In reply to Horace Heffner's message of Wed, 13 Apr 2005 19:09:02 -0800: Hi, [snip] > P = (rho*A/2) v^3 = (0.0891 kg/m^3) * ((314 m^2)/2) (90 m/s)^3 > > P = 10.2 MW > > F = (10.2 MW)/(90 m/s) = 113,300 N = > > F = 11,550 kg force = 11.5 metric tons of force > >So, even at 200 mi/hr jet stream speed, and light 20 km altitude air, we >are still looking at about a metric ton of force per megawatt. The >assumtion that all the energy can be extracted from the cross section area >A is also wildly optimistic. However, the less efficient the rotor at >extracting the energy fromthe air flow the less drag it is likely to exert >as well, so the ratios are not too so far off, but the energy recovery >would likely be worse than this predicts. > A Kevlar 49 cord just 1 mm in diameter would be strong enough, not counting the weight of the kite, but you would have 2 cords, and they would only contribute an extra 50 kg each to the weight. The cord would be about 50 km long. An aluminium wire with an area of 1 mm^2 would have a resistance of about 1.3 k ohm over that distance, but would only lose about 1.3% of the transmitted power if carrying 10 A at 1 MV (DC) (2.6% when taking the return wire into consideration). The Al wires would add an extra 270 kg to the weight.
This would seem to be feasible, and there is plenty of latitude in the choice of parameters, mostly due to the strength of kevlar 49, however generating 1 MV might be a bit difficult. Lower voltages are going to increase the weight of the conductors, though dropping to 0.5 MV might make a lot of difference in difficulty. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
