The most a disc can take

Wind arrives at a turbine carrying kinetic energy at a certain rate. The blades take some of it. The obvious question is how much, and the obvious answer — as much as the design is good enough to take — is wrong in a way that is worth an essay, because the ceiling is not set by the design at all.

It is set by a contradiction between two things the machine needs at once. To take energy from the air, the air has to be slowed down. To take energy from a lot of air, the air has to keep coming. Those pull in opposite directions, and where they balance is a number: 16/27, or 59.26 per cent.

The machine with nothing in it #

The model is called an actuator disc, and its definition is a list of things it does not have.

It is a surface across the stream that changes the pressure and not the velocity. It has no thickness, no blades, no rotation, no viscosity, no wake structure and no mechanism. It does not convert anything into anything. The only property it has is that a pressure drop occurs across it, and the only thing that is assumed about the fluid is that mass, momentum and energy are conserved in the tube of streamlines passing through it.

That sounds like the crudest possible model of a wind turbine, and it is. It is also unbeatable, which is the point. A ceiling derived from a model that contains no device is a ceiling for every device: a three-bladed rotor, a vertical-axis machine, a ducted fan, a sail, a bank of piezoelectric reeds, and anything nobody has thought of yet. The only way past it would be to violate one of the three conservation laws.

The streamtube is the whole geometry. Mass has nowhere to go: the same air that passes the disc passed some smaller circle far upstream and will pass some larger one far downstream, and since the air is slower at each of those in turn, the areas are fixed. That is the tube in the figure above, and its widening is not decoration — a turbine slowing the air by a third is drawing from a circle 82 per cent of its own area and pushing a wake twice its own area, and some of the wind that would have hit it goes round instead.

The streamtube is the whole of the model, so it is worth seeing the same object drawn where nothing is extracting anything from it.

One parameter, and it is a fraction #

Everything about the disc is set by one number: how much it slows the air. Write the speed at the disc as a fraction of the free stream,

$$u_{\text{disc}} = U(1 - a)$$

and $a$ is the axial induction factor, running from zero — a disc that does nothing — upwards.

The far wake speed is not a second free parameter. Writing the thrust twice, once as the rate at which the streamtube loses momentum and once as the pressure drop times the area, and requiring the two to agree, forces the disc speed to be the average of the two far-field speeds:

$$u_{\text{disc}} = \tfrac{1}{2}\left(U + u_{\text{wake}}\right) \quad\Longrightarrow\quad u_{\text{wake}} = U(1 - 2a)$$

That single line is the only non-obvious statement in the whole model, and everything downstream of it is arithmetic. Half the slowing happens in front of the disc and half behind it, which is a fact about the pressure field rather than about the machine.

Why the velocity is continuous across the disc and the pressure is not is worth a sentence, because it is the one place where the model’s abstraction does real work. The disc is a surface of zero thickness. Mass crossing it has nowhere to accumulate, so the normal velocity has to be the same on both faces; energy is being taken out of it, so the quantity Bernoulli’s theorem conserves is not the same on both faces, and the entire difference appears as a step in the static pressure. Bernoulli’s equation holds separately in front of the disc and separately behind it, and does not hold across it — which is the case the theorem’s hypotheses were written to exclude, and one of the five rows in the site’s own table of where it may and may not be carried.

Upstream, the air slows and its pressure rises, from ambient far away to a maximum just in front of the disc. Across the disc the pressure falls by more than it rose. Downstream it recovers to ambient again while the air slows further, and the recovery is what does the second half of the slowing. So the wake is still adjusting a long way behind the machine, and a turbine placed in another turbine’s wake is meeting air that is both slower and still changing — which is most of what the layout of a wind farm is about.

The thrust and the power then follow:

$$T = 2\rho A U^2 a(1-a) \qquad P = 2\rho A U^3 a(1-a)^2$$

and dividing each by the corresponding property of the undisturbed stream through the same area gives two coefficients with no units and no scale in them:

$$C_T = 4a(1-a) \qquad C_P = 4a(1-a)^2$$

What was computed, and how it was checked #

None of the four expressions above is used to draw the figures. The solver computes the mass flow at the disc, the areas at the other two stations from that mass flow, the thrust from the momentum lost and the power from the kinetic energy lost, and then compares its answers with the closed forms.

