Joule Thomson versus Carnot

Jul 2016
Joule Thomson cooling is part of many kryo-physical equipments.
I am wondering why this is the case, as Joule Thomson (=adiabatic, isenthalpic) cooling to me seems to be less efficient than the adiabatic, reversible (=isentropic) cooling like in a Carnot-Process:
During adiabatic, reversible cooling the expansion work, that the system is exercising on the environment on its way from p1 > p2 to p2 is p(V)dV. In Joule Thomson the work is p2dV. The second is less than the first. Joule Thomson can be considered a Carnot process with the 'excess work' being reapplied to the system as heat. This 'dissipative reheating' is what makes Joule Thomson inherently irreversible.
Most obvious: Joule Thomson cooling does not work for an ideal gas, while Carnot cooling does.
Is the technical simplicity (no moving parts, steady process as opposed to a Carnot-like 'steam engine') the only reason?

Kind regards

Apr 2015
Somerset, England

1) To implemement any form of refrigeration you need a gas with suitable real world properties in the temperature range of interest. Have you looked for one?

2) See my note in your other thread about real world gases.

3) Refrigeration cycles generally depend heavily on latent heat exchange. Latent heat is often significant compared to specific heat.
Jul 2016

I have not really considered the properties of any real gas. What I am thinking of is
the cooling of He4 to temperatures on the order of 1K by expanding the gas-phase in a Joule Thomson manner and then pumping the gas on the low pressure side.

I agree that the latent heat of the phase-transition liquid->gas is a contributer,
but if I remember correctly, that latent heat is rather low for He4 at those temperatures.
Anyway the cooling is generally referred to as 'Joule Thomson cooling' not 'Helium evaporation cooling' so I was expecting the latent heat to be less important.
In any case the end temperature would be lower, if the expansion was done in 'Carnot manner' rather than in "Joule Thomson' manner.