AutoC use of R14/Argon mixture explained
kang-China...
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In the lastest evaporator to used Ar lique, so R14 evaporation temperature is about -128℃
Unfortunately I think we are having a language translation problem :confused:
As Nol just stated: the Argon is not condensed.
But in order to have the evaporator get colder then -128C, especially when the evaporator pressure is above 0 psi, Argon is required. However due to the critical point of Argon, no amount of pressure would be able to condense it.
So what does the Argon do if it can not be condensed?
The exact mechanism is unknown, but as I stated in an earlier posted answer to your question, it is felt that two possibilities exist. The first one having to do with "Partial Pressure", where the presence of Argon in the evaporator, creates an effect on the evaporating R14, of being evaporated into a lower pressure than reality.
The use of Argon to create a "Partial Pressure" on the R14 is the same principle as used in Absorption Refrigeration Systems, and is best described by this excerpt from this WorldLingo article (for our scenario, substitute R14 for the Ammonia, and Argon for the Hydrogen)...
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The cooling cycle starts at the evaporator, where liquefied anhydrous ammonia enters. (Anhydrous means there is no water in the ammonia, which is critical for exploiting its sub-zero boiling point.) The "evaporator" contains another gas (in this case, hydrogen), whose presence lowers the partial pressure of the ammonia in that part of the system. The total pressure in the system is still the same, but now not all of the pressure is being exerted by ammonia, as much of it is due to the pressure of the hydrogen. Ammonia doesn't react with hydrogen - the hydrogen is there solely to take up space - creating a void that still has the same pressure as the rest of the system, but not in the form of ammonia. Per Dalton's law, the ammonia behaves only in response to the proportion of the pressure represented by the ammonia, as if there was a vacuum and the hydrogen wasn't there. Because a substance's boiling point changes with pressure, the lowered partial pressure of ammonia changes the ammonia's boiling point, bringing it low enough that it can now boil below room temperature, as though it wasn't under the pressure of the system in the first place. When it boils, it takes some heat away with it from the evaporator - which produces the "cold" desired in the refrigerator.
The 2nd possibility of what happens with the Argon and R14 mixture, is that the Argon gas will dissolve to a small degree into the condensed R14, thereby causing the R14 to evaporate at a temperature below it's norm when later expanded. The degree of Argon that goes into solution with the R14, is wholly dependent on the temperature of the condensed R14. So as it gets colder, and goes into a subcooled state, more Argon will tend to dissolve into it, thus the R14/Argon solution will evaporate even colder. This effect will continue until such time that an equilibrium is obtained, where the physical constraints of heat load versus evaporating pressure exert themselves (determined by limits of pressure drop through heat exchangers, and pumping ability of the compressor).
It is my belief that both of these theories are in effect with the R14/Argon mixture within an AutoC.
what about this question to answer thank you
hi mytek
in your AC-2,AC-3
I don't know the area
each cascade condenser
common temp about -20or-30 ,figure out the area is very easy.
But your AC-2,OR AC-3, condenser area is very very difficulty!
so you are a marvelous guy
could you tell about this '
