A few weeks ago, a question was submitted regarding adiabatic venting of a chlorine cylinder:
Issues with Chlorine Liquefaction

The member submitting the question was interested in why the stream exiting a chlorine tank became cold, particularly why the exit valve became cold. @poutnik presented and excellent answer to these questions, and also referred to the case of propane tanks which display similar behavior. I presented a similar discussion for the case of an ideal gas.

Subsequently, I became interested in how to quantitatively predict this kind of behavior, say for the typical case of venting a propane tank. Does anyone have any ideas on how to thermodynamically model such a situation? In such a model, the liquid propane and propane vapor would be gravitationally segregated, and only propane vapor would be able to exit the tank through a valve situated at the top. I was thinking of applying the open system (control volume) version of the 1st law of thermodynamics to solve for the temperature as a function of the mass of propane remaining in the insulated tank. I envision the model to include relative changes in the amounts of liquid and vapor in the tank as a result of liquid vaporization and vapor loss through the vent. I would also envision he model to treat the vapor as non-ideal because of the relatively high pressures initially (say, > 10 bars). So, in my judgment, use of thermodynamic tables on propane would be preferable over analytic representations of thermodynamic properties.



1 Answer 1


Since I have not yet received any answers to my question, I will post the answer I arrived at using the methodology I alluded to in the question.

The analysis makes use of the open system (control volume) version of the 1st law of thermodynamics, which, for the present transient adiabatic venting situation, reduces to $$d(mu)=h_Vdm=(u_V+Pv_V)dm\tag{1}$$where m is the mass of propane within the tank at any time, u is the average internal energy per unit mass (averaged over vapor and liquid), P is the equilibrium vapor pressure of the propane in the tank, and $v_V$ is the specific volume of the vapor. Eqn.1 tells us that the differential changes in internal energy of the liquid/vapor combined mass in the tank are equal to the differential amount of vapor enthalpy exiting the tank through the valve.

In addition to Eqn. 1, we have $$u=u_Lx+u_V(1-x)\tag{2}$$and $$v_Lx+v_V(1-x)=\frac{V}{m}\tag{3}$$where x is the mass fraction of liquid in the tank, the u's and v's are the specific internal energies and specific volumes, respectively, and V is the volume of the tank.

I combined the above 3 equations into a single differential equation (by eliminating the mass fraction x), and, after considerable mathematical manipulation, arrived at the following differential equation for the changes in propane mass in the tank along the equilibrium venting trajectory:$$d\ln{m}=\frac{dh_V}{\lambda v_V}+\frac{V}{mv_ V}d\ln{(\lambda/T)}-d\ln{(\lambda v_V)}\tag{4}$$where $$\lambda=\frac{h_V-h_L}{v_V-v_L}$$and where the h's are enthalpies.

I integrated Eqn. 4 numerically for the test case of a 33.4 liter insulated propane tank initially at 300 K and 10.0 bars and initially containing 8.4 kg propane, using the thermodynamic data presented in the following table (Thermopedia):
enter image description here

The results of the calculations are shown in the figure below. enter image description here

The beginning of venting is at 300 K and 8.4 kg propane, the upper right point of the line. The line shows the trajectory of temperature vs propane amount remaining in the adiabatic tank at any time. These results seem consistent with the description provided by @poutnik in the original thread.


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