A wind turbine site has its wind speed doubled by moving to a taller tower, everything else unchanged. The power available in the wind through the same swept area becomes

Sixteen times the earlier value

For an ideal actuator disc operating at the Betz optimum (axial induction factor 1/3), the air speed far downstream in the wake, relative to the undisturbed wind speed v∞, is

zero, since all the kinetic energy is removed

In the tip-speed ratio–power coefficient curve of a rotor, why does Cp fall when λ is raised far above the best value?

Blades work in the turbulent wake of the blades ahead of them and drag losses grow

Betz's law forbids a higher λ, since the power coefficient cannot exceed 16/27 beyond it

The generator cannot exceed its synchronous speed, which caps the rotor speed

The blades turn so slowly that wind passes between them unused, so the rotor extracts little

Which statement correctly contrasts the Savonius and Darrieus vertical-axis turbines?

Savonius is drag-driven and self-starting with low Cp; Darrieus is lift-driven with higher Cp but poor self-starting

Both are lift-driven and reach the same Cp, since both have a vertical axis and blades

Darrieus is drag-driven and Savonius is lift-driven, so Savonius has the higher Cp

Savonius is drag-driven and self-starting with low Cp; Darrieus is lift-driven with higher Cp but poor self-starting

Both need a yaw system to face the wind, as a horizontal-axis machine does

Above rated wind speed a pitch-controlled turbine holds its output at the rated power mainly by

turning the blades about their axes to reduce the angle of attack

raising the generator speed beyond synchronous speed so that more power is taken off

turning the blades about their axes to reduce the angle of attack

increasing the tower height during strong winds

raising the air density seen by the rotor

A good horizontal-axis blade is twisted and tapered along its length. The main reason is

the tip speed ωr and hence the flow angle vary with radius, and the shape holds the angle of attack near its best value along the span

it lowers the cut-in speed to zero so that the machine runs in any breeze, however light

it raises the rated power above the Betz limit by adding extra lift along the blade

it makes the blade easier to transport on a road and to lift onto the hub

What is the main advantage of variable-speed operation with a power converter over fixed-speed operation?

The rotor can stay near the best tip-speed ratio as the wind speed changes

No gearbox or generator is needed

The cube law no longer applies to the wind power

The rotor can stay near the best tip-speed ratio as the wind speed changes

Which of the following statements about the Weibull description of wind speed are correct?

With k = 2 it reduces to the Rayleigh distribution

A larger shape factor k means a narrower spread of speeds

The scale factor c has the units of speed

With k = 2 it reduces to the Rayleigh distribution

A larger shape factor k means a narrower spread of speeds

The mean power density equals ½ρ times the cube of the mean speed

The scale factor c has the units of speed

Which of the following statements about wind-turbine types are correct?

A vertical-axis machine accepts wind from any direction without a yaw mechanism

A Savonius rotor is driven mainly by drag

Modern three-blade horizontal-axis rotors operate at tip-speed ratios of roughly 6 to 8

A vertical-axis machine accepts wind from any direction without a yaw mechanism

A Savonius rotor is driven mainly by drag

A Darrieus rotor is generally self-starting at low wind speed

Modern three-blade horizontal-axis rotors operate at tip-speed ratios of roughly 6 to 8

Which of the following statements about the Betz limit are correct for an open rotor in unconfined flow?

No such turbine can extract more than 16/27 of the kinetic power of the wind passing through its swept area

At the optimum the thrust coefficient is 8/9

No such turbine can extract more than 16/27 of the kinetic power of the wind passing through its swept area

It comes from blade drag and wake rotation losses

At the optimum the thrust coefficient is 8/9

It is reached when the wake speed is zero

Wind Energy Conversion Systems: Wind Resource, Betz Limit, Turbine Types, Blade Design and Generators

This chapter belongs to the Engineering Sciences (XE) paper and sits in Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. It answers the heading “Wind energy conversion systems: Wind resource analysis, types and characteristics of wind turbines, Betz limit, wind turbine motor design considerations, blade profile, wind energy generators”. It teaches the power in the wind and power density, the Weibull and Rayleigh description of a site, scaling of speed with height, the momentum theory and the Betz limit, tip-speed ratio and the power coefficient, horizontal and vertical-axis machines and their power curves, blade profile and design choices, and the generators and drive trains. The numericals are power density, Weibull mean speed, height scaling, turbine output, rotor speed for a chosen tip-speed ratio and capacity factor; air density and every other constant is given in the question.

