Elevation angle

by Pierre Benhaïem in Airborne Windenergy Forum · · 3 Replies · View last reply Respond
Pierre Benhaïem
Pierre Benhaïem

So 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/portfolio-item/wes-100/ 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 Benhaïem
Pierre Benhaïem

http://www.kiteboardingevolution.com/wind-window.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.
 Generally a good compromise is expected when the average of elevation angle is around 30°, the cubed cosine being 0.65, so still not too high. In the other hand 30° can be seen as not high enough in regard to land use issue, the biggest part of the tether being close to the ground. And one km tether cannot be compared with planes landing.

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 Benhaïem
Pierre Benhaïem

In 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 Spilkin
Gordon Spilkin

My 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 scale-up 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 light-weight and easily scalable, but requires a larger lifter kite to account for the differential forces in the cable drive system.
The turbines do not directly face the wind due to the angle of the tether. By inserting universal joints at each turbine, the blades will be able to reorient to face the wind. Let us call this orientation the "alignment angle". The tension in the tether would partially prevent this reorientation. For example, if the tether angle is 45 degrees, and the alignment angle is half the tether angle, then by the cosine cubed rule, the efficiency would increases from 0.354 to 0.789. This is a considerable improvement and more than compensates for the loss in efficiency of the universal joints. Other values for efficiency at various alignment angles when the tether is at 45 degrees are as follows:
ALIGNMENT ANGLE        EFFICIENCY
           10                                   0.550
           20                                   0.744
           30                                   0.901
           40                                   0.989

If double universal joints are used at each turbine then the differential rotational speed caused by the joint angle is neutralized, and all the connecting shafts (or tensegrity structures) rotate at exactly the same speed. The big advantage of the universal joints is that it eliminates bending forces on the connecting shafts and the system can respond to wind gusts without stress. The size of the alignment angle increases when the ratio of the length of the joining shaft to the length of the double universal joint increases. This is advantageous because longer joining shafts will eliminate the shadowing effects of the front rotors to the rear rotors. Similarly an increase of wind speed will increase the alignment angle as the turbines try to face the wind. This is counteracted by the increase in tether tension because of increased wind which would orient the turbines along the tether axis. It is not clear which of these two effects predominates.