Joule thompson effect


Published on

Published in: Technology, Business
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Joule thompson effect

  1. 1. In thermodynamics, the Joule–Thomson effect or Joule–Kelvin effect or Kelvin–Joule effect describes thetemperature change of a gas or liquid when it is forced through a valve or porous plug while kept insulated sothat no heat is exchanged with the environment. This procedure is called a throttling process or Joule–Thomson process. At room temperature, all gases except hydrogen, helium and neon cool upon expansion bythe Joule–Thomson process.DescriptionThe adiabatic (no heat exchanged) expansion of a gas may be carried out in a number of ways. The change intemperature experienced by the gas during expansion depends not only on the initial and final pressure, butalso on the manner in which the expansion is carried out.If the expansion process is reversible, meaning that the gas is in thermodynamic equilibrium at all times, it iscalled an isentropic expansion. In this scenario, the gas does positive work during the expansion, and itstemperature decreases.In a free expansion, on the other hand, the gas does no work and absorbs no heat, so the internal energy isconserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but thetemperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.The method of expansion discussed in this article, in which a gas or liquid at pressure P1 flows into a region oflower pressure P2 via a valve or porous plug under steady state conditions and without change in kineticenergy, is called the Joule–Thomson process. During this process, enthalpy remains unchanged (see a proofbelow).A throttling process proceeds along a constant-enthalpy line in the direction of decreasing pressure, whichmeans that the process occurs from left to right on a T-P diagram. As we proceed along a constant-enthalpyline from high enough pressures the temperature increases, until the inversion temperature. Then as the fluidcontinues its expansion the temperature drops. If we do this for several constant enthalpies and join theinversion points a line called the inversion line is obtained. this line intersects the T-axis at some temperature,named the maximum inversion temperature. For hydrogen this temperature is -68°. In Vapour-compressionrefrigeration we need to throttle the gas and cool it at the same time. This poses a problem for substanceswhose maximum inversion temperature is well below room temperature. Thus hydrogen must be cooledbelow its inversion temperature if any cooling is achieved by throttling.Physical mechanismAs a gas expands, the average distance between molecules grows. Because of intermolecular attractive forces(see Van der Waals force), expansion causes an increase in the potential energy of the gas. If no external workis extracted in the process and no heat is transferred, the total energy of the gas remains the same because ofthe conservation of energy. The increase in potential energy thus implies a decrease in kinetic energy andtherefore in temperature.A second mechanism has the opposite effect. During gas molecule collisions, kinetic energy is temporarilyconverted into potential energy. As the average intermolecular distance increases, there is a drop in thenumber of collisions per time unit, which causes a decrease in average potential energy. Again, total energy isconserved, so this leads to an increase in kinetic energy (temperature). Below the Joule–Thomson inversiontemperature, the former effect (work done internally against intermolecular attractive forces) dominates, andfree expansion causes a decrease in temperature. Above the inversion temperature, gas molecules movefaster and so collide more often, and the latter effect (reduced collisions causing a decrease in the averagepotential energy) dominates: Joule–Thomson expansion causes a temperature increase.
  2. 2. The Joule–Thomson (Kelvin) coefficient Joule-Thomson coefficients for various gases at atmospheric pressureThe rate of change of temperature T with respect to pressure P in a Joule–Thomson process (that is, atconstant enthalpy H) is the Joule–Thomson (Kelvin) coefficient μJT. This coefficient can be expressed in terms ofthe gass volume V, its heat capacity at constant pressure Cp, and its coefficient of thermal expansion α as:See the Derivation of the Joule–Thomson (Kelvin) coefficient below for the proof of this relation. The value ofμJT is typically expressed in °C/bar (SI units: K/Pa) and depends on the type of gas and on the temperature andpressure of the gas before expansion. Its pressure dependence is usually only a few percent for pressures up to100 bar.All real gases have an inversion point at which the value of μJT changes sign. The temperature of this point, theJoule–Thomson inversion temperature, depends on the pressure of the gas before expansion.In a gas expansion the pressure decreases, so the sign of is negative by definition.With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas:If the gas temperature is then μJT is since is thus must be so the gasbelow the inversion temperature positive always negative negative coolsabove the inversion temperature negative always negative positive warmsHelium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of oneatmosphere are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up whenexpanded at constant enthalpy at typical room temperatures. On the other hand nitrogen and oxygen, the twomost abundant gases in air, have inversion temperatures of 621 K (348 °C) and 764 K (491 °C) respectively:these gases can be cooled from room temperature by the Joule–Thomson effect.For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded atconstant enthalpy.
  3. 3. ApplicationsIn practice, the Joule–Thomson effect is achieved by allowing the gas 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. No externalwork is extracted from the gas during the expansion (the gas must not be expanded through a turbine, forexample).The effect is applied in the Linde technique as a standard process in the petrochemical industry, where thecooling effect is used to liquefy gases, and also in many cryogenic applications (e.g. for the production of liquidoxygen, nitrogen, and argon). Only when the Joule–Thomson coefficient for the given gas at the giventemperature is greater than zero can the gas be liquefied at that temperature by the Linde cycle. In otherwords, a gas must be below its inversion temperature to be liquefied by the Linde cycle. For this reason, simpleLinde cycle liquefiers cannot normally be used to liquefy helium, hydrogen, or neon.[edit] Proof that enthalpy remains constant in a Joule–Thomson processIn a Joule–Thomson process the enthalpy remains constant. To prove this, the first step is to compute the network done by the gas that moves through the plug. Suppose that the gas has a volume of V1 in the region atpressure P1 (region 1) and a volume of V2 when it appears in the region at pressure P2 (region 2). Then thework done on the gas by the rest of the gas in region 1 is P1V1. In region 2 the amount of work done by the gasis P2 V2. So, the total work done by the gas isThe change in internal energy plus the work done by the gas is, by the first law of thermodynamics, the totalamount of heat absorbed by the gas (here it is assumed that there is no change in kinetic energy). In the Joule–Thomson process the gas is kept insulated, so no heat is absorbed. This means thatwhere E1 and E2 denote the internal energy of the gas in regions 1 and 2, respectively.Using the definition of enthalpy H = E + PV, the above equation then implies that:where H1 and H2 denote the enthalpy of the gas in regions 1 and 2, respectively.Derivation of the Joule–Thomson (Kelvin) coefficientA derivation of the formula for the Joule–Thomson (Kelvin) coefficient.The partial derivative of T with respect to P at constant H can be computed by expressing the differential ofthe enthalpy dH in terms of dT and dP, and equating the resulting expression to zero and solving for the ratioof dT and dP.It follows from the fundamental thermodynamic relation that the differential of the enthalpy is given by: (here, S is the entropy of the gas).
  4. 4. Expressing dS in terms of dT and dP gives:Using (see Specific heat capacity),we can write:The remaining partial derivative of S can be expressed in terms of the coefficient of thermal expansion via aMaxwell relation as follows. From the fundamental thermodynamic relation, it follows that the differential ofthe Gibbs energy is given by:The symmetry of partial derivatives of G with respect to T and P implies that:where α is the coefficient of thermal expansion. Using this relation, the differential of H can be expressed asEquating dH to zero and solving for dT/dP then gives:It is easy to verify that for an ideal gas the thermal expansion coefficient α is 1 / T, and so an ideal gas does notexperience a Joule–Thomson effect. The cooling of a gas by pure isentropic expansion is not Joule-Thomsoncooling, although it is sometimes erroneously called J-T cooling by some laboratory practitioners.