Wind Turbine Energy Calculator

Introduction to rotor swept area, hub-height wind and annual yield

This calculator answers the question a prospective turbine owner actually cares about: how many kilowatt-hours will this machine deliver at my site in a year? That is a different and much harder question than "how many watts does this rotor make in a 7 m/s breeze". A single-speed power figure is an instantaneous quantity; annual energy production, or AEP, is an integral of the machine's power curve against the frequency with which each wind speed occurs at the hub. Most quick wind calculators quietly conflate the two, multiply a rated or single-speed power by a number of hours, and produce an answer that can be wrong by a factor of three in either direction.

The tool on this page keeps the two ideas separate and shows both. It starts from the rotor geometry and the wind measured at whatever height your anemometer or reference dataset sits at, extrapolates that wind up to hub height with a stated shear exponent, corrects air density for site elevation and temperature, builds an idealised power curve with a cut-in speed, a cubic region, a rated plateau and a cut-out speed, and then integrates that curve against a Weibull distribution whose shape parameter you control. Setting the shape parameter to 2 reproduces the Rayleigh distribution, the reference distribution that the IEC 61400-12 power performance standard and the U.S. Department of Energy small wind literature both use when no site-specific distribution is available.

Alongside the correct answer, the result panel deliberately prints the two wrong answers people usually reach for: rated power multiplied by 8,760 hours, and the instantaneous power evaluated at the mean hub-height wind speed multiplied by 8,760 hours. Seeing the three numbers side by side is the fastest way to internalise why the cube law and the wind-speed distribution cannot be skipped.

How to use this hub-height and annual-energy estimator

Work down the form from geometry to site to machine:

1. Rotor diameter is the tip-to-tip circle the blades sweep, in metres. If a datasheet quotes a radius, double it. A 5 m rotor sweeps 19.6 m².
2. Mean wind speed is the long-term annual average at the height where it was measured. Airport observations, a met mast, or a wind-atlas grid cell are all valid sources provided you also enter the correct measurement height.
3. Reference height and hub height drive the shear extrapolation. Airport anemometers are usually at 10 m; the Department of Energy notes that wind speed at a typical small-turbine hub height of 24 m is 15% to 25% higher than at that anemometer height.
4. Shear exponent defaults to 0.14, the one-seventh power law for neutral stability over open terrain. Raise it toward 0.20 to 0.25 for wooded or built-up fetch, lower it toward 0.10 over water.
5. Site elevation and mean air temperature set the air density. Published power curves are referenced to 1.225 kg/m³, which corresponds to 15 °C at sea level.
6. Net power coefficient is the fraction of the wind's kinetic flux that reaches your meter, after aerodynamic, drivetrain, generator and conversion losses. It is capped at the Betz limit of 59.3%.
7. Cut-in, rated and cut-out speeds shape the power curve. Typical small turbines cut in at 2.5 to 3.5 m/s, reach rated output at 10 to 12 m/s, and furl or brake at 20 to 25 m/s.
8. Weibull shape factor $k$ controls how peaky the wind distribution is. Leave it at 2 for the standard Rayleigh assumption; use 1.5 for gusty, highly variable sites and 2.5 to 3 for steady trade-wind or monsoon regimes.

Select Calculate Annual Energy to populate the result panel, the chart and the bin-by-bin table. The what-if slider sweeps the reference-height mean wind speed so you can watch AEP and capacity factor respond in real time, and the CSV button exports the same per-bin numbers the chart draws so you can check them in a spreadsheet.

The wind power formula, the Betz limit and the air-density correction

A rotor of diameter $D$ sweeps the area

Formula: A = (π D^2) / 4

$$A=\frac{\pi D^2}{4}$$

Air of density $\rho$ crossing that disc at speed $v$ carries a mass flow of $\rho Av$ and a specific kinetic energy of $\frac{1}{2}{v}^2$. Multiplying gives the kinetic power flux through the disc, and multiplying again by the power coefficient $C_p$ gives the extracted electrical power that the Department of Energy small wind handbook writes as

Formula: P = C_p 1 / 2 ρ A v^3

$$P=C_p\frac{1}{2}\rho A{v}^3$$

The cube is the whole story of wind energy: doubling the wind speed multiplies the available power by eight, while doubling the diameter multiplies it only by four.

