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Temperature change in an isentropic flow of an ideal gas

  1. Apr 17, 2012 #1
    I am a bit confused by the definition of an isentropic process in the flow of an ideal gas.

    isentropic implies reversible & adiabatic.

    for a process to be reversible, there are no losses to friction (viscosity in this case),
    for a process to be adiabatic, there is no heat transfer with the surroundings.

    That being said, when a compressible gas flows through a nozzle, there is a temperature change. When this happens, a temperature gradient occurs. Where there is a temperature gradient, heat transfer occurs within the gas. I was always taught that, according to the 2nd principle of thermodynamics, a heat transfer due to a temperature gradient (from hot -> cold areas) is irreversible...which is directly contradicting the definition of a reversible & adiabatic flow...

    Hope this is a pertinent question...

    Thanks a lot for any help.
  2. jcsd
  3. Apr 17, 2012 #2
    Imagine a bunch of balloons containing gas. Imagine that they expand when they fly out of a window. The expansion of the gas cools the gas in the balloons. Yet, there is not any significant heat exchange between the balloons. That's because there's not enough time permitted for that exchange of heat to be significant. Thus, it is adiabatic.
  4. Apr 17, 2012 #3
    I do not have a problem with the fact that the process is adiabatic (no heat transfer with the surroundings). My problem is with the fact that the transformation is reversible, and yet there is a spatial temperature gradient which would cause heat transfer within the gas (not with the surroundings). Heat transfer due to a temperature gradient within the gas is irreversible...maybe I'm thinking too much...sorry
  5. Apr 17, 2012 #4
    If you count the separate balloons as members of the same "gas", you can easily see how there is not significant transfer of heat, not even between parts of the gas.
  6. Apr 17, 2012 #5
    That makes more sense ! Thanks a lot for clearing that up.
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