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Heat rejected between states

  1. Mar 27, 2013 #1
    0.05KG of steam at 15bar is contained in a rigid vessel of volume 0.0076 m3. What is the temperature of the steam? If the vessel is cooled, at what temperature will the steam be just dry saturated? Coolingis continued until the pressure in the vessel is 11 bar; calculate the final dryness fraction of the steam,and the heat rejected between the initial and final states.

    i have the answers as 250°C, 191.4°C, 0.857.

    How do i find the heat rejected between states?

    Im assuming it is the change in internal energy? but how do i get this from the steam tables for 2 states?
     
  2. jcsd
  3. Mar 27, 2013 #2

    SteamKing

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    Have you tried to apply the First Law to the system?
     
  4. Mar 27, 2013 #3
    I have tried the formula Q-W=change in internal energy.
    Assuming that work done = 0
    Q = change in internal energy
    I am not familiar with the steam tables and dont know how to obtain values for U as there are values for U with different subscripts and i do not know which on to use for the first and last state. Im assumign i would use Ug for the last state as it is dry saturated
     
  5. Mar 27, 2013 #4

    SteamKing

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    Your tables don't have enthalpy? Enthalpy is usually h.
     
  6. Mar 27, 2013 #5
    looking in the super heated steam at 15 bar, i have the specific volume as 0.0076/0.05, which has a corresponding enthalpy of 2925. Then looking in the saturated water and steam at 11 bar i have three enthalpys hf hg and hfg. I have the answer to be 21.5kj but none of them values for enthalpy would give me a difference of that?
     
  7. Mar 29, 2013 #6
    Can you help me with this?
     
  8. Mar 29, 2013 #7
    Why don't you tell us what the subscripts are in your steam table? To do this problem, you need to know the intermal energy per unit mass of the superheated steam in the initial state, and you need to know the internal energy per unit mass of the water and steam under saturated conditions in the final state. Can you determine these?
     
  9. Apr 8, 2013 #8
    Thanks for your reply chestermiller the information the steam tables are giving me are "hg" "hf" "hfg", i dont know which one to use for the internal energy for this particular question
     
  10. Apr 8, 2013 #9
    Let's check your understanding. How is the enthalpy defined in terms of internal energy, pressure, and volume?

    In the final state, you know how much saturated liquid water is present in your rigid vessel, and you know how much saturated water vapor is present in your rigid vessel. hf is the specific enthalpy of liquid water, and hg is the specific enthalpy of water vapor. This should be enough information to determine the enthalpy of the total contents of the vessel in the final state.
     
  11. Apr 9, 2013 #10
    ok so i have
    dryness fraction = 0.857
    hg = 2781
    hf = 781
    therefore
    final enthalpy = (2781 * 0.857) + ((1-0.857) * 781) = 2495

    For the initial enthalpy i have the formula H = U + pV
    but using the internal energy from the steam tables (pressure 15bar and temp 250) i read a value of u = 2697
    therefore H = 2697 + (1500000 * 0.0076) = 14097, but this is too big to be correct i assume
     
  12. Apr 9, 2013 #11
    This final enthalpy is per kilogram. You need to multiply by the mass. You also need to subtract pV (in appropriate units to get the final internal energy). The units of hg and hf in kJoules/kg? Watch your units.

    For the initial state, you already have the internal energy per unit mass. All you need to do is multiply by the mass.

    The focus is on determining the change in internal energy, not enthalpy.
     
    Last edited: Apr 9, 2013
  13. Apr 9, 2013 #12
    So to find specific final enthalpy then the resulting internal energy

    (2495 * 0.05) - (11,00000*0.0076) = 8235

    then the initial internal energy

    2697 * 0.05 = 134.85

    **The values are given in KJ
     
  14. Apr 9, 2013 #13
    (2495 * 0.05) - (11,00000*0.0076) = 8235
    Look at the above equation. Does something seem strange to you? Also, I told you to watch out for units. The conversion factor from bars to Pa is 100000. The conversion factor from bars to kiloPa is 100.
     
    Last edited: Apr 9, 2013
  15. Apr 9, 2013 #14
    I dont notice anything strange about the equation sorry? im not sure i understand your comment about units. Bars to joules? isnt one a unit of pressure and the other of energy? The only unit conversion i done was change 11 bar to 11,00000 pascal
     
  16. Apr 9, 2013 #15
    I meant Pa and kPa, not joules and kilojoules. I already changed that in my response. Regarding the equation, you are subtracting something from a positive number, and getting a larger positive number. Does that make sense?
     
  17. Apr 9, 2013 #16
    oh no, my mistake. that should be have a minus sign in front of it making it -8235. is this correct?
     
  18. Apr 9, 2013 #17
    I have the answer as 21.5kj but i cant get to it, do you know where i have gone wrong?
     
  19. Apr 9, 2013 #18
    No. You still didn't use the right units in evaluating pV.

    Determination of final internal energy:

    h = 2495
    Pressure = 11 Bars = 1100 kPa
    Volume = 0.0076

    Final internal energy = 2495 (0.05) -(1100)(0.0076)=124.75 kJ - 8.36 kJ = 116.39 kJ

    Initial internal energy = 134.85 kJ

    Heat removed = 134.85 - 116.39 = 18.46 kJ

    You have to be more careful with units, and you have to understand how the thermodynamic functions are related to one another.
     
  20. Apr 9, 2013 #19
    Oh i see, i have always used pascals as the Si unit for pressure, i can see you have arrived at the correct answer but why is it different in this instance?
     
  21. Apr 9, 2013 #20
    For the second part of this question (that i have already answered) was i right to take the specific volume of the vessel and look for the corresponding temperature of the specific volume in the superheated steam section then interpolate if not listed?
     
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