Betz's analysis of an ideal actuator disc bounds $C_p$. Writing the axial induction factor as $a$, so that the wind slows from $v$ upstream to $(1-a)v$ at the disc, the extracted fraction is

Formula: C_p = 4 a (1−a)^2

$$C_p=4a{(1-a)}^2$$

which is maximised at $a=\frac{1}{3}$, giving the Betz limit

Formula: C_p,max = 16 / 27 ≈ 0.5926

$$C_{p,\max }=\frac{16}{27}\approx 0.5926$$

Extracting more would require bringing the air to a standstill inside the disc, which would block the flow that is supposed to arrive behind it. Real rotors reach 0.25 to 0.45 before drivetrain losses; the handbook quotes exactly that range against the 59.3% ceiling.

Because power is linear in density, a site correction matters as much as a few tenths on $C_p$. Density follows the ideal gas law for dry air,

Formula: ρ = p / (R_d T)

$$\rho =\frac{p}{R_{d}T}$$

with $R_d=287.05$ J kg⁻¹ K⁻¹ and $T$ in kelvin. Pressure at elevation $z$ comes from the troposphere layer of the U.S. Standard Atmosphere,

Formula: p(z) = p_0 (1−(Lz)/T_0)^5.25588

$$p\left(z\right)=p_0{(1-\frac{Lz}{T_0})}^{5.25588}$$

with $p_0=101325$ Pa, $T_0=288.15$ K and a lapse rate $L=0.0065$ K/m. Feeding sea level and 15 °C back through returns 1.2250 kg/m³, the IEC reference density, which is a useful sanity check on the implementation.

Wind speed is lifted from the measurement height $z_{\mathrm{ref}}$ to hub height $z_{\mathrm{hub}}$ with the power law,

Formula: v_hub = v_ref (z_hub/z_ref)^α

$$v_{\mathrm{hub}}=v_{\mathrm{ref}}{\left(\frac{z_{\mathrm{hub}}}{z_{\mathrm{ref}}}\right)}^{\alpha }$$

The alternative logarithmic profile, $v\left(z\right)=\frac{u_{\ast }}{\kappa }\ln \frac{z}{z_0}$, with von Karman constant $\kappa =0.4$ and roughness length $z_0$, gives very similar answers over the 10 m to 50 m span that small wind cares about; this page uses the power law because a single measured exponent is the parameter site assessors actually report.

From instantaneous power to a distribution-weighted annual energy formula

The idealised power curve used here is piecewise:

Formula: P(v) = {0 v < v_in C_p 1 / 2 ρ A v^3 v_in ≤ v < v_r P_r v_r ≤ v ≤ v_out 0 v > v_out

The wind-speed distribution is Weibull with shape $k$ and scale $c$,

Formula: f(v) = k / c (v/c)^k−1 e^−(v/c)^k

$$f\left(v\right)=\frac{k}{c}{\left(\frac{v}{c}\right)}^{k-1}{e}^{-{\left(\frac{v}{c}\right)}^k}$$

with cumulative form $F\left(v\right)=1-e^{-{(v/c)}^k}$. The scale parameter is pinned to the mean you entered through the gamma function,

Formula: c = (v ―) / (Γ(1 + 1 /k))

$$c=\frac{\bar{v}}{\Gamma (1+1/k)}$$

At $k=2$ this collapses to the Rayleigh case, $c=2\bar{v}/\sqrt{\pi }$, because $\Gamma \left(1.5\right)=\sqrt{\pi }/2$. Annual energy then follows directly:

Formula: AEP = 8760 ∫ 0 ∞ P(v) f(v) d v

$$\mathrm{AEP}=8760\underset{0}{\overset{\infty }{\int }}P\left(v\right)f\left(v\right)dv$$

The calculator evaluates the cubic region with composite Simpson's rule on 2,000 subintervals between cut-in and rated,

Formula: ∫ a b g(v) d v ≈ h / 3 (g_0 + 4 g_1 + 2 g_2 + ⋯ + g_n)

$$\underset{a}{\overset{b}{\int }}g\left(v\right)dv\approx \frac{h}{3}(g_0+4g_1+2g_2+\cdots +g_n)$$

and closes the rated plateau analytically, because a constant times a Weibull density integrates to a difference of exponentials:

Formula: E_plateau = P_r (e^−(v_r/c)^k − e^−(v_out/c)^k)

$$E_{\mathrm{plateau}}=P_r(e^{-{(v_r/c)}^k}-e^{-{(v_{\mathrm{out}}/c)}^k})$$

Two derived numbers fall out of the same integral. Mean output power is $\bar{P}=\mathrm{AEP}/8760$, and the capacity factor is

