# Relationship between wind velocity, turbine rotor diameter, and power output

I'm trying to understand and identify the equations to use in defining the relationship between wind velocity, turbine rotor diameter, and power output for a wind turbine.

To simplify my question, let's use the following assumptions:

• Average wind speed of 8 m/s.
• Rotor length of 40m for an overall diameter of 80m.
• Ideal conditions, so we can neglect direction of wind with respect to the positioning of the turbine.
• This has to do with how much energy you can extract from the wind, which is an aerodynamics question.
– Daniel
Nov 14 '15 at 17:47
• Determined by Newton's laws of motion, the Bernoulli equation, and bounded by the Betz criterion. Nov 14 '15 at 17:51
• Thanks Brian, I have seen the equation:-power o/p=(0.652*pi*diameter^2)/4*wind_velocity^3...is it correct?
– Dwiparna Datta
Nov 14 '15 at 17:54

The Betz limit will give you a good theoretical maximum from your wind velocity and swept area. Your real world turbine will fall somewhere below this number based on its efficiency.

For measuring power production over a period of time, you need wind data for that particular area. Wind data can be iterated and summed for a particular turbine, but generally the Rayleigh distribution is used for a close approximation.

Like EnergyNumbers mentioned, it is not economically feasible to collect power from very high speed wind that only occurs once in a while, so turbines have braking and feathering systems to protect themselves in a power off state during these situations.

It's not a simple relationship.

First let's deal with the kinetic energy of the wind passing through the rotor.

The mass of air passing through it in one second, $m$, is equal to the the density of the air ($\rho$), times the surface area of the rotor (${\pi}r^2$), times the velocity of the air ($v$). i.e. ${\rho}{\pi}r^2v$.

The kinetic energy of that air is just $\frac{1}{2}mv^2$ which is $\frac{1}{2}{\rho}{\pi}r^2v^3$ in each second.

Now, an idealised wind turbine would capture as high a proportion of that energy as it possibly could: that proportion is given by the Betz limit, which is $\frac{16}{27}$.

However, in the real world, we don't achieve perfection, so at best the real proportion is a bit lower than that. And also, in the real world, we have to make trade-offs based on cost. And that means that at higher speeds, which happen less often, although we could capture much more power, it's just not worth the extra expense of uprating all of the electronics and the connection to the grid, for those times, because it wouldn't represent much extra energy over the lifetime of the turbine, but would be a lot of extra cost. So the blades get feathered at higher speeds - at any speed above what we define as the rated wind speed - and the power generated is capped at the rated power. And there gets to be some rare wind speed where, in the interests of a long and healthy lifetime for the turbine, it's most economic to shut it down completely until wind speeds drop again: that's the cut-out wind speed. And at very low wind speeds, there's virtually no energy in the wind - nothing worth harvesting, and not enough to get the blades moving. So below a certain speed - the cut-in wind speed - no power is generated.

All this gives us a curve of how generated power varies with windspeed that looks like this: (source)

The power generated depends on the product of two factors:

1. the windmill (wind turbine) power curve
2. the windspeed curve or histogram, % at each windspeed.

Here's a plot of 2 wind turbine types $$\times$$ 3 synthetic (Weibull-distributed) windspeed curves:

Let's compare this with the theoretical $$A \, v^3$$, as described e.g. in MacKay, Sustainable Energy -- without the Hot Air p 263 ff. Consider the effects of $$A$$ and $$v^3$$ on power separately:

$$A$$: the two wind turbines here have rotor diameters 101 and 141m, so $$A$$ ratio ~ 2 but max power ratio ~ 1.4. This doesn't tell us much: maybe one could build a wind turbine with twice the MW at $$2 A$$, but it would cost more than it's worth ?

$$v^3$$: the power ratios in table form

Ratios of MW at [6 7 8] / 6 m/s
E-101/3050 [100 132 161] %
E-141/4200 [100 131 159] %


show a factor ~ 1.4, not 2 $$\sim (7.5 / 6)^3$$ .

(Fine print: why synthetic windspeed curves ? Real data is hard to find (worth \$\$\\$). There are of course lots of histograms with average 8 m/s. Weibull give a family of curves or histograms with the same average, and a parameter $$k$$ that shifts less wind at lower / more at higher windspeeds.)

See also: wind-data.ch, wind-stats, globalwindatlas.info -- nice gui, legend re-scale

Summary: is there a simple model of wind turbine power$$( A, v, height, cost )$$ that fits power actually generated to within say 10 % ? Sounds unlikely.