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Buoyancy and Energy Conservation
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Zero999:

--- Quote from: NiHaoMike on May 20, 2020, 07:13:38 pm ---I think that if the practicality problems were somehow overcome, the limiting factor is the finite (even if very large) amount of hydrogen available to run the process with. Hence it would not be perpetual motion any more than tidal power is perpetual motion.

--- Quote from: Zero999 on May 20, 2020, 12:19:20 pm ---Underwater is a good example. Imagine the balloon is deflated and underwater. The deeper it is, the more pressure and thus work will be required to displace the same volume of water, so lift the same mass.

--- End quote ---
Underwater, a volume of gas can lift far more weight than it can in air, thus increasing the energy that can be extracted from lift. So electrolyze water deep in the ocean to lift bags that power a generator. Or actually, forget the bags and just have a pipe to the surface, using the pressure of the water to pressurize the gas. Something doesn't seem right since you could burn the hydrogen after using the pressure, getting back water to allow the process to not use up anything.

--- End quote ---
True, but water is heavier than air so it will take more work to displace the same volume, given the same depth, because the pressure will be higher. 10m head of water exerts a similar pressure to the earth's atmosphere at sea level. This is why diving with a hose to the surface for air, only works at shallow depths without a compressor: the lungs aren't strong enough to displace enough water to allow them to fill with air. The process you're using to produce the hydrogen will require more energy, at higher pressures.
T3sl4co1l:
The higher concentration of [H2] and [O2] will raise the voltage of electrolysis.  This isn't important under most circumstances (it takes 100s atm to get significant; voltages goes as log(concentration), but you're talking the kinds of pressure where gasses are as dense as liquids, and these can be important.

That said, it should only be a modest increase in voltage, while the pressure is exponentially more.  Not a huge barrier to the scheme.

Weird things can happen to density and solubility, too.  CO2 for example is liquid even at more modest depths (that is, compared to the average (abyssal plane) depth of the oceans), and is denser than water, so it sits there, dissolving and diffusing gradually.  O2 is likely soluble enough that you would at least need to bottle it; H2 I'm not sure, but I'm sure there's some kind of clathrate that forms down there, at those temperatures and pressures.  For sure, given the sheer amount of water column they have to rise through, very little if any need reach the surface.

Tim
splin:

--- Quote from: T3sl4co1l on May 20, 2020, 10:39:34 am ---Right.  The buoyant force isn't constant, because air density isn't constant; you'd have to integrate over the altitude to see what work has been performed.

Tim

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As others have said, the balloon would be designed to expand as the pressure reduces with altitude, so if the temperature were constant, the ideal gas law states that when the air pressure halves the pressure of the hydrogen also halves thus it doubles in volume displacing exactly the same mass of air as at ground level.

The drop in temperature will reduce the hydrogen volume proportionally to the absolute temperature which at 50km is around 270K so reducing the lifting capacity by about 7% from a ground temperature of 20C/291K. However it's actually colder at lower altitudes - around 205K at 18km, or 30% less lift than at launch.

https://personal.ems.psu.edu/~brune/m532/m532_theatmosphere.htm
splin:

--- Quote from: Domagoj T on May 19, 2020, 10:39:55 pm ---
Anyway, I started doing the math, and either I'm doing something wrong, or there might be something to it. Somebody double check my math?

The current balloon altitude record holder (BU60-1) reached 50 km, and had a volume of 60 000 m^3. The balloon in question had a mass of ~34,5 kg and the scientific payload + parachute of another ~5,5 kg.
If we can be generous and replace that science payload and parachute with wind turbine + battery of equal mass (total mass of 40kg) and keep the 50km altitude. To lift 40 kg you need 36 m^3 of hydrogen.

Potential energy of a 40kg object at 50km is 19.600 kJ. That's absolute maximum of energy available to extract.
On the other side, to generate 3,2 kg of hydrogen, at 50 kWh/kg, we need 512 kJ (160   kWh).
--- End quote ---

512kJ = 142Wh


--- Quote ---Sure, back of the envelope calculation, but that's quite a margin.