1. Wind resource analysis

Air of density ρ moving at speed v through an area A carries kinetic power P = ½ρAv³; per unit area this is the power density ½ρv³ (W/m²). With ρ = 1.225 kg/m³ and v = 8 m/s the power density is 0.5 × 1.225 × 512 = 313.6 W/m². The cube is the whole story of wind siting: doubling the speed gives eight times the power, and a 10 % error in speed is a 33 % error in power.

Wind speed at a site varies, so the site is described by a probability distribution. The Weibull form has probability density f(v) = (k/c)(v/c)^(k−1) exp[−(v/c)^k] and cumulative F(v) = 1 − exp[−(v/c)^k], with scale c (m/s) and shape k (dimensionless). The mean speed is v̄ = c Γ(1 + 1/k). The Rayleigh distribution is the special case k = 2, for which v̄ = c√π/2 = 0.886c, so c can be found from the mean alone. A larger k gives a narrower, more constant wind. The time the speed exceeds v is exp[−(v/c)^k]; with k = 2 and c = 9 m/s the fraction above 12 m/s is exp[−(12/9)²] = exp(−1.778) = 0.169.

⚠️ Mean speed cubed understates the power

Because P ∝ v³ and the wind varies, the mean power density is ½ρ⟨v³⟩, which is larger than ½ρv̄³. For a Weibull wind ⟨v³⟩ = c³Γ(1 + 3/k); for Rayleigh the ratio ⟨v³⟩/v̄³ is 6/π ≈ 1.91 (the energy pattern factor). Using the mean speed in the cube formula gives roughly half the true available power at such a site.

Height scaling. Speed rises with height because of surface friction. The power law is v₂/v₁ = (h₂/h₁)^α, with α ≈ 1/7 over open level ground (a value to be stated in a problem), larger over rough terrain. The logarithmic law is v₂/v₁ = ln(h₂/z₀)/ln(h₁/z₀) with z₀ the surface roughness length. Worked: 6 m/s at 10 m becomes 6 × 8^(1/7) = 8.08 m/s at 80 m, so the power density rises by 1.346³ = 2.44 times. Anemometer data are therefore always transferred to hub height before estimating energy.

The annual energy of a turbine is judged by its capacity factor = actual energy/(rated power × 8760 h). A 2 MW turbine delivering 5.256 GWh in a year has CF = 5.256 × 10⁶ kWh/(2000 kW × 8760 h) = 0.30.

2. Betz limit, power coefficient and tip-speed ratio

Treat the rotor as an ideal actuator disc in an unconfined stream tube: no drag, no wake rotation, uniform flow, incompressible air. Speed is v∞ far upstream, vd at the disc and vw in the far wake, and momentum theory gives vd = (v∞ + vw)/2. Define the axial induction factor a = 1 − vd/v∞, so vd = v∞(1 − a) and vw = v∞(1 − 2a). The power taken is P = ½ρA v∞³ · 4a(1 − a)², so the power coefficient is Cp = 4a(1 − a)². Setting dCp/da = 0 gives a = 1/3 and Cp,max = 16/27 = 0.593, the Betz limit. At that point the disc speed is 2/3 v∞, the wake speed is v∞/3, and the thrust coefficient CT = 4a(1 − a) is 8/9.

A real turbine output is P = ½ρA v³ · Cp · ηdrive, where Cp (actual, below 0.593) measures the aerodynamic extraction and ηdrive covers the gearbox and generator. Worked example. Rotor diameter 80 m, v = 10 m/s, ρ = 1.225 kg/m³, Cp = 0.42, ηdrive = 0.90. A = π × 40² = 5027 m². P = 0.5 × 1.225 × 5027 × 1000 × 0.42 × 0.90 = 1.164 × 10⁶ W, about 1164 kW. The Betz-limited aerodynamic power in the same wind would be 1825 kW.

