[petermcneeley]

* Not to clutter hn but here are my notes for that discussion:

http://www.sfu.ca/~mbahrami/ENSC%20461/Notes/Refrigeration%2... http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/watvap.ht... this shows that its quite large amount Efficiency fridge = 1 / ( Th / Tl - 1)

There is actually way more water in the air than 0.04 grams... i think this is because liters not m^3 for 1m^3 there is 30 grams at 30 degrees and 17.3 at 20 degrees. its at least ~15/10 - > 1.5 grams per Centigrade Say we do small diff of 4 degrees, 30 to 26. We gain 1.5 * 4 = 6 grams / m^3 To do this we need to cool 1m^3 water at 30 to 20. Now this is tricky because humidity but https://www.engineeringtoolbox.com/heating-humid-air-d_693.h... h = 1.01 (kJ/kg.oC) t + x [1.84 (kJ/kg.oC) t + 2502 (kJ/kg)] x is relative weight so its actually small like 0.03 so it changes the result very little h = 1.06 (kJ/kg.oC) t + x 2502 Kj/kg partial diff with t gives us 4.24kJ/kg change for cooling the air (sensible) partial diff with x gives us (0.006kg delta) 2500 - > 15kJ/kg for this 4 degree change for condensation with 6g/m^3 of water Since the cooling of air that didnt result in water is reclaimable energy lets ignore it Now efficiency of this heat transfer 1/ (303.15 / 299.15 - 1) = 75x heat transfer efficiency. Does this make sense? we know heat pumps are amazing :) so per gram its like 15kJ/6g = 2.5kJ/gram -> this is really expected as 2.3MJ/kg from original video!

However this heat (Q) is moved via work (W) at 75x rate This means to get 1kg (1L really) it only takes 30KJ So at 15watts (J/sec) -> 2000sec -> 33 min Perhaps this wont work because the MASSSIVE heat sink required :)

How much air needs to be moved? we get 6 grams for each meter cubed of air. Not great thats about 166m^3 of air that needs to filter. https://www.amazon.ca/Ultra-Strong-DC12V-Cooling-200CFM/dp/B... 200CFM -> 5.66 m^3/min 60 min/h = 339 ... this is double what we need. its about 1.6A 12v = 15 watt... but its more than we need. Half it?

[robotrout]

> Perhaps this wont work because the MASSSIVE heat sink required

Perhaps not! Thanks for running the numbers.