# Engineering thermodynamics/Joule-Thomson effect

## Wikipedia's proof that enthalpy is conserved This image used to depict the Joule-Thomson effect should be modified by someone capable of editing an svg file and uploading a new file with a different name that credits the original author. The plug should be replaced with a single small hole and arrows depicting fluid flow near the aperture.

The derivation found in Wikipedia:Joule-Thomson effect should be learned first. The equations superficially resemble those often introduced in a physics class for a single sealed piston that permits heat flow into or out of the system, as shown to the left. P is pressure, V is volume, T is temperature, and E is internal energy. To model throttling, we require two thermodynamic systems, on each side of the throttle. We use subscripts (1,2) to denote the state variables on each side of the throttle.

The two systems exchange matter, and our attention shall be focused on a specific amount of mass, $m$ that is transferred from region 1 to region 2. The state variables $(P_{1},V_{1},T_{1},E_{1})$ and $(P_{2},V_{2},T_{2},E_{2})$ , refer to this mass before and after it passes through the throttle.

Since the walls of the throttle don't move, they do no work on the gas. The two pistons do work on the larger systems at each end, and this work is somehow transferred to the mass, $m$ , as it transits the throttle. In spite of all the chaos inside the throttle, this net work is described by the simple equation:

$Work=P_{1}V_{1}-P_{2}V_{2}$ It is possible for a processes occur so rapidly that it may be conveniently described by the adiabatic approximation, meaning that there is not enough time for the transfer of energy as heat to take place to or from the system. We assume this to be the case for the fluid element of mass $m$ as it passes through the throttle. Borrowing from the first law of thermodynamics, we equate the work done in an adiabatic process to the change in total (internal plus kinetic) energy:

$P_{1}V_{1}-P_{2}V_{2}=(E_{2}+{\tfrac {1}{2}}m{\dot {\xi }}_{2}^{2})-(E_{1}+{\tfrac {1}{2}}m{\dot {\xi }}_{1}^{2})$ Rearrange to obtain,

$H_{1}+{\tfrac {1}{2}}m{\dot {\xi }}_{1}^{2}=H_{2}+{\tfrac {1}{2}}m{\dot {\xi }}_{2}^{2}$ where ${\vec {\dot {\xi }}}$ is the fluid's velocity, and the enthalpy is,

$H=E+PV$ Even though the kinetic energy can be large inside the throttle, the variables, ${\vec {\dot {\xi }}}_{1}$ and ${\vec {\dot {\xi }}}_{2}$ , refer to the motion while the fluid is in the two pistons, where kinetic energy can be small (if the pistons are sufficiently wide). While passing through the throttle, large external forces can accelerate the fluid to high velocity. These forces can be turbulent and cause irreversible changes.

It is remarkable that this complexity can be modeled by such simple formulas.

## Deriving the conservation of enthalpy from Euler's equations

Here we analyze the situation using Euler's equations for the fluid's motion. Although Euler's equations require vector calculus, they represent one of the simplest (and therefore least accurate) models for the motion of fluid. To avoid confusion, we adopt a somewhat unorthodox notation.

click to see the choice of symbols and the equations in slightly more convential notation

Avoid the use of:

• u, because it can denote either fluid velocity or energy per unit mass.
• v, because it can denote either velocity or volume per unit mass.

We shall use h as enthalpy per unit mass, but to avoid its use as height in gravitational potential, we set ${\vec {g}}=-{\vec {\nabla }}\Phi$ , so that gravitational potential energy is, $m\Phi$ .

{\begin{aligned}{\partial \rho \over \partial t}+\nabla \cdot (\rho \mathbf {\dot {\xi }} )=0&\rightarrow \quad \rho {\vec {\nabla }}\cdot {\vec {\dot {\xi }}}+{\vec {\dot {\xi }}}\cdot {\vec {\nabla }}\rho =0\\\rho {\frac {\partial \mathbf {\dot {\xi }} }{\partial t}}+\rho \mathbf {\dot {\xi }} \cdot \nabla \mathbf {\dot {\xi }} +\nabla p=\rho \mathbf {g} &\rightarrow \quad (\rho {\vec {\dot {\xi }}}\cdot {\vec {\nabla }}){\vec {\dot {\xi }}}+{\vec {\nabla }}p+\rho {\vec {\nabla }}\Phi =0\\{\partial E \over \partial t}+\nabla \cdot (\mathbf {\dot {\xi }} (E+p))=0&\rightarrow \quad {\vec {\nabla }}\cdot \left(\;(\rho e+{\tfrac {1}{2}}\rho {\dot {\xi }}^{2}+p)\,{\vec {\dot {\xi }}}\quad \right)=0\end{aligned}} The internal energy per unit mass is $e$ . At steady state, the compressible Euler equations in differential form are: 

