1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics: Heat Removed from Rigid Tank.

  1. Sep 6, 2015 #1
    1. The problem statement, all variables and given/known data
    In a rigid sealed tank, 0.32kg of saturated H2O vapour is cooled from 16 bar to 2.70 bar. Determine the heat removed in Kj.

    2. Relevant equations
    Q(12)-W(12)=U2-U1

    3. The attempt at a solution
    Assuming that we're only looking at the saturated vapour internal energies, and there is no work occurring in the system.

    State 1 is at 16 bar, and State 2 is at 2.70 bar.

    ug1 = 2596 kJ/kg,
    ug2 = 2540 kJ/kg.
    Ug1 = 0.32*2596 = 830.72kJ
    Ug2 = 0.32*2540 = 812.80kJ

    Therefore, Q12 = U2 - U1 = 812.80 - 830.72 = -17.92 kJ. But this is counter intuitive as when things are compressed they general go hotter.

    Is my logic on this correct or am I totally in the wrong ball park?

    Thanks for all your help.
     
  2. jcsd
  3. Sep 6, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    What's being compressed here? P1 > P2 ...

    Q is negative because you are removing heat from the vapor inside the tank.
     
  4. Sep 6, 2015 #3
    How do you know that you're only looking at saturated vapor? What is the specific volume of the vapor in the initial state, and what is the volume of the tank? What is the specific volume of the vapor in the final state? If it is only vapor in the final state, what is the mass of vapor in the final state? Has the mass of water in the tank actually changed?

    Chet
     
  5. Sep 6, 2015 #4
    @SteamKing Sorry mate, for some reason I wrote that thinking it was being compressed when it's actually expanding so it should be getting colder and as such will have a negative value of heat associated with it.

    Given the problem statement "In a rigid sealed tank, 0.32kg of saturated H2O vapour is cooled from 16 bar to 2.70 bar. Determine the heat removed in Kj." I believe that there is no need for specific volume of the final or initial state to be required as it is a rigid vessel and as such the volume remains constant.

    Also, I think we're looking at saturated vapour only because of the fact there is no dryness fraction to determine a mixture of liquid-vapour and given the problem states 'saturated H2O vapour' I am assuming that this means that we're using vapour only otherwise if it were to state 'saturated H2O liquid' then I'd refer to the internal energies of the saturated liquid at those pressures.

    If the tank is rigid then the mass at the initial state is equal to that of the final state, and this must be the case as I'm trying to apply the 1st law of thermodynamics for a closed system.

    Is it safe to assume that my assumptions are along the right lines?

    Cheers
     
  6. Sep 6, 2015 #5
    No! Some liquid water will form when you remove the heat. We call this condensation. Now, please go back up my questions on post#3 and answer these questions. This will get you started toward the correct solution.

    Chet
     
  7. Sep 6, 2015 #6
    At 16 bar the specific volume of the vapour is equal to 0.1237 m3/kg. The volume of the tank therefore is equal to: (0.1237m3/kg)*(0.32kg) = 0.0396 m3.

    At 2.7 bar the specific volume of the vapour is equal to 0.6686 m3/kg.

    The mass of the vapour in the final state is therefore equal to: (0.0396m3) / (0.6686 m3/kg) = 0.0592 kg.

    If there is 0.0592 kg of vapour in the final state this means that there is 0.32 kg - 0.0592 kg = 0.2607 kg of water in the final state?
     
  8. Sep 6, 2015 #7
    OK. Now you're on the right track. But you didn't determine the amounts of liquid water and water vapor in the tank quite correctly (although your results are close). Here's what you know:
    • At 2.7 bar, there is saturated liquid water and saturated vapor in the tank
    • The average specific volume of the liquid/vapor combination in the tank is 0.1237 m3/kg.
    • The specific volume of the vapor in the tank is 0.6686 m3/kg
    What is the specific volume of the saturated liquid in the tank?
    You know the formula for the average specific volume of the liquid/vapor combination in the tank in terms of the specific volume of the liquid, the specific volume of the vapor, and the mass fraction of liquid. What is that formula? Now, use that formula to determine the mass fraction of liquid in the tank. Then determine the mass of liquid and the mass of vapor in the tank.

    Chet
     
  9. Sep 6, 2015 #8
    vave = (mliquid*vf + mvapour*vg) / mtotal

    I know that:
    vave = 0.1237 m3/kg
    mvapour = 0.0592 kg
    mtotal = 0.32 kg
    vg = 0.6686 m3/kg

    mliquid*vf = vave*mtotal - mvapour*vg

    mliquid*vf = (0.1237*0.32)-(0.0592*0.6686) = 2.88*10^-6 m3

    This is where I get stuck.. Is this my mass fraction for the liquid in the tank? if it is, then I can multiply this by the total mass within the tank and I'll have the mass of the liquid, likewise if I do this with mvapour*vg ill get the mass of the vapour in the tank?
     
  10. Sep 6, 2015 #9
    The mass of water vapor in the tank is less than 0.0592 kg. It would only be 0.0592 kg if the entire tank volume (0.0396 m3) were filled with vapor. But, some of the tank volume has liquid in it, so the volume of water vapor in the tank is less than 0.0396 m3.

    Chet
     
  11. Sep 7, 2015 #10
    Okay so here is another take at the problem.

    Using: vave = (1-x)*vf + x*vg

    vave = 0.1237 m3/kg
    vf (2.7 bar) = 0.1070*10-2 m3/kg
    vg (2.7 bar) = 0.6686 m3/kg

    Therefore:

    0.1237 = (1-x)*(0.1070*10-2) + x*(0.6686), x = 0.1837

    mliquid = (1-x)*mtotal = (1-0.1837)*0.32 = 0.2612 kg
    mvapour = x*mtotal = 0.1837*0.32 = 0.0587 (less than 0.0592 thus doesn't fill the whole tank.

    Is this the correct approach?
     
  12. Sep 7, 2015 #11
    Yes!! Now all that's left is to determine the change in internal energy.

    Chet
     
  13. Sep 7, 2015 #12
    To determine the change in internal energy I assume this is the approach I must take.

    The internal energy at state 1 (16 bar) is equal to only the internal energy from the vapour phase and considering only vapour exists here we use the total mass 0.32kg. Thus:

    U1 = ug*mtotal
    U1 = 2596kJ/kg * 0.32 kg = 830.72 kJ

    At state 2 (2.7 bar) we have a mixture due to condensation of water. Thus:
    U2 = uf*mliq + ug*mvap
    U2 = 546 kJ/kg * 0.2612 kg + 2540 kJ/kg * 0.0587 kg = 142.62 kJ + 149.09 kJ = 291.72 kJ

    Change in internal energy is equal to U2 - U1, therefore:

    U2 - U1 = 291.72 kJ - 830.72 kJ = -539.00 kJ (Negative energy due to expansion).
     
  14. Sep 7, 2015 #13
    Your answer is right, but your interpretation is not. The decrease in internal energy is the result of removing the heat of condensation.
     
  15. Sep 7, 2015 #14
    I see! Thank you very much for your help Chet, very much appreciated as always.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thermodynamics: Heat Removed from Rigid Tank.
Loading...