Why is Joule-Thomson effect isenthelpic?

[ussername]

Let's have an experimental apparatus for realization of Joule-Thomson effect. It could be a thermally isolated pipe with a porous board separating two compartments inside. Different pressures can be set in the second compartment. The gas flows through the pipe and we measure temperature and pressure on the outlet.

Now applying a simple energy balance onto that pipe we get:
$$0=\dot{m}_{in}(h_{in}+\frac{v_{in}^{2}}{2}+g\cdot z_{in})-\dot{m}_{out}(h_{out}+\frac{v_{out}^{2}}{2}+g\cdot z_{out})$$
There is definitely a steady state ($\dot{m}_{in}=\dot{m}_{out}$) and there is no change of potential energy ($z_{in}=z_{out}$) so we get:
$$\Delta h=\frac{v_{in}^{2}-v_{out}^{2}}{2}$$
From the continuity equation we get:
$$\frac{v_{out}}{v_{in}}=\frac{\rho _{in}}{\rho _{out}}$$
Does it mean I will measure the same velocities $v _{in}$ and $v _{out}$ and the same densities $\rho _{in}$ and $\rho _{out}$ all the time?

 

[Chestermiller]

Let's have an experimental apparatus for realization of Joule-Thomson effect. It could be a thermally isolated pipe with a porous board separating two compartments inside. Different pressures can be set in the second compartment. The gas flows through the pipe and we measure temperature and pressure on the outlet.

Now applying a simple energy balance onto that pipe we get:
$$0=\dot{m}_{in}(h_{in}+\frac{v_{in}^{2}}{2}+g\cdot z_{in})-\dot{m}_{out}(h_{out}+\frac{v_{out}^{2}}{2}+g\cdot z_{out})$$
There is definitely a steady state ($\dot{m}_{in}=\dot{m}_{out}$) and there is no change of potential energy ($z_{in}=z_{out}$) so we get:
$$\Delta h=\frac{v_{in}^{2}-v_{out}^{2}}{2}$$
From the continuity equation we get:
$$\frac{v_{out}}{v_{in}}=\frac{\rho _{in}}{\rho _{out}}$$
Does it mean I will measure the same velocities $v _{in}$ and $v _{out}$ and the same densities $\rho _{in}$ and $\rho _{out}$ all the time?

The Joule Thomson effect is isenthalpic only if kinetic energy and potential energy effects are negligible. Otherwise, no.