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What Is the Betz Limit?

Updated 21 September 20267 min readWind & Hydropower

No wind turbine can extract more than 16/27 — about 59.3% — of the power in the wind passing through it. The reason has nothing to do with friction, blade design or generator losses. It follows from a single awkward fact: air that has given up all its energy has stopped moving, and stopped air cannot leave the rotor to make room for what is behind it.

Key takeaways

  • The limit comes from momentum and mass conservation alone: no assumption about blades, materials or losses enters the derivation.
  • Extraction is described by one number, the axial induction factor — how much the rotor has slowed the air by the time it reaches the disc.
  • Power coefficient works out as Cp = 4a(1−a)², which peaks at a = 1/3, giving 16/27 ≈ 0.593.
  • At that optimum the wind leaves the rotor at one third of its original speed. Slowing it further chokes the flow and captures less, not more.
  • The limit applies to a rotor in open flow. A machine that ducts or channels the flow is not bound by it, which is why hydro turbines are not.
On this page
  1. The question Betz asked
  2. The actuator disc: a rotor with the details removed
  3. Where 16/27 comes from
  4. What the limit does not say
  5. Why real turbines land below it
  6. Where the limit does not apply at all

The question Betz asked

Put a machine in a river of air and take energy out of it. How much can you have?

The tempting answer is "all of it, minus losses". Albert Betz showed in 1919 that this is not merely difficult but impossible, and that the ceiling is a clean fraction depending on nothing about the machine.

The argument needs no aerodynamics — only conservation of mass, conservation of momentum, and the willingness to follow them.

The actuator disc: a rotor with the details removed

Betz replaced the turbine with an actuator disc: a permeable surface that extracts energy from the flow through it, with no blades, no thickness, no rotation and no friction. No real rotor can do better than this idealisation.

Air approaches at speed v₁, passes through the disc at speed v, and continues downstream at v₂. Because the disc has taken energy out, v₂ < v₁ — and because the same mass of air per second must pass every cross-section of the streamtube, slower air needs more room. The streamtube expands as it passes through.

The streamtube through an actuator discA streamtube of air flowing left to right through an actuator disc. Upstream the tube is narrow and the air is fastest. At the disc the flow has already slowed and the tube has widened. Downstream the wake is slowest and widest. Because the same mass of air must pass every cross-section each second, slower air occupies a larger area, so extracting energy forces the streamtube to expand.actuator discarea Av₁free streamv = v₁(1 − a)at the discv₂ = v₁(1 − 2a)far downstreamnarrowwidestMass flow is the same through every section: ρ × area × speed. Slow the air and the tube must widen to carry it,which is the constraint that produces the limit. No blade, material or loss appears anywhere in the argument.
The same mass of air passes every section each second. Slower air needs a wider tube, so the streamtube expands through the rotor — and that expansion is the whole story.

That expansion is where the limit hides: slowing the air further means widening the wake further, and a wake cannot widen without limit.

Where 16/27 comes from

One variable carries the whole derivation: the axial induction factor, a, defined as the fraction by which the wind has already been slowed by the time it reaches the disc.

  • Speed at the disc: v = v₁(1 − a)
  • Speed far downstream: v₂ = v₁(1 − 2a)

The second relation is the one that does the work. The disc sits halfway through the deceleration: half of the slowing happens upstream of it, half downstream.

Now follow the two conservation laws.

Mass flow through the disc of area A is ρAv = ρAv₁(1 − a).

Force on the disc is the rate of change of momentum: F = ṁ(v₁ − v₂) = ρAv₁(1 − a) · 2av₁.