The tip-speed ratio is λ = ωR/v = (blade-tip speed)/(wind speed). Cp depends on λ (and pitch angle) through a hump-shaped curve with one best λ. Too low a λ lets wind pass between slowly turning blades unused; too high a λ makes blades work in each other's turbulent wake and raises drag losses. For a stated λ = 7, R = 45 m and v = 10 m/s: ω = λv/R = 1.556 rad/s, i.e. N = 60ω/2π = 14.9 rpm. A large rotor must therefore turn slowly, which is why a gearbox (or a many-pole generator) is needed.

⚠️ Betz is a ceiling on the wind, not on the machine

The limit says that an open rotor cannot extract more than 59.3 % of the kinetic power of the wind flowing through its swept area; it does not say a good turbine reaches it. Real Cp values of good machines sit well below, as drag, wake rotation and finite blade count take their share. Extraction of 100 % would stop the wind behind the disc, which would block the flow.

3. Types and characteristics of wind turbines

Common wind-turbine types
Type	Principle	Features
Horizontal-axis (HAWT), 2 or 3 blades	lift on aerofoil blades	highest Cp; needs a yaw system to face the wind; high λ of about 6 to 8 for modern three-blade rotors
Multi-blade (water-pumping type)	lift and drag, high solidity	high starting torque, low λ near 1; low Cp
Savonius (vertical-axis)	drag-driven cups	accepts wind from any direction, self-starting, low λ below 1, low Cp
Darrieus (vertical-axis)	lift on curved blades	no yaw needed; generally not self-starting; good Cp at moderate λ; cyclic loads

A machine is characterised by its power curve, power against hub-height wind speed. Below the cut-in speed nothing is delivered; between cut-in and rated speed the output follows Cp·½ρAv³ and rises as v³; above rated speed the output is held at the rated power by pitching blades or by stall; above the cut-out speed the rotor is stopped for safety. Note that rated speed is where the rated power is first reached, and for the best use of the machine it is chosen well above the mean speed of the site.

Pitch control turns each blade about its axis above rated speed to reduce the angle of attack and feather the blades; it holds power steady and gives a safe stop.
Stall control uses fixed blades whose aerofoil stalls progressively as the speed rises; it is simpler and cheaper but gives less precise limiting.
Yaw control turns the whole nacelle into the wind on a horizontal-axis machine; a vertical-axis machine needs none.

4. Blade profile and design considerations

A blade is an aerofoil: relative flow at the blade is the vector sum of the wind speed v and the blade's own motion ωr, so the relative speed and flow angle change along the span. The lift-to-drag ratio of the aerofoil at its best angle of attack governs Cp; a high L/D is needed to reach a high Cp at a high λ. Because ωr grows with radius, the flow angle at the root differs from that at the tip, so a good blade is twisted (a larger pitch angle near the root) and tapered (a wide chord near the root, narrow near the tip) to hold the angle of attack near the best value along the whole span. Solidity, the blade area fraction of the swept disc, is low for fast rotors with few blades and high for slow, high-torque ones.

Blade count: fewer blades give a higher optimum λ and a lighter rotor; three blades are the usual compromise for smooth loading and appearance.
Tip speed: it is limited by blade stress, erosion and noise, so a larger rotor turns at fewer rpm (N = 60λv/2πR).
Hub height and rotor diameter: power grows with D² and with the cube of the speed at hub height, which grows with the tower; hence the trend to tall towers and large rotors.
Loads and materials: blades are light composites designed for fatigue under cyclic gravity and gust loads; the tower, foundation and extreme-wind survival speed are design drivers.

