I was watching Thunderf00t’s self-congratulatory video about the failure of the Fontus self-filling water bottle, and it occurred to me to do some calculations.
Looking at a psychrometric chart we can see that under the 21° 32% relative humidity conditions the dude was testing the dehumidifier under, the dewpoint is about 4°; this means you need to cool the air by 17 K before you start getting any water out. To a great extent, you could do this with a heat exchanger such as a recuperator or regenerator which cools the air on the way into the condensing chamber and warms it back up on the way out; the total energy involved is the 1.012 J/g/K specific heat of air times 17 K, or about 17 J/g. That’s because the humidity ratio horizontal line on the psychrometric chart shows us that we have about 0.005 g of water per g of dry air under these conditions, which doesn’t change as we cool the air down to saturation; this 0.005 ratio means this is about 3.4 kJ per g of water contained in the air, which we hope to conserve in the heat exchanger.
Water’s enthalpy of vaporization is about 44 kJ/mol, 2.4 MJ/kg, or 2.4 kJ/g, and that’s the energy we have to remove in order to condense the water. You also have to remove more energy to cool the air (and, less significantly, the water) further. It probably isn’t practical to cool it below 0°, because then you have to waste more cold on forming ice, but (as you can see on the psychrometric chart) at 0° the saturation concentration is still .0037 grams of water per gram of air, so at that point you’ve only gotten about a quarter of the water out.
So in summary, you cool a gram of air down from 21° to 0°, spending 21.3 J to cool it down and another 3.1 J to extract 1.3 mg of water from it. Then you use this 0° air to cool down the incoming air through a heat exchanger as much as possible, getting back most of that 21.3 J. To quantify, I think a typical energy recovery heat exchanger can recover 85% of the “lost heat”, so we probably only “lose” 3.2 J in this scenario — that is, have to actively remove 3.2 J of heat from each gram of input air in addition to the 3.1 J spent on condensation.
Thermoelectric coolers like those used in the Fontus are unfortunately rather inefficient, with typical coefficients of performance of around 0.5, though perhaps as high as 1.5 for the small 20 K temperature difference we’re talking about here, according to Meerstetter’s chart. But there are other refrigeration technologies with better coefficients of performance.
So, we have to remove about 3.2 J + 3.1 J = 6.3 J of energy per gram of air, and we may have to spend 12.6 J to do that with a Peltier cooler. And we get 1.3 mg of water. This works out to about 9.7 megajoules per kilogram or liter of water produced. This is higher than Thunderf00t’s number of 2.3 MJ/ℓ, which was merely an approximation of the enthalpy of vaporization of water.
(Thunderf00t’s original video was working from the “energy factors” of dehumidifiers, which are required to be at least 1.2 ℓ/kWh to 1.6 ℓ/kWh to qualify for an “ENERGY STAR” label. These work out to 2.25 to 3 MJ/ℓ, but maybe that’s measured on a more humid or hotter day.)
To support one person with four liters of water per day, then, we would need about 39 MJ, which is an average of about 450 watts. This is feasible for a person to carry a solar panel for, but not to exert.
But that isn’t all we can do.
There’s supposedly a wilderness survival trick for distilling water using a sheet of clear plastic. You dig a hole in the ground, put a bowl in the center of the hole, cover the hole with plastic, weight it down with rocks around the edge, and put another rock in the center to make a dimple over the bowl. In the sun, the air under the plastic heats up like a greenhouse, evaporating water from the dug surface of the soil, until its dewpoint rises above the temperature of the plastic. Water then condenses on the plastic and runs down the slope to the dimple in the center, where it drips into the bowl.
To keep the air under the plastic saturated with moisture, it is often recommended to pile foliage in the hole, perhaps smashed up a bit in order to allow faster evaporation.
That is:
If you can raise the air temperature to 30° in your greenhouse, which is fairly easy, and get the air to 100% saturation by blowing it through smashed-up leaves or moist dirt or whatever, then you have about 27 mg of water per gram of air. If you can cool that back down to 20° — a purely passive process, and again one that can be facilitated with a recuperative heat exchanger — you can recover about 12 mg per g of it.
Not only is this almost an order of magnitude more humidity per volume of air, reducing the problems associated with device size, the bigger deal is that it’s powered almost entirely by low-grade thermal energy, like what you can get from sunlight. Some fans would be useful to circulate air between the condenser and the evaporator, and maybe some water pumps, but the mechanical power needed will be orders of magnitude smaller — less than 10 W, probably less than 1 W.
The thermal power still needs to be adequate to evaporate the water; if the thermal collector is 50% efficient (a common figure for inexpensive glazed solar thermal collectors, much better than the 16% of photovoltaic panels) then you need 4 kg/day · 2.4 MJ/kg / 50% = 220W on average. You probably need 660W during the day to average 220W for all 24 hours, so at 1000 W/m² you need almost a square meter of thermal collector per person.
