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Q & A: evaporation rates

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Most recent answer: 02/13/2012
Q:
I am working on a desalinization project and need technical advice. These are the basics: If I have a 1,000 sq ft pool of heated water, covered with a chilled metal baffle, then enclosed the entire unit in a greenhouse, how much liquid water can I produce through condensation in one day? Any ideas or feedback would be helpful. Thanks, Rick Mannoia
- rick (age 59)
sayville NY -USA
A:
Probably the limiting factor will be how much heat energy you can get to the liquid, since there's a latent heat of vaporization required for evaporation. That's 2260 kJ/kg. So it just depends on how much total heat you can pump in. A Joule is a watt-second, so for each kW of power you get about 86,000 kJ/day. That could evaporate about 38kg of water.

By the way, to maximize efficiency your chiller for the metal baffle should dump its waste heat into the saltwater. You probably already knew that.

Mike W.

(published on 02/09/2012)

Follow-Up #1: solar desalination output

Q:
Hello Mike W. and thank you for your reply. If my math is correct 38 kg of water is about 12 gallons. Not enough. Picture saltwater being fed through black (swimming-pool) pipe and solar heated. This hot water will fill a 10 x 100 ft pool. Additionally, saltwater will be cooled (running through canvas evaporation hoses) and dripped onto the tent-like 10 x 100 ft metal baffle. This entire 10 x 100 ft unit will rest inside an enclosed greenhouse. The cooled saltwater will drain into a spillway, while fresh water will be collected as it runs off the baffle. In a hot climate (Africa, Asia, South America) this greenhouse will generate a great amount of heat and chilled water. 12 gallons per day is not enough. Is my math correct? Do you have other suggestions? Any feedback would be appreciated. Rick Mannoia
- rick (age 59)
sayville, NY USA
A:
38kg is just over 10 gal. Of course, that's the approximate expected daily output for a 1kW source of heat. You may be able to get a lot more solar heat than that. A typical average solar energy input is around 280 W/m2. If you're in some sunny desert you may get more. If your gathering area is comparable to your pool area (~100m2), then you could get around 30kW. That would get you around 300 gal/day.

Depending on what your scarcest resources are (start-up money, land, salt water, ...) you may be able to do better. The thermodynamic limit is certainly a lot better, because when you let the input solar energy just heat up water you throw out quite a lot of its free energy, the parts due to its frequency spectrum and to its narrow range of directions. There are ways of harnessing those parts (photovoltaic cells, solar concentrators via mirrors) to use the solar power to generate electricity. That electricity can then be used to drive a heat pump, heating up your salt water and cooling the metal baffle. So long as the waste heat of the electrical generation process is used to also heat the saltwater, you'd get more water out per amount of land used, although with higher initial capital costs.

Mike W.

(published on 02/10/2012)

Follow-Up #2: solar desalination limits

Q:
Hello Mike W., thank you again for your input. I wish I could send you a drawing of this “Aqua-Pod” I am working on. Unfortunately this website does not allow attachments. I would like to show you an illustration. Yes, I would like to build this water farm in undeveloped countries. Yes it will be hot, and yes, I have access to unlimited salt water. What makes this project unique is that it does NOT require outside energy (actually nil energy) I may have to pump the saltwater up into a holding tank. Using solar power and gravity, as the heated water evaporates, hits the chilled baffle, it will condensate and be piped into a holding tank. 300 gallon a day? Even by my estimation that seems high. If we figure 10 hours a day sunlight heating the pipe & heating the greenhouse, dripping cool water on the baffle, do you really foresee 300 gallons? Thank you for your help, I would like to include you in this project. Rick Mannoia
- Rick (age 59)
Sayville, NY USA
A:
Hi Rick- That 300 gal/day assumed that your average sun intensity over the course of the day was equal to the world-wide average, that all of the incoming power over 100 m2 was used to evaporate water, and that all of the evaporated water was gathered. The last two assumptions are obviously a bit optimistic- some heat will leak out and some of the evaporated water will condense somewhere besides your gathering system. The first assumption might be a little pessimistic, since it sounds as if you may be working in sunnier-than-average locations near the equator.

So maybe 300 gal/day is a little optimistic, but if you really use most of your area for sun-absorbing material and get that heat into the water, it shouldn't be way off.  Would 200 gal per day be ok?

Mike W.

(published on 02/13/2012)

Follow-up on this answer.