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Ideal gas law combined with vapor pressure

  1. Mar 1, 2015 #1
    I am working on a DIY project and want to understand the system/mathematical model of a traditional steam espresso machine.

    How I have started to think about the problem is a closed container with liquid and gas. At room temperature I know the vapor pressure of the water. As I increase the temperature the volume will stay the same. When the temperature increases the vapor pressure will also increase but at the same time the pressure of the gas should follow the ideal gas law. I am assuming for simplicity reasons that the liquid volume is not changing (the liquid to gas change is insignificant and the liquid won't compress).

    Once the pressure/temperature reaches a set level the valve is opened. At this stage the temperature remains constant but the pressure decreases and volume increases.

    Can I get help finding the equations during these different states to describe and calculate the system of the liquid and gas in the closed container (temperature and pressure based on the volume)?
     
  2. jcsd
  3. Mar 1, 2015 #2
    Is there air in the head space of your closed container?

    I don't understand the situation when the valve is opened. Is heat supplied to the liquid to maintain its temperature? Is the gas exposed to the atmosphere, so that the pressure drops suddenly to atmospheric? How is the volume increasing; is this just that the gas is escaping from the container into the air?

    Chet
     
  4. Mar 1, 2015 #3
    Chet,

    Thanks for the quick reply!

    Yes there is air in the head on the container. For example the container is 500mL and only 300mL of water is poured in before being closed.

    Once the valve is opened the pressure from the gas pushes the hot water through a tube at the bottom of the container. This way the gas isn't exposed to the atmosphere and as the water is pushed out the volume of the gas increases and the pressure decreases.
     
  5. Mar 1, 2015 #4
    It still isn't clear to me exactly what you want to calculate after the valve is opened. And, in the first part, it would be better if you specified the temperature that the chamber is heated to. Then we could put some numbers on it.

    Chet
     
  6. Mar 1, 2015 #5
    I am sorry for not being clear.

    Step 1-
    When I put the water in and begin to increase the temperature, I need to calculate the pressure to ensure everything is operating at safe levels. My main confusion here is as the temperature increases the vapor pressure increase but following PV=nRT the gas pressure will also increase. IF the gas pressure increases does this effect the rate of the vapor pressure increasing?

    Step 2-
    Once the temperature is increased to a set/threshold level (this level is also dependent on the associated pressure) and the valve is opened, I would like to be able to track the pressure after the valve is opened.

    Ideally the temperature before the valve is opened would be 90-95 degrees C however if the pressure at 95 C is too much I will have to decrease the threshold temperature to ensure the valve is opened at a lower temperature so the pressure is lower.
     
  7. Mar 1, 2015 #6
    OK. You're going to help me do part 1. The pressure in the container is going to be determined by the pressure in the head space. The pressure in the head space is going to be the sum of the "partial pressure" of the water vapor and the partial pressure of the air. Initially, the temperature is going to be 20 C (293 K). The partial pressure of water vapor at this temperature is negligible, and the partial pressure of the air is going to be 1 atm (100 kPa). At 95C, the partial pressure of the water vapor is going to be equal to the equilibrium vapor pressure of water at that temperature. Look up the equilibrium vapor pressure of water at 95C. What is it? The partial pressure of the air in the head space is going to be determined by the ideal gas law. If the pressure of the air at 20C is 100kPa, what is the pressure of the air at 95C (from the ideal gas law)? What is the sum of the water vapor and air partial pressures at 95C?

    I'll wait to hear back from you with these answers?

    Chet
     
  8. Mar 1, 2015 #7
    the pressure do to water vapor at 95 C is .83421 atm

    the pressure of the atmospheric air at 95 C is 1.256 atm

    this results in a total pressure of 2.09 atm

    Does the increase in pressure of the atmospheric air have any effect on the vapor pressure of water? Also with the increase of pressure the boiling point of the water increases does this effect the vapor pressure?
     
  9. Mar 1, 2015 #8
    No.
    No.

    What units do you want to work in: atm, psi, Pa, kPa?

    Do you want to use gauge pressure or absolute pressure. Right now, 2.09 atm. is absolute pressure.

    Chet
     
  10. Mar 1, 2015 #9
    any units are fine to work in, converting across them is not too complex. Absolute pressure is also good. I am simply using these calculations to estimate/monitor the pressure across the temperature to make sure I don't exceed the system limits.
     
  11. Mar 2, 2015 #10
    What do you mean by wanting to track the pressure. Do you want it as a function of volume of water expelled, or as a function of time?

    Chet
     
  12. Mar 2, 2015 #11
    Tracking pressure as a function of volume. Another question I have is how does the boiling point of the water change over the increase of temperature/pressure of the system? Because what I have read is the water will boil when the vapor pressure reaches the atmospheric pressure but in the closed system the atmospheric air pressure continues to increase with the temperature. Is there a temperature where the water will begin to boil?

    Thanks for all the help so far!!
     
  13. Mar 2, 2015 #12
    No.

    If you are going to track pressure as a function of volume, then are you permitted to allow the water to escape very slowly from the bottom? If not, you may need to include the fluid mechanics in the channels beneath the chamber.

    Chet
     
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