![]() | ||
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 real gas or liquid (as differentiated from an ideal gas) 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; these three gases experience the same effect but only at lower temperatures.
Contents
- History
- Description
- Physical mechanism
- The JouleThomson Kelvin coefficient
- Applications
- Proof that the specific enthalpy remains constant
- Throttling in the T s diagram
- Derivation of the JouleThomson coefficient
- Joules second law
- References
The throttling process is commonly exploited in thermal machines such as refrigerators, air conditioners, heat pumps, and liquefiers.
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.
History
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.
Description
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.
The temperature change produced during a Joule–Thomson expansion is quantified by the Joule–Thomson coefficient,
Physical mechanism
There are two factors that can change the temperature of a fluid during an adiabatic expansion: a change in internal energy or the conversion between potential and kinetic internal energy. Temperature is the measure of thermal kinetic energy (energy associated with molecular motion); so a change in temperature indicates a change in thermal kinetic energy. The internal energy is the sum of thermal kinetic energy and thermal potential energy. Thus, even if the internal energy does not change, the temperature can change due to conversion between kinetic and potential energy; this is what happens in a free expansion and typically produces a decrease in temperature as the fluid expands. If work is done on or by the fluid as it expands, then the total internal energy changes. This is what happens in a Joule–Thomson expansion and can produce larger heating or cooling than observed in a free expansion.
In a Joule–Thomson expansion the enthalpy remains constant. The enthalpy,
where
For an ideal gas,
The ratio of the value of
For liquids, and for supercritical fluids under high pressure,
The physical mechanism associated with the Joule–Thomson effect is closely related to that of a shock wave, although a shock wave differs in that the change in bulk kinetic energy of the gas flow is not negligible.
The Joule–Thomson (Kelvin) coefficient
The rate of change of temperature
See the Derivation of the Joule–Thomson (Kelvin) coefficient below for the proof of this relation. The value of
All real gases have an inversion point at which the value of
In a gas expansion the pressure decreases, so the sign of
Helium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one atmosphere are very low (e.g., about 45 K (−228 °C) for helium). Thus, helium and hydrogen warm 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,
Applications
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 cooling produced in the Joule–Thomson expansion makes it a valuable tool in refrigeration. In vapor-compression refrigeration, as used in common refrigerators and air conditioners, it produces cooling in the throttling valve. 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). A gas must be below its inversion temperature to be liquefied by the Linde cycle. For this reason, simple Linde cycle liquefiers, starting from ambient temperature, cannot be used to liquefy helium, hydrogen, or neon. However, the Joule–Thomson effect can be used to liquefy even helium, provided that the helium gas is first cooled below its inversion temperature of 40 K.
Proof that the specific enthalpy remains constant
In thermodynamics so-called "specific" quantities are quantities per unit mass (kg) and are denoted by lower-case characters. So h, u, and v are the specific enthalpy, specific internal energy, and specific volume (volume per unit mass, or reciprocal density), 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 "flow work" done on the amount of gas by the rest of the gas is: W1 = m P1v1. In region 2, the work done by the amount of gas on the rest of the gas is: W2 = 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 the amount of gas is, by the first law of thermodynamics, the total heat supplied to the amount of gas.
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 77.2 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 coefficient
It is difficult to think physically about what the Joule–Thomson coefficient,
The first step in obtaining these results is to note that the Joule–Thomson coefficient involves the three variables T, P, and H. A useful result is immediately obtained by applying the cyclic rule; in terms of these three variables that rule may be written
Each of the three partial derivatives in this expression has a specific meaning. The first is
and the third is the inverse of the isothermal Joule–Thomson coefficient,
This last quantity is more easily measured than
This equation can be used to obtain Joule–Thomson coefficients from the more easily measured isothermal Joule–Thomson coefficient. It is used in the following to obtain a mathematical expression for the Joule–Thomson coefficient in terms of the volumetric properties of a fluid.
To proceed further, the starting point is the fundamental equation of thermodynamics in terms of enthalpy; this is
Now "dividing through" by dP, while holding temperature constant, yields
The partial derivative on the left is the isothermal Joule–Thomson coefficient,
where α is the cubic coefficient of thermal expansion. Replacing these two partial derivatives yields
This expression can now replace
This provides an expression for the Joule–Thomson coefficient in terms of the commonly available properties heat capacity, molar volume, and thermal expansion coefficient. It shows that the Joule–Thomson inversion temperature, at which
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).This rule was originally found by Joule experimentally for real gases and is known as Joule's second law. More refined experiments of course found important deviations from it.