I Enthalpy in Throttle Process (Joule–Thomson expansion)

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1. Dec 28, 2017

MisterX

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. Dec 28, 2017

Staff: Mentor

Are you familiar with the derivation of the open-system (control volume) version of the first law of thermodynamics?

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