Joule Thomson coefficient

[Matthew99]

Dear community!

Obviously, the Joule Thomson coefficient is a function of pressure and temperature, as inversion temperature plots show. Thus, I was wondering why this is the case. I found an explanation that says that depending on the pressure, either repelling forces or attractive forces predominate and according to conservation of energy, when the gas expands without a change in enthalpy the thermal energy and therefore the temperature has to decrease when the potential energy is increasing and vice versa -  resulting in a positive JT-coefficient when attractive forces predominate and a negative JT-coefficient when repelling forces predominate. This seems reasonable. However, inversion temperature plots show that the JT-coefficient changes substantially when temperature is changed. So far, I have not figured out why the temperature of the gas has that much of an importance here. Anyone got an intuitive explanation? Thank you in advance for your help!

[studiot]

The JT coefficient is zero for an ideal gas.

To estimate this coefficient you need at least a simplified Van Der Waals equation of state

[math]\mu  = \frac{{\frac{{2a}}{{RT}} - b}}{{{C_p}}}[/math]

Where mu is the JT coefficient, a and b ar the VDW coefficients and the rest have their usual significance.

I'm not sure how you will get an 'intuitive' feel from that.

Edited May 25, 2020 by studiot

[joigus]

Ok. I'm more trained in statistical mechanics, but from what I remember about real gases, and taken the cue from Studiot's post with the definition of the Joule-Thomson coefficient --which I had forgotten completely, I must confess--. b is the volume of the molecules (or 3x#molecules, I don't remember). So what's playing a role there is the size of the molecules as little hard impenetrable balls.

It you take a look at the Van der Waals potential energy, it looks something like this:

So it is slightly attractive at long distances but becomes strongly repulsive at short distances. Now, as the temperature from a statistical point of view is the average energy per degree of freedom, large temperatures imply that your molecules get to the range in their collisions that looks like the steep repulsive force you're seeing on the curve. The more average kinetic energy (temperature) the closer you approach the "hard ball" regime, so to speak. That's the only intuitive explanation I can think of. And as the b (the effective size of the molecules in the hard-ball model) is playing a role in your coefficient, I would say that could be the reason.

I hope it helps.

[Matthew99]

Thank you very much for your answers!

Considering the Van der Waals potential, it seems very reasonable that the JT coefficient changes with temperature and pressure. However, the formula that studiot posted suggests that the JT coefficient is a function of temperature only, as the other coefficients are constants and as far as I know the VDW coefficients do not change with pressure. How does that match?

Edited May 26, 2020 by Matthew99

[studiot]

25 minutes ago, Matthew99 said:

Thank you very much for your answers!

Considering the Van der Waals potential, it seems very reasonable that the JT coefficient changes with temperature and pressure. However, the formula that studiot posted suggests that the JT coefficient is a function of temperature only, as the other coefficients are constants and as far as I know the VDW coefficients do not change with pressure. How does that match?

You did ask for an intuitive answer, and I offered a simplified 'ball park' answer.

The full VDW analysis leads to more insight, including the fact that there are two inversion temperatures at a given pressure and above a certain pressure the gas is permanently inverted.

But the treatment is fully mathematical.

[math]\frac{{2a}}{{R{T_I}}} - \frac{{2abP}}{{{R^2}T_I^2}} - b = 0[/math]

 

 

Edited May 26, 2020 by studiot

[studiot]

Two folks have said thanks, but no one has said more.

I am quite happy to post the maths.

[joigus]

12 hours ago, studiot said:

Two folks have said thanks, but no one has said more.

I am quite happy to post the maths.

More.

[studiot]

On 5/27/2020 at 12:54 AM, joigus said:

More.

 

Part 1

The analysis if the Joule-Thompson or Joule-Kelvin flow or throttling is interesting because it demonstrates so many points in a successful thermodynamic analysis.

1. Appropriate system description
2. Distinction from similar systems
3. Identification of appropriate variables
4. Correct application of states
5. Distinction between the fundamental laws and the equations of state and their application

JT flow is a continuous steady state process. The system is not isolated or necessarily closed, but may be treated as quasi-closed but suitable choice of variables.
It cannot be considered as closed, for instance, if we consider a 'control volume' approach, common in flow processes, since one of our chosen variables (volume) varies.
 

By contrast, the Joule effect is a one off or one shot expansion of an isolated system.

So to start the analysis here is a diagram.

 

1 mass unit eg 1 mole of gas within the flow enters the left chamber between adiabatic walls and equilibrates to the V₁, P₁, T₁ and E₁.
Since P₁  >  P₂ the flow takes this 1 mole through the porous plug into the right hand chamber where it equilibrates to  V₂, P₂, T₂ and E₂.

