What is the rate of evaporation per cm² or m² of water at 80°C and 5%
salt content given that atmospheric pressure is normal and temperature
is 80°C?

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Request for Question Clarification by hedgie-ga on 08 Sep 2005 23:42 PDT

You cannot expect a quantitave answer when saying 
 'wind is minimal'.  
 You can specify closed space, water and water vapour in a (closed)
testtube. In this case an equilibrium between saturated vapour
and water will be established - and then rate will be zero.
To appreciated some of the factors, you may look at
http://answers.google.com/answers/threadview?id=512638

Also depends on humidity.

In addition to the ambient humidity (as racegar has pointed out), the
evaporation rate will also depend on the windspeed.  Wind has the
effect of both (1) reducing the thickness of the atmospheric boundary
layer, which steepens the humidity gradient near the air-water
interface, and (2) causing raves/ripples (and in the extreme case,
detached water droplets), which increases the effective surface area
of the water.

The humidity will be 50% at 80°C there is minimal wind as it is in an
enclosed space.

Buyn-ga,
Since the subject is "Water desalination", were you considering
drawing off and condensing the water evaporated at 80°c?

Yes evaporated water will be condensed and will be withdrawn from the chamber.

Okay, so we want detailed information about low temperature
desalination processes.  I would imagine that a statement that a
device produced a certain amount of distilled water per hour, at 80°c,
and gave the surface area of the container, then the rate of
evaporation could be calculated.
From what I have seen, 80°c appears hotter than temperature mentioned
some places, and there are probably other variables.

This is a lengthy site that may have some information.

http://www.usbr.gov/pmts/water/media/pdfs/report007.pdf