Here are the following problems. Firstly, the sheer scale. One of the major producers makes 150 million gallons a year. Or half a million gallon a day, or roughly an olympic sized swimming pool. Or... lets go all out.
Half a million gallons is roughly 2 million liters, or 2 million kilograms of water. Assume air temperature of 20 degrees. To boil off the water, its going to take 4 kilojoules per kilogram per degree celcius.... so thats 640 GJ already. Ignore heat of vaporization.
Incident solar radiation at like ... 1000W per m^2. 12 hours -> 43200 seconds per day, so 1 m^2 receives 44MJ is a day... which means you need about 15000m^2 of surface area to evaporate off a day's worth of production in a day. Optimistically.
Actually, I think the heat of vaporisation is more than the energy to heat the water to boiling: 2.26 MJ/kg, which takes us to ~4TJ/day.
The 1000W solar radiation is also for the sun directly overhead without clouds, which obviously can't happen for 12 hours a day. I've found a source saying Bangkok gets something like 20MJ/m^2/day, and higher latitudes presumably have less. With those figures we're at 200000m^2 of evaporation area, and it's still optimistic (what about rain?).
Which really doesn't seem like that big a deal for a farm or large factory -- it's a 100m x 150m pad. Presumably you could keep it sanitary by using plastic sheeting over a concrete pad.
Half a million gallons is roughly 2 million liters, or 2 million kilograms of water. Assume air temperature of 20 degrees. To boil off the water, its going to take 4 kilojoules per kilogram per degree celcius.... so thats 640 GJ already. Ignore heat of vaporization.
Incident solar radiation at like ... 1000W per m^2. 12 hours -> 43200 seconds per day, so 1 m^2 receives 44MJ is a day... which means you need about 15000m^2 of surface area to evaporate off a day's worth of production in a day. Optimistically.