Betz limit and power available in the wind

[PierreB]

@dougselsam

And speaking of power generated, let’s recall that if a parachute is moving downwind at half the windspeed, it is only exposed to 1/8th the available power, from what is an airborne version of a Savonius turbine, already the least efficient concept commonly pursued, and which is only pursued because of the fact that “any idiot can understand how it works”…

@PierreB

“AWE drag power systems can harvest up to 16/27 of the power available in the wind.
AWE lift power systems can harvest up to 4/27 of the power available in the wind.”
A drag power system has the turbines aloft like a Makani’s wing.
A lift power system has a ground-based generator and is a pumping kite system (yo-yo, jojo, reel-out/in…).

@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.

@PierreB

Page 3:

The power P that can be generated with a tethered airfoil operated either in drag
or in lift mode had under idealized assumptions been estimated by Loyd [8] to be
approximately given by P = 2/27ρAvw³CL (CL/CD)²
where A is the area of the wing, CL the lift and CD the drag coefficients, and vw the
wind speed.

Hi Doug,
Let’s forget for a moment the questionable terminology of “lift” and “drag” power here.

The power available in the wind for reel-out/in (“lift power”) = yo-yo mode is 4/27, so 1/4 Betz limit (quote 2), so two times more than “1/8th” (quote 1). Indeed the idealized reel-out speed remains 1/3 wind speed (quote 3). The apparent wind speed is 2/3 real wind speed (quote 3) rather than “half the windspeed” (quote 1).

The power is the traction force multiplied by the reel/out speed (quote 3), and can also be reached by using the formula (quote 4) but with simplification, without “CL(CL/CD)²” which is applicable to crosswind kites, but not to our surface, which can be a parachute going downwind. Using the same figures (quote 3), this gives us: 2/27 x 1.2 x 100 x 10³ = 8888 W, so exactly the same result.

Precision about 4/9 (quote 3) of the force (and also the power in the end, the reel-out speed being 1/3 wind speed), resulting from apparent wind speed of 2/3 real wind speed (quote 3):

@PierreB

1/3 wind speed downwind leads to (1/3)², so 4/9 available power. It is due to the lower apparent wind speed², 10² becoming ((2/3)10)².