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Compressing steam and water together

  1. Jun 29, 2016 #1
    I want to describe here an imaginary experiment. First. take a thick walled cylinder with an open end and pour boiling water into it. The steam coming out will replace the air and then put a piston over the open end and fix it in an airtight way. But, the piston can move up and down.
    Now, as the steam will cool to atmospheric level and pressure inside will reduce and the piston will move down. Now, lets move the piston further to compress the steam inside with the remaining water. I want to know what will happen then?
    In case, if there is pure steam only inside, the steam will quickly get superheated and began to act like a gas. But, in this case, the steam will always be in contact with water here and have to be saturated and can never act like a gas. I am guessing that in such a case the steam can be compressed with far less energy consumption than pure steam.
    How? In this case, the steam will always be saturated and can't be superheated and therefore no need for extra effort to compress superheated steam that will act like a gas.
    Am I right?
     
  2. jcsd
  3. Jun 29, 2016 #2
    If the piston is frictionless and weightless, the outside pressure will be 1 atm. So as soon as the temperature is dropped below 100C and the system is allowed to equilibrate, all the steam will condense to liquid. The piston will be sitting right on the top of the liquid.

    Maybe, instead of allowing the system to cool, you want to consider the case where the system is adiabatic (insulated)?
     
    Last edited: Jun 29, 2016
  4. Jun 29, 2016 #3
    Right! Now, lets assume some points more. The piston can move freely and frictionless, but we have some control over it so that some space is allowed over the water. And also, there is some hidden mechanism inside the walls of the cylinder that can be switched on and off. In short, we can make the process adiabatic or isothermal at our will.
    Now, after the water being cooled, as there is some space available over the water, that should be filled with steam and the pressure of the steam will the same as the saturated steam pressure at that atmospheric temperature level. Now, we have made the system adiabatic and started to move the piston. In short, we started to compress the steam adiabatically. What will happen then?
    To be precise, I want to know what will happen if both steam and water are compressed together adiabatically. In such case, the steam will always be in touch with water and therefore can't be superheated and have to remain saturated. In such case, what should be the power consumption?
     
  5. Jun 29, 2016 #4
    I couldn't understand your first paragraph at all, but the 2nd paragraph seems clear. You have liquid water and water vapor in a cylinder, starting at 100 C and 1 atm. You then compress the system adiabatically and reversibly, and you would like to know the amount of work done. The answer is going to depend on the fraction of the cylinder filled with liquid water to begin with. For example, if there is no liquid water present initially, the answer is going to be very different from the case in which there is say 50% liquid water in the volume initially. Please choose an initial volume fraction of liquid water.

    Chet
     
  6. Jun 30, 2016 #5
    Do you want to say that amount of power consumption will vary with amount of water. In my opinion, the presence of water will be enough. The water will keep the steam saturated and it can never be superheated. I hope you can understand my point.
    And I don't want to mean that the water inside will be at 100C. I clearly want to say that the water inside will be at the same temperature level as the surrounding atmosphere. But, the water will have space over it filled with steam and the temperature of the steam will be the same as the water and pressure of the steam will be the saturated steam pressure at that temperature.
    My one and only concern is that whether such an arrangement can reduce the power consumption for steam compression as the steam will always be saturated.
     
  7. Jun 30, 2016 #6
    Suppose there is only one drop of liquid water in the container to start with. Do you still maintain that the drop of water won't evaporate and the system will be saturated throughout the process?
     
  8. Jun 30, 2016 #7
    If the process is reversible/idea, then just one single drop will be enough. Whatsoever, we all know that there is nothing irreversible in this universe and hence it's just time wasting to quarrel on whether one drop is enough or two. Lets assume that there is at leas half-a-centimetre of water there. Then?
     
  9. Jun 30, 2016 #8
    Just pick a mass percentage of liquid water in the container to begin with. I will then solve the following problem for you, if that is OK:

    Insulated Container
    Liquid water and water vapor initially in container at equilibrium at 20 C
    Adiabatic reversible compression of the contents (no heat exchange with atmosphere outside container).
    Calculate change in internal energy, work, and changes in amounts of liquid water and water vapor.

    Would you prefer that I solve this, or would you prefer to do the calculation.
     
  10. Jun 30, 2016 #9
    I will prefer both. I want to do it in my way and you do it in your own way. Lets see whether that will match or not. In fact, my little knowledge of physics tell me that the power consumption would be just enough to keep the steam saturated while raising both its temperature and pressure.
     
  11. Jun 30, 2016 #10
    OK, but I still need you to specify an initial mass fraction for the liquid water. The pressure vs volume relationship for the contents of the container is going to depend on the initial mass fraction of the liquid water. And this determines the work required per unit mass of cylinder contents.
     
  12. Jun 30, 2016 #11
    Kindly tell me how the pressure vs volume relationship will depend on the mass fraction and how much water is necessary for 1 kg of steam (suppose).
     
