Unintuitive to the uninitiated

[TimberWolf]

Surely pitch of blade would be a factor too?
an increase in blade pitch would vary the ratio and thus the output?

Pitch and the number of blades plays a part plus a whole load of other factors shunted into the efficiency factor (which itself can only be less than 0.59 due to Betz' law). Blades can interfere with themselves too

[Bear]

Surely pitch of blade would be a factor too?
an increase in blade pitch would vary the ratio and thus the output?

I assume that in the design of the blades they have calculated the optimum pitch for the aero dynamics of the blade. Any change in pitch would therefore increase the drag coefficient (reduce the effectiveness).

So you are right in that it would change the output - but probably downwards

[Diver]

Surely pitch of blade would be a factor too?
an increase in blade pitch would vary the ratio and thus the output?

[Pickle2]

For example why doesn't it help to build higher - in a wind farm - in an attempt to minimise land use?

Is it something to do with cleaning, ladders and health and safety?

[EternalOptimist]

Because the higher you go, the weaker is the Earths magnetic field, which means there is less electricity to extract from the air flow. You can prove this through extrapolation. If the turbine was very high, say 100 miles high, there would be no air flow at all, because it would be in space hence no electricity.
Plus if you made them too tall they would bend over in a strong wind, and effect known as blowing over.

[Xenophon]

TimberWolf is awarded +5 Xeno Geek Points.

[TimberWolf]

I'm still fascinated by this book http://www.withouthotair.com/ (thanks Cliphead)

Wind turbines

I think it can easily be seen that the power that a wind turbine can extract from the wind will be related to the length of the blades in some way (the square in fact), and perhaps less intuitively that it's also related to the cube of the velocity of the air passing through it (i.e. the wind speed). In fact the formula for calculating the power that can be extracted from the wind is 0.5 * efficiency * airDensity * PI * r^2 *v^3. Or roughly r^2*v^3. So a 1m radius turbine in a 6 m/s wind can produce up to around 200 Watts.

Here's the initially unintuitive bit. In a wind farm, if you pack turbines too close together they become inefficient and it turns out you need at least 5 diameters separation between them to avoid the worst of this effect. The resulting algebra means that the power you can extract from the wind (in a wind farm) then becomes independent of the radius of the blades (cough)! It's just related to wind speed cubed and air density, which by the magic of algebra is also a unit of Watts per square metre. The author (PhD from Caltec, professor at Cambridge, la de da) concludes that this area relates to the amount of land land used and is a constant for a given wind speed. See http://www.withouthotair.com/ page 332 of the PDF book.

At first glance this doesn't seem right. There appears to be a change of interpretation involved here with what the algebra represents (in the first case the area swept by the blades in a vertical plane and now to the area of land used) and it doesn't seem to make sense that power extracted can be independent of the blade radius. For example one might envisage a wind turbine 1 km high that takes up not much more land space than a more vertically challenged version but which has a much larger cross sectional area. Can you see how this works out? (it does) For example why doesn't it help to build higher - in a wind farm - in an attempt to minimise land use?