5. Wind energy generators and drive train

The rotor turns slowly (10 to 20 rpm for a large machine) while a conventional four- or six-pole generator wants about 1000 to 1500 rpm, so a gearbox raises the speed. Synchronous speed is Ns = 120f/p; for p = 6 and f = 50 Hz, Ns = 1000 rpm, so a rotor at 12.5 rpm needs a gear ratio 1000/12.5 = 80. A many-pole direct-drive generator removes the gearbox at the price of a large, heavy machine.

Generators used in wind turbines
Generator	Speed	Features
Squirrel-cage induction	nearly fixed, just above synchronous speed	rugged and cheap; draws reactive power; fixed-speed operation cannot hold the optimum λ
Doubly fed induction (DFIG)	variable, with a converter on the rotor circuit sized for a fraction of the power	a limited speed range, so the rotor tracks the best λ; the grid is supplied at fixed frequency
Synchronous with permanent magnets (PMSG)	fully variable, full-power converter	can be geared or direct-drive; high efficiency; needs magnets and a full converter

🎯 Why variable speed pays

Cp peaks at one λ, so for the rotor to stay at that λ the speed ω must follow the wind: ω = λopt v/R. A fixed-speed machine sits at its best λ at only one wind speed. A variable-speed machine with a power converter tracks the peak over the range, capturing more energy and reducing gust loads on the drive train by letting the rotor speed up and down.

