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We'll be cooling 10 comps with a 550W pump with 50m max head.
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The bong will be used outside. how much water would be lost?
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Math time!
10 comps and a "550w pump" probably works out to less then 1kw IME. Don't believe the pump power ratings, typical is generally half that, if as much. (Like my 24w Eheim 1250 that uses 9w). Lets say 1000w.
Enthalpy of evaporation is 22kj/mol for water at 101.3kpa and (IIRC) 20C. That works out to one mol lost every 22 seconds. A mole of water is 18g and a liter has a mass of 1kg, so you'd loose a liter every 1222 seconds or about every 20.5 minutes. That means 70.7 liters (about 18 gal) evaporated per day.
Plus the evaporated water gets replaced with cold water from the tap, so you get some "free" cooling right there. Lets use the above figure for simplicity of 70.7 l. Thats 70700g at say 10C below bong temp so you'd get almost 50 minutes of free cooling. Of course if you want sub ambient temps you'd loose that much trying to cool incoming water . . .
Not too bad compared to AC.
Edit: Just checked and according to my dad a cubic foot of water (~7.5gal) is 10 cents or so here in Arizona. If thats correct, then you'd be looking at just
24 cents a day assuming you can run the coolant at outside airtemp. Thats way less then a tenth of what AC would cost. I think I see why swamp coolers are so popular here.
You may want to check my math though.