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Thermodynamic Reversible/Irreversible Closed Systems

  1. Feb 2, 2013 #1
    I am completely lost with this problem. How would I go about breaking it up into manageable components that so I won't get lost in the problem?

    1. The problem statement, all variables and given/known data

    A 2 m3 rigid storage tank containing 10 kg of steam is heated from 300 to 400 C by transferring heat from a hot reservoir at 500 C. Determine the total amount of heat transferred to accomplish this change. Determine the entropy change of the steam, the surroundings, and the universe. Assume steam is a real fluid following the steam tables in Appendix A.III.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 3, 2013 #2
    This problem is a bit vague. Is the steam saturated or superheated at 300 C? If you assume it is saturated, you can determine its internal energy and its specific volume. Because the container is rigid, the specific volume remains constant when you heat because both mass and volume are constant. Once again go to the superheat tables for 400 C and determine the pressure and internal energy at the known specific volume. The difference in internal energies is the heat transfer. This should get you started.
     
  4. Feb 3, 2013 #3
    Thanks for the response. There seems to be no indication in the problem statement whether or not the steam is saturated or superheated. The topic I am currently covering in class is Reversible/Irreversible Closed Systems and Unsteady Flow Systems. Do you know what part of the problem statement indicates reversiblity or irreversibility and how that concept applies here?
     
  5. Feb 3, 2013 #4
    The situation is irreversible because there is no way you can get the heat from the rigid container back into the reservoir (returning container and reservoir to initial states)without supplying work. Think of two blocks at different temperatures coming into contact and are isolated. They come to a common temperature. But they cannot be returned to their initial states without the addition of work. Heat will not move from cold to hot without work (heat pump).

    You can determine the entropy change of the steam when heated from 300 C to 400 C from steam tables. You can also determine the heat transfer needed from internal energy changes based on steam tables. Now assuming the reservoir temperature does not change (big reservoir), you can determine the entropy change (decrease) of the reservoir. Comparing that to the increase in entropy of the container contents, you can determine the change in the entropy of the universe for this process. As you know it must increase because the process is irreversible.
     
  6. Feb 3, 2013 #5

    rude man

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    This is not my specialty but look at the following:

    By looking at a T-v diagram (T = 300C and 400C and v = 0.2m^3/kg) for H2O you can see that you're always in the superheated steam region. From those two data points you can look up the change in internal specific energy Δu which, since W = 0 gives you Δq (1st law). Multiply by your actual specific volume gives you your total ΔQ.

    Changes in entropy and enthalpy are read right off your steam tables.

    My steam tables list T and v for a certain p, spaced 0.2 MPa apart, so I would have to interpolate a bit but I got p(300C) ~ 1.2 MPa and p(400C) ~ 1.6 MPa from which two tables I got Δu ~ 160 kJ/kg. Use the 1st law to get Δq and then scale to your actual specific volumer.

    The reservoir at 500C loses entropy ΔS = ΔQ/Tr J/K where Tr is reservoir temperature. The universe ΔS is ΔS of the steam - loss of reservoir entropy.
     
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