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Non-flow process with steam (Constant Pressure) Part II

  1. Feb 18, 2013 #1
    In a non-flow process, 0.6kg of steam at a pressure of 10 bar and a temperature of 235°C is cooled at a constant pressure in a cylinder. After 45 minutes the cylinder contains 0.12 kg of saturated water and 0.48kg of dry saturated steam. Calculate
    (a)the change in internal energy, the work energy and the heat energy transferred during the process.

    My working for u2
    From the steam table:
    @ 10bar 0.12kg of saturated water uf = 762
    @ 10bar 0.48kg of dry saturated water ug = 2584
    Hence u2=2584+762=3346.

    Putting this value into the formula together with the value of u1, I am unable to get the right answer. I believe my approach for u2 is wrong.


    Please advice, thanks.
     
  2. jcsd
  3. Feb 18, 2013 #2
    Aren't you supposed to multiply the specific energies by the mass of the respective component?
     
  4. Feb 18, 2013 #3
    You mean 762x0.12 + 2584x0.48 = u2 ?
     
  5. Feb 18, 2013 #4
    Yep.
     
  6. Feb 18, 2013 #5
    voko, I cannot get the answer, can you help me take a look at my working.

    Stage 1, Finding u1:
    @ 10bar, 235°C

    Doing interpolation:

    Temp Ug
    250 2711
    235 u1
    200 2623

    250-200/2711-2623 = 235-200/u1-2623
    50/88=35/u1-2623

    50u1=131150=3080
    u1=2684.6

    u2 working:
    u2=762x0.12 + 2584x0.48
    =1331.76


    U2-U1=m(u2-u1)
    =0.6(1331.76-2684.6)
    =-811.704 (Ans wong)

    Book answer is -279kJ
     

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  7. Feb 18, 2013 #6
    What you computed as u2=762x0.12 + 2584x0.48 is in fact U2. Can you see why?
     
  8. Feb 18, 2013 #7
    Because it's not J/kg anymore but I'm confused why do I need to multiply the specific energies by the mass of the respective component first if the mass is not the full value (0.12 saturated, 0.48 dry saturated).

    For example if the question is changed to "after 45 minutes the cylinder contains 0.6kg of dry saturated steam"

    @ 10bar 0.6kg of dry saturated water ug = 2584
    For this instance how come I don't have to multiply by the mass first? I mean I will do it when I'm using U2-U1=m(u2-u1) later on.

    But from post #1
    @ 10bar 0.12kg of saturated water uf = 762
    @ 10bar 0.48kg of dry saturated water ug = 2584
    I have to mutiply by the specific mass first.
     
  9. Feb 18, 2013 #8
    0.12 saturated, 0.48 dry saturated means there are two components, and you do not know the specific energy of the mix. So you compute the full internal energy of each component then sum them, getting the full internal energy of the entire mix. THEN you can divide the full internal energy by the total mass and get the specific energy of the mix, but that would be doing more than necessary.
     
  10. Feb 18, 2013 #9
    Yes you are right, thank you for the explanation.
     
  11. Feb 18, 2013 #10
    Think of this as a two step process. You start out with 0.6 lb of superheated steam at 10 bars and 235C. You first cool this to the saturation temperature (dew point) of 179.9 C and 10 bar. The total internal energy change for this is 101 J/kg, or 60.1 J for 0.6 kg. Then you condense out 0.12 kg of liquid water at the same temperature and pressure. The change in internal energy is 2584 - 762 = 1822 J/kg. Since the amount of liquid water condensed out is 0.12 kg, the change in internal energy is 219 J. Therefore the total change in internal energy is 219 + 60 = 279 J.
     
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