing being demonstrated.

And nothing here is a temperature. Dissipation is a rate of energy destruction per unit volume; turning it into a temperature rise needs a conduction problem and a boundary condition, and in an unbounded flow at speed the answer is a small fraction of a kelvin. Where it is not small is in a confined film, and that is a different geometry with a different answer.

One number, for scale #

It is worth putting an actual temperature on all of this, because the phrase “becomes heat” invites a picture that is wrong by several orders of magnitude.

A car at thirty metres a second has a drag of about four hundred newtons, so it is depositing twelve kilowatts into the air behind it — comparable with its own cabin heater and considerably more than its engine’s radiated heat. That energy is spread through a wake perhaps three metres across, and the air in it is passing through at thirty metres a second, so the mass flow is of order three hundred kilograms a second and the temperature rise is a fortieth of a kelvin.

Twelve kilowatts, and four hundredths of a degree. Nothing about the flow is warm, and everything about the accounting is enormous. That ratio is generic: mechanical power is easy to make large and temperature rises are hard, because the specific heat of air is a thousand joules per kilogram per kelvin and there is a great deal of air.

The places in this collection where the rise is not negligible are all places where the fluid cannot escape — a film in a bearing, where the same power density is confined to a gap of microns — or where the speed is high enough for the kinetic energy itself to be a temperature, which is the wall that heats itself and is a compressible problem rather than a viscous one.

Who found it, and when #

The equality of drag power and total dissipation is old enough to have no single owner; it is a corollary of the mechanical-energy equation and appears in Lamb. The measurement that matters here is newer and is usually attributed to the wind-tunnel practice of the 1920s: Betz’s wake survey, which extracted a drag from a downstream traverse and had to be careful about exactly this distinction — the momentum deficit gives the drag, the energy deficit does not, and confusing them gives an answer that is wrong by about a fifth.

The surprising connection is with the induced drag of a wing, which is an inviscid drag and which this argument nevertheless covers. A lifting wing in an ideal fluid leaves kinetic energy behind it in a vortex pair, and the power it takes to do so is exactly the induced drag times the speed. There is no dissipation anywhere in that flow — and yet the bookkeeping is identical, because the third term in the balance does not care whether the first is zero. The reaction to a wing’s lift is the momentum half of the same statement.

Where the ladder goes next #

Above this rung is the question of whether a flow can choose how much to dissipate — the cheapest shape a set of walls allows — which turns the bookkeeping into a prediction. Beside it is the third thickness, which is where the 78.6 per cent comes from and where it stops being 78.6.

Below it is the price of a gradient, which is where the function being integrated came from, and the Reynolds number, which is the axis every comparison here is made along.