[mannykannot]

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.h...

[jjk166]

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.

[mannykannot]

> 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.

I Don't think that can be right - for example, at 18 °C, the isobaric specific heat of water is 4.18 kJ/(kg.K), while the rate of change of the latent heat with temperature is only 2.4 kJ/(kg.K), and at 100 °C, the figures are 4.22 kJ/(kg.K) and 2.7 kJ/(kg.K) respectively.

> 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).

Only up to a point: you cannot condense steam at 100 °C in a condenser where the incoming coolant is also at 100 °C. Using only passive methods (heat exchangers) you cannot, even assuming perfect efficiency, recover all of the heat needed in a distillation process for reuse within that distillation process.

While distilling at higher temperatures need not be anywhere near as inefficient as it seems if you don't include the use of heat exchangers, the numbers given above don't seem different enough to justify distilling at a higher temperature than necessary, under realistic assumptions of efficiency, which is not surprising, given that it does not seem to be done.

[jjk166]

Vaporizing at 18 C produces vapor at 18 C, vaporizing at 100 C produces vapor at 100 C. Once you account for cooling the vapor back down to ambient temperature (specific heat pretty close to constant varying from 1.86 to 1.89 kJ/(kg.K) from 18 to 100 C) you get the same amount.

You use the cold incoming water to condense the hot vapor, thus pre-heating it prior to distillation. You get out all the heat you put in, the issue is that vapor has higher entropy than liquid water so you can't use that heat efficiently enough to vaporize the same quantity of water. That's in the efficiency term we are handwaving away.

[trashtester]

One way to bring the recovered energy higher could be to use some of the steam energy to drive a steam engine (with most still used in a heat exchanger). The steam engine, in turn, can provide at least some of the power needed to take the sea water in the last stage past 100°C.