Joule-Thomson Effect CHRISTOPH WINDMEIER, Linde AG, Pullach (M€unchen), Germany RANDALL F. BARRON, Louisiana Tech University, Ruston, United States A fluid flow encountering an obstruction like, e.g., an expansion valve, an orifice plate, a porous plug, or a capillary tube, will experience a pressure drop. Since outer heat transfer and changes in kinetic energy are usually negligible, and because there is no transfer of work from and to the fluid flow, the resulting expansion can be considered to be isenthalpic. A non-ideal fluid will hence experience a change in temperature. This fundamental phenomenon is called the Joule–Thomson (J–T) effect and it enables the production of cold temperature levels by isenthalpic expansion of highpressure fluids. The origin of the J–T effect lies within the occurrence of intermolecular forces in nonideal fluids: A pressure reduction is accompanied by an increase of specific volume. Consequently the average intermolecular distances also increase. The internal work done on attractive or repulsive interactions will result to a respective change of system temperature. The constant enthalpy lines on the temperature– pressure plane are shown in Figure 1. The effect of temperature change for an isenthalpic change in pressure is represented by the Joule–Thomson coefficient, mJT, defined by mJT ¼ @T @p h The Joule–Thomson coefficient is equal to the slope of the isenthalpic lines in Figure 1. In the region where mJT < 0, isenthalpic expansion results into an increase in temperature, whereas in the region where mJT > 0, respective expansion results into a temperature decrease. The curve that separates the two regions is called the inversion curve. The J–T coefficient is zero along the inversion # 2015 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim 10.1002/14356007 curve, because the slope of the isenthalpic line is zero. Thermodynamics show that the J–T coefficient may be expressed in terms of volumetric and caloric properties as follows: mJT ¼ " # 1 @v T v cp @T p ð1Þ where v is the specific volume and cp the specific heat at constant pressure of the material. Since an ideal gas does not possess any intermolecular forces one can expect a value of zero for its J–T coefficient. Substituting the equation of state of an ideal gas v ¼ (RT)/p into Equation 1 gives: @v @T p ¼ R v ¼ p T where R is the gas constant and p is the pressure. Substitution into Equation 1 gives mJT ¼ 0 for an ideal gas. For a real gas, the J–T coefficient may be negative, zero, or positive, depending on the initial temperature and pressure of the material. The simplest equation of state for real gases is the van der Waals equation of state: pþ a ðv bÞ ¼ RT v2 where the constant a gives a measure of the intermolecular forces between the molecules, and the constant b provides a measure of the finite size of the gas molecules. Constants a and b are thus dependent upon the species and can be determined using the first and second derivative of the isothermal line at the critical point. 2 Joule-Thomson Effect The inversion curve is formed by all points at which the J–T coefficient is zero. The inversion temperature Ti from Equation 2 is given by: Ti ¼ 2a ð1 b=vÞ2 bR The maximum inversion temperature for a van der Waals fluid is the temperature on the inversion curve where p ¼ 0 (or b/v ¼ 0) or Ti, max ¼ 2a/bR. The maximum inversion temperature (in K) for several gases is as follows: Figure 1. General temperature–pressure diagram for a real gas The expression for the J–T coefficient for a van der Waals gas is given by Equation 2: mJT ¼ ð2a=RT Þð1 b=vÞ2 b h i cp 1 ð2=vRT Þð1 b=vÞ2 ð2Þ For large values of the specific volume, as typically occurring in gas phases, Equation 2 can be approximated by: mJT ¼ 1 2a b cp RT This expression shows that the J–T coefficient is positive when T < 2a/bR and negative when T > 2a/bR. Helium [4He] Carbon monoxide Hydrogen Neon Argon Air Oxygen Nitrogen Methane 45 652 205 250 794 603 761 621 939 The maximum inversion temperatures for helium, hydrogen, and neon are below ambient temperature. When starting out from ambient temperature, liquefaction systems that use J–T expansion alone to produce low temperature cannot be used to liquefy these gases: Additional means such as external sources of refrigeration or expansion engines must be used in order to liquefy these gases.
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