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Thermodynamics; Finding Work of a Piston cylinder Device

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    An insulated piston-cylinder device contains 5 kg of water at 100oC with a quarter of the mass is vapor. The water is compressed reversibly to 900 kPa. Find the work done during this process. Let the ambient temperature to be 27oC.


    2. Relevant equations
    I can use the energy balance equation

    3. The attempt at a solution
    I am not sure how to continue and find the work of an irreversible process. What equation shall i take to find the work in a piston device?
     
  2. jcsd
  3. Apr 30, 2012 #2

    rude man

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    If you have a mixture of liquid water and water vapor at 100 deg C, what must p be?
    Since you're told p is increased to 900 kPa, what does that imply about the liquid-vapor mixture?

    Add the fact that liquid water is incompressible so that ∫pdV = 0 once the "quality" of the mixture reaches 0%, think latent heat of vaporization.

    Or, you can do pΔV.

    (Obviously, the ambien temp. is immaterial since the piston is insulated.)
     
    Last edited: Apr 30, 2012
  4. Apr 30, 2012 #3
    How do i use the latent heat to find the work? Do i calculate to find the h or perhaps find the specific volume then multiply by the mass to find the volume. then i can use the equation of work which is:
    w=p(v2-v1)
     
  5. Apr 30, 2012 #4

    rude man

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    You'll have to pardon me here. I need to figure this out correctly myself first. I apologize and hope someone else will stepin here in the meantime.nI will get back as soon as I can.
     
  6. Apr 30, 2012 #5

    rude man

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    Never mind using enthalpy. That was a blunder on my part. Enthalpy doesn't change since no heat is added or subtracted from the mixture.

    So, compute the volume of the steam, given its mass and temperature and knowing it's saturated, and use pΔV. Of course, V = 0 when you're done appplying the work - why?
     
    Last edited: Apr 30, 2012
  7. Apr 30, 2012 #6
    why will the volume be equal to zero?
     
  8. Apr 30, 2012 #7

    rude man

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    By volume V I meant the volume of steam. Well, think about what's happening. You're apllying work to squish the steam until it's gone. The clue as to why it's gone rests in the fact that the final pressure is upped to 900 kPa. What is the pressure when you just removed the last bit of steam?
     
  9. Apr 30, 2012 #8
    Won't the final pressure be 900?
     
  10. Apr 30, 2012 #9

    rude man

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    Yes, but remember I pointed out that water in liquid form is incompressible? If it's incompressible, what is the work done in increasing pressure on it?
     
  11. Apr 30, 2012 #10
    Oh, so shall i use the equation Q=m*cv*(T2-T1) instead? The t1 is given and i can find the t2 from the pressure=900kpa
     
  12. Apr 30, 2012 #11

    rude man

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    Sara, I have to apologize again. I'm not sure what the exact siuation is. I thought p amd T stayed cosntant while the steam was compressed until it was gone. Now I have grave doubts about that. And my statement that H stays constant would then be invalidated, sinve H can change isentropically if p changes, which I'm beginning to think it does.

    I apologize again, and suggest you keep the post & hopefully somebody smarter will help us both! Meanwhile I am not giving up & will let you know asap.

    .
     
  13. Apr 30, 2012 #12

    I like Serena

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    I'm still working through the problem myself and I was kind of hoping to see the solution here. :smile:

    Since no heat is exchanged, I'm drawing the conclusion that the total entropy must remain the same.

    From the steam tables, the entropy and the internal energy can be calculated in the initial state.

    In the final state, we're looking at either another liquid-vapor combination, or a condensed liquid.
    Either way, the entropy must be the same, from which the final temperature and masses can be deduced.

    With the internal energy in the final state, you can calculate the work done, which is the difference in internal energies.
     
    Last edited: Apr 30, 2012
  14. Apr 30, 2012 #13

    rude man

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    That sounds much better! Thanks! I will muse on that.
     
  15. Apr 30, 2012 #14

    rude man

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    The problem as I see it is that we know the final p, but not T. If the water is supercooled then there is no T to go along with p as there would be in the unsaturated region.

    At this point, if I had to give an answer I would argue that p and T don't change until the vapor has all been compressed into liquid, and after that there is no more work done since the liquid is incompressible. That would reduce the problem to finding the volume of the vapor at onset of compression (easy), then calling that V and going with W = Vp, p being cosntant at the initial p.

    I'm just worried about enthalpy "not changing". Seems like the enthalpy of the water-vapor mixture should be higher than the enthalpy of just the water, even with T still constant. That would dictate a change in p, otherwise H couldn't change, since dH = dQ - Vdp and dQ = 0.

    .
     
  16. Apr 30, 2012 #15

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    Assuming that entropy is constant (dQ=TdS=0), my calculations show that the system ends up in the liquid-vapor region with a much smaller volume and T=175 oC.

    So pressure and temperature increase, while volume decreases.
    Furthermore entropy remains constant, while internal energy and enthalpy both increase.
     
    Last edited: Apr 30, 2012
  17. Apr 30, 2012 #16

    rude man

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    My problem with that is that condensation represents a loss rather than a gain in U (or Q, but there is no Q in this situation). Heat is not absorbed by the vapor, it's liberated, isn't it?

    Consider the part of the refrigeration cycle where the compressed vapor at the high-p point, and saturated, is reduced in V at constant T and p (isothermally). During this part of the cycle, heat is expelled to the higher T reservoir as the vapor is compresed to a pure (saturated) liquid. So should you worry about that?

    I'd say that if in your analysis came to an end-point isentropically where p = 900 kPa and there was still some vapor left, then you've solved the problem, using ΔU = W. I would not agree with that were there no vapor left. Thank you for your willingness to jump in!
     
  18. Apr 30, 2012 #17

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    Yes, I realized that heat was released by the condensation instead of absorbed, so I was looking at it the wrong way around.
    That's why I edited my previous post.
     
  19. Apr 30, 2012 #18

    rude man

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    Well, I think you went the right way with using isentropy to get to (p=900kPa, T=175C) with quality > 0. I think that was very clever.
     
  20. Apr 30, 2012 #19

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    Thanks. :)
     
  21. May 1, 2012 #20
    I figured out how to solve it as well. Since the process is reversible, then the piston is adiabatic, then the process is isentropic. Thus, s1=s2. We can Find the work using the equation -W=m(U2-U1)
    The u1 is determined from state one from the given information. To find U2, we must realize that s1=s2 so at the given pressure and s2 we can find the state, and then find the corresponding u2.
    Then we calculate and solve for work.
     
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