Prove me wrong, please?
edit: thanks for all the great comments, this is really helpful. My main take-away is that it does work, but requires dry air. In humid conditions it doesn’t really do anything.
Spouse bought this thing that claims to cool the air by blowing across some moist pads. It’s about as large as a toaster, and it has a small water tank on the side. The water drips onto the bottom of the device, where it is soaked up by a sort of filter. A fan blows air through the filter.
- Spouse insists that the AIR gets cooled by evaporation.
- I say the FILTER gets cooled by evaporation.
- Spouse says the cooled filter then cools the air, so it works.
- I say the evaporation pulls heat (and water) from the filter, so the output is actually air that is both warmer and wetter than the input air. That’s not A/C, that’s a sauna. (Let’s ignore the microscopic amount of heat generated by the cheap Chinese fan.)
By my reckoning, the only way to cool a ROOM is to transport the heat outside. This does not do that.
We can cool OURSELVES by letting a regular fan blow on us = WE are the moist filter, and the evaporation of our sweat cools us. One could argue that the slightly more humid air from this device has a better heat transfer capacity than drier air, but still, it is easier to sweat away heat in dry air than in humid air.
Am I crazy? I welcome your judgment!
it’s not technically a net loss- some of that energy is lost as it escapes the system, but conservation of energy generally means that as the air cools down, the water gets warmer. it’s just that the water has extremely excellent thermal mass, meaning that the air appears to cool much more than the water gains heat. This is especially true if the water itself is cold to begin with. (ie, blowing it over ice cubes.)
This is not the dominant factor, and this system will still work even if the water is warmer than room temperature. The primary energy sink is the vaporization of water.
Vaporizing 1g of water takes 2257 Joules. https://en.wikipedia.org/wiki/Enthalpy_of_vaporization
The water does not change temperature during that process.
Heating 1 g of water 1C takes 4.184 J. To heat it from from 0C to 25C (about freezing to room temp) it thus takes104.6 J, much less than the 2257 Joules required to vaporize it. These numbers could be modified to properly account for the variace with temp, but the effect actually gets larger then I believe.
No energy leaves the system; it goes into changing the state of water and is stored in the water.