Buoyancy and Energy Conservation - Page 1
Forum section: electronics / general

author: msuffidy
author_rank: Frequent Contributor
timestamp: May 19, 2020, 07:01:03 pm
content: I have seen some persons attempt to extract power from the constant force of gravity. This however usually fails because power output seems to require a one time motion, and a potential change in a falling object. So I came up with a situation that potentially breaks conservation, or it is more subtle where the inputs are.So I was thinking that say you have a hydrogen balloon. So it goes up to as high as it can go and then releases the hydrogen. But say you have propellers on the sides that charge a battery as the object falls. Is it possible you would have more charged energy than is required to make more hydrogen from water? Or is this somehow fundamentally flawed? Could buoyancy be a potential changing displacement that breaks conservation? Also there may be a vacuum balloon version of this that requires pumping, but that may be too heavy to actually do.

author: Zero999
author_rank: Super Contributor
timestamp: May 19, 2020, 08:42:18 pm
content: The answer is no.The balloon can only lift an object equal to the mass of the air inside it. Sucking the gas out and replacing it with a vacuum would require the same amount of energy as lifting the same mass of air to outer space. This is easier to calculate for a tank under water, because water doesn't compress like air does.

author: nctnico
author_rank: Super Contributor
timestamp: May 19, 2020, 08:56:46 pm
content: Well... if you want to make use of gas rising up... you could make a system where you let water evaporate at a low place, make it condense again at a high place and make the liqiuid water stream drive a turbine + generator. The sun would be an ideal heat source for such a system.

author: Domagoj T
author_rank: Frequent Contributor
timestamp: May 19, 2020, 10:39:55 pm
content: Quote from: Zero999 on May 19, 2020, 08:42:18 pmOP is suggesting to generate hydrogen, not vacuum.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, at50 kWh/kg, we need 512 kJ (160 kWh).Sure, back of the envelope calculation, but that's quite a margin.

author: Gregg
author_rank: Super Contributor
timestamp: May 20, 2020, 12:38:33 am
content: Quote from: nctnico on May 19, 2020, 08:56:46 pmThat has already been done and examples are really cheap.

author: cdev
author_rank: Super Contributor
timestamp: May 20, 2020, 01:11:38 am
content: That's basically what hydroelectricity is. It works really quite well, even on a small scale.Quote from: Gregg on May 20, 2020, 12:38:33 amCool beak, warm body, and I think some air has been evacuated from the bird's body and the liquid in the bird's body is likely alcohol.The bird is a bit like a lava lamp, which is made with naptha and wax, dye, and plain water.

author: Circlotron
author_rank: Super Contributor
timestamp: May 20, 2020, 07:16:57 am
content: Quote from: Domagoj T on May 19, 2020, 10:39:55 pmAt sea level perhaps. As you get higher the atmosphere gets less dense and the balloon would loose its ability to lift. But actually the balloon would gradually expand and become less dense itself. What would that do to the lifting capability vs altitude?

author: Circlotron
author_rank: Super Contributor
timestamp: May 20, 2020, 07:20:19 am
content: Quote from: Zero999 on May 19, 2020, 08:42:18 pmI would think it could lift a weight equal to the weight of the air the balloon displaces minus the weight of the gas or vacuum inside the balloon. Example - the balloon will lift a whole lot more if you place it underwater.

author: Circlotron
author_rank: Super Contributor
timestamp: May 20, 2020, 07:23:56 am
content: What if you used electrolysis to make Hydrogen and Oxygen bubbles several miles below the surface of the ocean and harnessed the force of these rising bubbles somehow?

author: Domagoj T
author_rank: Frequent Contributor
timestamp: May 20, 2020, 07:45:15 am
content: Quote from: Circlotron on May 20, 2020, 07:16:57 amI'm not a balloon expert, but from what I've seen, high altitude balloons are not filled completely on the ground and the balloons are much bigger than the volume of lifting gas exactly for that reason. As the balloon climbs, the lifting gas expands into the empty part of the balloon.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 20, 2020, 10:39:34 am
content: 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

author: Zero999
author_rank: Super Contributor
timestamp: May 20, 2020, 12:19:20 pm
content: Quote from: Circlotron on May 20, 2020, 07:20:19 amYes, I should have said displacement.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.

author: msuffidy
author_rank: Frequent Contributor
timestamp: May 20, 2020, 05:44:10 pm
content: Wow thanks for all the good analysis. One other way of altering buoyancy may be mechanically compressing the balloon to make it less buoyant.

author: james_s
author_rank: Super Contributor
timestamp: May 20, 2020, 06:21:38 pm
content: You're falling into the same trap as all these other free energy/over unity guys. Adding additional complexity until you reach a point where you don't understand or can't keep track of all the factors and then pondering whether the system would actually work. It won't, the laws of physics will always win, even if you manage to obfuscate it from yourself, there will always be some factor that will stop it from working the way you imagine it might.

author: NiHaoMike
author_rank: Super Contributor
timestamp: May 20, 2020, 07:13:38 pm
content: 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 pmUnderwater, 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.

author: Zero999
author_rank: Super Contributor
timestamp: May 20, 2020, 09:18:08 pm
content: Quote from: NiHaoMike on May 20, 2020, 07:13:38 pmTrue, 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 20, 2020, 10:10:45 pm
content: The higher concentration of  and  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

author: splin
author_rank: Frequent Contributor
timestamp: May 20, 2020, 10:26:28 pm
content: Quote from: T3sl4co1l on May 20, 2020, 10:39:34 amAs 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

author: splin
author_rank: Frequent Contributor
timestamp: May 21, 2020, 01:26:20 am
content: Quote from: Domagoj T on May 19, 2020, 10:39:55 pm512kJ = 142WhQuoteThe numbers do seem to stack up but there are a few pratical issues!First problem:=========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^3Therefore 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:========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!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! 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.

author: Domagoj T
author_rank: Frequent Contributor
timestamp: May 21, 2020, 06:52:12 am
content: Quote from: splin on May 21, 2020, 01:26:20 amOops, 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.

author: tszaboo
author_rank: Super Contributor
timestamp: May 21, 2020, 09:02:02 am
content: Quote from: NiHaoMike on May 20, 2020, 07:13:38 pmWhat makes you think, that it requires the same amount of energy to do electrolysis at high pressure than at 1 bar?

author: Zero999
author_rank: Super Contributor
timestamp: May 22, 2020, 09:18:12 am
content: Quote from: NANDBlog on May 21, 2020, 09:02:02 amIt depends on what you mean by the same amount?The higher the water pressure, the more work will be required to displace the same volume of water. If the pressure is higher, then more moles of hydrogen and oxygen will need to be produced, to displace the same volume of water. Another issue is, more of the oxygen and hydrogen produced will simply disolve in the water, rather than emerging as gasses.