Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to estimate the amount of water condensing from air on a surface, given the air's temperature and relative humidity and how they change over time, the surface temperature, material's thermal properites, roughness and whatever else needs to be given about the air and surface?

For my purpose, we may assume the surface starts off dry, but the more general situation would be as interesting. If enough water condenses, it'll form drops and run off - can we account for that?

Are there some fairly simple formulas or rules of thumb? High accuracy is not needed; I'll be happy to get grams per sq meter per second (or whatever) to within a factor of two. (What if we wanted higher accuracy?)

share|cite|improve this question
I'm not sure about this one... it almost seems more like chemistry than physics. Then again, I have no idea how to answer the question, so I'll leave it alone. – David Z Nov 16 '10 at 5:01
@David I would say this is in the blurry interface of physics and chemistry, as you can see in the link, the basis for it is statistical mechanics. And actually, this is very much the kind of question I would like to see here. – Bernardo Kyotoku Nov 16 '10 at 9:00
I agree with Bernardo Kyotoku. To be quite honest, I'm sick of all the high energy / GR questions and would prefer a bigger spread of questions. – j.c. Nov 18 '10 at 2:56
up vote 1 down vote accepted

So you want to know how much water a certain surface adsorbs. This is really dependent on the surface material/conditions. Check adsorption and relative humidity on Wikipedia. To where I have analyzed, it seems that there is about enough information in the two articles. I am not a specialist on the subject so I might be missing some important factor.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.