  13. Jun 30, 2016 #12
    You'll see that when we work the problem. Please, just be patient. I am suggesting that we assume 10 mass percent liquid water and 90 mass percent water vapor initially at 20 C. Is that OK with you?
     
  14. Jul 1, 2016 #13
    No problem. That means you have taken 900 gm steam and 100 gm water. Right?
    So far, this isn't a problem. But, question is, do you consider the steam to be saturated always or not.
     
  15. Jul 1, 2016 #14
    Yes.
    Good.
    It depends on how much the system gets compressed (adiabatically and reversibly). Certainly for small volume compression ratios, the system will be saturated.
    Before we begin, I have some questions for you too.
    1. In this adiabatic reversible compression, do you think that the mass fractions of liquid water and water vapor will (a) remain constant or (b) change.
    2. Do you think that, if the proportions change, the amount of work will depend on that?
    3. Can you apply the usual ideal gas adiabatic compression equations to the vapor if the number of moles of vapor change?
     
  16. Jul 1, 2016 #15
    In reply to your questions:
    1. In my opinion, I think the ratio will change. The work done by the piston will change the enthalpy of both the steam and the water and the temperature of water will rise. Therefore, too much water means more power consumption. What is necessary is to limit the amount of water to an optimal level so that most of the work will go to steam to change its enthalpy.
    2. Certainly. More water means more work necessary that will add to the as heat.
    3. Never! In this case, the steam will remain saturated always and the idea gas compression equations can't be applied here. We just have to calculate the change of enthalpy of steam by subtracting the gross enthalpy of steam at initial stage from the final stage.
    In short, the system will be just like a reverse process of multiple cylinder steam engine. Inside multiple cylinder steam engines, steam will go from one cylinder to another (from small diameter to large diameter) and being in the saturated almost always.
     
  17. Jul 1, 2016 #16
    Okay. I think we are ready to begin. Do you want to go first or shall I?
    Incidentally, in this closed adiabatic system, the work will be equal to the change in internal energy, not the change in enthalpy.
     
  18. Jul 1, 2016 #17
    Lets start together. My process is very simple. By this process, as water will always be saturated, therefore the power necessary would be enthalpy of saturated steam at a given lower temperature subtracted from the gross enthalpy at a given higher temperature. As for example, 80C and 100C which we have discussed in another thread. Though the actual consumption will be a little higher just like any process, but unlike other steam compression process, the steam will never get superheated and idea gas laws can't be applied to it.
    I beg to differ. "Internal energy" means only temperature, but enthalpy means both pressure-volume and internal energy. In this process, the pressure and volume wouldn't be the same. Therefore, even in an adiabatic process, it's change in enthalpy, not just internal energy.
     
  19. Jul 1, 2016 #18
    I guess we have a fundamental disagreement here. For a closed system, the first law of thermodynamics is given by:
    $$\Delta U=Q-W$$
    If the process is adiabatic, Q = 0, and we are left with:
    $$\Delta U=-W$$
    If we are unable to agree on this, we will never be able to agree on the remainder of the solution.

    In my judgment, you need to refresh your background on even the most fundamental aspects of thermodynamics. Here is a mini tutorial I prepared for Physics Forums Insights which presents a quick review of the 1st and 2nd laws of thermodynamics. Maybe this will help: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

    I have obtained a rough estimate of the solution to this problem using the Graphical Method on a pressure enthalpy diagram:

    Enthalpy pressure.PNG
    The thick black line in the figure shows the series of states that the cylinder contents passes through as they are compressed. We start off at the lower left end of the line with a saturated mixture of liquid water and water vapor at 20 C and 0.0234 bar, featuring 90 % water vapor and 10% liquid water. As the contents are compressed, the temperature and pressure rise in tandum (according to the saturation vapor pressure relationship) and the liquid water gradually evaporates while the amount of water vapor increases. The total entropy of the contents of the container does not change, because the process is adiabatic and reversible. To satisfy this condition, the amount of liquid water has to be decreasing, while the amount of water vapor has to be increasing. When the contents reaches about 70 C and 0.30 bar, all the liquid will have been evaporated and we will have 100% saturated water vapor in the container. If we compress the contents of the container further than that, the vapor will enter the superheated region. If the contents were compressed to 1 bar, the superheated vapor will be at a temperature close to 200 C.

    I will fine tune these results using the data in the steam tables, if your are interested.

    Chet
     
    Last edited: Jul 1, 2016
  20. Jul 1, 2016 #19
    From the perspective of !st law of thermodynamics, ΔU means gross enthalpy of the system. It doesn't denote only the internal energy. Can you explain why the work done is added only to the internal energy of the system only if both pressure, volume and temperature has been changed?
    Wikipedia (https://en.wikipedia.org/wiki/Enthalpy) says something different and what you have said above is true is saturated steam behaves like an ideal gas. But, that isn't the fact here.
     
    Last edited: Jul 2, 2016
  21. Jul 2, 2016 #20
    I stand by everything I said. Please see my private conversation comments where I feel like I can speak more frankly.

    Chet
     
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