At 3:17 PM 4/13/5, Stephen A. Lawrence wrote: >Suppose the air is >stationary, and the turbine is being dragged along by the cable. Then >it's obvious where the energy is coming from: It's coming from the >cable, and the generated power must be no larger than F*V where F is the >force on the cable and V is the velocity at which the generator is being >dragged. Unless you've got some way of cooling the breeze passing over >the fans and turning the extracted heat into electricity, there's just >no other energy source in the picture. > >So, to turn that around, the force on the cable must presumably be, at a >minimum, the power generated, divided by the wind speed.
Yes, I think you are on the right track for getting simple a ballpark estimate of the parameters. The energy E_m carried by a mass m of air at velocity v is E_m = (1/2) m v^2 Given a cross section of area A of air flow covered by a rotor we have an air volume flow rate of A*v and mass flow rate of rho*A*v. We thus, by substitution of mass flow rate for m in the above, have an available power P given by: P = (1/2) (rho*A*v) v^2 = (rho*A/2) v^3 If as you suggest, the wind is operated on over a distance D, we have roughly E = F*D E/t = F*(D/t) = F*(v) P = F*v F = P/v = (rho*A/2) v^2 We thus see roughly that power is proportional to v^3 and resistance is proportional to v^2, which are well known facts. In a "standard atmosphere" (see CRC Handbook) air at altitude 0 has rho = 1.225 kg/m^3 at temperature 288 K, but at 20 km rho = 0.0891 kg/m^2 at 216.6 K. Assume the jet stream is moving at about 200 mi/hr, or about 90 m/s. We would be fantastically lucky to build a rotor with 10 m radius, for A = 314 m^2. This gives: 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. I think even harder than achieving working windmills of this kind may be keeping track of where the jet stream is moving and getting windmills there in time. It strikes me as far better to build windmills on mountain top ridges in places like Alaska where the wind rose data show fantastic energy densites. The air density is 10 times higher as well. The problem has been overcoming the demanding engineering problems associated with arctic temperatures and ultra high winds, as well as the formidable logistics problems associated with a highly remote mountain environment. Engineering of windmills and arctic lubricants has greatly improved since the 1970's when the State of Alaska last looked into moutntain top windmill citing. Windmills now operate successfully in the antarctic environment. The cold of the arctic environment might also assist in conversion of that energy into storable and transportable form. Regards, Horace Heffner