Power extracted is that force times the speed at which it acts, which is the speed at the disc:

P = F·v = 2ρAv₁³ a(1 − a)²

Divide by the power available in the undisturbed wind through the same area, ½ρAv₁³, and the machine disappears entirely:

Cp = 4a(1 − a)²

Everything about the turbine has cancelled. What remains is a curve with a single peak. Differentiate, set to zero, and the maximum sits at a = 1/3, giving

Cp,max = 4 · (1/3) · (2/3)² = 16/27 ≈ 0.593

Power coefficient against axial induction factorChart of the power coefficient four a times one minus a squared, plotted against the axial induction factor from zero to one half. The curve rises from zero, reaches its maximum of sixteen twenty-sevenths, about 0.593, at an induction factor of one third, and falls back to 0.5 at an induction factor of one half where the wake has been brought to a standstill.Power coefficient, Cp = 4a(1 − a)²Axial induction factor, a00.150.300.450.6000.10.20.30.40.5a = 1/3, Cp = 16/27 ≈ 0.593the wind leaves at one third of its speed1/3a = 0.5: the wake has stopped,and capture has fallen to 0.5take too littleand most of the wind sails pastExact curve, not an illustration: 4a(1 − a)² plotted directly. Every property of theturbine cancelled out of the derivation.
The whole result in one curve. Extract too little and you leave energy in the wind; extract too much and the flow refuses to cooperate.

At that optimum the wind leaves at v₂ = v₁(1 − 2/3) = one third of its original speed. Taking more would mean slowing it further, and the curve says plainly what happens then: Cp falls.

The curve as numbers
Induction factor aSpeed at the discWake speedPower coefficient Cp
0100% of v1100% of v10 — nothing extracted
0.190%80%0.324
0.280%60%0.512
1/367%33%0.593 — the maximum
0.460%20%0.576
0.550%0%0.500 — the wake has stopped, and capture has fallen

Calculated directly from v = v1(1 − a), v2 = v1(1 − 2a) and Cp = 4a(1 − a)². Nothing about a turbine enters these figures; they follow from the two conservation laws alone.

The last row is the one worth staring at. Bringing the wake to a standstill — apparently extracting everything — delivers *less* power than the optimum, because the flow through the disc has fallen further than the extra energy per kilogram makes up for.

Velocity through the rotor at three loadingsThree cases of flow through a rotor. Lightly loaded, the wind barely slows and most of its energy passes through unused. At the Betz optimum, the wind is at two thirds of free stream at the disc and one third in the wake. Over-loaded, the wake velocity approaches zero and the flow begins to spill around the rotor instead of passing through it.Lightly loaded: a is smallthe wind barely slowsmost of the energy carries straight on past the rotorCp is low because little has been takenAt the optimum: a = 1/3two thirds of the speed remains at the disc,one third in the wakeenough flow still leaving to make room for what followsOver-loaded: a above 1/2the wake has nearly stopped, so the flow goes aroundthe momentum model breaks down here and real rotorsavoid the regime entirely
At the optimum, two thirds of the speed remains at the disc and one third in the wake. Push harder and the air starts going around the rotor rather than through it.

What the limit does not say

Four misreadings are worth heading off, because each one turns a clean result into nonsense.

  • It is not about blades. No blade count, aerofoil, material or rotational speed appears anywhere in the derivation. Why turbines settle on three blades is a separate engineering question, answered in why wind turbines have three blades.
  • It is not about friction or generator efficiency. The disc is frictionless and the derivation has no losses in it at all. Real losses come *after* this ceiling, not instead of it.
  • It is not a design target. A machine that hits a = 1/3 at one wind speed is off the optimum at every other, which is why modern rotors vary their speed to hold the induction factor near the sweet spot as the wind changes.
  • It does not cap a whole wind farm at 59.3%. It applies to the flow through one rotor's swept area; turbines downstream work in disturbed air, which is a wake problem, not a Betz problem.

Why real turbines land below it

A good modern rotor reaches a power coefficient somewhere below the Betz value, and the gap is made of ordinary engineering:

  • Blade drag. Aerofoils generate lift and drag together; drag is a torque tax at every radius.
  • Finite blades. The actuator disc is continuous. A real rotor is three slender blades sweeping a mostly empty circle, so the flow is never uniformly loaded.
  • Tip losses. Air escapes around each blade tip from the high-pressure side to the low, doing no useful work.
  • Wake rotation. A real rotor applies torque to the air, so the wake spins. That rotational energy is taken from the flow and not converted.
  • Drivetrain and generator. Everything downstream of the hub takes its share before electricity leaves the nacelle.
From the Betz limit down to a real power coefficientA descending step chart starting at the Betz limit of 0.593 and falling through successive engineering losses: wake rotation, finite blade number and tip losses, blade profile drag, and drivetrain and generator losses. What remains at the bottom is the power coefficient a real machine delivers. The proportions are illustrative; the ordering and the mechanism of each step are not.Power coefficient00.150.300.450.60Betz limit, 0.593 — momentum theory, no lossesidealdiscwakerotationfinite bladesand tip lossesbladedragdrivetrain andgeneratorwhat reaches the gridIllustrative proportions: the size of each step varies by machine and wind speed. What is not illustrative is the order —the ceiling comes from momentum theory, and everything below it is engineering that can in principle be improved.
Illustrative proportions. The ceiling is set by momentum theory; everything below it is engineering that can, in principle, be improved.

