3 ms·

Technically evaporation takes less energy than boiling as your product is ambient temperature water vapor instead of hot water vapor. You also don't have to rep

by jjk166 3y ago

Technically evaporation takes less energy than boiling as your product is ambient temperature water vapor instead of hot water vapor. You also don't have to replace heat lost to the environment by the hot water and steam during the boiling process. The difference can be lessened by good insulation and heat regeneration, but they still can't be perfect. Of course on the flip side, your goal is fresh liquid water, so you need to condense the vapor. Condensing hot vapor is easy, just expose it to cooler ambient conditions. Condensing ambient vapor is harder, and will require you to run something like a refrigeration cycle or a chemical desiccant system which will need energy to be regenerated. Most commercial systems use vacuum distillation which boils water at low temperatures and pressures, which has its own drawbacks but is generally more efficient.

AlexandrB 3y ago
Fair enough. I guess what I'm getting at is that the heat of vaporization sets a lower bound for how much energy you need to add, regardless of whether you boil the water or evaporate it.
- DoctorOetker 3y ago
  You are correct that at ambient temperatures there is an ultimate lower bound for energy. You are incorrect that at ambient temperature the lower energy bound is set by the latent heat of vaporisation, as others have pointed out this is theoretically recoverable. At ambient temperature there is however still a fundamental physical limit: the solvation energy of the salt in the water: https://en.wikipedia.org/wiki/Solvation#Solvation_energy_and_thermodynamic_considerations https://en.wikipedia.org/wiki/Solvation#Solvation_energy_and... Now the whole globe does not have the same ambient temperature, and as you know about global warming it would be great to shed some energy in the form of heat. There are many forms of desalination. Another way to desalinate is freezing: when salty water freezes, it pushes out the salt, so while desalinated water ice forms, the liquid water surrounding the ice will increase in salinity and become brine. One could then use simple nets or grills to separate ice from brine. Suppose one has a space elevator, or even a tether from a balloon, but capable of carrying significant weight. The temperature falls roughly adiabatically with height. Above the tropopause the atmosphere is essentially cloud-free, CO2 free and below freezing point (say -60 deg C). Hence the latent heat of fusion (freezing) can be shed to outer space. So it should be possible to lift salty water up an elevator, allow it to freeze over, separate brine from ice at the top, then lower the separated brine and ice. The energy required to lift the brackish water is compensated by the energy released by lowering the freeze-distilled water and brine. What comes up must go down, so simplistically speaking a pulley in equilibrium, so that the only energy intentionally exerted is lost to pulley and air friction. Then one would be cooling the planet and receiving frozen freeze-distilled water at the same time. The law of conservation of misery is typically not a fundamental law of nature, but imposed by reluctance to study of those who dictate artificial laws.
mannykannot 3y ago
I just found out that the latent heat of evaporation itself decreases with increasing temperature and disappears at the critical temperature [1]. With regard to the energy expenditure for reaching that temperature: if we were merely raising the temperature of the water and then cooling it down again, I think it would be correct to say that with a completely efficient contra-flow heat exchanger, perfect insulation, and no pumping losses, the steady-state heat input could be arbitrarily low. If we now modify this system to evaporate and then condense some of the water at the point of highest temperature, we would have to supply, and then extract, the latent heat at whatever temperature the evaporation is performed at. Once that has been performed, the outflow would comprise of the same amount of water as before (and at the same temperature), and it would be equally available to warm up the incoming stream as in the initial scenario (though now we would need two heat exchangers in order to keep the fresh water separate.) Of course, both the heat exchanger and the insulation will have losses, but we are rejecting quite a bit of heat in the condenser, and it is at the highest temperature in the system. Would that, in principle, be available to make up for any losses elsewhere? This makes me wonder if, counter-intuitively, it could be more efficient to do the distillation at higher temperatures, at least up to the point where the diminishing latent heat can no longer compensate for the losses of running at a higher temperature? I'm leaving out some considerations that I don't know how to handle (and probably others that have not occurred to me.) For one thing, there's the question of what happens if the evaporation occurs into a chamber containing some air, rather than just steam (my guess is that the relevant temperature is determined by the water vapor partial pressure.) For another, what difference does having salt dissolved in the water make? And this may all be moot, as this system has no moving parts, so the pressure is probably atmospheric (or somewhat below, if the condensation can be exploited to create a partial vacuum.) I have no idea if any of this makes the slightest bit of sense, and it's probably wrong - as you say, most systems run at reduced pressure. [1] https://www.engineeringtoolbox.com/water-properties-d_1573.html https://www.engineeringtoolbox.com/water-properties-d_1573.h...
- jjk166 3y ago
  The latent heat of vaporization decreases, but the amount it decreases by is the same amount of heat you need to put in to raise the water to that temperature. You're still trying to do the same thing - overcome the intermolecular bonds, but at higher temperatures you've already invested most of the energy you need to do so. With perfect insulation and heat recovery (zero loss) all that matters is the change in entropy between the starting and end products. Both the energy for raising the temperature and for the vaporization is theoretically recoverable (when you condense a vapor back to a liquid it releases the same amount of heat that it took to vaporize it). But you can't have perfect insulation and heat recovery in practice, and the losses become worse with increasing temperature - or more accurately increasing temperature difference, so trying to cool things down below ambient won't help you either.
hinkley 3y ago
If your water source is cooler than ambient air, is that still a problem? If it works less well on cool days, well you don't need as much water on cool days as you do on hot ones.
- jjk166 3y ago
  Ambient here is referring to the temperature of the water. The hotter the air is, the higher the partial pressure of water can be, meaning you need to take even more heat out to get it to condense.
  - trashtester 3y ago
    As long as you use the sea water to cool the steam, this is not really an issue. The more energy you extract from the steam, the less additional energy you need to evaporate the water.

by jjk166 3y ago

AlexandrB 3y ago

Fair enough. I guess what I'm getting at is that the heat of vaporization sets a lower bound for how much energy you need to add, regardless of whether you boil the water or evaporate it.