Three things are asserted before any disc is drawn, and they fail differently.

The streamtube carries one mass flow. Evaluated at the upstream station, at the disc and in the far wake, the three come out equal to the last bit. A figure whose tube did not do that would be drawing three unrelated circles.

The power computed from the energy lost equals the thrust times the speed at the disc. Nothing in the arithmetic forces that. It is true only because the disc velocity is the mean of the far-field speeds, so asserting the equality asserts the model’s one interesting statement — on every figure, rather than once in a derivation nobody rereads.

The coefficients match their closed forms to ten decimal places, which catches a transcription error in either direction.

The maximum comes out at $a = 1/3$ and $C_P = 16/27 = 0.5925925926$, and the search that found it knew nothing about either number.

Why there has to be a maximum #

The algebra gives the answer and hides the reason. The reason is that the two things a turbine wants are the same thing measured in opposite directions, and it is clearest on an axis of what leaves rather than what is taken.

At the right-hand end the air leaves at the speed it arrived and the disc has taken nothing per kilogram. At the left-hand end the air leaves at rest and the disc has taken everything per kilogram — but air at rest does not leave, the streamtube has closed off, and the mass flow is zero. A perfect extractor extracts nothing, which is the sentence the folk version of this subject never reaches.

Between two zeros there is a maximum, and its position is not obvious in advance. The energy per kilogram goes as the difference of two squares and the kilograms go linearly, so the product is a cubic, and a cubic pinned at zero and one has its maximum at a third of the way along.

Past the optimum the audit still balances, and the power has fallen — which is the part of the result that is easiest to state and hardest to believe.

The other curve, which is what the tower has to carry #

The power coefficient is the one everybody quotes. The thrust coefficient is the one that decides how much steel is in the ground.

At the best induction factor $C_T = 8/9 = 0.889$, and it is still climbing: the thrust peaks at $a = 1/2$, where $C_T = 1$ and the power has already fallen to a quarter of its maximum. The two curves are not maximised at the same place and they are not close. Running a rotor harder than Betz’s point buys less power and more load, which is why a large turbine pitches its blades to shed lift in high winds rather than simply letting them take what is offered.

The size of that load is worth putting in units. A rotor of 100 metres diameter in a 12 m/s wind at its design point carries a thrust of about $\tfrac{1}{2}\rho A U^2 C_T$, which is roughly half a meganewton — fifty tonnes of steady horizontal push at the top of a tower a hundred metres tall. It is not a gust load and it does not go away; it is the reaction to the momentum the machine is taking out of the air, and it is present whenever the machine is working.

The same balance reappears in a shape this site has already drawn. A control volume gives the force on whatever is inside it without going near the thing, and the answer does not depend on where the box is drawn — an argument made in full for an aerofoil, on six boxes of different sizes, with the surface integral agreeing to ten decimal places. The disc is the same theorem with a simpler interior, and the fact that it works for a rotor whose blades nobody has specified is exactly the property that makes the limit universal.

Where the model stops, and it stops sharply #

At $a = 0.5$ the far wake is at rest. Past it the momentum theory says the air behind the disc is travelling backwards, which is not a slow wake but a different flow: the streamtube the whole derivation is written on has broken open and the assumption that it is closed has failed.

The solver refuses induction factors of a half and above rather than clamping them, and the figure’s curve stops there rather than continuing. That refusal is checked in the site’s gate, because the alternative is a smooth curve continuing into a region where the model contradicts itself — which is exactly the failure this site is built to prevent, and it looks like nothing at all on the page.

Real rotors do run past it. The state is called the turbulent-wake state, it happens in high winds with the blades heavily loaded, and what it does is measured, not derived: the empirical thrust curve there is a fit, and momentum theory has nothing to say about it. Everything on the right-hand side of the figures in this essay is a derivation. Nothing past $a = 0.5$ would be.

Three other things the disc does not contain, each of which costs a real machine something:

• The wake spins. A rotor that extracts power applies a torque, and the reaction leaves angular momentum in the wake that the axial theory has no place to put. That is the whole of the next rung but one, and it is worth one to two per cent of the available power at the tip-speed ratios turbines actually run at.
• Blades are finite. The tips shed vorticity for the same reason a wing with ends does, and the loss has the same character.
• Blades have drag. The disc has no viscosity in it at all, so the friction on a real blade is entirely outside the model.

