bash my idea

[Ares]

alright guys, I need you all to bash my idea. yes Im a newbie I was reffered here.

I have a design, and would like it put to the test, any and all possible problems pointed out.

I first looked at the idea of evaporative cooling and identified the problems. this is what I came up with

size.
and
mist output.

thats all I saw as a problem.

so we'll start with size, the standard method I saw was shower head blowing down.

now the 2 points of this... time it stays in the air being "evaporated", this directly correlates to speed of descent or length of descent.

the second being, the size of the droplets, thus smaller droplets, more surface area, doesnt need to fall as far. but if they get too small, they wont fall. no good.

so... rather then just showering the water down, trickle it down through a brillo pad type material, this makes it fall slower.

alright, so I THINK thats somewhat helped, if it falls twice as slow, it would have the efficiency of 2 or 3x as tall of a chamber.

next problem, the output, now you want the air coming out to have 100% humidity, but vapor mist and humidity are different... it can be 100% humidity and it can be raining. not the same.

so though we dont want to recondense the humid water, we do want the mist caught up in the wind back... so route the output back down and bubble it through a VERY fine aquarium bubble stone into the holding tank. this should remove excess water particles from the mix. bubbling only the 100% humidity air out(also you get the smaller bubble effect from it.

heres a photo to help.

so please comment on what is wrong with this, or just tell me Im doing stuff thats been tried before.

[Ares]

yeah I see a problem... no fan

forgot about that, thats the "problem" ideally the fan would be inline... which Im not sure a water proof fan exists for the application.

since its open on both sides, you cant have a 3rd point to blow in as it would just blow down into the holding tank, not up through the mesh.

now you could seal it up top between the evaporating chamber and the holding tank. and blow into the holding tank, and up the chamber, and out. but then you need a seperate bubbler tank outside it.

[bigben2k]

You can recover some humidity from the exhausting air through a rock filter, but you'll need a significant amount of air pressure to do that.

The other issue is that you're not getting rid of all the heat: you're putting it back in the loop (warmer water).

[Ares]

not sure if thats true...(though might be). as I saw it, the 2 things coming out of the top of the "bong" are humid air, and an actual vapor mist. the mist is not warm, it should infact be cold, with the energy going into the water that truely evaporated into the gaseous form. I have a feeling like 50% of the water loss is not being evaporated at all, like a waterfall having the huge water cloud, thats not really all evaporated water, and it is infact extremely cold.

bubbling it through more water shouldnt pull the evaporated water out of the air, infact it should just further increase the cooling effects, but the mist should more or less drop out.

your right though, air pressure is key, it would need to be a sealed chamber, with a strong fan and huge tubing, along with multiple VERY large diffusing stone.