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State of water in the system

  1. Jan 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A rigid container of volume 0.5 m3 contains 1.0 kg of water at 120oC (vf= 0.00106 m3/kg ,vg=0.8908 m3/kg )
    v-specific volume (vf is v of saturated liquid and vg is v of saturated vapour)
    The state of water is
    (a) compressed liquid
    (b) saturated liquid
    (c) a mixture of saturated liquid and saturated vapor.
    (d) super-heated vapor

    2. Relevant equations
    v = vf + x(vg- vf)
    x- dryness fraction
    3. The attempt at a solution
    In order to know whether the water can exist in vapor state in the container at 120oC , i need to find the pressure? because pressure decides the boiling point.

    If i consider dryness fraction of zero then the volume occupied will be 0.00106 m3 right ? given that the volume of the container is 0.5 m3 certainly the water is not compressed so option (a) is ruled out.

    In order to rule out option (b) and (d) , i need to know the pressure right ? i don't know how to proceed.

    The answer given to me is (c) but i want to know how to rule out (b) and (d).
     
  2. jcsd
  3. Jan 13, 2016 #2

    Bystander

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    0.00106 m3 is how much mass?
     
  4. Jan 13, 2016 #3

    ehild

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    The volume of liquid water is negligible with respect to the volume of the container. You are given the specific volume of the vapour at 120 °C. How much is the mass of the vapour in the container? Is it less or greater than the total mass of water?
     
  5. Jan 13, 2016 #4
    According to the values given, the total amount of water in the container is 1kg ,if i consider dryness fraction as zero , i get specific volume as 0.00106 m3/kg ,so one kg of saturated liquid water has a volume of 0.00106 m3 right ?

    At 120oC i don't think it is possible for only liquid water to exist in saturated form in such a large volume of the system because the rest of it will be vacuum then right ? can i rule out option(b) this way ?
     
    Last edited: Jan 13, 2016
  6. Jan 13, 2016 #5
    The dryness fraction is not given and i don't know how to find it. The total mass of the water (either liquid or vapour or both) in the system is given as 1kg.
     
  7. Jan 13, 2016 #6

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    Is 0.00106 m3 more or less than 1 kg?
     
  8. Jan 13, 2016 #7
    I am not able to get what you are asking , i said 1kg of saturated liquid water occupies a volume of 0.00106 m3 (assuming dryness fraction as zero)
     
  9. Jan 13, 2016 #8

    ehild

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    "A rigid container of volume 0.5 m3 contains 1.0 kg of water at 120oC (vf= 0.00106 m3/kg ,vg=0.8908 m3/kg )
    v-specific volume (vf is v of saturated liquid and vg is v of saturated vapour)"

    Forget the dryness factor.
    vf = 0.00106 m3/kg is the specific volume of the water. So 1 kg water would occupy 0.00106 m3, much less than the volume of the container.
    So the volume of 0.5 m3 is mostly filled in by the saturated vapour. Knowing the specific volume of the vapour, vg=0.8908 m3/kg, what is the mass of 0.5 m3 of it?
    Then how much water is in liquid state?
     
    Last edited: Jan 13, 2016
  10. Jan 13, 2016 #9
    Why don't you just solve for x????

    Chet
     
  11. Jan 13, 2016 #10

    ehild

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    The OP said x was the dryness factor. I read that the dryness factor is something characterizing the steam, the amount of water droplets in it.
     
  12. Jan 13, 2016 #11
    x is the mass fraction of the water that is in the vapor phase, and v is the mass-average specific volume of the liquid and vapor.
     
  13. Jan 13, 2016 #12

    ehild

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    Is
    There is no liquid water phase on the bottom of the container? The whole water is in the form of wet steam?
     
  14. Jan 13, 2016 #13
    If there's gravity, then there is liquid water in the bottom of the container and water vapor in the head space (at equilibrium). If there is no gravity (or if all the liquid water hasn't coalesced yet), then there can be liquid water drops in the vapor phase. But the vapor is gas, and the liquid is liquid. For that equation to apply, it doesn't matter where the liquid water and the water vapor are situated in the container.

    Chet
     
  15. Jan 13, 2016 #14
    I don't think the rest of the volume is filled by saturated vapour ,because the total mass in the system itself is 1 kg only , if 1 kg of saturated liquid water is sitting at the bottom occupying a volume of 0.00106 m3 ,there is nothing else left to occupy the remaining volume in the container!!
     
  16. Jan 13, 2016 #15
    This is not correct. Just solve for x with your equation and tell us what you get. This will tell you the precise mass of water in the vapor phase and the precise mass of water in the liquid phase.
     
  17. Jan 13, 2016 #16
    I got x=0.5607 ,so the answer is (c) ,is that all ?
     
  18. Jan 13, 2016 #17
    You don't get off that easily now. Now that you have this result, tell us the mass of liquid, the mass of vapor, the volume of vapor, and the volume of liquid in the 0.5 m3 container.
     
  19. Jan 13, 2016 #18
    Mass of vapour =0.56 kg
    mass of liquid =0.44 kg
    Volume of vapour = 0.0005936 m3
    volume of liquid = 0.391952 m3

    the volume numbers don't sound right but that's what i got.
     
  20. Jan 13, 2016 #19
    The volume numbers are not right. Show us how you got these numbers. They have to add up to 0.5 m3.
     
  21. Jan 13, 2016 #20
    I just multiplied the mass of the liquid and the specific volume vf to get the volume of the liquid ,i did similarly for the volume of the vapour with vg.
     
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