Every one of those subtracts. None of them can be arranged to add.

The hypothesis the limit is quietest about #

The streamtube expands, and it expands into an unbounded stream. That hypothesis is used at every step and stated at none, and taking it away raises the ceiling.

Put the same disc in a channel — a duct, a river, a tidal strait — and the tube can no longer widen freely. The flow it displaces has to go somewhere, and the only place available is past the machine through a fixed remaining area, so the bypass flow accelerates. That raises the pressure difference across the disc and the power with it, and the limit becomes $16/27$ divided by the square of the fraction of the channel left open. A machine filling a tenth of the section gains about a quarter; one filling half of it more than doubles the ceiling.

So a tidal fence across a narrow strait can exceed Betz’s number, and it does so without violating anything: the argument that produced $16/27$ assumed the wind could go round, and in a channel it cannot.

And for a whole channel the right calculation is a different calculation entirely. A tidal strait is not driven by the kinetic energy passing through it; it is driven by the head difference between the seas at its two ends, and the flow through it is whatever that head can force against the channel’s resistance. Adding turbines adds resistance, which reduces the flow — so the naive estimate of multiplying the undisturbed kinetic flux by a power coefficient is a calculation about a stream that stops existing once the machines are installed.

Done properly, the maximum extractable power from a tidal channel comes out proportional to the driving head amplitude times the peak undisturbed flow, with a coefficient near a fifth, and it depends scarcely at all on how many turbines are used or how good they are — a fence of poor machines and a fence of excellent ones reach nearly the same ceiling, because both are limited by the same head.

Which is Betz’s own lesson applied one level up: the ceiling belongs to the flow rather than to the machine, and choosing the right flow to draw the box round is the whole of the work.

The number as a piece of arithmetic about the world #

16/27 is 59.26 per cent, and it is worth converting into the units an argument about energy policy is conducted in.

A rotor of diameter $D$ in wind of speed $U$ has an available power of $\frac{1}{2}\rho \frac{\pi D^2}{4} U^3$, of which at most 59.26 per cent is collectable. The cube is the striking part: doubling the wind speed multiplies the available power by eight, so a site’s average wind speed matters far more than anything about the machine, and an average taken over a distribution of speeds is not the same as the speed of the average — a site with gusty wind is worth more than a steady one of the same mean.

Modern large rotors reach power coefficients of 0.45 to 0.50 in their design band. Against a ceiling of 0.5926 that is 76 to 84 per cent of what is physically available, which is a very different statement from “they only get half of it” and is the sentence this essay’s refutation exists to replace.

Who found it, and when #

Albert Betz published the result in 1920, at Göttingen, in the same institute and the same decade that produced the boundary layer and the lifting-line theory. Frederick Lanchester had reached it independently in England in 1915, and Nikolai Zhukovsky in Russia in 1920; the tradition of calling it Betz’s limit is a convention rather than a judgement, and “the Betz–Joukowsky limit” is the more careful name.

What is worth noticing is the date. It precedes any wind turbine worth the name by half a century. The limit was found by people thinking about propellers — the actuator disc had been Rankine’s and Froude’s since the 1860s, built to answer a question about ships — and the entire argument is indifferent to which way the energy is flowing. That is the next rung: the same disc, run backwards, gives the efficiency of every propulsor ever built.

Where the ladder goes #

Three rungs, and the middle one is the surprise.

Rung two runs the disc the other way. Instead of taking energy out it puts energy in, and the same momentum balance produces Froude’s propulsive efficiency — a number that decides the shape of every aircraft engine and that also contains no engine.

Rung three puts back the angular momentum. A disc cannot apply a torque, and a rotor must, so Glauert’s optimum sits below Betz at every finite tip-speed ratio. That gap is computed here and it never quite closes.

Both are still control-volume arguments, and both share the property that makes this whole field different from the eight before it: the box does not need to know what is inside it. That is why a model with no machine in it can set a limit on every machine, and it is why these results are still exact in a subject where the equations cannot be solved and the flow will not stay laminar.