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I Enthalpy in Throttle Process (Joule–Thomson expansion)

  1. Dec 28, 2017 #1
    Why is it written that enthalpy (##H##) is constant for Joule-Thompson expansion? It seems the essence of this process is to convert from one pressure to another with no heat loss. How does that connect with enthalpy being constant? When I learned about enthalpy it seemed to be most relevant to systems at constant pressure. However in the Joule-Thompson effect the pressure is surely changing. Sources mention that the "flow work" is ## P_2V_2 - P_1V_1##. This leads to constant ##H## by conservation of energy. I guess I don't understand why the flow work is ##\Delta(PV)## and not ## \int P dV##. Maybe it also seems strange for me to think of a fluid doing work on itself or something. Anyway, I feel confused about this.
     
  2. jcsd
  3. Dec 28, 2017 #2
    Are you familiar with the derivation of the open-system (control volume) version of the first law of thermodynamics?
     
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