Any ideas? Thanks.
As a kid we used to make snowballs in winter and put them in the freezer. Come mid-summer when we went to the frig to grab them and surprise everyone, they were gone. From a kid’s perspective, one could assume good ol’ Mom tossed them. In reality the snowballs sublimated into water vapor. In a closed container such as the freezer that could only mean the freezer air was dry.
If you take an empty frig, turn it on and let it cool for three days, the only moisture in the frig comes from the air in it that was cooled where the moisture in the stagnant air condensed out of the vapor. The frig does not introduce moisture on its own.
Wouldn’t refigerators dry the air in them? The air is drawn over the cooling coils which are below freezing. I would think that this condenses the moisture out onto the coils as ice crystals.
Makes sense. Thanks.
I would think though, that opening the door, or storing things like lettuce in the fridge could make the air moister than it would be if the fridge remained empty and closed? 
Of course, but the air is continually drawn over the cooling coils and so the added moisture is removed.
I used to be the janitor at a theater pre-WWII. One of my jobs was to operate the airconditioning system. It was a fancy chilled water system in which there was a huge duct with a rain field in it. The rain was chilled by the system and the air from the theater was pulled through the rain. The net effect was that the air was cooled below its dew point so that the moisture condensed out. Although the air went through a water spray, the air was dryer when it left the spray than when it entered. In the midwest, in August, that was an important feature.
It’s worth noting that:
a) many dehumidifiers work by running the air over a refrigeration couild to condense out moisture;
b) many car defrosters dehumidify the air with the A/C coils to reduce fogging and improve defrosting
c) “frost free” freezers dry the interior air so thoroughly with super-cold coils that any frost dimplu sublimates
d) in many home units, the refrigerator is cooled by regulated airflow to/from the freezer rather than having a separate cooling coil. Air that’s been dried enough to keep a 25F (-4C) freezer frost-free is incredibly dry at 41F (5C). I don’t have the saturated vapor pressure (or density) for water in air at 4C, but I do know that air will hold three times as much water at 5C vs -10C, so even 100% humidified (saturated) air at tyhe surface of a -10C coil will be a dry 33% Relative Humidity in a 5 C fridge. (I’d imagine that the small surface of the freezer coils would be colder than -10C to efficiently cool the much larger freezer to -4C. IIRC, heat transfer per time increases with the fourth power of the temperature gradient)
I believe that it’s only radiative heat transfer that T^4 holds- IIRC, the equation for Watts output is W= sigmaAT^4, sigma being the stephan-boltzmann constant (and that’s for blackbody radiation, which is idealized anyway). The heat transfer in a refrigeration system would be convective, as far as I know: W=hA (T1-T2). *
- it’s been a while since I did any actual math on heat transfer, but I believe I am recalling this correctly.
regards,
karinne
Wallacek is right - well, pretty right - it’s spelled “Stefan”. (*it’s only been about 12 hours since I did any actual math on heat transfer.) Heat transfer in a refrigerator is practically all going to be convective. In this case it’s about proportional to the first power of differential temperature, and also proportional to about the 0.8 power of air speed.
Oops- right on the Stefan/Stephan. I actually remember the constant, tho- it’s one of the easier engineering constants to recall, as it’s 5.67E-8. Much more mnemonic than, say, Avogadro’s or Planck’s constant.
Interestingly enough, I work in heat transfer engineering (computers / ICs) but I don’t do much by-hand calculating.
regards,
Karinne