How does the Claude cycle Helium liquifier work?

In summary, the conversation discusses a problem in thermal physics involving a system where gas is compressed, cooled, and then expanded to produce liquid helium. The liquid helium is removed and not replaced, raising questions about the process and the calculations needed to determine the values of (Pc, Te) given (Pin, Tin). Further discussion explores the possibility of returning the boiled off liquid helium to the cycle or finding another way to replace it.
  • #1
TMFKAN64
1,126
22
I've been working on problem 12-3 in Kittel and Kroemer's "Thermal Physics", and I'm not sure I fully understand what is going on here.

As far as I can tell, the system being described is something like this:

Gas leaves a compressor at (Tc, Pc) and enters a heat exchanger. The gas leaves the heat exchanger at (Te, Pc) and enters an expansion engine, which it leaves at (Tin, Pin). The cooled gas then enters a J-T valve, which liquifies [tex]\lambda[/tex] of it, leaving [tex]1-\lambda[/tex] to leave the chamber at (Tin, Pout). This gas is then cycled through the heat exchanger, which raises its temperature to Tc. This gas at (Tc, Pout) is fed back into the compressor.

The liquidification coefficient, [tex]\lambda[/tex], is given, as are the values Tin and Pin. An equation is given for the work that must be extracted by the expansion engine: [tex]W_e = \frac{5}{2}\lambda R (Tc - Tin)[/tex]. The compressor is assumed to operate isothermally, and the expansion engine isentropically. 1 mol/s of helium leaves the compressor. Helium is assumed to be an ideal gas, and the heat exchanger is also ideal.

My first question is simply how is the Helium withdrawn as a liquid replaced? Do we simply add new Helium (or the Helium boiled off by the load being cooled) to the compressor input? (The expression given for the expansion engine work seems to me to imply that we are not sending this replacement Helium through the heat exchanger, since it is assuming that [tex](1 - \lambda)[/tex] moles of gas are being returned to the compressor through this path... or am I misunderstanding what is going on here?)

Secondly, the problem itself asks for a calculation of (Pc, Te) given the values of (Pin, Tin). If the work done by an expansion engine is [tex]\frac{5}{2}R(Te - Tin)[/tex], we can equate this to the expression for [tex]W_e[/tex] given above and find an expression for Te in terms of Tc, Tin and [tex]\lambda[/tex]. However, I don't see how to eliminate Tc from this equation. (The fact that the process is isentropic enables the calculation of Pc given Pin, Tin, and Te, but I don't see how this can help eliminate Tc.)

Any help would be appreciated, thanks in advance!
 
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  • #2
TMFKAN64 said:
My first question is simply how is the Helium withdrawn as a liquid replaced? Do we simply add new Helium (or the Helium boiled off by the load being cooled) to the compressor input? (The expression given for the expansion engine work seems to me to imply that we are not sending this replacement Helium through the heat exchanger, since it is assuming that [tex](1 - \lambda)[/tex] moles of gas are being returned to the compressor through this path... or am I misunderstanding what is going on here?)
It appears that this engine is designed to make liquid Helium. It appears that the liquid He is removed and not replaced.

Secondly, the problem itself asks for a calculation of (Pc, Te) given the values of (Pin, Tin). If the work done by an expansion engine is [tex]\frac{5}{2}R(Te - Tin)[/tex], we can equate this to the expression for [tex]W_e[/tex] given above and find an expression for Te in terms of Tc, Tin and [tex]\lambda[/tex]. However, I don't see how to eliminate Tc from this equation. (The fact that the process is isentropic enables the calculation of Pc given Pin, Tin, and Te, but I don't see how this can help eliminate Tc.)
I would suggest that you show us your work first to see if we can help you.

AM
 
  • #3
Well, if [tex]\frac{5}{2}R(Te - Tin) = \frac{5}{2}R\lambda(Tc - Tin)[/tex], then [tex]Te = \lambda Tc + (1 - \lambda)Tin[/tex]. But that's almost besides the point... the big question is where is there another relationship that can be used to eliminate Tc from this equation?

I thought that perhaps it could come from the heat exchanger... the energy added to the return gas is [tex]\frac{3}{2}R(1 - \lambda)(Tc - Tin)[/tex], and this must equal the energy removed from the compressed gas, [tex]\frac{3}{2}R(Tc - Te)[/tex]. However, this simplifies to the exact same expression as above.

If we did have a value for Te, then we could use the fact that for an isentropic process [tex]\frac{T^{5/2}}{P}[/tex] is constant to find Pc.
 
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  • #4
Andrew Mason said:
It appears that this engine is designed to make liquid Helium. It appears that the liquid He is removed and not replaced.

Oh, I forgot to address this... I don't think that this is possible. If 1 mol/sec enters the Linde stage as the problem specifies, and [tex]1 - \lambda[/tex] mol/sec of this flow is sent back to the compressor through the heat exchanger, how could we maintain the 1 mol/sec coming out of the compressor? Don't we have to either return the boiled off liquid helium to the cycle somehow or otherwise replace it?
 

Related to How does the Claude cycle Helium liquifier work?

1. What is a Claude cycle helium liquifier?

A Claude cycle helium liquifier is a device used to liquefy helium gas by employing the principles of thermodynamics and the Joule-Thomson effect. It is named after its inventor, Georges Claude.

2. How does a Claude cycle helium liquifier work?

The Claude cycle helium liquifier works by compressing the helium gas and then cooling it down using a heat exchanger. The cooled gas is then expanded through a valve, causing it to cool further due to the Joule-Thomson effect. This cooled gas is then used to cool down the incoming helium gas, creating a cycle that eventually results in the liquefaction of helium.

3. What is the purpose of a Claude cycle helium liquifier?

The main purpose of a Claude cycle helium liquifier is to produce liquid helium, which is used in various scientific and industrial applications, such as in cryogenics, MRI machines, and superconductors.

4. What are the advantages of using a Claude cycle helium liquifier?

Compared to other methods of liquefying helium, such as the Hampson-Linde cycle, the Claude cycle helium liquifier is more efficient and requires less energy. It is also a continuous process, making it suitable for large-scale production of liquid helium.

5. Are there any limitations to using a Claude cycle helium liquifier?

One limitation of a Claude cycle helium liquifier is that it can only be used for helium gas with a high purity level, as impurities can affect the efficiency of the liquefaction process. It also requires a substantial initial investment and maintenance costs, making it less suitable for small-scale or occasional use.

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