7 replies to this topic

[Himanshu Sharma]

Dear all

I am writing this post after reading lot many threads on Low Metal temperature while HC depressurization due to Joule Thomson effect.

My question pertains to Hydrogen Depressurization as we all know that hydrogen experiences a reverse joule Thomson effect. Hydrogen Temperature should increase while depressurising.

I simulated this using HYSYS depressurising facility and the results predict that initial few minute temperature goes up and then vessel fluid’s temperature starts reducing and ultimately metal wall temperature also drops down.

How is this possible !!!

I thought on the lines that energy utilised while heating of the gas is coming from vessel fluid cooling down ,but then how do you explain the normal joule thomson cooling.

I know i am missing a link,where i need your help !

This will be a common scenario while depressurising H2 rich gas comprsesors,can anybody please explane from operational experience.

[Zauberberg]

Yes, that is true for Hydrogen (or any other component) - but only if you are above the maximum inversion temperature. See the literature on this topic (e.g. Perry).

Each gas has its own maximum inversion temperature after which throttling causes heating rather than cooling. Hydrogen is known as the gas with the lowest MIT. It means in practice that the gas has to be cooled below this temperature if any further cooling is to be achieved by throttling.

[Himanshu Sharma]

Sir

I should have provided the Temperature and pressure levels.Hydrogen in the subject case is at 74 deg c and 180 k/cm2 g pressure ,well above the inversion temperature of Hydrogen thats -222 deg c.(Right side of the inversion point )

so the above logic fails to answer my concern and as per thermodynamics Hydrogen should heat up.

Lets try and find the missing angle !

[Zauberberg]

For Hydrogen, the maximum inversion temperature is -64C, not -222C; however you should be above that value in any case when having 74 degC fluid temperature. Inversion temperature depends on the pressure as well - see attached chart.

Let me look into the model and I'll come back if I get any conclusions.

http://en.citizendiu...ion_temperature

Attached Files

•  Locus T-P.bmp   306.33KB   52 downloads

[Himanshu Sharma]

was reading semiconductors ,typed the wrong figure for MIT !

anyways sir ,any conclusions???

i still need the help.

[AHT]

Himanshu:

I might be wrong, but I think that the point here is that the H2 mas inside the vessel has decreased, therefore decreasing the collisions between the H2 molecules, and consequently the temperature (and internal energy). The incease in H2 temperature should be only for steady-state flow (enthalpy) situations. For the vessel itsel that it is no the case.

Why don't you try to check the simulator solution by doing an energy balance with the aid of a H2 thermodynamic table? You already know the initial state, and you can arbitrarily stablish any subsequent case (if your are depressuring, fix an arbitray time and pressure, and as you know the vessel volume, then you can work out the total mass inside the vessel). To estimate the temperature at this "final" estate, you just need the specific volume (or density) that you have already estimated.

Hope this helps!!!

[Zauberberg]

AHT,

I'm giving you a star for this post because you are probably right. I can't think of some other solution to the problem except if the Joule-Thomson effect has not been properly integrated in Hysys thermodynamics, or there are some other issues if the gas is in supercritical conditions such is Hydrogen at ~180 barg and 74 degC.

I tried to model isenthalpic flash across the valve and indeed there is no temperature decrease when Hydrogen is flashed from 180 barg to atmospheric pressure. The same happens if the stream is flashed from 100, 50, and 10 bar, the last being below the critical pressure of Hydrogen (13.16 bara). I would like to hear thermo-expert opinion on this subject, perhaps also to contact Aspen Tech Support and ask for explanation. I'm thinking to do that next week.

[Himanshu Sharma]

A close look at simulation will reveal that while the source fluid temperature decreases (the one in vessel) ,still the Hydrogen coming out of RO has higher temperature than source (joule Thomson effect) .

AHT a well deserved star !

I could also think on a parallel track but couldn’t post being busy in other things .

Vessel is a constant volume system and Temperature has to decrease as simple as P=const*T.While pressure decreases, temperature has to decrease. There is no expansion in the system and hence Joule Thomson effect does not apply here.

While passing thru a restricted area gas has to undergo expansion; if it’s a free adiabatic expansion, enthalpy is conserved and hence joule Thomson effect applies here. Temperature may increase or decrease depending upon initial stage of system.

Even if the volume of the system is not constant as if imagining the case of Hydrogen coming out of a bursting tyre(God Forbid ). The hydrogen that is expanding inside the tire is also doing work by pushing hydrogen out of the tire. In this case, the first law reduces to: dU = Hout. Similarly, we can draw a control surface around some finite volume inside the tire which we might see as being an isentropic expansion in which the entropy change is zero. Either way, the process is the same, and the hydrogen will cool off because it must do work as it follows a line of constant entropy .

The Hydrogen passing through the valve as the tire goes from a region of higher pressure to a region of lower pressure. The air in this case is not doing work, and so the first law reduces to dU = 0 or Hin = Hout.Joule Thomson effect applies here !
For Hydrogen (unlike other gases) MIT lies below room temperature that’s why gas gets heated up!

@Zauberberg
Sir, expansion thru the valve should come in category of JT effect and while i simulated the same i got a temperature increase.

There is one more point i would like to make that if the gas undergoes a reversible expansion that is entropy is conserved for the entire path, gas will be always cooled! This is not JT effect and you can relate that simply to Kinetic theory.

As a gas expands, the average distance between molecules grows. Because of intermolecular attractive forces (see Van der Waals force), expansion causes an increase in the potential energy of the gas. If no external work is extracted in the process and no heat is transferred, the total energy of the gas remains the same because of the conservation of energy. The increase in potential energy thus implies a decrease in kinetic energy and therefore in temperature.

A second mechanism has the opposite effect. During gas molecule collisions, kinetic energy is temporarily converted into potential energy. As the average intermolecular distance increases, there is a drop in the number of collisions per time unit, which causes a decrease in average potential energy. Again, total energy is conserved, so this leads to an increase in kinetic energy (temperature). Below the Joule–Thomson inversion temperature, the former effect (work done internally against intermolecular attractive forces) dominates, and free expansion causes a decrease in temperature. Above the inversion temperature, gas molecules move faster and so collide more often, and the latter effect (reduced collisions causing a decrease in the average potential energy) dominates: Joule–Thomson expansion causes a temperature increase.

PS:I realise that i wrote a lot ,but may be it will be worth reading !