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Vapor pressure of water between two heated plates on a press

  1. Mar 15, 2017 #1
    Hi, I'm trying to do some figuring on vapor pressures. If one were to take two plates heated to 220F and press water between them in a press that exerts 100psi, what would the resulting vapor pressure of the water be?

    Now, what temperature would you heat the plates to if you were using a press that exerts 625psi and wanted to achieve the same vapor pressure as with 220F and 100psi scenario? Is this possible or am I going down a dead end? Is there a formula like (x) psi results in an increase of (y) temperature?

    Lastly - does longer time exposure to a constant heat and pressure continue to build vapor pressure, or does the vapor pressure remain constant once achieved?

    Any answers or formulas that might be used to solve this would be very greatly appreciated :-) I hope I didn't post this in the wrong section. Thank you!
  2. jcsd
  3. Mar 15, 2017 #2
    You seem to be confused over the concept of "vapor pressure." What is your understanding of the definition of this term?

    If a water is pressurized at 220 F and 100 psi, what do you think the state (of aggregation) of the water would be? (a) liquid, (b) vapor (c) solid, (d) combination of two of these
  4. Mar 15, 2017 #3
    My understanding is it should be in vapor at that point, that as temp rises, so does the vapor pressure of the steam escaping. My understanding is also that something pressed with great force between two plates will result in an increase of temperature. Trying to reconcile these two understandings (possibly misunderstandings), thank you for your help with grasping this concept!
  5. Mar 15, 2017 #4
    Unfortunately, both are misunderstandings.

    If you have a pure vapor and a pure liquid of the same substance in contact with one another in equilibrium, the temperature of the combination determines the pressure of the combination. Under these circumstances, the pressure is called the "vapor pressure at the specified temperature."

    Squeezing a liquid does not cause its temperature to rise, because water is nearly incompressible, and no work is done on the liquid.

    If you have water at 100 psi and 220 F, you will have all liquid water and no water vapor. The equilibrium vapor pressure of water at 220 F is less than 20 psia. This is way below the imposed pressure of 100 psia. So no vapor could be present.
  6. Mar 16, 2017 #5
    It makes very much sense that you way water is nearly incompressible, because I do notice that the vast majority of the water simply runs out, but it seems there is some that is trapped between the plates that evacuates as steam and reaches very much higher temps than what the plates are set at. Does that change what you are saying? If the water is allowed to escape the system as pressure and temps rise? If not my question may really be much simpler.
    It may just be: can the conditions within the environment of two heated plates at 220 F and 100 psi of applied pressure be replicated using an alternate temperature with 625 psi? (I see 220 F and 100 psi work for others in my industry; I have control over my temps, however my pressure source is 625 psi) I assumed there was a curve that follows along, as psi was increased, temperature could be decreased to achieve the same result. Simply trying to balance an equation so to speak, get as close to the environment that 220 F and 100 psi creates as possible using my press that puts out 625 psi.Thank you so much for helping me wrap my head around this! I have been feeling like I'm banging my head against a wall.
  7. Mar 16, 2017 #6
    If the water is able to run out, then its pressure can't rise.
    I can't follow what you are saying here.
  8. Mar 16, 2017 #7
    Two plates are heated at 220 F and pressed together at 100 psi. This creates a different environment between the two plates than if the plates were heated at 220 F and pressed together at 10 psi, correct?
  9. Mar 16, 2017 #8


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    Staff: Mentor

    If the plates are not sealed together then it sounds like you are just pressing together to wet metal plates (at atmospheric pressure). I would not say you are "creating an environment" in between them. Any water remaining would be in cavities caused by imperfections in the flatness of the surface and would not be pressurized. Then when you heat the plates, you evaporate/boil the water off at atmospheric pressure.

    Can you explain in more detail what exactly you are trying to accomplish here?

    In any case, to find out what vapor pressure (saturation pressure) boiled water at various temperatures is, you need to look at a saturated steam table:
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