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Reversible processes and carnot cycle

  1. Mar 8, 2009 #1
    i want to ask why heat transfer is considered ti be an irreversibility and why in carnot carnot cycle heat addititon is at constant temperature to make this process reversible
     
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  3. Mar 8, 2009 #2

    Mapes

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    Any smoothing of a gradient (in temperature, charge, matter, etc.) is thermodynamically irreversible because entropy increases; the universe can't be restored to exactly its former condition. In a Carnot cycle, heat transfer is assumed to occur without a gradient and from infinitely large thermal reservoirs. Although this arrangement is impossible in practice, it is useful to consider in theory because it enables a heat engine with maximum possible efficiency. Does this answer your question?
     
  4. Mar 9, 2009 #3
    do u mean that any mprocess cuases an increase in entropy is an irreversible process?is it general rule?
    thx for ur reply
     
  5. Mar 9, 2009 #4

    Mapes

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    Yes, any process that involves an increase in total entropy is irreversible.
     
  6. Mar 13, 2009 #5
    ok can u tell me a website or textbook that explains this relation between entrop and irreversibilty?because in carnot syscle in the heat addition process there is increase in the entropy of system however it is cosidered to be reversible process .
    thx
     
  7. Mar 13, 2009 #6

    Mapes

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    Zemansky's Heat and Thermodynamics is pretty good. If you're near a university library, browse for immediately readable thermodynamics books (actually flip through the selection and think about which ones you'd most enjoy reading) and compare their treatment. I learn best by scanning different books on a subject and seeing what they all mention and how it's expressed in different ways. Perhaps this approach works for you too.

    Whenever flow occurs in response to a gradient, entropy is not just transferred but also created. Consider heat transfer between two bodies. If we assume one is very large (this is the reservoir), then the energy removed won't affect its temperature. If we also assume the temperature difference is very small, then entropy creation is minimal. Take these assumptions to their extreme and we have the Carnot cycle, where no entropy is created at all. There is still entropy transfer: the energy associated with heat transfer carries its own entropy. But the transfer is perfectly efficient, and what's lost by the hot reservoir is exactly gained by the engine. Later, when sending energy to the cold reservoir, what's lost by the system is exactly gained by the cold reservoir.
     
  8. Mar 13, 2009 #7
    i laso ve another question i can understand the concept of irreversibility that there is energy dissipated in the process but i can uget that just for friction burt for other reasons of irreversibilities for example heat transfer i cant get whre is the energy dissipated here.thx
     
  9. Mar 13, 2009 #8

    Mapes

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    My advice is to link irreversibility in your mind not with "energy dissipation" but with "flow in response to a gradient that tends to erase that gradient." There are a couple advantages. First, gradient-induced flow is easy to quantify, by taking the dot product of the flow vector (in this case, heat flow and direction) with the mathematical gradient of the potential (in this case, the temperature). This dot product is actually proportional to the rate of entropy increase. Second, as you've pointed out, it's not easy to visualize "energy dissipation" when a hot object heats a cold object. It's an amorphous term (how is energy dissipated here?). But the gradient-induced flow view encompasses both friction (matter deforms in response to stress gradients, energy moves in response to temperature gradients) and simple heating/cooling. Swapping these descriptions in my mind was a key part of moving from beginner to advanced thermo.
     
  10. Mar 13, 2009 #9
    nice description mapes:smile:
     
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