# Calculating temperature change due to Joule-Thomson effect

I have a 10 L compressed argon cylinder with filling pressure of 10 MPa. According to Perry's Chemical Engineer's Handbook, argon's J-T coefficient (unit: K/MPa) is 3.7 at 0.1 MPa, 2.6 at 10 MPa. (Temperature is 300 K.)

When I open up the cylinder valve, 10 MPa argon in the cylinder will expand into 0.1 MPa atmosphere. So the $\Delta P$ is 9.9 MPa.

How much will the temperature change?
The initial temperature is 300 K.

1. $9.9 \times 3.7 = 36.63$

(Which means the temperature of argon outside the cylinder will be 263.37 K)

1. $9.9 \times 2.6 = 25.74$
2. Something in between (like when the J-T coefficient in effect changes continuously as the gas expands).

## 3 Answers

It is going to be something in between. The definition of the JT coefficient is $$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$If the temperature-dependence can be neglected, then this gives $$\Delta T=-\int_{0.1}^{10}{\mu(P)dP}$$

• Unless I'm mistaken, won't they need to evaluate this numerically (thus presumably needing more points then they have) since there isn't a known functional form for the JT coefficient in terms of pressure for a real gas? – Tyberius Sep 7 '17 at 14:30
• For a large change like this, it can be done using generalized corresponding states correlations for the 2nd virial coefficient to determine the residual enthalpy at the initial and final states. See Smith and Van Ness, Chapter 7, Throttling Processes. – Chet Miller Sep 7 '17 at 14:48

The initial state of the argon gas is completely defined by the given temperature and pressure:

• Initial temperature $T_1=300\ \mathrm K$
• Initial pressure $p_1=10\ \mathrm{MPa}$

The corresponding values of other properties of the gas can be looked up in so-called steam tables. For example, in REFPROP – NIST Standard Reference Database 23, Version 9.0 we find for the given initial temperature and pressure:

• Initial density $\rho_1=167.60\ \mathrm{kg\ m^{-3}}$
• Initial specific internal energy $u_1=78.313\ \mathrm{kJ\ kg^{-1}}$
• Initial specific enthalpy $h_1=137.98\ \mathrm{kJ\ kg^{-1}}$
• Initial specific entropy $s_1=2.8716\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
• Initial Joule–Thomson coefficient $\mu_{\mathrm{JT},1}=2.6066\ \mathrm{K\ MPa^{-1}}$

The final state of the gas is not completely defined by the given values since only the final pressure is given

• Final pressure $p_2=0.1\ \mathrm{MPa}$

which is not enough to look up other values in steam tables.

However, we know that Joule–Thomson expansion is an isenthalpic process; i.e. the specific enthalpy $h$ remains constant. Thus

• Final specific enthalpy $h_2=h_1=137.98\ \mathrm{kJ\ kg^{-1}}$

Now we have a second data point that can be used to look up other parameter values, for example

• Final density $\rho_2=1.8105\ \mathrm{kg\ m^{-3}}$
• Final specific internal energy $u_2=82.747\ \mathrm{kJ\ kg^{-1}}$
• Final specific entropy $s_2=3.8155\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
• Final Joule–Thomson coefficient $\mu_{\mathrm{JT},2}=4.5226\ \mathrm{K\ MPa^{-1}}$

and also

• Final temperature $T_2=265.65\ \mathrm K$

Since the pressure is not changing gradually, in the sense that it is either $p_1$ or $p_2$, not in between (it is not a reversible process), the temperature drop should be $\Delta T=\mu_\mathrm{JT}(p_2)\cdot p_2-\mu_\mathrm{JT}(p_1)\cdot p_1$.

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