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Old 06-20-2014, 05:32 PM   #20 (permalink)
mort
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Quote:
Originally Posted by P-hack View Post

I guess the problem can be simplified a bit (and water is probably a better test medium)

Hello P-hack,
Naturally the design could be optimized to demonstrate the viability of the idea. For instance, make the gas envelope a torus with the wind generator-propeller in the opening and the various mechanicals underslung of the gas bag.
Consider the Goodyear blimp, Spirit of Goodyear. It has a gas capacity of about 200,000 cu ft which in helium provides 14,000 lbs. of lift (at sea level etc.)
The amount of power you can get out of the wind, that's during rapid descent, is:
P = 1/2 A*d*v**3
where v is the wind speed
d is air density
and A is the swept area of the blades
For modern designs, efficiency of the actual wind power mechanism is around
45% - that is 45% of the kinetic energy in the wind can be recovered as electricity.
A regular torus 120 feet wide with a hole 30 feet wide (R=45' and r=15') has a volume of 200,000 cu ft. The hole in the center has an area of 700 sq ft
Air density is 0.0023769 slugs/cu ft (sea level, 70 degrees F, etc.) The power produced is about:
P = .45 * .832 * v**3 (efficiency times area times density times speed cubed)
Now this gets a bit scary-
If you could descend at 125 mph (183 ft/sec) that would develop 9000 hp for 54 seconds (falling 10,000 ft) Assume a compressor-expander with round trip efficiency about 80% 9000 hp now is the difference between the power needed to compress the gas and the power you get back expanding the gas. You could run a compressor at 45000 hp for .91 minutes and compress 200,000 cu ft of helium in that time (3 stage compressor 15 psi to 215 psi helium adiabatic expansion coefficient=1.67). By the way that's about 8 Tesla S batteries worth of juice. That's about 100% of your gas bag volume. Assume the torus is actually streamlined for descent, and the Cd is just about the Cd of the windmill -- .5 at 700 sq ft. The drag on the airship falling at 183 ft/sec is 14,000 lb, so the ship is at terminal velocity for free fall.
The point is that you can fall fast enough to generate the power needed to recharge the battery. You don't need to completely deflate the bag to descend, but the faster you fall the better.

The airship could be scaled up to quite huge size so the load capacity would be sufficient to carry that big compressor and battery.
-mort
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