Betz limit and power available in the wind

[PierreB]

@PierreB

Let us try to check this with an example, wind speed = 10 m/s, air density = 1.2.
The power of a HAWT is determined by the disk swept by the rotor. Supposing the disk area is 100 m².
Power at Betz limit: 1/2 x 100 x 1.2 x 1000 x 16/27 = 35555 W.
Now let us take a surface of 100 m². Traction (force): 1/2 x 100 x 1.2 x 100 = 6000 N. Let us reel-out this surface in the optimal speed of 1/3 wind speed, considering the wind speed on the surface is now only 2/3 wind speed, leading to the result: 20000 x 4/9 = 8888 W, so exactly 1/4 of 35555 W.

Recall: traction 6000 N X 1/3 wind speed = 20000.
Now assuming the elevation angle is 30°, leading to a cosine of 0.8660.
The reel-out speed in the horizontal becomes 2.955 m/s.
And the apparent wind speed becomes 10 - 2.955 = 7.045 m/s instead of 6.66 m/s.
7.045² = 49.63. Then 49.63/100 = 0.4963 instead of 4/9, so a bit more (10%).