Where the limit does not apply at all

The derivation assumes the rotor sits in an open flow that is free to go around it. Remove that assumption and the ceiling goes with it.

A turbine in a duct, pipe or penstock has the flow delivered to it, with nowhere else to go. There is no streamtube expanding into open air and no requirement to leave room for what follows, which is why hydro turbines routinely convert far more than 59.3% of the energy available to them. The same is true of a tidal barrage, which works on head rather than on the kinetic energy of an open stream — a distinction that separates two whole technology families in tidal energy versus wave energy.

Where the Betz limit applies and where it does notTwo cases. On the left, a rotor in open flow: streamlines can divert around it, so loading the rotor too heavily pushes flow around rather than through, which is the constraint that produces the limit. On the right, a turbine inside a penstock or duct: the water has nowhere else to go, the full flow must pass through the machine however much energy is extracted, and the open-flow argument does not apply.Open flow: the limit appliesflow can go aroundLoad the rotor too heavily and the airtakes the easier route past it. That isthe constraint that produces 16/27.Ducted: it does notnowhere else to goA penstock delivers the whole flow to themachine. Nothing has to be left moving tomake room, so hydro turbines routinelyconvert far more than 59.3%.The limit is a statement about unbounded flow, not about turbines.
The limit assumes the flow can go around. Remove that escape route, as a penstock does, and the argument that produced 16/27 no longer applies.

Diffuser-augmented wind turbines sit in between and are the usual source of "we beat Betz" claims. A shroud does let a rotor produce more power than a bare rotor of the same diameter, by drawing in air from a wider upstream area. Measured against *that* area, the limit is intact. The claim is really about which area you put in the denominator.

The same caveat applies to a tidal stream turbine, which is a wind turbine in water: open flow, same momentum argument, same ceiling, in a fluid roughly eight hundred times denser.

One last thing the limit does not govern: what a machine does when there is far more energy available than it can use. Above its rated wind speed a turbine deliberately stops trying to extract the optimum and sheds the surplus instead — a control decision rather than an aerodynamic one, and the subject of why wind turbines stop in strong wind.

Frequently asked questions

Has any turbine ever exceeded the Betz limit?

Not in open flow. Claims usually come from measuring against the wrong area — for instance quoting power against the rotor's swept area while a duct or diffuser draws air from a larger upstream area. Measured against the area the flow actually comes from, the limit holds.

Is the Betz limit the reason turbines are only about 45% efficient?

It is the ceiling, not the explanation for the gap. Real machines sit below 16/27 because of blade drag, finite blade count, tip losses, wake rotation and drivetrain losses. Betz says where the roof is; engineering decides how close you get.

Does the limit apply to vertical-axis turbines?

Yes. The derivation says nothing about the geometry of the machine, only that it extracts energy from an unbounded flow passing through a defined area. Vertical-axis rotors are bound by the same ceiling and typically operate further below it.

What happens if a rotor tries to slow the wind more than the optimum?

Less power, not more. Beyond roughly a = 0.5 the simple momentum model breaks down entirely: the flow no longer passes cleanly through and instead spills around the rotor in a turbulent wake, a regime real machines avoid.

Why is the limit sometimes written as 59.3% and sometimes as 16/27?

They are the same number. 16/27 is the exact result of the derivation; 0.5926 is its decimal form, usually rounded to 59.3%.

Sources

Named organisations whose published material underpins this article. Where no link is given, the source is named rather than linked.

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Last reviewed 21 September 2026. How we research and review