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Capillary Action

  1. Feb 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Kindly view the attached.

    2. Relevant equations

    ΔP = γ(1/R1 + 1/R2)

    3. The attempt at a solution

    I've began the topic of fluid mechanics, capillary pressure, surface tension and such and was given this question to try. Now from my limited knowledge it seems to me that in the case of the liquid being water, a concave meniscus will be formed due to the water's adhesion to the inner plate walls (due to the water's polarity) and this will result in upwards capillary action. As for the case in which the liquid is mercury, the greater cohesive force between the mercury and the walls will cause the height of the mercury between the plates to drop lower than the surface of the mercury outside, and the mercury will form a convex meniscus. Is this correct?

    For the second part of the question, the young-laplace equation as given is in terms of two radii of curvature, but since we are dealing with plates, would I be right in thinking that it only has a single radius of curvature (the distance between the plates) and the other would be undefined (due to their being no definite end to plates in the direction parallel to them)? How this comes into play, and how to calculate the pressure above and below the meniscus, I'm not too sure on though. Can anyone please give some help?

    Much appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2015 #2
    Anyone?
     
  4. Feb 21, 2015 #3

    SteamKing

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    http://en.wikipedia.org/wiki/Meniscus
     
  5. Feb 22, 2015 #4
    Awesome, so from what I read there I'm more or less on the right track. As for the next part of the question, how is it that I am meant to proceed? Is it just that I rearrange the young Laplace equation for the pressures above and below? But then that answer won't be in terms of what I know from the question.
     
  6. Feb 22, 2015 #5
    :oldconfused:
     
  7. Feb 22, 2015 #6
    Anybody? :)
     
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