DoctorOetker 3y ago
You are correct that at ambient temperatures there is an ultimate lower bound for energy. You are incorrect that at ambient temperature the lower energy bound is set by the latent heat of vaporisation, as others have pointed out this is theoretically recoverable. At ambient temperature there is however still a fundamental physical limit: the solvation energy of the salt in the water: https://en.wikipedia.org/wiki/Solvation#Solvation_energy_and_thermodynamic_considerations https://en.wikipedia.org/wiki/Solvation#Solvation_energy_and... Now the whole globe does not have the same ambient temperature, and as you know about global warming it would be great to shed some energy in the form of heat. There are many forms of desalination. Another way to desalinate is freezing: when salty water freezes, it pushes out the salt, so while desalinated water ice forms, the liquid water surrounding the ice will increase in salinity and become brine. One could then use simple nets or grills to separate ice from brine. Suppose one has a space elevator, or even a tether from a balloon, but capable of carrying significant weight. The temperature falls roughly adiabatically with height. Above the tropopause the atmosphere is essentially cloud-free, CO2 free and below freezing point (say -60 deg C). Hence the latent heat of fusion (freezing) can be shed to outer space. So it should be possible to lift salty water up an elevator, allow it to freeze over, separate brine from ice at the top, then lower the separated brine and ice. The energy required to lift the brackish water is compensated by the energy released by lowering the freeze-distilled water and brine. What comes up must go down, so simplistically speaking a pulley in equilibrium, so that the only energy intentionally exerted is lost to pulley and air friction. Then one would be cooling the planet and receiving frozen freeze-distilled water at the same time. The law of conservation of misery is typically not a fundamental law of nature, but imposed by reluctance to study of those who dictate artificial laws.

mannykannot 3y ago

I just found out that the latent heat of evaporation itself decreases with increasing temperature and disappears at the critical temperature [1]. With regard to the energy expenditure for reaching that temperature: if we were merely raising the temperature of the water and then cooling it down again, I think it would be correct to say that with a completely efficient contra-flow heat exchanger, perfect insulation, and no pumping losses, the steady-state heat input could be arbitrarily low. If we now modify this system to evaporate and then condense some of the water at the point of highest temperature, we would have to supply, and then extract, the latent heat at whatever temperature the evaporation is performed at. Once that has been performed, the outflow would comprise of the same amount of water as before (and at the same temperature), and it would be equally available to warm up the incoming stream as in the initial scenario (though now we would need two heat exchangers in order to keep the fresh water separate.) Of course, both the heat exchanger and the insulation will have losses, but we are rejecting quite a bit of heat in the condenser, and it is at the highest temperature in the system. Would that, in principle, be available to make up for any losses elsewhere? This makes me wonder if, counter-intuitively, it could be more efficient to do the distillation at higher temperatures, at least up to the point where the diminishing latent heat can no longer compensate for the losses of running at a higher temperature? I'm leaving out some considerations that I don't know how to handle (and probably others that have not occurred to me.) For one thing, there's the question of what happens if the evaporation occurs into a chamber containing some air, rather than just steam (my guess is that the relevant temperature is determined by the water vapor partial pressure.) For another, what difference does having salt dissolved in the water make? And this may all be moot, as this system has no moving parts, so the pressure is probably atmospheric (or somewhat below, if the condensation can be exploited to create a partial vacuum.) I have no idea if any of this makes the slightest bit of sense, and it's probably wrong - as you say, most systems run at reduced pressure. [1] https://www.engineeringtoolbox.com/water-properties-d_1573.html https://www.engineeringtoolbox.com/water-properties-d_1573.h...

jjk166 3y ago
The latent heat of vaporization decreases, but the amount it decreases by is the same amount of heat you need to put in to raise the water to that temperature. You're still trying to do the same thing - overcome the intermolecular bonds, but at higher temperatures you've already invested most of the energy you need to do so. With perfect insulation and heat recovery (zero loss) all that matters is the change in entropy between the starting and end products. Both the energy for raising the temperature and for the vaporization is theoretically recoverable (when you condense a vapor back to a liquid it releases the same amount of heat that it took to vaporize it). But you can't have perfect insulation and heat recovery in practice, and the losses become worse with increasing temperature - or more accurately increasing temperature difference, so trying to cool things down below ambient won't help you either.

hinkley 3y ago

If your water source is cooler than ambient air, is that still a problem? If it works less well on cool days, well you don't need as much water on cool days as you do on hot ones.

jjk166 3y ago
Ambient here is referring to the temperature of the water. The hotter the air is, the higher the partial pressure of water can be, meaning you need to take even more heat out to get it to condense.
- trashtester 3y ago
  As long as you use the sea water to cool the steam, this is not really an issue. The more energy you extract from the steam, the less additional energy you need to evaporate the water.