Issues with Chlorine Liquefaction

A common problem that occurs when releasing chlorine from cylinders during industrial chlorination processes is that of liquefaction of the gas and ice deposition in and around the cylinders which slow the rate of withdrawal of chlorine from the cylinders. Thermodynamically, its a legit process when chlorine will undergo expansion, with the help of Joule Thompson coefficient, it will cool down.


My issue with the reasoning is that since the expansion is occurring at the valve itself and adiabatically, there shouldn't be cooling inside the cylinder yet I do have pictures of a cylinder with ice deposited all over its surface due to expansion via cooling. How do we explain the fact that the cooling effect is travelling back through the expansion nozzle and more importantly, how do we calculate the cooling rate and total energy removed from the cylinder via expansion through nozzle?




You have it backwards. The major part of the cooling occurs within the cylinder itself, with very little cooling (if any) occurring in the valve due to Joule Thomson.


The gas remaining within the cylinder at any time has expanded adiabatically and reversibly to do work in forcing the gas exiting the cylinder ahead of it into the valve. On the other hand, the gas passing through the valve is experiencing both expansion cooling and viscous frictional heating, so that the net effect is much smaller.


To calculate the cooling effect within the cylinder, one merely needs to apply the open system version of the 1st law of thermodynamics to this adiabatic expansion into the valve. For an ideal gas, this would read $$nC\_vdT=RTdn$$ where n is the number of moles of gas within the cylinder at any time.


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Gases, that at room temperature condense before reaching sufficient storage pressure, are provided in the liquified form with gaseous headspace, what allows much better storage capacity.


Liquid chlorine has the vapor pressure [$\pu{5830 mm Hg}$ at $\pu{25^{\circ}C}$](https://webwiser.nlm.nih.gov/substance?substanceId=338&identifier=Chlorine&identifierType=name&menuItemId=42&catId=48), what is about $\pu{7.7 atm}$.
Having cylinders with just few atmospheres of gaseous chlorine would not make much sense.


Adiabatic or Joule-Thompson expansion cooling has minor impact here. Additionally, considering what the output chlorine pressure may be, there may not be much expansion at all, depending on application.


The major effect is evaporation cooling of liquid chlorine, based on its evaporation enthalpy. The similar effect is observed on the outdoor camping LPG burners. When cooking, you can feel how cold the thin wall LPG cartriges become.


* The [chlorine boiling point](https://en.wikipedia.org/wiki/Chlorine) is $\pu{−34.04^{\circ}C}$.
* The [propane boiling point](https://en.wikipedia.org/wiki/Propane) is $\pu{−42.25}$ to $\pu{−42.04^{\circ}C}$.


Therefore chlorine is even more easily liquified than propane. With an intense output flow, the boiling chlorine may reach deep frost temperature.


If the pressure regulator has equal or lower temperature than the liquid chlorine - e.g. for the minor expansion effect, then the gaseous chlorine will condense there. Similarly as water vapor condenses on glasses, if you enter a vapor chamber in a spa, or if you enter a warm room from cold outside during winter. Mild heating of regulators may help.


Dew or frost deposition on cylinders indicates another issue, as low cylinder temperature significantly decreases the gas pressure. (LPG cartridges are often kept in warm sleeping bags overnight, as butane boiling point is near $\pu{0^{\circ}C}$.) It would help if there was applicable some technical way to keep cylinders at room temperature.


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