| > 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. |
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.