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Claude cycle Helium Liquifier

  1. Jun 26, 2006 #1
    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    My first question is simply how is the Helium withdrawn as a liquid replaced?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!
    Last edited: Jun 26, 2006
  2. jcsd
  3. Jun 29, 2006 #2

    Andrew Mason

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    It appears that this engine is designed to make liquid Helium. It appears that the liquid He is removed and not replaced.

    I would suggest that you show us your work first to see if we can help you.

  4. Jun 30, 2006 #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.
    Last edited: Jun 30, 2006
  5. Jun 30, 2006 #4
    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?
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