Joule-Thomson Expansion

[Kakashi]

A gas flows along an insulated pipe (q=0) through a porous plate that separates two sections of the pipe at different constant pressures P1 and P2.

In Thermodynamics we study systems at equilibrium but here the gas is not in equilibrium because there is a pressure difference across the porous plate.If this were not the case, the pistons would accelerate and the pressures would change with time. The pistons are externally constrained so that P1 and P2 are held constant.

I read that gases flow from high pressure to lower pressure to equalize pressure. If gas flows from the left chamber to the right chamber through the porous plate, wouldnt this initially cause the gas pressure in the left chamber to drop so that $$ P_{gas,left}<P_{1} $$ causing the left piston to move inward and compress the gas until the pressure is restored to P1? Similarly, gas entering the right chamber would tend to increase the pressure so that $$ P_{gas, right}>P_{2} $$ causing the right piston to move outward and expand the volume until the pressure returns to P2. After this adjustment, there would still be a pressure difference between the left and right chambers. Does this mean that piston motion continues indefinitely, eventually pushing all the gas from the left chamber through the porous plate so that all the gas ends up on the right side?



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[Chestermiller]

Who says that in thermodynamics, we study gases that are in thermodynamics equilibrium? The states of the gas in irreversible processes do not have to be equilibrium states; only the initial and final status have to the thermodynamic equilibrium states.

There are 2 thermodynamic equilibrium states being considered here. 1. The gas significantly upstream of the porous plug (which is the initial thermodynamic equilibrium state) and 2. The same gas significantly downstream of the porous plug (which is the final thermodynamic equilibrium state). If we apply the open-system version of the first law of thermodynamics to the process causing this irreversible change in the gas from thermodynamic equilibrium state 1 to thermodynamic equilibrium state 2, we obtain $$\Delta H =0$$where H is the gas enthalpy per unit mass. So $$H(T_1,P_1)=H(T_2,P_2)$$where 1 and 2 signifiy the upstream thermodynamic equilibrium state and the downstream thermodynamic equilibrium state, respectively.

 

[Vincf]

I think you have a misconception about Joule-Kelvin expansion, which is sometimes distorted by the diagrams used. It's not a small upstream reservoir with gas expanding into a small downstream reservoir via a porous wall (or expansion capillary). In true Joule-Kelvin expansion, there's a compressor that constantly maintains the upstream and downstream pressures. Therefore, these pressures don't change. You don't need to imagine pistons upstream and downstream of the porous wall: it's the upstream gas that pushes and the downstream gas that resists.
Consider a control surface containing the porous wall. Gas constantly enters on one side and constantly exits on the other: it's an open system. (In applied thermodynamics, we almost always deal with open systems).

To apply the first law of thermodynamics to an open system, we must consider the closed system within the control surface at time $t$ and observe what happens to it between $t$ and $t+dt$. For Joule Kelvin expansion, we are in steady state, and ultimately, everything behaves as if a mass of gas dm enters and exits later without having exchanged heat, but having received work from the pressure forces. Its thermodynamic states at the inlet and outlet are well-defined and are equivalent to equilibrium states. Its specific enthalpy remains unchanged.