Enthelpy in Throttle Process (Joule–Thomson expansion)

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Enthalpy is considered constant during the Joule-Thomson expansion because the process occurs at constant enthalpy under adiabatic conditions, where no heat is exchanged with the surroundings. This expansion involves a change in pressure without heat loss, leading to the conservation of energy principle, which maintains constant enthalpy. The flow work is expressed as P2V2 - P1V1, illustrating the relationship between pressure and volume changes during the process. The confusion arises from the distinction between flow work and the integral form of work, as well as the concept of a fluid doing work on itself. Understanding the open-system version of the first law of thermodynamics can clarify these concepts.
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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.
 
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Are you familiar with the derivation of the open-system (control volume) version of the first law of thermodynamics?
 

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