Key takeaways

Wind power P = ½ρAv³; power density ½ρv³. Speed enters as a cube, and mean power uses ⟨v³⟩, which exceeds v̄³ (Rayleigh factor 6/π).
Weibull: F = 1 − exp[−(v/c)^k], v̄ = cΓ(1 + 1/k); Rayleigh is k = 2 with v̄ = 0.886c. Height scaling: (h₂/h₁)^α with α ≈ 1/7 (stated), or the logarithmic law.
Betz: Cp = 4a(1 − a)², maximum 16/27 = 0.593 at a = 1/3 (disc speed 2/3 v∞, wake v∞/3, CT = 8/9). It is a ceiling for open rotors, not a target.
λ = ωR/v. Cp has one best λ; large rotors turn slowly; variable speed tracks the optimum. Pitch and stall control limit power above rated speed; HAWTs are lift-driven with a yaw system; Savonius is drag-driven, Darrieus lift-driven.
Blades are twisted and tapered aerofoils working near the best angle of attack. Generators: squirrel-cage (fixed speed), DFIG (partial converter), PMSG (full converter, gearbox optional); Ns = 120f/p.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A wind turbine site has its wind speed doubled by moving to a taller tower, everything else unchanged. The power available in the wind through the same swept area becomes
1. Two times the earlier value
2. Eight times the earlier value
3. Four times the earlier value
4. Sixteen times the earlier value
Show answer
Answer: B — Eight times the earlier value
P = ½ρAv³ varies as v³, so doubling v gives 2³ = 8 times. Four times would follow from a square law, which holds for the thrust force rather than the power.
Taking ρ = 1.225 kg/m³, the power density of a steady wind of 8 m/s, in W/m², is nearest to
1. 39
2. 196
3. 627
4. 314
Show answer
Answer: D — 314
½ρv³ = 0.5 × 1.225 × 512 = 313.6 W/m². The value 39 uses v² (0.5 × 1.225 × 64 = 39.2), and 627 forgets the factor one-half.
A Rayleigh wind-speed distribution is the Weibull distribution with
1. shape factor k = 2
2. shape factor k = 1
3. scale factor c = 1 m/s
4. shape factor k = 3
Show answer
Answer: A — shape factor k = 2
Rayleigh is the k = 2 member of the Weibull family, with v̄ = 0.886c. The k = 1 case is the exponential distribution, which is much broader.
At a site the wind follows a Weibull distribution with k = 2 and c = 9 m/s. The fraction of the time during which the speed exceeds 12 m/s is nearest to
1. 0.33
2. 0.83
3. 0.17
4. 0.06
Show answer
Answer: C — 0.17
The exceedance fraction is 1 − F = exp[−(12/9)²] = exp(−1.778) = 0.169. The value 0.83 is the cumulative F (the time below 12 m/s) and 0.33 is the ratio 12/9 minus 1 that has no meaning here.
For an ideal actuator disc operating at the Betz optimum (axial induction factor 1/3), the air speed far downstream in the wake, relative to the undisturbed wind speed v∞, is
1. v∞/2
2. 2v∞/3
3. zero, since all the kinetic energy is removed
4. v∞/3
Show answer
Answer: D — v∞/3
vw = v∞(1 − 2a) = v∞(1 − 2/3) = v∞/3. The speed 2v∞/3 is the speed at the disc, not in the far wake. A zero wake speed would mean no flow through the disc and no power.
In the tip-speed ratio–power coefficient curve of a rotor, why does Cp fall when λ is raised far above the best value?
1. Blades work in the turbulent wake of the blades ahead of them and drag losses grow
2. Betz's law forbids a higher λ, since the power coefficient cannot exceed 16/27 beyond it
3. The generator cannot exceed its synchronous speed, which caps the rotor speed
4. The blades turn so slowly that wind passes between them unused, so the rotor extracts little
Show answer
Answer: A — Blades work in the turbulent wake of the blades ahead of them and drag losses grow
At high λ the rotor is nearly a solid disc: each blade meets turbulent, slowed air from the previous one, and drag losses and tip losses rise. The first description applies to a λ that is too low, not too high.
Which statement correctly contrasts the Savonius and Darrieus vertical-axis turbines?
1. Both are lift-driven and reach the same Cp, since both have a vertical axis and blades
2. Darrieus is drag-driven and Savonius is lift-driven, so Savonius has the higher Cp
3. Savonius is drag-driven and self-starting with low Cp; Darrieus is lift-driven with higher Cp but poor self-starting
4. Both need a yaw system to face the wind, as a horizontal-axis machine does
Show answer
Answer: C — Savonius is drag-driven and self-starting with low Cp; Darrieus is lift-driven with higher Cp but poor self-starting
A drag-driven rotor has λ below 1 and is self-starting but inefficient, while a lift-driven Darrieus rotor works at higher λ with better Cp, but at low speed its torque is small. Neither needs a yaw system, since the axis is vertical.
Above rated wind speed a pitch-controlled turbine holds its output at the rated power mainly by
1. raising the generator speed beyond synchronous speed so that more power is taken off
2. turning the blades about their axes to reduce the angle of attack
3. increasing the tower height during strong winds
4. raising the air density seen by the rotor
Show answer
Answer: B — turning the blades about their axes to reduce the angle of attack
Pitching the blades towards feather reduces the angle of attack and the lift, and with it the torque. A faster generator would take more, not less, power, and neither tower height nor air density is a control variable.
