Engineering thermodynamics/Joule-Thomson effect

Wikipedia's proof that enthalpy is conserved

State variables defined for a sealed piston

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 Template:Nowrap 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 [[w:adiabatic process|Template:Nowrap]], 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:

$P_1{V}_1-P_2{V}_2=(E_2+\frac{1}{2}m{\dot{\xi }}_2^2)-(E_1+\frac{1}{2}m{\dot{\xi }}_1^2)$

Rearrange to obtain,

$H_1+\frac{1}{2}m{\dot{\xi }}_1^2=H_2+\frac{1}{2}m{\dot{\xi }}_2^2$

where $\overrightarrow{\dot{\xi }}$ 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, ${\overrightarrow{\dot{\xi }}}_1$ and ${\overrightarrow{\dot{\xi }}}_2$, 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.

Template:Cot 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 $\overrightarrow{g}=-\overrightarrow{\mathrm{\nabla }}\mathrm{\Phi }$, so that gravitational potential energy is, $m\mathrm{\Phi }$.

$\begin{array}{cc}\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \dot{\mathit{\xi }})=0& \to \rho \overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{\dot{\xi }}+\overrightarrow{\dot{\xi }}\cdot \overrightarrow{\mathrm{\nabla }}\rho =0\\ \rho \frac{\partial \dot{\mathit{\xi }}}{\partial t}+\rho \dot{\mathit{\xi }}\cdot \mathrm{\nabla }\dot{\mathit{\xi }}+\mathrm{\nabla }p=\rho 𝐠& \to (\rho \overrightarrow{\dot{\xi }}\cdot \overrightarrow{\mathrm{\nabla }})\overrightarrow{\dot{\xi }}+\overrightarrow{\mathrm{\nabla }}p+\rho \overrightarrow{\mathrm{\nabla }}\mathrm{\Phi }=0\\ \frac{\partial E}{\partial t}+\mathrm{\nabla }\cdot (\dot{\mathit{\xi }}(E+p))=0& \to \overrightarrow{\mathrm{\nabla }}\cdot ((\rho e+\frac{1}{2}\rho {\dot{\xi }}^2+p)\overrightarrow{\dot{\xi }})=0\end{array}$ Template:Cob

The internal energy per unit mass is $e$. At steady state, the compressible Euler equations in differential form are:

$\begin{array}{cc}\rho \overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{\dot{\xi }}+\overrightarrow{\dot{\xi }}\cdot \overrightarrow{\mathrm{\nabla }}\rho & =0\\ (\rho \overrightarrow{\dot{\xi }}\cdot \overrightarrow{\mathrm{\nabla }})\overrightarrow{\dot{\xi }}+\overrightarrow{\mathrm{\nabla }}p+\rho \overrightarrow{\mathrm{\nabla }}\mathrm{\Phi }& =0\\ \overrightarrow{\mathrm{\nabla }}\cdot ((\rho e+\frac{1}{2}\rho {\dot{\xi }}^2+p)\overrightarrow{\dot{\xi }})& =0\end{array}$

Here, $\rho$ is the fluid mass density, $\overrightarrow{\dot{\xi }}$ is the fluid velocity vector, and $p$ is the pressure, and $\mathrm{\Phi }$ is gravitational potential. The total energy density (total energy per unit volume) is:

$E=\rho e+\frac{1}{2}\rho {\dot{\xi }}^2$

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:

$\overrightarrow{\mathrm{\nabla }}\cdot (\rho \overrightarrow{\dot{\xi }})=0$

$\overrightarrow{\mathrm{\nabla }}(p+\frac{1}{2}\rho {\dot{\xi }}^2+\rho \mathrm{\Phi })=$ terms that are not always zero.

$\overrightarrow{\mathrm{\nabla }}\cdot ((h+\frac{1}{2}{\dot{\xi }}^2)\rho \overrightarrow{\dot{\xi }})=0$

The second equation is the familiar Bernoulli's equation, but extra terms are not shown that Template:Nowrap. The first and third equations can be combined using a well-known identity involving the Template:Nowrap:

$\rho \overrightarrow{\dot{\xi }}\cdot \overrightarrow{\mathrm{\nabla }}(h+\frac{1}{2}{\dot{\xi }}^2)=0=\hat{s}\cdot \mathrm{\nabla }(h+\frac{1}{2}{\dot{\xi }}^2)$

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 Template:Nowrap). Along this direction,

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

Other items that might be added to this page

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