There is an interesting discussion to be had about the possible efficiency of wind turbines which presents opportunities for the instructor to look at several different kinds of computations. The discussion ends up with an optimization problem which could be approached by the classical methods of calculus – if the students have that available to them – or could be approximated simply by drawing a graph of one critical function.

We start by asking: how much energy is available to a wind turbine? Two key parameters govern this: the area \( S\) swept out by the turbine blades, and the speed \(v\) of the incoming wind – the turbine should rotate so that the plane of the blades is perpendicular to the wind. If \(\rho\approx 1.2 \text{kg}/\text{m}^3\) denotes the density of air, then in one second a cylinder of air of cross-section \(S\) and area \(v\) approaches the machine. The mass of this amount of air is \(\rho S v\) and the kinetic energy it contains is half the mass times the square of the speed, that is

\[ {\textstyle \frac12} \rho S v^3. \]

This is the energy per unit second, i.e. the *power* (in watts), theoretically available in the wind incident on the turbine.

That is the first “theory” part of the lesson and one ought now do to some specific calculations. For example, a modern wind turbine might have a blade length of 30 meters; what is the theoretically available power at a wind speed of 10 meters per second? Using the formula \(S = \pi r^2 \) for the area of a circle, \(r \) being 30 meters in this example, the formula above gives

\[ 0.5 \times 1.2 \times \pi \times 30^2 \times 10^3 \approx 1.8 \times 10^6 W, \]

or nearly two megawatts. One should work things out for a few other examples.

Now comes the second theoretical kicker, though: it is impossible to extract *all* that kinetic energy. Why? Remember that kinetic energy is the energy of air (or something) *in **motion; *if you extract all its kinetic energy, it will stop. So to extract all the kinetic energy from the incoming wind we’d have to imagine the air just stopping when it got to the turbine and piling up without going anywhere. This is clearly ridiculous.

To get a better theoretical picture, we have to imagine what happens to a column of air as it moves through the turbine. As the air moves through the turbine it must slow down some (that is what extracting kinetic energy means, even if – as we have seen – we can’t slow it down to zero!) As it slows down it must spread out, so that there is the same total volume of moving air. (Remember, the volume of air moving per second is the area times the speed – if the speed decreases, the area must increase.) Therefore we get a picture like this: the wind arrives at a high speed \(v_1\) and departs at a slower speed \(v_2\). It passes through the turbine at a speed \(v\) which is between \(v_1\) and \(v_2\); in fact, we will assume that

\[ v = {\textstyle\frac12} (v_1+v_2). \]

(Though we treat this as an assumption, it is in fact a consequence of the physical laws of conservation of energy and momentum; students with sufficient background in physics would be able to follow the proof, which can be found in many places.) If \(V=Sv\) is the volume flow rate then the kinetic energies of the incoming and outgoing air are \(\frac12 Vv_1^2 \) and \(\frac12 Vv_2^2\) respectively; the difference, which is the energy available at the turbine, is (remembering the definition of \(V\))russ

\[ {\textstyle\frac14} S(v_1+v_2)(v_1^2-v_2^2) = {\textstyle\frac14}Sv_1^3 (1+\lambda-\lambda^2-\lambda^3), \]

where \(\lambda=v_2/v_1 \in (0,1)\). In this equation, \(S\) and \(v_1\) are given quantities, but the ratio \(\lambda\) can be changed by adjusting the blade pitches and so on. It therefore is reasonable to ask what choice of \(\lambda\) will maximize the energy output. Using either elementary calculus or graph-plotting estimates one discovers that the maximum value of \(1+\lambda-\lambda^2-\lambda^3\) occurs when \(\lambda=\frac13\); at this point

\[1+\lambda-\lambda^2-\lambda^3 = \frac{32}{27}\]

and the maximum power available at the turbine is \( {\displaystyle\frac{16}{27}} \approx 59\% \) of the naive estimate \({\frac12}S v_1^3\). This is called the *Betz limit* on turbine efficiency. Thus in the example we used for calculation above, only about 1 megawatt of the theoretical 1.8 megawatts can actually be extracted by the turbine and turned into useful energy.

Two occurrences of 14 should be \tfrac{1}{2}

Thank you for pointing this out.

There was a typo preventing typesetting of the fraction. It should be 1/4, and has now been corrected.

Thanks again!

Russ deForest