Formula: CF = (P ―) / P_r

$$\mathrm{CF}=\frac{\bar{P}}{P_r}$$

The quantity that explains why single-speed shortcuts fail is the energy pattern factor, the ratio of the mean cube to the cube of the mean:

Formula: EPF = E[v^3] / v―^3 = (c^3 Γ(1 + 3 /k)) / v―^3

$$\mathrm{EPF}=\frac{E\left[v^3\right]}{{\bar{v}}^3}=\frac{c^3\Gamma (1+3/k)}{{\bar{v}}^3}$$

For Rayleigh wind this evaluates to a clean constant:

Formula: EPF = 6 / π ≈ 1.9099

$$\mathrm{EPF}=\frac{6}{\pi }\approx 1.9099$$

So a rotor with no cut-in, no rated cap and no cut-out would produce 91% more energy than a naive "power at the mean speed" estimate. Add a cut-out and a rated plateau and that surplus is clipped; add a cut-in and the low tail is lost. Which way the naive estimate errs therefore depends on the site, and only the integral resolves it.

Finally, a useful independent check. The Department of Energy small wind handbook publishes the rule of thumb $\mathrm{AEO}=0.01328D^2{V}^3$, with diameter in feet, mean speed in miles per hour and output in kWh per year. Running this page's integrator with a Rayleigh distribution, sea-level density and no curve limits reproduces that rule to within 0.1% at a net power coefficient of 0.199: a 10 ft rotor in a 10 mph mean wind gives 1,329 kWh against the handbook's 1,328 kWh. The handbook's constant therefore bakes in roughly a 20% net system efficiency, which is a good reality check on what "efficiency" means once every loss is counted.

Worked example: a 5 m rotor on a 30 m tower at 500 m elevation

Take a 5.0 m diameter machine on a 30 m lattice tower. The nearest long-term wind record is an airport anemometer at 10 m reporting a 5.5 m/s annual mean. The site sits 500 m above sea level with a 12 °C annual mean temperature and open farmland fetch, so $\alpha =0.14$. The turbine is specified at a 30% net power coefficient, cutting in at 3 m/s, rated at 11 m/s and cutting out at 25 m/s, and we assume Rayleigh wind, $k=2$.

1. Swept area: $A=\pi \times 5^2/4=19.63$ m².
2. Pressure at 500 m: $p=101325\times {(1-0.0065\times 500/288.15)}^{5.25588}=95461$ Pa.
3. Density: $\rho =95461/(287.05\times 285.15)=1.1663$ kg/m³, which is 95.2% of the 1.225 reference.
4. Hub-height wind: $v_{\mathrm{hub}}=5.5\times {(30/10)}^{0.14}=6.41$ m/s, a 17% uplift for tripling the height.
5. Rated electrical power: $P_r=0.30\times 0.5\times 1.1663\times 19.63\times {11}^3=4572$ W, so about 4.6 kW.
6. Power at the mean hub-height speed: 907 W. This is the number a naive calculator reports, and it is not the average.
7. Distribution-weighted mean power: 1,376 W, giving AEP = 12,055 kWh per year and a capacity factor of 30.1%.

Now compare the shortcuts. Multiplying rated power by 8,760 hours gives 40,050 kWh, a 3.3-fold overestimate, because the machine spends only a small slice of the year at or above 11 m/s. Multiplying the 907 W single-speed figure by 8,760 hours gives 7,941 kWh, a 34% underestimate, because the cube law rewards the windy tail far more than the calm centre of the distribution penalises it. Only the integral, 12,055 kWh, respects both the curve and the distribution.

Scenario comparison and site-sensitivity tables

The first table holds the example turbine fixed (5 m rotor, 30% net $C_p$, 3/11/25 m/s curve, Rayleigh wind) at sea-level standard density, where rated power is 4,802 W, and varies only the hub-height mean wind. It shows how badly the single-speed shortcut tracks the truth, and in which direction.

Two things stand out. First, AEP is nowhere near linear in mean wind speed: moving a site from 5 to 7 m/s more than doubles the yield. Second, the naive single-speed estimate is not merely wrong, it is wrong by an amount that itself depends on the wind resource, so it cannot be repaired with a fixed fudge factor.

The second table shows how much the shear exponent alone moves the hub-height wind for a 5.5 m/s reference reading at 10 m. Because power goes as the cube, the 0.10 to 0.25 spread at 45 m is a factor of ${(8.01/6.39)}^3\approx 1.97$ in available power.

The third table isolates air density, the linear term that altitude-blind calculators silently get wrong.

