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I Water evaporation in semi-open system

  1. May 17, 2016 #1
    Hello guys!
    I'am having trouble getting my head around the physics of evaporation in quite specific situation.

    So i have a closed container with opening in the bottom (cross-section in picture below). Lets say that water has a constant temperature of 80°C and container has a temperature 90°C, therefore the water cannot condense on the inner edges of the container. Outside the container is air with sea-level pressure.
    cont.jpg
    So water evaporates and since vapor is lighter than air and it rises up. As it builds up in the container it should create some pressure (right?). Then due to the pressure the vapor should start escaping through the bottom (right?). So the first question is how much pressure is needed for the vapor to escape through the bottom?
    And since vapor pressure builds up, that should slow the rate of evaporation. Is there any way to calculate this?

    I will appreciate any help or links to related problems. Thanks!
     
  2. jcsd
  3. May 17, 2016 #2
    What is the equilibrium vapor pressure of water at 80 C? If the pressure in the room is 1 atm, what do you think the total pressure in the "vapor space" within the container above the liquid water is?
     
  4. May 17, 2016 #3
    At 80 C vapor pressure is 0.46 atm.
    Pressure in the "vapor space" space should still be 1atm since it's an open system.
    What does that tell me?
     
  5. May 17, 2016 #4
    That tells you that the most that the average partial pressure of the water vapor in the head space can be at any time is 0.46 atm. The rest of the gas has to be air.
     
  6. May 17, 2016 #5

    ogg

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    If the pressure outside the partially enclosed space is 1 atm, (assuming the opening isn't extremely small), then the pressure inside is 1 atm. I am not sure how the water remains 10° cooler than the container it is in, could you explain that (apparent) violation of the 0th Law of Thermodynamics? (Two objects in contact at equilibrium will reach the same temperature). While pressure is one way, and often used, to characterize a gas's composition, it is better here to consider that the key question is about the mixing of the air water gas with the atmosphere - which will occur around the opening. In real world conditions, the air-water mix will cool and water will condense into fog droplets and fall (there will also be some water vapor diffusion - unless the air is at 100% rh)
     
  7. May 18, 2016 #6
    So if I get this right, then the only way vapor starts flowing through the bottom opening into atmosphere is if the vapor temperature is at least 100 C, i.e., water boils ??

    The container is being heated. Therefore water will be few degrees cooler that container.
     
  8. May 18, 2016 #7

    russ_watters

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    Staff: Mentor

    No, they certainly were not telling you that. The gases act independent of each other: as long as the partial pressure of water vapor in your container is higher than the partial pressure of atmospheric water vapor, water vapor flows out.
     
  9. May 18, 2016 #8
    So far we have established that the composition of the vapor in the overhead space involves a mole fraction of water vapor of ≤ 0.46 and a mole fraction of air of ≥0.54 (all at 1 atm.). In the air outside, the mole fraction of water vapor is much less than 0.46 and the mole fraction of air is close to 1. So there is a concentration difference between the gas outside and the gas in the overhead space. Therefore, there is a mass transfer driving force for water vapor to exit the head space. Part of this the mass transfer (probably not much because the higher temperature is on top) can be by convection, so most will likely occur by diffusion in the tube. Knowing the diffusion coefficient, Fick's law can be used to determine the mass flow rate of water vapor. There is also heat transfer occurring by conduction occurring simultaneously within the tube. All this can be modelled.
     
  10. May 18, 2016 #9
    And should not that be when vapor temperature is atlest 100 C?

    Edit: ok no. Water vapor pressure in atmosphere is 0.023 bar at 20 C
     
  11. Jul 21, 2016 #10
    I have a related question but a bit different. Suppose I have a bottle of liquid (other than water - similar or low volatility). The bottle has a small opening (say 10 mm) on its top and wind blows over the top of the bottle. So, there will be pressure drop through the orifice or opening.
    Is there any way to calculate for a given pressure drop what will be the evaporation rate?

    Or let me rephrase the question -

    Is there any method to determine rate of evaporation in a open system with surrounding pressure less than atmospheric (or maybe vaccum) ?
     
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