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Thread: The Joule-Thomson Coefficient

  1. #1 The Joule-Thomson Coefficient 
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    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.

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  3. #2  
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    Join Date
    Dec 2006
    The Joule-Thomson (Kelvin) coefficient

    As for how the Joule-Thomson effect is achieved in practice:

    The real gas is allowed to expand through a throttling device (usually a valve) which must be very well insulated to prevent any heat transfer to or from the gas.
    There must be no external work extracted from the gas during the expansion (the gas must not be expanded through a turbine, for example).
    The effect is applied in the Linde technique as a standard process in the petrochemical industry for example, where the cooling effect is used to liquefy gases, and also in many cryogenic applications (e.g. for the production of liquid oxygen, nitrogen and argon). Only when the Joule-Thomson coefficient for the given gas at the given temperature is greater than zero can the gas be liquefied at that temperature by the Linde cycle. In other words, a gas must be below its inversion temperature to be liquified by the Linde cycle. For this reason, simple Linde cycle liquifiers cannot normally be used to liquify helium, hydrogen and neon.

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