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I Thermodynamics -- Entropy of an Isometric Process

  1. Jun 26, 2017 #1
    In order to better explain my question let me give a precise situation and then state my question

    Say I have a well insulated rigid container containing some mass m of a saturated liquid-vapor mixture of water at some pressure P1. Initially it's at some quality x1. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized.

    To find the entropy change of this process I'm told to

    1) Look up the values for water in a thermodynamic steam table for H2O and use the given (observed) property values (for the given property at the given pressure P1) of specific entropy from s1=(1-x1)s1,f + x1*s1,g

    2) Assume the specific entropy of the water after it's done heating and further pressurizing to its fully vaporized state to be s1,g=s2

    3) Multiply the mass by the change in specific entropy m(s2 - s1) thus obtaining total entropy change of the process.

    Now that I have established where this curiosity originated from let me ask my question

    During step 2 I assumed (without knowing why) that it is simply O.K. to treat s2 as s1,g this seems highly odd to me since I'm heating a rigid, insulated, container increasing the content's temperature as well as its pressure. By the time the water reaches state two it is no longer the same information on the table for sg at some new pressure and temperature P2 and T2 respectively..

    Won't using the entropy values for state 2 from state 1 in the table result in inaccurate results since specific entropy should be different for a higher pressure and temperature?
     
  2. jcsd
  3. Jun 26, 2017 #2
    They expect you to be assuming that the system pressure is being held constant (not the volume).
     
  4. Jun 26, 2017 #3
    Rigid container problems are typically constant volume whereas moving boundary problems are constant pressure.
     
  5. Jun 26, 2017 #4
    OK. So the constraint on this problem is that the volume of the container is constant. But, in that case, ##s_2## is not equal to ##s_{1g}##. Do you know how to find the final state of the water vapor in the case where the total volume is constant using the steam tables? In terms of the initial specific volumes of the liquid and vapor, what is the volume of the container? What is the final specific volume of the vapor if it now fills the container?
     
    Last edited: Jun 26, 2017
  6. Jun 26, 2017 #5
    Edit:
    Oh man was I wrong. (and so were the online solution help sites)

    I can determine s2 from the specific volume by looking at the table and finding where the specific volume vg matches v1.

    _____


    I wouldn't be able to find the final state of the saturated liquid-vapor mixture (as far as I know) since there aren't two intensive properties of which I can determine about it.

    I can determine the initial volume and final volumes since they are the same and are based off of the vf and vg values given at the first state. Since neither mass nor volume are changing the specific volume of the system must be the same at both states.

    I thought about this last night and I suppose the process could be isothermal if we suggest that all of the heat entering the system is used to convert the liquid part of the mixture to vapor, this same suggestion may be extended to the pressure not increasing as this energy is used to change the phase of the (pure)substance and not increase the temperature or the pressure.

    If these suggestions are correct for this model, which I suspect they are, then I have no problem seeing the 2nd state at the same temperature and pressure and thus the conclusion sg,1=s2 ...

    Would suffice for me but it would be interesting to see if there's another angle to this or if anyone can confirm these suggestions etc.
     
    Last edited: Jun 26, 2017
  7. Jun 26, 2017 #6
    What do the words "until all the liquid in the tank is vaporized" mean to you?
     
  8. Jun 26, 2017 #7
    Since those words are used in the context of "saturated liquid-vapor mixture of water" I would predict the words "until all the liquid in the tank is vaporized" to mean that the saturated liquid-vapor mixture with some given quality between 0 and 1 is now at quality of 1 where it is purely a saturated vapor.
     
  9. Jun 26, 2017 #8
    Oh man was I wrong. (and so were the online solution help sites)

    I can determine s2 from the specific volume by looking at the table and finding where the specific volume vg matches v1.

    Pressure is increasing several times and temperature by around 15% (abs scale)
     
    Last edited: Jun 26, 2017
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