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I Water evaporation by negative pressure and 0% of RH

  1. Nov 9, 2016 #1

    I'm trying to evaporate water in some samples. I'm trying 3 methods.
    1. Baking samples with 100°C, normal air pressure, and normal RH.
    2. Put the samples in 0% RH, 23°C, and normal air pressure.
    3. Put the samples in container of -15 psi air pressure, 23°C, and normal RH.
    The total amounts of water evaporation from each tests are different.
    Is there any equation to quantify how much water evaporate in each condition?
    I'm looking for an answer like:

    [70°C, 14.7 psi, 30% RH] = [23°C, 14.7 psi, 0% RH] = [23°C, -15psi, 30% RH].

    I hope this is a just simple question :)

    Thank you
  2. jcsd
  3. Nov 9, 2016 #2


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    What's the total air flow?
  4. Nov 9, 2016 #3
    The air of each container should be refreshed by air from outside. However, I don't know the flow rate. Could you suppose each container has the same air flow?
  5. Nov 9, 2016 #4


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    Depends on how accurate a model you need. For a very rough model, maybe this will be enough:
    Look up the vapor pressure of water versus temperature.
    Calculate the partial pressure of water in the air.
    (surface area) * ((Vapor pressure) - (partial pressure)) should be approximately proportional to the evaporation rate. Of course, flow rate matters too, but you'll be getting into complicated models. Actively heating the water will speed up the evaporation, since it takes energy to evaporate.

    Case 1 will be different, since you are very close to boiling, which is faster than evaporation, since bubbles can form, greatly increasing the effective surface area.
  6. Nov 10, 2016 #5
    Wow this is tough. I'll keep studying about it.
    Thank you very much.
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