{\begin{aligned}\rho {\vec {\nabla }}\cdot {\vec {\dot {\xi }}}+{\vec {\dot {\xi }}}\cdot {\vec {\nabla }}\rho &=0\\(\rho {\vec {\dot {\xi }}}\cdot {\vec {\nabla }}){\vec {\dot {\xi }}}+{\vec {\nabla }}p+\rho {\vec {\nabla }}\Phi &=0\\{\vec {\nabla }}\cdot \left(\;(\rho e+{\tfrac {1}{2}}\rho {\dot {\xi }}^{2}+p)\ {\vec {\dot {\xi }}}\;\right)&=0\end{aligned}} Here, $\rho$ is the fluid mass density, ${\vec {\dot {\xi }}}$ is the fluid velocity vector, and $p$ is the pressure, and $\Phi$ is gravitational potential. The total energy density (total energy per unit volume) is:

$E=\rho e+{\tfrac {1}{2}}\rho {\dot {\xi }}^{2}$ with $e$ being the specific internal energy (internal energy per unit mass). The specific enthalpy (i.e., per unit mass) is:

$h=e+{\frac {p}{\rho }}={\frac {E}{\rho }}-{\frac {1}{2}}{\dot {\xi }}^{2}+{\frac {p}{\rho }},$ The three compressible Euler equations at steady state can be written as:

${\vec {\nabla }}\cdot (\rho {\vec {\dot {\xi }}})=0$ ${\vec {\nabla }}\left(p+{\tfrac {1}{2}}\rho {\dot {\xi }}^{2}+\rho \Phi \right)=$ terms that are not always zero.
${\vec {\nabla }}\cdot \left(\;(h+{\tfrac {1}{2}}{\dot {\xi }}^{2})\,\rho \,{\vec {\dot {\xi }}}\ \right)=0$ The second equation is the familiar Bernoulli's equation, but extra terms are not shown that do not always vanish. The first and third equations can be combined using a well-known identity involving the product of a scalar and a vector:

$\rho {\vec {\dot {\xi }}}\cdot {\vec {\nabla }}(h+{\tfrac {1}{2}}{\dot {\xi }}^{2})=0={\hat {s}}\cdot \nabla (h+{\tfrac {1}{2}}{\dot {\xi }}^{2})$ Here, we have used defined ${\hat {s}}$ to be a unit vector parallel to the motion of the fluid (called the tangent vector in the Frenet-Serret coordinate system). Along this direction,

$(h+{\tfrac {1}{2}}{\dot {\xi }}^{2})$ is a constant.

click to view
1. These images might be introduced.
2. Ideas from http://de.wikipedia.org/wiki/Joule-Thomson-Effekt can translated into English and used. Isenthalpic contours for temperature versus pressure showing inversion curve Failed effort by XSteam to find inversion curve for water

Footnotes and references

1. The only good source I could find using a cursory internet search was Wikipedia. But I am 90% confident that these formulas are correct, and 100% confident that this argument has been presented elsewhere if it is correct. In fact, these arguments have probably been presented elsewhere even if they are wrong.
2. The walls do no work in the reference frame of the throttle; the fluid element "sees" a moving wall compress it, but we are not working in that reference frame.
3. Note how the work is "transferred" from the piston to the mass, $m$ , even though the mass is on the opposite side of the piston.
4. The choice of minus sign is dictated by the assumption that the gas is flowing from 1 to 2
5. Bailyn, M. (1994). A Survey of Thermodynamics, American Institute of Physics Press, New York, ISBN 0-88318-797-3, pp. 52–53. Text from http://en.wikipedia.org/w/index.php?title=Adiabatic_process&oldid=651370206