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Pdv or Vdp

  1. Oct 12, 2014 #1
    • Warning! Posting template must be used for homework questions.
    Workdone= integral Pdv
    a kg of saturated water was compressed isentropically from 1 bar to 10 bar.
    I solved it in the following logic: Since water is incompressible dv=0 , work =0

    But my answers was wrong.

    The solution was integral Vdp where V is Vf at 1 bar

    My doubt is , is workdone =integral {(Pdv) or (Vdp)}...? Or both?
    If both, how work=integral pdv and vdp ?
    If it is equal to integral Vdp, then how is it?
     
  2. jcsd
  3. Oct 12, 2014 #2

    haruspex

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    If water were completely incompressible then you would be right that no work is done, but it is very slightly compressible.
    Beyond that, I agree with you that work done is PdV. Without knowing the compressibility coefficient there is no way to deduce the work done. Using VdP makes no sense to me.
     
  4. Oct 12, 2014 #3

    jack action

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    The relationship between work and energy:
    In this regard:

    [tex]\delta W = d\left(pV\right) = pdV + Vdp[/tex]
    The first part is know as boundary work, where there is a variation of volume. The second part is any other form of work (often called shaft work) where there is no variation of volume.

    The shaft work may not do actual work on the system (i.e. displacement), but it changes the energy level of the system. If the process is reversible and adiabatic, then work had to be done somewhere for this to happen. So, in a sense, «work done» is still technically correct.

    But, if someone wants to argue, an isentropic process is not necessarily reversible and adiabatic. :p
     
  5. Oct 12, 2014 #4
    Was it saturated liquid water or saturated water vapor? Either way, the work is PdV.

    Chet
     
  6. Oct 12, 2014 #5
    Saturated liquid water
     
  7. Oct 12, 2014 #6
    Then your original answer is definitely correct, and it's not vdp.

    Chet
     
  8. Oct 13, 2014 #7
    But the textbook solution is vdp only.
     
  9. Oct 13, 2014 #8

    haruspex

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    Can you post the whole of the textbook solution? That might shed more light on it.
     
  10. Oct 13, 2014 #9
    vdp is definitely incorrect. But there is a more accurate solution to this problem than simply saying that dw = pdv = 0. This solution involves the coefficwient of volumetric thermal expansion ##\alpha## and the bulk compressibility ##\beta##. If you are interested, I can lead you through how to obtain the correct solution.

    Chet
     
  11. Oct 13, 2014 #10

    jack action

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    @ Chestermiller:

    I don't understand why you say that it is incorrect. Here's an example of what is taught at Ohio university about steam power plants:
     
  12. Oct 13, 2014 #11
    Hi Jack Action,

    The original problem statement didn't say anything about an open system (flow process). In fact, it seems to imply a closed system of 1 kg.

    For an open system, your description is certainly appropriate, with the change in enthalpy per unit mass of water passing through the system (at steady state) equal to minus the shaft work. For a closed system, of course, the development you provided does not apply.

    It is not clear whether this homework assignment was designed to give the student experience with applying the open system version of the first law.

    In any event, in my previous post, I though it might be interesting to develop the closed system solution for compression of 1 kg of water at constant entropy. Any interest?

    Chet
     
  13. Oct 13, 2014 #12
    It is the isentropic compression by pump in RANKINE CYCLE .....I'll post the problem statement with solution soon. (Unable to post now due to some error)
     
  14. Oct 13, 2014 #13
    Now, don't you think it would have been pretty important to let us know that from the outset?

    Chet
     
  15. Oct 14, 2014 #14
    the solution is integral Vdp where V is Vf at 1 bar.
    dp= p2-p1 , where p1= 1 bar and p2=10bar.
     
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