Solar Wind Energy Tower Inc. (fka SWET)

[dog turd cannon]

Yes... to completely absorb 1 kg of water in perfect conditions requires 3.34 kg of dry air as I stated.

In the simplest terms the evaporation of 1kg of water liberates 2257 kJ of energy. The potential energy of 1kg of water at 800m is 1kg*9.8*800 = 7840 kJ. Now the potential energy at the ground is 0.

Now at 100 degrees F the saturation vapor density is 55 grams per cubic meter so 1 kg will absorb into 18 cubic meters. Now 18 cubic meters of “wet” air weighs 2.34 kg + 1kg (the water).

So the potential energy of this is 3.34 kg * 800m * Acc. Assuming it reaches 50 mph (22.35 m/s) at the bottom that is an acceleration of 0.63 m/s2.

That is an energy of 1668 kJ.

So it is 75% loss of energy.

~7840kJ up... and you get back ~1668kJ.