Quote:
Originally posted by nicozeg
Ben: I don’t know the precise physics explanation, but an evaporative cooling system just take out the heat because that’s the energy needed to turn water into vapor. That energy is stored in the vapor and is released when it condensates and return to liquid state. It’s just the same as in a phase change cooling loop, where cold is generated at the evaporator, and heat is released at the condenser. Only difference is that here you forget about the condensation part, and dispose off the coolant in its vapor state; and refill it with liquid one. You just give Mother Nature the job of the condensation stage. 
Volenti: How much water evaporates daily that setup? I bet that several liters a day
And how did you solder the vertical pipes? I don’t see the tees.
Very efficient design! You don’t waste pump power with a big height.
OT: Ben, when I first read the forum in the morning, the radius thread was dead on the 10K reads mark.
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Ok, here is what actually happens. First you must realize, heat cannot be destroyed, only transferred.
First the water that drips is at room temp. This water when it drips evaporates, and the air becomes humid (and hot). The reason it becomes hot is because the hottest bits of water will evaporate first (this is the same as when you hop out of the pool, and it feels cold). So the water will eventually reach an equilibrium with the air temperature, a temperature where the water wont get any colder, and the air wont be heated up as much, if the air gets cooled by the water, the water then evaporates and heats the air up again. So to cool air, you would need to run air through pipes in the water at the bottom.
Or, in short, if the water gets cooled by the air normally, and you want the water to cool the air (when it gets cold) where will the heat go? As it cannot be destroyed.