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According to textbooks, the Joule Thomson coefficient requires a isenthalpic process ($H = const.$). I found much related information about the throttling process. But what happens, if a pipe system for not-ideal gas (e.g. hydrogen) at some process pressure has a leakage and the environment is at atmospheric pressure at room temperature?

Is that an isenthalpic process, and why? The environment has an unlimited volume (gas volume remains the same though), therefore should the process not be isenthalpic? -> Joule Thomson does not apply? -> How can the temperature change be calculated?

The question somehow relates to Why does Joule-Thomson expansion of a gas do no (external) work?, however I am not sure of the implications. I know this is propably very trivial, but bear with me.

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  • $\begingroup$ You are aware that the JT expansion relates to continuous steady flow through an open system, right? In your case, the open system would be the wall of the pipe. And, there is supposed to be some significant flow resistance component within the system, like a porous plug, for example. $\endgroup$
    – Chet Miller
    Commented Sep 25, 2020 at 21:46
  • $\begingroup$ Therefore the question. My (limited) understanding is, that JT does not apply here but one could argue that a leakage is a flow resistance (the obstacle where the pipe is broken) and the environment outside of the pipe somehow and that there is as continuous steady flow, at least in the beginning. But, if JT does indeed not apply, how can the temperature change in this leakage process be described? $\endgroup$
    – Irenaius
    Commented Sep 26, 2020 at 5:01
  • $\begingroup$ Treat it as choked flow. The expansion is not free if going into an already occupied volume. $\endgroup$
    – Jon Custer
    Commented Sep 26, 2020 at 15:40
  • $\begingroup$ @JonCuster As far as I understand, choked flow treats the gas velocity and its limitation once sonic speed is reached. My question was if a) a pipe leakage is an isenthalpic process and b) how the temperature change can be described/calculated. $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 6:30
  • $\begingroup$ @ChetMiller I assume you are saying that we do not have constant enthalpy, therefore JT does not apply, is that right? Do you have a hint on the involved temperature dependency? I would highly appreciate an answer with background. $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 8:17

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Enthalpy is constant when a fluid is squeezed slowly through a restriction with no heat exchange. The reason is that when some mass $M$ of fluid flows into the restriction, it brings its internal energy $U_1$ with it, and the fluid or piston behind it (whatever is pushing it along) does work $p_1 V_1$ to push it through, where $V_1$ is the volume of the mass $M$ of fluid and $p_1$ is its pressure. This same mass $M$ then emerges on the other side of the restriction and pushes back whatever fluid or piston is there, doing work $p_2 V_2$ in order to make room for the final volume $V_2$ of the fluid under discussion. It emerges with internal energy $U_2$ and by conservation of energy we must have $$ U_1 + p_1 V_1 = U_2 + p_2 V_2 $$ assuming that no heat has flowed either in or out. There may be further effects going on, such as chemical reactions or phase changes, but these are all included in $U_1$ and $U_2$.

So you see in the case of a leak into the atmosphere, you don't need to worry about the total volume of the atmosphere. All you need to know is that the leaking gas pushing some of the atmosphere out of the way---an amount $V_2$ to be specific.

Approximations

The above assumes there is no net heat transfer into or out of each parcel of gas that leaks out (during the leaking, that is). This will be an approximation in practice. The name "Joule Thomson process" (also called "Joule Kelvin process") refers to the idealized process with no heat transfer. A real process will be some approximation to this. In the case of a leak the approximation is usually quite good if there is a reasonable amount of pressure on both sides of the leak, and it is not too slow.

If the expansion were very slow then there would be time for non-negligible heat flow. In this case, if the temperature inside and outside the pipe were initially the same then I guess the process would be isothermal. In practice it will be some sort of intermediate case with no simple description.

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  • $\begingroup$ But would there not be exactly such a heat exchange in a real system? Thus enthalpy is not constant? $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 11:57
  • $\begingroup$ The heat transfer in JT is assumed to be negligible, mostly because of the short residence time involved. $\endgroup$
    – Chet Miller
    Commented Sep 28, 2020 at 15:54
  • $\begingroup$ @Irenaius The name "Joule Thomson process" (also called "Joule Kelvin process") refers to the idealized process with no heat transfer. A real process will be some approximation to this. In the case of a leak the approximation is usually quite good if there is a reasonable amount of pressure on both sides of the leak, and it is not too slow. $\endgroup$
    – Andrew Steane
    Commented Sep 28, 2020 at 16:45
  • $\begingroup$ So you say, it can be concluded that in good approximation the temperature change in a leak such as described in the question CAN in fact be described by a Joule Thomson process. What kind of process do you get, if the process is too slow? How can the temperature dependence be described in that case? $\endgroup$
    – Irenaius
    Commented Sep 28, 2020 at 16:55
  • $\begingroup$ @Irenaius It depends on the conditions. If the temperature inside and outside the pipe is initially the same, and the expansion were very slow, then I guess it would be isothermal. In practice it will be some sort of intermediate case with no simple description. $\endgroup$
    – Andrew Steane
    Commented Sep 28, 2020 at 18:43

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