http://www.awec2017.com/presentations-main.html then click on Fort Felker: a lot of informations.
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Fort Felker (please see the attachment below or the link above): “Our largest kite to date, it has a wingspan of 26 m, and has eight onboard rotors that are each 2.3 m in diameter. For comparison, our previous prototype, Wing 7, was a 20 kW system with a 8m wingspan and with four rotors 0.7 m in diameter.”
The turbines aloft should add 1/2 drag of the wing alone in order to reach an optimal efficiency, the speed becoming 2/3. I checked this with my FlygenKite:
(please see also my comments as corrections).
But the drag of “eight onboard rotors that are each 2.3 m in diameter” (33 m²) seems to be largely above 1/2 the wing alone drag. Have you some explains please? Is it the large drag of the support of the turbines?
ch01_awec94_Felker.pdf (2.2 MB)
Another flygen AWES:
The rotors have comparable relative dimensions as M600 rotors, but there are four rotors instead of eight.
Dave Santos’ answer on https://groups.yahoo.com/neo/groups/AirborneWindEnergy/conversations/messages/25436 :
“High M600 rotor disc area is needed for E-VTOL, including safety factor. Full drag for harvesting is not the design necessity.”
Indeed E-VTOL requirement seems plausible but even half of rotors area looks to be high in regard to the wing alone, by taking into account of its high L/D ratio, so its low D.
See also on http://www.kitekraft.de/#4_System_Development with still higher values (rotor area/wing area) as mentioned previously, perhaps because of the high wing lift allowing larger rotors aloft:
“Kite
Wing Span: 4 m
Wing Area: 1 m²
Weight: 30 kg
Rotor Diameter: 0.4 m
Number of Rotors: 8
Total Rotor Area: 1 m²”
In my opinion the turbines area is very large in regard to the wing area, above all for the M600, probably due to E-VTOL requirement as previously discussed.
So during the flight lowering the thrust coefficient of the turbines is needed in order to keep a L/D ratio high enough. The used means can be the pitch or stall like for wind turbines working at high wind speed. In the same time that can also avoid an excess of rpm. In the other hand generators have to be bigger and heavier due to the required higher diameter of the turbines.
Hypothesis for a calculation of M600: 2/27 x 1.2 x 1.2 x 50 [supposed area] x 1000 [cubed wind speed] x (10.6)² [(L/D ratio)²] = roughly 600 kW before cosine loss.
L/D ratio wing alone = 10.6; optimized L/D ratio with turbines = 7.0666666; wing area estimated at 50 m². With a lift coefficient CL of 1.2, wing area x CL = 60; so wing area x CD should be 5.66, leading to a drag coefficient of 0.1132. The thrust (drag) of the turbines should add 50% of 5.66, so 2.83. The thrust coefficient of the turbines is 2.83/33 = 0.0857575.
Verification: 33 x 1.2/2 x 70.666666 x70.666666 x 70.666666 x 0.0857575 = roughly 600 kW.
It looks that the turbines are (10 times too) large due to E-VTOL requirement, then the thrust coefficient is lower during generation. And a relatively important diameter of the turbines leads to possible relatively slow and heavy generators aloft: these large turbines are not optimized for the generation phase, due to E-VTOL requirement.
IMHO Makani M600 looks to be by far the most advanced and efficient crosswind AWES.
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What’s the base formula for this? Might try to confirm your results at some point.
http://homes.esat.kuleuven.be/~highwind/wp-content/uploads/2013/08/Diehl2013a.pdf : equation (2) page 3, Loyd’s formula : [P = 2/27 aD A w3 CL (CL/CD)² ] . I used this simplification with not still cosine cubed inclusion. aD is air density, w is wind speed. A is the kite area.
Concerning the verification with the turbines, I applied (a little wrongly because it is secondary turbines) the wind power formula before Betz limit: https://www.raeng.org.uk/publications/other/23-wind-turbine : [P = 1/2 aD A w3]. This time A is the swept area. But such a verification is not essential. I don’t use a coefficient of power or it is at an ideal value of 1.
The rule is increasing the kite drag by 50% with the turbines aloft. As the Makani turbines look to be huge in regard to the wing area, the only possibility I saw is a low thrust coefficient during generation, being high only for take-off. But I can be wrong.
Note: as I indicated I don’t know M600 area, thinking about 50 m² from the pictures, and the given 26 or 28 m span.
Bro how it comes 70.6666 in verification part?
So wind speed is assumed to be 10 m/s. Then L/D ratio with turbines is assumed to be 7.06666.
10 X 7.06666 = 70.6666. The 7.06666 value is 2/3 of 10.6 which is the L/D ratio without turbines aloft.
I think there are a lot of mistakes in my now old calculation:
Not, the result would be roughly 6 MW, although it is 600 kW but with other numbers.
No, it was what I thought, but the turbines are well sized.
So I would prefer take another basis of hypothetical calculation on:
The usual formula of wind energy by swept area is applied for this time.
Now from the supposed 40 m² of the wing, assuming a lift coefficient of 2, a L/D ratio of 12 (with tether drag), an efficiency of 0.65 after cosine cube loss: 40 x 1.2 x 2/27 x 2 x 10³ x 12² x 0.65 = about 660 kW. Derived Loyd’s formula is used.
So the value of the wing roughly matches the value of the onboard turbines.
These are only assumptions and I expect that some other possible mistakes can be corrected.