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.

 

[Kakashi]

I am quite confused by the way my textbook presents the Joule–Thomson experiment. A gas flowing through a throttling valve from a high pressure Pi to a lower pressure Pf is an open system. However, the book treats the gas as if it were a closed system and imagines the upstream side acting like a piston compressing the gas at pressure Pi, while a downstream piston resists at pressure Pf.

Is the book effectively tracking a small mass of gas that passes through the valve, such that at some time the mass occupies a region upstream and the surrounding upstream fluid (represented by the left piston) pushes it into and through the valve? Later, the same mass occupies a region downstream, so relative to that moving mass the surrounding fluid or the left pistonsweeps across the region it previously occupied.

How does this viewpoint justify the use of the closed-system first law?

Is fluid pushing fluid mechanically equivalent to a piston applying the same pressure in a piston–cylinder process?

 

[Vincf]

The first law of thermodynamics, written in the form $U_2 - U_1 = W + Q$ , only applies to a closed system. That's why we isolate a closed system at a given instant and track it over time.

In the explanation provided, the closed system contains the porous zone that maintains the pressure gradient.
You can imagine two (immaterial) walls that delimit this closed system upstream and downstream. The external gas exerts pressure on the walls. Upstream, it pushes, and downstream, it resists.

Now, imagine the gas moving forward. The two walls move simultaneously with the gas. If the upstream wall has swept through the volume $V_i$, the work done is $P_i V_i$. The work done downstream is $-P_f V_f$. The work done is the same as if there were pistons.

Since there is no heat exchange, the first law of thermodynamics can be written as: $$U_2 -U_1 = P_i V_i - P_f V_f$$ But if you look at how the closed system has evolved, you see that the central part has not changed because we are in steady state (The fluid flows continuously, but it remains in the same state in this central region).
If we denote $U_c$ the internal energy of the central part, we have: $U_2=U_c+U_s$ and $U_1=U_c+U_i$ if we denote $U_i$ the internal energy of the gas which, upstream, had the volume $V_i$ and $U_f$ the internal energy of the gas which, downstream, had the volume $V_f$. $$U_2 -U_1 = U_f -U_i$$
In the end, we are left with: $$U_f -U_i=P_i V_i - P_f V_f$$ or:
$$U_f +P_f V_f=U_i+P_i V_i $$ or, with enthalpy: $$H_f=H_i$$
In this expression, $H_f$ is the enthalpy of the downstream gas of volume $V_f$ and $H_i$ is the enthalpy of the upstream gas of volume $V_i$.

Since the mass flow rate is constant (steady state), the mass corresponding to $V_f$ and $V_i$ are identical, equal to $M$. You can therefore divide by this common mass, and we obtain that the upstream specific enthalpy is equal to the downstream specific enthalpy.

Note that this reasoning can sometimes be a little confusing because the fluid considered upstream and downstream is not the same. However, when defining specific enthalpy, it's simply a matter of taking a quantity of fluid and dividing by its mass.
Also note that if you want to track a closed mass of fluid passing through the porous region, it will initially occupy the volume $V_i$ and have the enthalpy $H_i$ . Then it will pass through the porous region and many complicated things will happen to it. Finally, after a certain time, it will emerge and occupy the volume $V_f$ with the enthalpy $H_f=H_i$: so in the end, a closed mass of fluid does indeed have the same enthalpy at the start and at the finish.

In applied physics, this kind of reasoning is constantly encountered, and the first law is written once and for all, applied to an open system. It's not very complicated. It amounts to stating that, for an open system, the internal energy can be changed by supplying heat and work, just as with a closed system, but it can also be changed by introducing matter into the system.