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Archived Fluid statics and relative humidity problem

  1. Aug 8, 2014 #1
    1. The problem statement, all variables and given/known data
    A tube (both ends open, height h, diameter d) is immersed into water so that length h1 of the tube is above water. Then the top of the tube is covered tightly and the tube is pulled out of the water so that length h2 is above water. The relative humidity of air is [itex] \phi [/itex] and the temperature is T. What is the length of the air column in the tube after it has been pulled out? Water evaporation due to the humidity has to be taken into account.


    2. Relevant equations

    [itex] p_1 V_1 = p_2 V_2 [/itex]

    3. The attempt at a solution

    I know how to solve the problem without taking the evaporation into account:
    [itex] p_1 V_1 = p_2 V_2[/itex],where p1 (atmospheric pressure) and p2 are the pressure of the gas (air) before and after pulling the tube out of the water, and V1 and V2 are volumes of the air. We can rewrite the equations
    [itex] p_a h_1 = p_2 h [/itex]
    Then another equation of fluid equilibrium when the tube is pulled out
    [itex] p_2 + \rho g (h_2 - h) = p_a [/itex]. Solve that for h and that's it, except for the water evaporation part. How is it supposed to be taken into account? I know what relative humidity is basically, but failed to find any useful mathematical approach in this situation. Any help appreciated!
     
  2. jcsd
  3. Feb 4, 2016 #2
    In the initial state, the partial pressure of the water vapor in the air is φpsat, where psat is the saturation vapor pressure of water, and the initial partial pressure of the air is pa-φpsat, where pa is the atmospheric pressure. In the final state, the partial pressure of the air in the head space is (pa-φpsat)h1/h2, but the partial pressure of the water vapor in the head space will be psat (because the gas in the head space will become saturated with water vapor). So the total pressure of the gases in the head space will be $$P_{tot}=(p_a-\phi p_{sat})h_1/h_2+p_{sat}$$
    This must match the pressure at the top of the water column ##p_a-\rho g (h-h_2)##. So,
    $$(p_a-\phi p_{sat})h_1/h_2+p_{sat}=p_a-\rho g (h-h_2)$$

    This equation can be solved for h2.
     
    Last edited: Feb 4, 2016
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