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The numbers do seem to stack up but there are a few pratical issues!  :-DD

First problem:
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The air density at 50km is so low (0.08% ground level) that an enormous turbine would be needed to extract any useful energy. I tried a few cacluations:

To minimise the blade diameter the turbine would need to descend as fast as possible (power out is proportional to wind speed cubed). But this is limited by the maximum tip speed which needs to be less than the speed of sound.

Assuming a two blade prop, the optimum tip speed ratio (blade tip speed/wind speed) is around 6 (athough a single blade prop would likely be better for this application being lighter).

Arbitrarily limiting tip speed to 300m/s (Mach 0.88), the wind speed, ie. descent speed, should be <= 50m/s. At that speed the turbine would be losing potential energy at m x g x h = 40x50x9.8 = 19.6kW. Thus the wind energy input to the turbine is 19.6kW.

The kinetic energy in the wind passing throughy the turbine = 1/2 m v^2 where m is the mass of air through the turbine each second and v is the air velocity.

m = air density * blade swept area x wind velocity.

Thus power in = 1/2 x density x pi x r^2 x v^3

Therefore to capture 19.6kW of wind (input) power at 50m/s at 50km altitude the blade diameter would need to be 19.7m. That seems a bit unlikely, even for carbon fibre blades, given the available weight budget after the battery and generator mass.

However the air density improves dramatically at slightly lower altitudes - 4x greater at 40km and 18x at 30 km requiring blade diameters of 10m and 4.7m respectively. 4.7m may be achievable in 10kg or less (just guessing).

2nd problem:
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The weight of the battery. The potential energy of 40kg at 50km is 5444Wh. The theoretical maximum efficiency of a wind turbine is 59% (Betz limit). Assuming an overall efficiency of 40%, allowing for drag, turbine, generator, battery charger and battery charging efficiency losses the battery would have to store 2178Wh.

Current Li-ion batteries have an energy density of less than 300Wh/kg meaning >=  7.3kg of batteries. That isn't unreasonable so the batteries might not be a problem after all. The charge rate isn't a problem either - descent time at 50m/s from 50km = 1000s, thus charge rate is only 3.6C which is modest for most batteries.

Third problem:
=========
Weight of generator. Given 40% efficiency the generator has to be around 9kW. Perhaps somene else can provide a reasonable weight for a state of the art 9kW generator?

Fourth problem:
=========
How much fuel and manpower would be required retrieving this thing (what should we call it? Balloony Mc. Balloon-Face?) and returning it to it's start point after spending time in the jetstream? Realistically some means of controlling the descent would be needed to return it reasonably close to it's launch position. Much heavier units, descending much faster would emeliorate this problem somewhat but this is likely the killer issue.

Fifth problem:
=========
What happens to all the hydrogen dumped at the top of the stratosphere? How much will escape into space rather than recombining with oxygen? How long would the oceans last if this system were to supply most of mankinds' energy needs? What would be the maximum altitude to ensure most of the hydrogen is recaptured in the atmosphere?

Scaling up will help considerably - 40kg is a tiny power station but ultimately there will be severe limits to the number of locations where governments will tolerate large masses pluming earthwards at 100mph+ when in control - and considerably faster when they aren't!  :-DD

I like this idea - I reckon I could watch it all day! Lots of automation would be needed to capture the returning units and swap the batteries. Aviation in the vicinity might be somewhat hazardous however!

[EDIT] Forgot to add that as the air density increases as the unit descends the optimum rotor diameter and descent speed will change so the net efficiency of a (necessarily) fixed diameter rotor will likely vary considerably during it's descent.
Domagoj T:

--- Quote from: splin on May 21, 2020, 01:26:20 am ---512kJ = 142Wh
--- End quote ---
Oops, the error probably snuck in while I was playing with all the numbers and forgot to update a cell in excell. Nice catch.
The 160 kWh number is correct (3,2 kg of hydrogen x 50 kWh/kg).
The 512 kJ is wrong. It should have been 576 000 kJ (160 x 3600), which brings us short of 19 600 kJ of potential energy.
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