Interpreting AEP, capacity factor and the energy chart

The headline number to act on is AEP in kilowatt-hours per year. Divide your annual household or farm consumption by it to get the fraction of load the machine could offset, and multiply it by your retail or export tariff for a first-pass revenue figure. A 12,055 kWh/yr result covers roughly the annual electricity use of an average U.S. home, though matching generation to consumption hour by hour is a separate and harder problem.

Capacity factor is the honesty check. It is mean output divided by rated output, and it tells you how hard the nameplate is working. Small turbines on modest inland sites commonly land between 10% and 25%; a 30% figure implies a genuinely good wind resource or an optimistically low rated speed. If your capacity factor comes out above about 45%, suspect an input error: either the rated speed is set too low, the shear exponent is too generous, or the mean wind speed is a windy-month average rather than an annual one.

The chart plots three things against wind speed. The dark line is the power curve in kW, flat once rated speed is reached. The pale filled area is the probability density of the wind, which peaks well below rated speed at most sites. The orange bars are the product of the two, the annual energy contributed by each 1 m/s bin. That bar chart is the single most instructive picture in wind resource assessment: the tallest bars almost never sit at the modal wind speed, and they almost never sit at rated speed either. They sit between the two, in a band where the wind is still reasonably common and the cube law has already made it valuable. The exported CSV gives the same bins numerically, including the hours per year spent in each bin.

Assumptions and limitations of a Rayleigh-weighted yield estimate

This is a screening tool, not a due-diligence model. Its assumptions are worth stating plainly.

• Idealised power curve. A real curve is measured, not assumed, and is neither perfectly cubic below rated nor perfectly flat above it. IEC 61400-12 exists precisely because the shape matters; a manufacturer curve measured to that standard will shift AEP by several percent relative to this model.
• One distribution, no seasonality or diurnal cycle. A single Weibull fit hides the fact that winter and night-time winds are usually stronger. It also hides directional variation, so it cannot represent a site whose good wind arrives from a blocked sector.
• Steady, uniform inflow. Turbulence intensity, wind veer across the rotor, and the difference between the hub-height speed and the rotor-equivalent speed are all ignored. High turbulence typically costs energy and always costs fatigue life.
• No wake, obstacle or tower effects. Buildings, tree lines and neighbouring turbines all reduce and disturb the flow. The rule of thumb is to clear the tallest nearby obstacle by 9 m within a 150 m radius, which this calculator has no way to check.
• No availability, soiling, icing or electrical losses beyond your $C_p$. If you enter a rotor-only coefficient you will overstate delivered energy by the drivetrain and inverter losses, typically 10% to 20% combined.
• Dry-air density. Humidity lowers density slightly, by well under 1% at normal conditions, and is neglected. The standard-atmosphere pressure model is valid only in the troposphere, so elevations above about 11,000 m are rejected.
• The shear exponent is an input, not a measurement. It is the largest single source of uncertainty in most small-wind estimates, and a guessed value of 0.14 against an actual 0.22 will understate energy by roughly a third.

Treat the output as a bounded estimate. If the numbers only work at the optimistic end of your input ranges, the project does not work.

Common questions about wind turbine energy estimates

Sources and further reading

Sources. U.S. Department of Energy, Wind Energy Technologies Office / WINDExchange, Small Wind Guidebook (Small Wind Electric Systems Handbook) — the power equation $P=C_p\frac{1}{2}\rho A{v}^3$, the swept area $\pi D^2/4$, the Betz maximum of 16/27 (59.3%), the practical $C_p$ range 0.25–0.45, the 15 °C sea-level rating condition, the 15–25% hub-height wind uplift over airport anemometer height, and the rule of thumb AEO = 0.01328 D²V³ (energy.gov/eere/wind/windexchange/small-wind-handbook). IEC 61400-12 series, Wind energy generation systems — Part 12: Power performance measurements of electricity producing wind turbines — measured power curve, clause 9.1.1 density correction to the 1.225 kg/m³ reference, clause 9.3 annual energy production, and the use of the Rayleigh cumulative distribution F(V) as the reference wind-speed frequency distribution. NOAA / NASA / USAF, U.S. Standard Atmosphere, 1976 — sea-level pressure 101,325 Pa, temperature 288.15 K, tropospheric lapse rate 0.0065 K/m and the barometric exponent 5.25588 used for the density correction. NREL, Wind Shear Characteristics at Central Plains Tall Towers (NREL/CP-500-40019) and the NREL Annual Technology Baseline — power-law shear exponents and representative capacity factors. Betz limit derivation: $C_p=4a{(1-a)}^2$ maximised at $a=1/3$.