The Joule-Thomson coefficient (Chemical engineering)

Discussion in 'Chemical | Metallurgy' started by mbeychok, Sep 20, 2007.

  1. mbeychok

    mbeychok Author, Consultant

    When a real gas, as differentiated from an ideal gas, expands at constant enthalpy (i.e., no heat is transfered to or from the gas, and no external work is extracted), the gas will be either cooled or heated by the expansion. That change in gas temperature with the change in pressure is called the Joule-Thomson coefficient and is denoted by µ, defined as:

    µ = (dT/dP) at constant enthalpy

    The value of u depends on the specific gas, as well as the temperature and pressure of the gas before expansion. For all real gases, µ will equal zero at some point called the "inversion point". If the gas temperature is below its inversion point temperature, µ is positive ... and if the gas temperature is above its inversion point temperature, µ is negative. Also, dP is always negative when a gas expands. Thus:

    If the gas temperature is below its inversion temperature:
    -- µ is positive and dP is always negative
    -- hence, the gas cools since dT must be negative

    If the gas temperature is above its inversion temperature:
    -- µ is negative and dP is always negative
    -- hence, the gas heats since dT must be positive

    "Perry's Chemical Engineers' Handbook" provides tabulations of µ versus temperature and pressure for a number of gases, as do many other reference books. For most gases at atmospheric pressure, the inversion temperature is fairly high (above room temperature), and so most gases at those temperature and pressure conditions are cooled by isenthalpic expansion.

    Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at atmospheric pressure are very low (e.g., about -222 °C for helium). Thus, helium and hydrogen will warm when expanded at constant enthalpy at atmospheric pressure and typical room temperatures.

    It should be noted that µ is always equal to zero for ideal gases (i.e., they will neither heat nor cool upon being expanded at constant enthalpy).

    By contrast, when external work is extracted during the expansion of a gas (as when a high-pressure gas is expanded through a turboexpander), the expansion is isentropic (i.e., occurs at constant entropy) rather than isenthalpic as in a Joule-Thomson expansion. For an isentropic gas expansion, the gas temperature always cools and the temperature drop is more than would be achieved by an isenthalpic Joule-Thomson expansion.
  2. Pedrosbier

    Pedrosbier Newbie

    Hello crazy engineers!

    My name is Pedro, I do not have a formal engineering degree, but I am definitely a crazy inventor since many years. My current project is the development of a nitrogen liquefier for a research and development institution in Brazil. This thread came to my attention as I am intending to use helium for the last stage of the cascade refrigeration system I am trying to design.

    Well, nitrogen liquefies at arround -195 °C (1atm), so I need to reach at least -190°C in order to liquefy the pressurized nitrogen gas. Therefore, my plan is to pressurize the helium to 250 PSI, cool it, and then expand it in a counter-flow heat exchanger where the nitrogen gas would liquefy.

    Very good so far, but now I need to know how much do I need to cool the pressurized helium so that when it is expanded it will be able to liquefy the nitrogen gas. So I started to research the properties of helium. Oh boy! Internet research can be very missleading sometimes! So I got very confused by the different lines of thinking arround helium. There seems to be a lot of mysticism arround the thermodinamic properties of this gas.

    Two things came to my concern as when using helium as a refrigerant in a normal refrigeration cycle:

    1. The critical point of helium4 is 5.19 K at 0.227 MPa:
    Normal refrigerants evaporate and condense arround the refigeration cycle, takind and giving latent heat. For my aplication, given the target temperature of -190°C, helium would most likely remain as a gas or super-critical fluid, but never liquid.
    Question: Would that represent a barrier for achieving cooling?

    2. Helium has a negative Joule thompson coeficient @ 1atm.
    Question: What implication does this have in achieving colling effect from helium when expanding it from 250 PSI at temperature X to 30 PSI trying to achieve -190°C? What would X most likely be?

    I would apreciate any inputs.

    Best regards,
    Pedro
  3. gpadamwar

    gpadamwar Certified CEan

    Dear sir,
    I wanted to know that if synthesis gas at 115 kg/cm2g and 20°C is expanded using throttle valve to 2 kg/cm2g how to calculate
    - new temp of synthesis gas ?
    - wheather partial pressure of individual gases should be used?
    - if any graph is to be use what is the procedure to use it?
    - if incoming synthesis gas is cooled using expanded gas how much temp can be achieved of incoming gas?
    - Is it possible to calculate the time required to achieve that temp after cooling using expanded gas?

    synthesis gas analysis is H2= 66%, CH4=7.5%, Ar=2.5%,and rest is nitrogen


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