How Does Rotor Diameter Affect Wind Power Output?

[The Guv.]

Ok, doing a project at the moment and this is one of the issues I have to overcome;

Exercise: Electricity Generation

Investigate the use of wind generation. The unit provided has a rotor diameter of 3m and an electrical output of 500W at a wind speed of 10.0m/s.

The power output of a wind generator is proportional to the cube of the wind speed.

Instead of relying on one wind generator, calculate the necessary rotor diameter if two smaller wind generators were to be used to provide the same total output.

The information I have researched is as follows:

A typical turbine with a 600 kW electrical generator will typically have a rotor diameter of some 44 metres (144 ft.). If you double the rotor diameter, you get an area which is four times larger (two squared). This means that you also get four times as much power output from the rotor.

Source: http://www.windpower.org/en/tour/wtrb/size.htm [/size]

If this is the case then a 88m diameter rotor should generate 2400kW? But the below diagram from the same page shows a 80m diameter rotor generates 2500kW!

http://www.windpower.org/res/rotorsk.gif

When I put this data into a graph I end up with a curve rather than a straight uniformed line?

http://img221.imageshack.us/img221/1328/powergraphmi7.jpg

Can anyone help me understand what I am doing wrong and guide me through what I am meant to be doing?

I'm not doing physics or maths I am doing Arch & Construction Management.

Many thanks!

 

[The Guv.]

Oh and I don't want the answer to the given question I want the formulas involved in understanding the relationship between swept area and power generation.

Just read the sticky and I think think I need to put it in the homework bit do i?

 

[andrevdh]

You can use the statement ... "If you double the rotor diameter, you get an area which is four times larger (two squared)." This comes from the fact that the swept area is in the same ratio as the power produced due to the amount (volume per second) of air that flows through the turbine blades (if the airflow is kept constant) we have that

P \propto A

where A is the swept area and P is the power produced. Which in the case of the problem boils down to

\frac{d_x ^2}{250} = \frac{d_{500} ^2}{500}

The volume of air that flows throught the turbine per second will be

Av

where v is the wind speed. A turbine converts (a fraction of) the kinetic energy of the air that flows through it to electrical energy. Which means that the power output will be (somewhat) directly proportional to this product. The reason why this will be only approximately true is that the turbine is more effective at some wind speeds due to the blade construction and orientation.

 

[The Guv.]

Thanks, this helped a lot with my calculations!