A good horizontal-axis blade is twisted and tapered along its length. The main reason is
1. the tip speed ωr and hence the flow angle vary with radius, and the shape holds the angle of attack near its best value along the span
2. it lowers the cut-in speed to zero so that the machine runs in any breeze, however light
3. it raises the rated power above the Betz limit by adding extra lift along the blade
4. it makes the blade easier to transport on a road and to lift onto the hub
Show answer
Answer: A — the tip speed ωr and hence the flow angle vary with radius, and the shape holds the angle of attack near its best value along the span
Relative flow is the vector sum of v and ωr, so at each radius the flow angle differs. Twist and taper keep the angle of attack near the best L/D. No shape can pass the Betz limit.
What is the main advantage of variable-speed operation with a power converter over fixed-speed operation?
1. No gearbox or generator is needed
2. The Betz limit rises to unity
3. The cube law no longer applies to the wind power
4. The rotor can stay near the best tip-speed ratio as the wind speed changes
Show answer
Answer: D — The rotor can stay near the best tip-speed ratio as the wind speed changes
Since ω = λopt v/R, the rotor speed has to follow the wind for Cp to stay at its peak, and a converter lets the speed differ from the grid-fixed one. The Betz limit and the cube law are properties of the wind and are unaffected.
Which of the following statements about the Weibull description of wind speed are correct?
1. With k = 2 it reduces to the Rayleigh distribution
2. A larger shape factor k means a narrower spread of speeds
3. The mean power density equals ½ρ times the cube of the mean speed
4. The scale factor c has the units of speed
Show answer
Answer: A — With k = 2 it reduces to the Rayleigh distribution; B — A larger shape factor k means a narrower spread of speeds; D — The scale factor c has the units of speed
k = 2 is Rayleigh; a larger k concentrates the speeds; c is in m/s. The mean power density is ½ρ⟨v³⟩ and exceeds ½ρv̄³ for any wind that varies.
Which of the following statements about wind-turbine types are correct?
1. A vertical-axis machine accepts wind from any direction without a yaw mechanism
2. A Savonius rotor is driven mainly by drag
3. A Darrieus rotor is generally self-starting at low wind speed
4. Modern three-blade horizontal-axis rotors operate at tip-speed ratios of roughly 6 to 8
Show answer
Answer: A — A vertical-axis machine accepts wind from any direction without a yaw mechanism; B — A Savonius rotor is driven mainly by drag; D — Modern three-blade horizontal-axis rotors operate at tip-speed ratios of roughly 6 to 8
The statements on yaw, on the Savonius drag drive and on tip-speed ratios of 6 to 8 are standard properties. A Darrieus rotor generally produces little starting torque and usually needs help to start, so the self-starting statement is wrong.
Which of the following statements about the Betz limit are correct for an open rotor in unconfined flow?
1. No such turbine can extract more than 16/27 of the kinetic power of the wind passing through its swept area
2. It comes from blade drag and wake rotation losses
3. At the optimum the thrust coefficient is 8/9
4. It is reached when the wake speed is zero
Show answer
Answer: A — No such turbine can extract more than 16/27 of the kinetic power of the wind passing through its swept area; C — At the optimum the thrust coefficient is 8/9
The derivation assumes an ideal disc with no drag and no wake rotation, so those losses lower Cp further and are not its origin. At a = 1/3, CT = 4a(1 − a) = 8/9, and the wake speed is v∞/3, not zero.
Wind at a site follows a Rayleigh distribution (Weibull with k = 2) with scale factor c = 9 m/s. Using Γ(1.5) = 0.8862, the mean wind speed in m/s, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 7.98
v̄ = cΓ(1 + 1/k) = 9 × Γ(1.5) = 9 × 0.8862 = 7.976, i.e. 7.98 m/s. Verified with Python (7.9760).
The wind speed at 10 m height is 6 m/s. Using the power law v₂/v₁ = (h₂/h₁)^α with α = 1/7, the wind speed at a hub height of 80 m, in m/s and to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 8.08
v₂ = 6 × (80/10)^(1/7) = 6 × 8^0.142857 = 6 × 1.3459 = 8.075, i.e. 8.08 m/s. Checked in Python (8.0754).
A turbine of rotor diameter 80 m operates in a steady 10 m/s wind with air density 1.225 kg/m³, power coefficient 0.42 and combined gearbox and generator efficiency 0.90. The electrical power output in kW, rounded to the nearest whole number, is ____.
Numerical answer — type the value.
Show answer
Answer: 1164
A = π × 40² = 5026.5 m²; P = 0.5 × 1.225 × 5026.5 × 10³ × 0.42 × 0.90 = 1.1638 × 10⁶ W = 1164 kW. Verified in Python (1163.77).
A rotor of radius 45 m is to run at a tip-speed ratio of 7 in a 10 m/s wind. The rotor speed in rpm, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 14.9
Tip speed = λv = 70 m/s; ω = 70/45 = 1.5556 rad/s; N = 60ω/2π = 14.85, i.e. 14.9 rpm. Verified in Python (14.854).
A 2 MW (2000 kW) turbine delivers 5.256 GWh (5.256 × 10⁶ kWh) of electrical energy in a year of 8760 h. Its capacity factor, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.30
CF = 5.256 × 10⁶/(2000 × 8760) = 5.256 × 10⁶/1.752 × 10⁷ = 0.30.