You might think that you could avoid this solar evaporation nonsense, since temperature changes between day and night are often sufficiently large to cross the dewpoint by themselves. And that’s true, but even in moist climates, that is not dependable every day, and in dry climates it never happens.
Here in Buenos Aires, it’s currently 21°, and the dewpoint is 18°; we should hit 18° for a couple of hours at about 4 AM. This should happen again on Thursday, where the dewpoint will rise to 19° during the day (with a temperature of 22° to 24°) and then the temperature will fall to 19° during the night and morning.
At a dewpoint of 18°, the air holds about 0.013 grams of moisture per gram of air, of which you can condense about 0.001 g/g per kelvin of cooling. So to condense 4 kg of water by cooling the humid Rioplatense air by a single degree below its dewpoint, you need to slightly cool 4 tonnes of air, extracting about 4 megajoules of heat from it, plus the 9.6 megajoules (4 kg · 2.4 MJ/kg) for condensing the water. The most reasonable way to do this is probably to cool down a cold reservoir at night by some 15 megajoules. If your cold reservoir is just water, this is going to require minimally 3.5 tonnes of water and probably 7 or 10 tonnes. On the bright side, 10 tonnes of water can be stored a lot more easily than 4 tonnes of air.
Phase-change materials, like the recently-discussed saturated aqueous solution of sodium hydroxide, probably allow you to do this with only a few hundred kg of mass.
However, all of this is overlooking the fact that it wouldn’t work very well this week, because in both cases the low temperature comes after the high dewpoint. You’d have to store the cold from now until Thursday afternoon, four days. Doable, but a major undertaking.
Consider, by contrast, Albuquerque. Currently the temperature there is 17° and the dewpoint is -8°. Despite its 15° temperature swing from day to night, you are just not going to have any luck condensing water out of that air with a passive cool reservoir stored on a timescale of less than months; the highest the dewpoint gets is -1°.
If you send a balloon up, it gets cold. The environmental lapse rate of the lower troposphere, up to about 11 km, is about -6.5 K / km, so at 11 km the temperature is usually about 70 K colder than at sea level. In almost all cases, that brings it below the dewpoint at ground level.
If a balloon goes up and then comes back down, it has done no net work, so need not necessarily have dissipated any energy. By bringing heat from the ground up, or bringing cold from the upper atmosphere down, it has not done any work yet either — the thermodynamic free energy there comes from the sun heating the earth’s surface, energy which is eventually dissipated by winds forming convection cells. It’s just speeding up the convection a bit locally.
In practice, you need some energy difference to induce the balloon-atmosphere system into motion in the desired direction, but it could be very small compared to the heat you harvest. A net force of 1 N should be enough to induce a 100 kg phase-change reservoir payload to rise at an acceptable rate. If you let it rise 3200 meters, dissipating that 1 N in drag, it will have dissipated 3.2 kJ. Having cooled the payload to below 0° and dropping a little gas at that altitude so as to reduce lift by 2 N (about 170 liters at 1.2 kg/m³), it will then return to the ground dissipating another 3.2 kJ. A payload of 100 kg of water ice at 333 kJ/kg will have lost 33.3 MJ of enthalpy of fusion, which was bought with 6.4 kJ, about 0.02% of the total. You could probably even afford 10 N or 100 N.
That said, it is nontrivial to construct and fly a balloon of the 83 m³, 5.4 m in diameter, needed to lift 100 kg. (I’m disregarding the weight of the hydrogen or helium within as smaller than my margin of error.)
Your 100 kg of ice is enough to condense almost 14 ℓ of water at 2.4 MJ/kg.
Typical vapor-compression heat pumps like residential air conditioners and normal electric refrigerators have coefficients of performance around 2, which is largely a result of having to pump heat against a large heat gradient, typically 15° to 25° of difference. (Note, however, that this is already several times better than Peltier coolers typically reach.)
(Thunderf00t implicitly claims that it would violate thermodynamic law for a refrigerator to have a CoP over 1, but he is mistaken.)
We can’t do much about the ΔT (we have to pump heat from the water-condensing chamber into the surrounding environment) but if we can reach that COP of 2, then instead of spending 12.6 J per gram of air to remove 6.3 J of heat, we can spend 3.2 J, which works out to 2.4 MJ/kg of water, coincidentally exactly the same as the enthalpy of vaporization of water.
If it so happens that your air is already at 100% humidity, a heat pump could in theory condense water with an arbitrarily small ΔT, just by replacing the air very rapidly. With an arbitrarily small ΔT, a heat pump could theoretically have an arbitrarily high coefficient of performance. In practice, water-source heat pumps used in Japan for residential climate control often reach a CoP of 6, which would allow you to condense water at 2.4 MJ/kg ÷ 6 = 0.4 MJ/kg.
However, as Thunderf00t points out, under those circumstances, water is probably already abundant.
