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Thermodynamics: reversible process, equilibrium and entropy

  1. Dec 17, 2006 #1
    Hi, everyone. i have several questions about thermodynamics which I just read.

    1. Why does a reversible process require equilibrium of the system all the time?

    2. Is every process reversible if it's very very slow so that the system is in equilibrium all the time( and there's no friction or some other factors that may dissipate the energy) ?

    3. When we calculate the entropy (clausius definition of change in entropy)Why do we use the thermal energy transfered when the system had followed a reversible path? Because of the equilibrium state?

  2. jcsd
  3. Dec 17, 2006 #2
    The entropy is an state function: it doesn't depend on the path, just on the initial and final states. But you can measure volume, pressure, temperature, etc.. whe the process finishes. Entropy is not measurable in such a way. But if the process is slow, infinitely slow, in a way that you can say it is in permanent equilibrium, then you can measure the change in entropy as thermal exchange under specific conditions. If the process is spontaneous, i.e., not in equilibrium, entropy will have increased in the universe, that is, in the system+environment. Under this circumstances, thermal transfer will not represent change in entropy, and more important, the process is irreversible just because entropy will have increased in a greater extent that what thermal transfer is responsible of. So there is no possibility to reverse the process as ther is less heat available than entropy to decrease.
    By the way, reversible processes are just a mental model. If you want a process to happen, you have to push it a bit, and this bit carries irreversibility.
    A different this is a battery, that you can charge thousands of times. But you always spend more energy to reverse it that the energy you get from it. And the entropy of the universe continues to grow...
  4. Dec 17, 2006 #3
    Thanks! vivesdn
    And I want ask another question.
    Between Otto engine and Diesel engine, which is more efficient?
    Is there any mathematical proof for that?
  5. Dec 18, 2006 #4
    In my opinion, it cannot be said that one is better than the other (in terms of models).
    Otto cycle is modelized as follows:
    1) Adiabatic compression
    2) Constant volume heating (that's the phase where fuel is burning)
    3) Adiabatic expansion, where work is produced
    4) Constant volume cooling (that's the phase where gases are exhausted)
    At the end, efficiency is 1-1/(r^(k-1)).
    k is the relation between Cp and Cv (specific heat). r is the compression ratio.
    It is important that on an otto engine, the compression phase includes the fuel. So it is not possible just to increase r as needed to achieve better efficiency.

    Diesel cycle is modelized as follows:
    1) Adiabatic compression
    2) Constant pressure heating (that's the phase where fuel is burning), with work being produced
    3) Adiabatic expansion, where more work is produced
    4) Constant volume cooling (that's the phase where gases are exhausted)
    The efficiency of this cycle is again a function of the compression ratio (I do not have it at hand). The important difference is that the compression phase is performed without fuel.

    Working with the same ratios, I think that otto produces higher efficiencies, but diesel cycle will work at much higer compression ratios, so at the end, produces higher efficiency with the same heat flow. 4 to 10 compression ratios are usual for otto engines while 12 to 20 are common for diesel engines. If you build a stronger engine, you can increase compression ratio using a diesel cycle, but it is not easy to accomplish it on an otto cycle, as fuel will ignite before the spark.
    Another difference: on the paper, you will get higher pressures but lower temperatures for the diesel cycle, as expansion starts when ignition starts.

    On real life, the real performace of diesel cycle is higher than that of an otto cycle, and much lower that those of the models.
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