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Basically I have a large Evaporative Cooler (one of those fans that you can put water in the bottom of). It has a really large container so it can last a day or two. The problem is once the water gets hot its doesn't work as effectively.

I was wondering if there's some type of cooling element I can purchse that could be rigged up in the water? I'm just not sure what I should be Googling for, does something like this exist?

Using a commercial refrigerator would be too big and too expensive. Is their some form of electronic device which can be used to apply refrigeration or cooling to reduce the water temperature, and how well would this work ?

Russell McMahon
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Sean Bannister
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  • I don't see what this has to do with electronics design... – Majenko Nov 09 '11 at 09:25
  • This can be marginally related to Peltier elements (http://en.wikipedia.org/wiki/Thermoelectric_cooling). However, without a more specific rephrasing it is quite off topic. – Konsalik Nov 09 '11 at 09:34
  • you want a way to cool your .. cooler? this is really a physics question. – JustJeff Nov 09 '11 at 11:05

1 Answers1

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You can use a "Peltier effect" cooler of the sort that you get in 12V powered portable refigerators / chilly bins/ eskys / ... . You can ither use an existing bin as the source or you can but Peltier effect devices from surplus and electronics suppliers. As you would then need a fan + heatsink + 12V power supply a complete bin may be cheaper and would be easier.

HOWEVER - it seems likely that the loss of effectiveness you are perceiving may not be real or may not be major. Another effect may be occurring.
I say this because the very major part of the cooling from such a device is supplied by the "latent heat of vaporisation of the water and the actual heating of the water as a liquid should be a minimal part of its cooling effect.

Energy required for heating water or for vaporising it is measured in Joule = Watt seconds.

The energy involved is

  • About 4.2 Joule per cc of water per degree C heated.

  • About 2260 Joule per cc of water when evaporated.

If you heat 1 cc of water fron 0c to 100 C you require only 4.2 x 100 = 420 Joule but you need 2260 to vaporise it.

If you heat your water by say 30C then you remove 30cc x 4.2 J/cc = 126 Joule/CC
but this is far less (~= 6%)
than the 2260 Joule needed to evaporate 1 cc of water

INTEREST:

It requires 2260 Joule to vaporise 1cc of water.
As there are 3600 seconds in an hour it takes
3600/2260 =~ 1.6 Watt to vaporise 1cc per hour or 1600 Watt to vaporise 1 litre/hour.

Russell McMahon
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