In thermodynamics, the Joule–Thomson effect (also known as the Joule–Kelvin effect, Kelvin–Joule effect, or Joule–Thomson expansion) describes the temperature change of a gas or liquid when it is forced through a valve or porous plug while kept insulated so that 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 by the Joule–Thomson process.
The effect is named after James Prescott Joule and William Thomson, 1st Baron Kelvin, who discovered it in 1852. It followed upon earlier work by Joule on Joule expansion, in which a gas undergoes free expansion in a vacuum and the temperature is unchanged, if the gas is ideal.
The throttling process is of the highest technical importance. It is at the heart of thermal machines such as refrigerators, air conditioners, heat pumps, and liquefiers. Furthermore, throttling is a fundamentally irreversible process. The throttling due to the flow resistance in supply lines, heat exchangers, regenerators, and other components of (thermal) machines is a source of losses that limits the performance.
- 1 Description
- 2 Physical mechanism
- 3 The Joule–Thomson (Kelvin) coefficient
- 4 Applications
- 5 Proof that the specific enthalpy remains constant
- 6 Throttling in the T-s diagram
- 7 Derivation of the Joule–Thomson (Kelvin) coefficient
- 8 Joule's second law
- 9 See also
- 10 References
- 11 Bibliography
- 12 External links
The adiabatic (no heat exchanged) expansion of a gas may be carried out in a number of ways. The change in temperature experienced by the gas during expansion depends not only on the initial and final pressure, but also 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 is called an isentropic expansion. In this scenario, the gas does positive work during the expansion, and its temperature decreases.
- In a free expansion, on the other hand, the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but the temperature 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 of lower pressure P2 via a valve or porous plug under steady state conditions and without change in kinetic energy, is called the Joule–Thomson process. During this process, enthalpy remains unchanged (see a proof below).
A throttling process proceeds along a constant-enthalpy curve in the direction of decreasing pressure, which means that the process occurs from right to left on a Temperature-Pressure diagram. If the pressure starts out high enough, the temperature increases as the pressure drops, until an inversion temperature is reached and a phase transition occurs, called the inversion point. After this, as the fluid continues its expansion, the temperature begins immediately to drop. If we measure this point, using a specific gas, for several constant enthalpies, and join the inversion points, a curve called the inversion line is obtained. The inversion line intersects the T-axis at some temperature, called the maximum inversion temperature, because inversion cannot occur at that temperature, regardless of pressure.
In vapor-compression refrigeration the gas must be throttled and cooled at the same time. Hydrogen must be cooled below its inversion temperature if any cooling is to be achieved by throttling. For hydrogen this temperature is −68 °C. This poses a problem for substances whose maximum inversion temperature is well below room temperature.
Thermodynamic Interpretation of the experiment:
Assume that a certain amount of gas passes through the porous plug. Let the pressure and temperature on the left side of the porous plug (the initial state) be P1 and T1 with volume V1. Let the pressure and temperature on the right (the final state), be P2 and T2 with a volume V2.
As the gas is compressed, the work done on the gas is P1V1, whereas the work done by the gas during the expansion is P2V2. This yields for the net work done by the gas W = P2V2 −P1V1.
The Joule-Thomson effect depends crucially on the small deviation from an ideal gas given by intermolecular forces, specifically both the attractive and the repulsive parts of the Van der Waals force as approximated (for example) by the Lennard-Jones potential.
As a gas expands, the average distance between molecules grows. Because of the attractive part of the intermolecular force, expansion causes an increase in the potential energy of the gas. If no external work is extracted in the process and no heat is transferred, the total energy of the gas remains the same because of the conservation of energy. The increase in potential energy thus implies a decrease in kinetic energy and therefore in temperature.
A second mechanism has the opposite effect. During gas molecule collisions, kinetic energy is temporarily converted into potential energy (corresponding to the repulsive part of the intermolecular force). As the average intermolecular distance increases, there is a drop in the number of collisions per time unit, which causes a decrease in average potential energy. Again, total energy is conserved, so this leads to an increase in kinetic energy (temperature).
Below the Joule–Thomson inversion temperature, the former effect (work done internally against intermolecular attractive forces) dominates, and free expansion causes a decrease in temperature. Above the inversion temperature, gas molecules move faster and so collide more often, and the latter effect (reduced collisions causing a decrease in the average potential energy) dominates: Joule–Thomson expansion causes a temperature increase.
The Joule–Thomson (Kelvin) coefficient
The rate of change of temperature with respect to pressure in a Joule–Thomson process (that is, at constant enthalpy ) is the Joule–Thomson (Kelvin) coefficient . This coefficient can be expressed in terms of the gas's volume , its heat capacity at constant pressure , and its coefficient of thermal expansion as:
See the Derivation of the Joule–Thomson (Kelvin) coefficient below for the proof of this relation. The value of is typically expressed in °C/bar (SI units: K/Pa) and depends on the type of gas and on the temperature and pressure of the gas before expansion. Its pressure dependence is usually only a few percent for pressures up to 100 bar.
