To estimate the energy in wind, let's imagine holding up a hoop with area A, facing the wind whose speed is v. Consider the mass of air that passes through that hoop in one second. Here's a picture of that mass of air just before it passes through the hoop:

oop oop

And here's a picture of the same mass of air one second later:

The mass of this piece of air is the product of its density p, its area A, and its length, which is v times t, where t is one second.

The kinetic energy of this piece of air is

So the power of the wind, for an area A - that is, the kinetic energy passing across that area per unit time - is jtnv

This formula may look familiar - we derived an identical expression on p255 when we were discussing the power requirement of a moving car.

What's a typical wind speed? On a windy day, a cyclist really notices the wind direction; if the wind is behind you, you can go much faster than

I'm using this formula again: mass = density x volume

miles/ |
km/h |
m/s |
Beaufort |

hour |
scale | ||

2.2 |
3.6 |
1 |
force 1 |

7 |
11 |
3 |
force 2 |

11 |
18 |
5 |
force 3 |

13 |
21 |
6 |
force 4 |

16 |
25 |
7 | |

22 |
36 |
10 |
force 5 |

29 |
47 |
13 |
force 6 |

36 |
31 |
16 |
force 7 |

42 |
68 |
19 |
force 8 |

49 |
79 |
22 |
force 9 |

60 |
97 |
27 |
force 10 |

69 |
112 |
31 |
force 11 |

78 |
126 |
35 |
force 12 |

Figure B.1. Speeds.

Figure B.1. Speeds.

Figure B.2. Flow of air past a windmill. The air is slowed down and splayed out by the windmill.

normal; the speed of such a wind is therefore comparable to the typical speed of the cyclist, which is, let's say, 21 km per hour (13 miles per hour, or 6 metres per second). In Cambridge, the wind is only occasionally this big. Nevertheless, let's use this as a typical British figure (and bear in mind that we may need to revise our estimates).

The density of air is about 1.3 kg per m3. (I usually round this to 1 kg per m3, which is easier to remember, although I haven't done so here.) Then the typical power of the wind per square metre of hoop is

2 Pv

Not all of this energy can be extracted by a windmill. The windmill slows the air down quite a lot, but it has to leave the air with some kinetic energy, otherwise that slowed-down air would get in the way. Figure B.2 is a cartoon of the actual flow past a windmill. The maximum fraction of the incoming energy that can be extracted by a disc-like windmill was worked out by a German physicist called Albert Betz in 1919. If the departing wind speed is one third of the arriving wind speed, the power extracted is 16/27 of the total power in the wind. 16/27 is 0.59. In practice let's guess that a windmill might be 50% efficient. In fact, real windmills are designed with particular wind speeds in mind; if the wind speed is significantly greater than the turbine's ideal speed, it has to be switched off.

As an example, let's assume a diameter of d = 25 m, and a hub height of 32 m, which is roughly the size of the lone windmill above the city of Wellington, New Zealand (figure B.3). The power of a single windmill is efficiency factor x power per unit area x area

Indeed, when I visited this windmill on a very breezy day, its meter showed it was generating 60 kW.

To estimate how much power we can get from wind, we need to decide how big our windmills are going to be, and how close together we can pack them.

How densely could such windmills be packed? Too close and the upwind ones will cast wind-shadows on the downwind ones. Experts say that windmills can't be spaced closer than 5 times their diameter without losing significant power. At this spacing, the power that windmills can generate per unit land area is power per windmill (B.4) land area per windmill

This number is worth remembering: a wind farm with a wind speed of 6 m/s produces a power of 2 W per m2 of land area. Notice that our answer does not depend on the diameter of the windmill. The ds cancelled because bigger windmills have to be spaced further apart. Bigger windmills might be a good idea in order to catch bigger windspeeds that exist higher up (the taller a windmill is, the bigger the wind speed it encounters), or because of economies of scale, but those are the only reasons for preferring big windmills.

This calculation depended sensitively on our estimate of the windspeed. Is 6m/s plausible as a long-term typical windspeed in windy parts of Britain? Figures 4.1 and 4.2 showed windspeeds in Cambridge and Cairngorm. Figure B.6 shows the mean winter and summer windspeeds in eight more locations around Britain. I fear 6m/s was an overestimate of the typical speed in most of Britain! If we replace 6m/s by Bedford's

f |
f | |

h |
d f |
T |

5d | ||

Figure B.4. Wind farm layout.

Figure B.4. Wind farm layout.

Power per unit area wind farm 2 W/m2 (speed 6 m/s)

Table B.5. Facts worth remembering: wind farms.

winter

Stornoway Kirkwall Kinloss Leuchars Dunstaffnage Paisley Bedford St Mawgan winter

Stornoway Kirkwall Kinloss Leuchars Dunstaffnage Paisley Bedford St Mawgan

100 km

Figure B.6. Average summer windspeed (dark bar) and average winter windspeed (light bar) in eight locations around Britain. Speeds were measured at the standard weatherman's height of 10 metres. Averages are over the period 1971-2000.

100 km

4 m/s as our estimated windspeed, we must scale our estimate down, multiplying it by (4/6)3 ~ 0.3. (Remember, wind power scales as wind-speed cubed.)

On the other hand, to estimate the typical power, we shouldn't take the mean wind speed and cube it; rather, we should find the mean cube of the windspeed. The average of the cube is bigger than the cube of the average. But if we start getting into these details, things get even more complicated, because real wind turbines don't actually deliver a power proportional to wind-speed cubed. Rather, they typically have just a range of wind-speeds within which they deliver the ideal power; at higher or lower speeds real wind turbines deliver less than the ideal power.

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