The 'system' is just this 1 mole of gs, not the whole flow. The system thus passes from state1 to state 2.
The First Law can thus be applied to the change.
Since the process is adiabatic, q  =  0 and the work done at each state is PV work.

E₂ - E₁  =  P₁V₁  - P₂V₂

since the system expands and does negative work.

Rearranging gives

E₂  + P₂V₂ =  E₁ + P₁V₁ 

But E + PV =  H or enthalpy.

So the process is one of constant enthalpy or ΔH  =  0.
Note this is unlike ΔE which is not zero.

Since ΔE is not zero, P₁V₁ is not equal to P₂V₂

More of this later.

Since H is a state variable and ΔH = 0

[math]dH = 0 = {\left( {\frac{{\partial H}}{{\partial T}}} \right)_P}dT + {\left( {\frac{{\partial H}}{{\partial P}}} \right)_T}dP[/math]

[math]{\left( {\frac{{\partial H}}{{\partial T}}} \right)_P}dT =  - {\left( {\frac{{\partial H}}{{\partial P}}} \right)_T}dP[/math]

[math]{\left( {\frac{{\partial T}}{{\partial P}}} \right)_H} =  - \frac{{{{\left( {\frac{{\partial H}}{{\partial P}}} \right)}_T}}}{{{{\left( {\frac{{\partial H}}{{\partial T}}} \right)}_P}}}[/math]

 

Where    [math]{\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}[/math]  is defined as the Joule-Thompson coefficient, μ,

However we actually want our equation to contain measurable quantities to be useful so using

[math]H = E + PV[/math]            again

[math]dH = PdV + VdP + dE[/math]

[math]0 = TdS - PdV - dE[/math]

add previous 2 equations

[math]dH = Tds + VdP[/math]

divide by dP at constant temperature

[math]{\left( {\frac{{\partial H}}{{\partial P}}} \right)_T} = T{\left( {\frac{{\partial S}}{{\partial P}}} \right)_T}dP + V[/math]

But

[math]{\left( {\frac{{\partial S}}{{\partial P}}} \right)_T} =  - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}[/math]

so

[math]V = T{\left( {\frac{{\partial V}}{{\partial T}}} \right)_P} + {\left( {\frac{{\partial H}}{{\partial P}}} \right)_T}[/math]

combining this with our fraction for μ we have

[math]\mu  = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H} = \frac{{T{{\left( {\frac{{\partial V}}{{\partial T}}} \right)}_P} - V}}{{{C_P}}}[/math]

Which give a practical form with quantities that can be measured.

[math]\Delta T = \frac{{T{{\left( {\frac{{\partial V}}{{\partial T}}} \right)}_P} - V}}{{{C_P}}}\Delta P[/math]

Joule and Thompson found that the change in temperature is proportional to the change in pressure for a range of temperature restircted to the vicinity of T.

 

The next stage is to introduce the second law and the connection between different equations of state and their meanings or implications.

 

Edit I think I've ironed out all the latex now but please report errors to the author.

 

Edited May 30, 2020 by studiot

[joigus]

Thank you, Studiot. +1. This motivates me a lot to review my thermodynamics.

I studied this ages ago, and it all has been ringing a bell while reading. The JT process was announced to us back then as something definitely important. Then I started studying statistical mechanics, which is great, because you get to see how it all works from the atoms to the thermodynamic variables, but you lose sight of many things because of the oversimplification.

[joigus]

2 hours ago, studiot said:

Edit I think I've ironed out all the latex now but please report errors to the author.

 

I've just seen one dP hanging there that you must clear out when dividing by dP. At,

\[\left(\frac{\partial H}{\partial P}\right)_{T}=T\left(\frac{\partial S}{\partial P}\right)_{T}+V\]

[studiot]

9 hours ago, joigus said:

I've just seen one dP hanging there that you must clear out when dividing by dP. At,

 

(∂H∂P)T=T(∂S∂P)T+V

 

Yes you are right thank you, though I thought I had spotted it and fixed it once.

I hate LaTex.

[Matthew99]

Very interesting! It's great to see when application of some theoretical maxwell relations actually gives us a practical result which is of great importance in many industries for sure. Given that walls with perfect heat insulation are rather a matter of theoretical considerations than of a real world system, would the result change significantly if we allow some heat transfer with the surroundings?

Thanks a lot for your effort!

[studiot]

3 minutes ago, Matthew99 said:

Thanks a lot for your effort!

That was part 1  - it was taking so long I though I'd split it.

Part 2 is a bit less mathematical, but I'm gald to see you can cope with partial diffs.