# Why will dew point matter?

I don't understand why evaporative cooling can't go to temperatures as low as fridges (like 3°C)? For a link to the design I'm talking about: https://engineering.stackexchange.com/questions/35452/how-effectively-will-this-evaporation-setup-cool-down-the-water

As you see, the accepted answer tells me it is NOT possible to bring down the temperature to 3°C because of "dew point consideration".

I calculated the dew point for 5°C at a relative humidity of 85% and the dew point is 2°C. How does that mean that the water mist cannot cool the container that far?

EDIT: I think one reason why the guy answered it's impossible (in the link above) is that he thought a constant water spray volume is going on. That would mean that the relative humidity goes to 100% eventually, as the water-holding capacity of air goes down with temperature. The thing is, I gradually decrease the water spray volume, so that the relative humidity stays constant at 85%.

• Since the original question was on Engineering SE, perhaps this follow up might be better there as well? – Jon Custer May 8 at 14:37
• Yes, but this is more theory than practical... – El Flea May 8 at 14:37
• The engineering link provides good advice. In addition, note that the closer your cold temperature approaches the dew point temperature, the lower the rate of heat transfer will become. Once that rate of heat transfer matches the rate at which heat leaks in from the environment, you will reach an equilibrium temperature that you can't go below. Note that because a temperature difference is the driving force for heat transfer, this equilibrium temperature will take quite a long time to reach. – David White May 8 at 15:38
• @DavidWhite So in short, you're saying that it IS possible to cool something down from 40°C to 3°C in atmospheric pressure? – El Flea May 9 at 6:10
• @ElFlea, it's theoretically possible, if you have PERFECT insulation that prevents all heat flow from the outside world. You will, of course, be unable to find insulation that can do that, so you will find that it is practically impossible to cool something down to the dew point via evaporative cooling. Why are you so concerned with proving that such a feat can be done? – David White May 9 at 17:00

Assuming you did your calculation correctly, evaporative cooling should occur if the air temperature is above 2$$^0$$C. As far as I know unless the relative humidity is 100% at 3$$^0$$C there should be no reason for evaporative cooling to cease at 3$$^0$$C.
The rate of evaporation is going to be roughly proportional to the "driving force," determined by the difference between the equilibrium vapor pressure of the water and the partial pressure of water in the gas phase. In terms of relative humidity, this reduces to: $$driving force=P_{vap}\left(1-\frac{RH}{100}\right)$$So, higher RH is going to reduce the driving force. But, even more important, lowering the temperature reduces the equilibrium vapor pressure, and thus the driving force. As an example, near 0 C, the equilibrium vapor pressure is about 611 Pa, while at 25 C, the equilibrium vapor pressure is 3170 Pa. So, for the same RH, the evaporation rate at 25 C is about 5X the evaporation rate near 0 C.