What Should Hl Be in a Joule-Kelvin Helium Liquefier?

[sachi]

We have a helium liquefier which takes compressed He gas at 14 K, liquefies a fraction alpha and rejects the rest at 14K and atmospheric pressure.
I've got to the stage alpha = (Hf - Hi)/(Hf - Hl)
where Hf, Hi, Hl are the final, initial and liquid enthalpies respectively. I know that the liquid and the final enthalpies are determined, and that we have to find the smallest possible Hi to maximise alpha.
The only problem is determining what Hl should logically be. All we know that it must be in equilibrium with the vapour in the liquefier. Any hints would be greatly appreciated.

Sachi

 

[Jhero]

-Maria:

Hi there,

Thank you for sharing your progress and asking for help. It sounds like you are working on a very interesting project.

To determine the liquid enthalpy in this scenario, it would be helpful to consider the thermodynamic properties of helium. Helium is a noble gas with a low boiling point of 4.2 K at atmospheric pressure. This means that at 14 K and atmospheric pressure, helium will be in a supercritical state, meaning that it is neither a gas nor a liquid.

In order to liquefy helium, the pressure must be increased and the temperature must be lowered. This can be achieved by using a compressor to increase the pressure and then a heat exchanger to cool the gas. The heat exchanger is where the gas is cooled to 14 K and a fraction of it is liquefied.

So, in order to determine the liquid enthalpy, you will need to consider the properties of supercritical helium and how it behaves when it is cooled and compressed. You may also need to consider the properties of the heat exchanger and how it affects the liquefaction process.

I hope this helps guide you in the right direction. Good luck with your research!