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Thermodynamics pure substances problem

  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Hellooo, so this is the question that i have:

    A 0.5-m3 rigid tank initially contained a saturated liquid-vapor mixture of water at 140 °C is now heated until the mixture reaches the critical state. Determine the mass and the volume of liquid before the heating process

    2. Relevant equations
    v = vf + x*vfg
    x = mvapor/mtotal
    v = V/m

    3. The attempt at a solution
    I do not understand how to find the mass before the heating process :(
    I found the X which is 0,004086
    vf at 140C is 0,001080m3/kg
    vg at 140C is 0,5089m3/kg

    the final answer should be mass = 157,83 kg
     
  2. jcsd
  3. Feb 8, 2016 #2
    There seems to be information missing. Have you provided the complete problem statement?
     
  4. Feb 8, 2016 #3
    yes everything is there, unless if you do not have access to the saturated water tables! :O For the critical state, the value of v is 0,003155m3/kg
     
  5. Feb 8, 2016 #4
    If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

    EDIT: Wait a second. If you determined X, then you must already have done this. X is the mass fraction of water vapor. What is the mass fraction of liquid water? Given that you know the total mass of water in the tank and the mass fraction of liquid water, what is the mass of liquid water?

    Chet
     
  6. Feb 8, 2016 #5
    Yes i got 158,48kg :) but what i am suppose to do after finding this value? Is this the total mass of water in the tank after or before the heating process ?
     
  7. Feb 8, 2016 #6
    I don't know :( In the solution book they wrote:

    mf = (1-xf) * mtotal
    where total = 158,48 kg but i don't know where they took 1-xf and what is xf
     
  8. Feb 8, 2016 #7
    It is both. The mass of water in the tank has not changed.
     
  9. Feb 8, 2016 #8
    xf is what you called X in your original post. It is the mass fraction of water vapor in the tank. So, 1-xf is the mass fraction of liquid water in the tank. So, if you know the total mass of water in the tank and the mass fraction of this water that is liquid, you can find the mass of liquid water.
     
  10. Feb 8, 2016 #9
    but where does the 1 come from? I thought the equation was X = mvapor/mtotal so if my X is equal to 0,004086 and my mtotal is 158,48kg how do they get 157,83kg ?
     
  11. Feb 8, 2016 #10
    The sum of the mass fractions has to add up to 1. $$\frac{mvapor}{mtotal}+\frac{mliquid}{mtotal}= \frac{(mvapor + mliquid)}{mtotal}=1$$If the mass fraction of vapor is 0.004086, then the mass fraction of liquid must be 1-0.004086.
     
  12. Feb 8, 2016 #11
    OHHHH okay!!! I understand now!! :D Thank you very much! You are the best!!!!!!
     
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