# Wind power generation help please

1. Dec 13, 2006

### The Guv.

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

The information I have researched is as follows:

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!

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

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!

2. Dec 13, 2006

### 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?

3. Dec 13, 2006

### 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.

Last edited: Dec 13, 2006
4. Dec 13, 2006

### The Guv.

Thanks, this helped a lot with my calculations!