Elevation angle
by Pierre Benhaïem in Airborne Windenergy Forum · · 4 Replies · View last reply Respond
Pierre BenhaïemSo here is a start for a later study. Wind speed 12 m/s. Air density 1.2 . A kite of type rotor sweeping 7800 m² as its disk, not as the frontal airspace. Efficiency Betz limit/2. Ground conversion like rotating reel or Daisy's: transferable torque supposed to be maximal = 1. With elevation angle 30°; cubed cosine = 0.65 . Power = 2.38 MW.With elevation angle 60°; cubed cosine = 0.125. Power = 0.45 MW. Now a static kite, its area is 7800 m². Lift is (supposing lift coefficient is 1.2) 808700 N. (roughly 80 tons). Drag (supposing drag cofficient is 0.12) is 80870 N (roughly 8 tons). L/D ratio is 10. Elevation angle is 81° without WECS. So the question is how much WECS vertical area the kite can carry in order to have an elevation angle being 30°, then 60°. Efficiency of WECS is supposed to be 80% Betz limit. Taking account of mass and drag of WECS. Supposing 18 m diameter and 1.5 ton and 100 kW rotors like http://windenergysolutions.nl/portfolioitem/wes100/ are carried. To obtain 0.45 MW 4.5 turbines are required for 6.75 tons. Drag is swept area 254 m² x 4.5 x wind speed² x 8/9 = 146300 N (14.6 tons). So 741200/227170 = L/D ratio = 3.26 . Elevation angle is 62°, a not so different value. Of course the result is not the same with wind speed < 12 m/s. 

Pierre Benhaïemhttp://www.kiteboardingevolution.com/windwindow.html here is a kite window. Everybody really involved in AWE, in kite, or in science, knows that the maximal power point of the kite window is around elevation angle = 0° (not 45°) and facing the wind. On this maximal point there is 0 loss by cosine. But as the kite flies the optimal power zone is between 0 and 45° of elevation angle (where cubed cosine losses are not too high) and also of sides angles. And an elevation angle (of 30° for example) concerns the kite in energy production; for a static kite carrying WECS, the things are a bit different. In the short study above I try to compare a rotor sweeping at 30° (more power) and 60° (less power) with a static kite having the same area and carrying WECS (turbines) in order to try knowing if higher elevation angle can be obtained with a same power. As example the elevation angle of Altaeros AWES can be higher, allowing the turbine (WECS) to be high in altitude. By the same a lifting kite (see the example of the static kite in the first message) can have also a higher elevation angle; and as it carries WECS, a question is how much area of WECS it can carry, keeping an elevation angle high enough. It is a complex topic as an airborne wind energy system faces to two contradictory concerns; lift(towards the top) and wind energy production (towards wind direction). 

Pierre BenhaïemIn the previous message elevation angle = 0° is considered as the maximal power point by geometry, not in practical considerations, the wind velocity gradient being not included as such. I try to see how different elevation angles with their respective advantages (power for a low value, lift/drag ratio before conversion for a high value) can involve in possible new AWES. I have some idea about it. 

Gordon SpilkinMy system consists of multiple turbines on a common axis similar to Doug's Superturbine(R). The axes of the turbines are part of the tether. The length of the torsion shafts or tensegrity elements between turbines is relatively short and so scaleup problems are not too severe. The problem is the transfer of energy to the ground. I favor using a rope driving system combined with a generator on the ground. This system is lightweight and easily scalable, but requires a larger lifter kite to account for the differential forces in the cable drive system. 

Pierre Benhaïemhttp://www.energykitesystems.net/GordonSpilkin/BalloonKite2byGprdpmSpilkin382017.html : this description can help for the message above. 