All real gases have an inversion point at which the value of changes sign. The temperature of this point, the Joule–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 is||since is||thus must be||so the gas|
|below the inversion temperature||positive||always negative||negative||cools|
|above the inversion temperature||negative||always negative||positive||warms|
Helium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one atmosphere are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up when expanded at constant enthalpy at typical room temperatures. On the other hand nitrogen and oxygen, the two most 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, is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.
In 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 external work is 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, 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 liquefied by the Linde cycle. For this reason, simple Linde cycle liquefiers cannot normally be used to liquefy helium, hydrogen, or neon.
Proof that the specific enthalpy remains constant
In thermodynamics so-called "specific" quantities are quantities per kilogram and are denoted by lower-case characters. So h, u, and v are the enthalpy, internal energy, and volume per kilogram, respectively. In a Joule–Thomson process the specific enthalpy h remains constant. To prove this, the first step is to compute the net work done when a mass m of the gas moves through the plug. This amount of gas has a volume of V1 = m v1 in the region at pressure P1 (region 1) and a volume V2 = m v2 when in the region at pressure P2 (region 2). Then in region 1, the work done on the amount of gas by the rest of the gas is = m P1v1. In region 2, the work done by the amount of gas on the rest of the gas is m P2v2. So, the total work done on the mass m of gas is
The change in internal energy minus the total work done on (IUPAC convention) the amount of gas is, by the first law of thermodynamics, the total heat supplied to the amount of gas (here it is assumed that there is no change in bulk kinetic energy). In the Joule–Thomson process the gas is insulated, so no heat is absorbed. This means that
where u1 and u2 denote the specific internal energies of the gas in regions 1 and 2, respectively. Using the definition of the specific enthalpy h = u + Pv, the above equation implies that
where h1 and h2 denote the specific enthalpies of the amount of gas in regions 1 and 2, respectively.
Throttling in the T-s diagram
A very convenient way to get a quantitative understanding of the throttling process is by using diagrams. There are many types of diagrams (h-T diagram, h-P diagram, etc.) Commonly used are the so-called T-s diagrams. Figure 2 shows the T-s diagram of nitrogen as an example. Various points are indicated as follows:
- a) T = 300 K, p = 200 bar, s = 5.16 kJ/(kgK), h = 430 kJ/kg;
- b) T = 270 K, p = 1 bar, s = 6.79 kJ/(kgK), h = 430 kJ/kg;
- c) T = 133 K, p = 200 bar, s = 3.75 kJ/(kgK), h = 150 kJ/kg;
- d) T = 77.2 K, p = 1 bar, s = 4.40 kJ/(kgK), h = 150 kJ/kg;
- e) T = 77.2 K, p = 1 bar, s = 2.83 kJ/(kgK), h = 28 kJ/kg (saturated liquid at 1 bar);
- f) T = 77.2 K, p = 1 bar, s = 5.41 kJ/(kgK), h =230 kJ/kg (saturated gas at 1 bar).
As shown before, throttling keeps h constant. E.g. throttling from 200 bar and 300 K (point a in fig. 2) follows the isenthalp (line of constant specific enthalpy) of 430 kJ/kg. At 1 bar it results in point b which has a temperature of 270 K. So throttling from 200 bar to 1 bar gives a cooling from room temperature to below the freezing point of water. Throttling from 200 bar and an initial temperature of 133 K (point c in fig. 2) to 1 bar results in point d, which is in the two-phase region of nitrogen at a temperature of 72.2 K. Since the enthalpy is an extensive parameter the enthalpy in d (hd) is equal to the enthalpy in e (he) multiplied with the mass fraction of the liquid in d (xd) plus the enthalpy in f (hf) multiplied with the mass fraction of the gas in d (1 − xd). So
With numbers: 150 = xd 28 + (1 − xd) 230 so xd is about 0.40. This means that the mass fraction of the liquid in the liquid–gas mixture leaving the throttling valve is 40%.
Derivation of the Joule–Thomson (Kelvin) coefficient
In this section a derivation of the formula
for the Joule–Thomson (Kelvin) coefficient is given.
The partial derivative of T with respect to P at constant H can be computed by expressing the differential of the enthalpy, dH, in terms of dT and dP, and solving for the ratio of dT and dP with dH = 0. The differential of the enthalpy is given by:
Here, S is the entropy of the gas. Expressing dS in terms of dT and dP gives:
(see Specific heat capacity) we can write:
The symmetry of partial derivatives of G with respect to T and P implies that:
where α is the cubic coefficient of thermal expansion. Using this relation dH can be expressed as
Solving for by equating dH to zero gives:
Joule's second law
It is easy to verify that for an ideal gas defined by suitable microscopic postulates that αT = 1, so the temperature change of such an ideal gas at a Joule–Thomson expansion is zero. For such an ideal gas, this theoretical result implies that:
- The internal energy of a fixed mass of an ideal gas depends only on its temperature (not pressure or volume).
- Critical temperature
- Ideal gas
- Enthalpy and Isenthalpic
- Reversible process (thermodynamics)
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- Figure composed with data obtained with RefProp, NIST Standard Reference Database 23
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- "Inversion Curve of Joule-Thomson Effect using Peng-Robinson CEOS". Demonstrations Projects of Wolfram Mathematica.