The Joule-Thomson coefficient for methane can indeed be determined from the corresponding enthalpy-temperature diagram. A particularly comprehensive presentation of this subject (but not specific for methane) can be found at: V.A. Kirillin, V.V. Sychev, A.E. Sheindlin, "Engineering Thermodynamics", Mir Publishers, Moscow, 1987 (3rd printing), transl. from Russian by: S. Semyonov, Section 7.6 (pp. 253-264).

Thanks Carlos , I'm searching for methane diagram to compare with the CFD.

Thanks Mykhailo, I tried to search in your appreciated work  but I couldn't find the formula.  Please could you guide me to find ? or send privately the formula.

Hello Mohammed,

The Joule Thomson (dT/dP) at constant H  coefficient is defined for an isenthalpic pressure change, for instance through a porous medium. In order to compute the JT coefficient you need the molar volume V, the molar heat capacity at constant pressure Cp, absolute temperature T, thermal expansivity alpha, and the coefficent muJT is obtained through :

muJT= V*(T*alpha - 1.00)/Cp

In this expression V and alpha can be obtained from an equation of state or NIST data.The toatl molar heat capacity Cp is obtanined as the sum of the ideal gas heat capacity, which you can obtain from tables, and the residual heat capacity that is usually taken from EOS calculations or Monte Carlo simulations when pressure effects cannot be neglected.

Note that the JT effect cools down the fluid when depressurizing it at moderate pressure. It heats up the fluid at very high pressure (typically above 500 bars) when P is decreased (counterintuitive, isn't it ) ? .

Hello Philiippe,

Thanks very much for highly appreciated answers.

Please could you help me by answerIng more questions to eliminate a lot of confusions related to J_ T coefficient.

Regards to dT /dP, could I through methane T_ H diagram from T1 and P1 to T2 and P2 ,  get the temperature _pressure gradient and hence the coefficient , is it constant along this process for fixed intiaI state?

Regards to the expansion coefficient I couldn't find it for methane gas, only for liquid and solid states ,how can I find it for gas  phase, as I havn't any simulation programs.

Finally if I have the temperature and pressure distributions between an initial and final expansion process , how can I plot the J_T coefficient along this process?

Hello Mohammed,

Thanks for your kind words !

Here is a short answer : in a Joule-Thomson expansion, you start from an initial state at T1, P1. You decompress your fluid with an given pressure drop (for instance by pushing a piston upstream to maintain P1 and withdraw the piston downstream where pressure is P1+deltaP = P2 . delta P is negative. The flow rate depends then on the permeability of the porous medium. In a true JT expansion, the output temperature T2 is T1+deltaT, and deltaT is given by the JT coefficient : muJT=deltaT/deltaP . Thus you can't impose T2, it's a result of the JT expansion, as muJT is given by fluid properties.

In an isenthalpic diagram you must find T2 such that enthalpy does not change  H(P1, T1) = H(P2, T2) , you cannot impose T2 and P2 and H simultaneously.

Surprisingly, the JT coefficient in the ideal experiment is not a transport property, as thermodynamics tells us that it depends just on equilibrium properties. The main reason is that it assumes a permanent adiabatic regime.

At ambient temperature for pure methane in moderate pressure conditions (say below 100 bars) the JT coefficient is positive (pressure drop --> cooling) . In very high pressure conditions (for instance 1000 bars) the behavior of the Joule Thomson coefficient is opposite (pressure drop --> heating) because you are above the JT inversion pressure, ie the pressure for which the JT coefficient is zero.

I hope it answers your questions, please ask for more details if required !

PhU

Hello Phillipe

Again great thanks as I become less confused due to your support.

But I'm sorry ,I was in dark as I meant P- H diagram instead of T-S and I will not hesitated to ask you for more explanation regards this topic.

Thanks