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Constant Temperature Process (Steam)

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    In an isothermal process, where the dryness fraction of steam is given - Does the law of P1V1 = P2V2 still stand? Mass is also given here.

    I know that I should not be using PV = mRT for steam.

    Thanks

    3. The attempt at a solution
     
  2. jcsd
  3. Mar 3, 2014 #2
    Hi Jameseyboy. Welcome to Physics Forums!!!

    No. The law does not still apply. If you want some more help, you need to tell us more about your specific problem, and what your thinking is about how to approach it. Please try to "dope out" your ideas for us.

    Chet
     
  4. Mar 3, 2014 #3
    Thanks for getting back to me with my question!

    Basically, I am away from the actual question at the moment:

    - Constant temperature process
    - The process begins with a wet/dry mix
    - Dryness fraction given
    - P1 is given
    - V2 is given

    I found out U1 by using steam tables u = uf + x(ug - uf)
    I also found V1 by using S.T with the given P1
    I found T1 by S.T at P1

    I multiplied both of these by the mass to find the actual values, not the per KG value.

    However I am not sure how to go about finding the final value of U or H.

    The value of V2 does not indicate it is superheated (with the same temp given by P1), so I am basically stumped.

    Sorry this is vague but I can only remember a couple of actual values.
     
  5. Mar 3, 2014 #4
    You should have found V1 by using the same kind of equation as you used for u.

    Now to get the new set of conditions, you need to determine whether V2/m lies between the specific volumes of the liquid and the saturated vapor at T1. If it is, then the only thing that changes is x. The pressure is still P1. Does it lie between these values?

    Chet
     
  6. Mar 5, 2014 #5
    So to gain V1 I have used "Futher properties of water and steam" using 5 bars of pressure i.e. V1 = vf + x(vg - vf) (obviously multiplying this by the mass).

    How am I to assume that pressure is still 5 Bar for the second part of the process?

    Edit: The V2/m does not lie at the same T1 and is actually smaller than Vg at T1. I assumed this means its superheated, however after checking the tables the new V1 does not correspond with the value. Really not sure what I am supposed to make of this.

    Thanks again for the help
     
    Last edited: Mar 5, 2014
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