Cold blowdown of vessel through orifice - thermodynamical explanation 1

[runsaep]

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Hi, could someone in a very simple way explain why:

the process upstream the orifice influencing the vessel is so-called "isentropic" and downstream the orifice (flare tailpipe, headers) the process is "isenthalpic", is this correct interpreted ?

The purpose with this query is to understand what happens in the vessel.

Can you first start with explaining in an easy manner these two processes ? In Wikipedia etc. I feel that it is described in a complex, detailed way, therefore
I challenge you to make it comprehendible!

 

[MortenA]

it is _complicated_ What is it that you dont understand? The main difference between the two processess is that over an orifice as it says its isenthalpic - e.i. then enthalphy [corrected] (in the gas) remains constant. This means that the work performed when the pressure is let down results in a temperature increase in the gas. Whereas in a isentropic process the entrophy remains constant - but the work is "taken out" of the process. So the expansion in the vessel is similar to you having a piston and decreasing the pressure by increasing the chamber volume. The work in this prossess "leaves" the gas part of the system (and is thus not heated). Note that when say "heated" here i disregard the temperature decrease resulting from the expansion. Its the same for a liquid, it will get warmer across a valve because the cooling from expansion is limited.

Best regards, Morten

 

[LittleInch]

This post covered similar topics

https://www.eng-tips.com/viewthread.cfm?qid=448359

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[chicopee]

It is time for the OP to take a course in thermodynamics. Meanwhile go to your nearest library and get an appropriate text book or either a mechanical or chemical engineering handbook.

 

[runsaep]

My understanding increases hopefully Chicopee , I have now read several other explanations in forums elsewhere and none describe it as work Morten A. Virtually all, describes it as an isentropic process in vessel- and over the valve as an isenthalpic process, as you said.

Isenthalpic over orifice: I cite from the Campbell Handbook; " ΔH = Q - W. Expansion across a valve or choke: Since Q and W are zero, equation reduces to ΔH = O. There is no change in H in this isenthalpic process. This is the basis for temperature drop when expanding across a valve". I presume this is valid also for an orifice.

Isentropic in vessel: Campbell Handbook chapter 7, "The Second Law of Thermodynamics" sets up the entropy formula this way: ΔS = ∑Saccum + Q/T + Sp +/- Sc where Sp is entropyproduction, dSsccum is accumulated entropy over the system and Sc is for any atomic transformations:

It shortens to ΔS = 0 since due to adiabatic Q/T=0, ∑Saccum=0 (no accum. of entropy during process) and Sc = 0 (no atomic transformations). Further it says; "friction loss in fluid flow systems is a function of Sp. If Sp = 0, there is no friction loss."

Little friction, more system-order, closer to "perfectness".

So,in vessel, Q=0 since no exchange of heat of gas with surrounding air and negligible friction in leaving the vessel (Sp = 0) - therefore ΔS = 0. Isentropic.

If someone could interpret the formula ΔH = T*Ds + VΔP in view of the depressurization process it would be interesting.

 

[Latexman]

Isenthalpic and isentropic processes are ideal models that we all learned in thermodynamics. A long time ago in the early days of thermodynamics, someone, I don't know who, must have either rationalized in their mind, or collected data in an experiment that proved, that flow through an orifice and flow through a valve is closest to isenthalpic. That knowledge has been propagated through the years to a LOT of thermodynamic textbooks. It sounds like you don't believe/trust this, or you want a deeper understanding. May I suggest your research take a deep dive and find the original person who postulated or determined this working model. If you would be so kind as to let us know what you find, we would appreciate it very much.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[runsaep]

Latexman! I wrote "I presume this is valid also for an orifice". i.e. isenthalpic. It is in the VESSEL to be blown down the process may be said to be close to isentropic.
Over the valve/orifice it is isenthalpic ΔH = 0, but look here: In Perry Handbook chapter 6-22 "Fluid and particle dynamics" an isentropic flow through a nozzle is depicted/sketched and the section is called " adiabatic frictionless nozzle flow". It is said: "The figure illustrates adiabatic discharge of a perfect gas through a frictionless nozzle from a large chamber where velocity is effectively zero. "

 

[Compositepro]

https://en.wikipedia.org/wiki/Joule–Thomson_effect

 

[runsaep]

Thanks Compositepro, but I do not think the explanation of J-T in Wikipedia is clearcut and easy to understand, do you ?

 

[Compositepro]

No, I cannot say I have ever had a very thorough understanding of it but know when it is relevant. If you want to try to understand it, study reversible versus irreversible expansion of gasses in a thermodynamics textbook. Joule-Thompson coefficients of gasses can be positive or negative depending on the gas and on temperature. This means that some gasses actually get warmer when when they expand through an orifice, but most will cool. It is a complicated topic and can only be simplified so much.

 

[MortenA]

Latexman allow me to disagree, flow over an orifice and a valve is close to similar and both isenthalpic (or close to) whereas flow over an expander is close to isentropic. Inside the vessel the expansin (due to mass leaving the vessel) is very close to isentropic - thus the gas in the vessel will end up being significantly colder at P== atmosphering compared to the very first gas coming from the valve/orifice (same dP from start to end) (if there is no condensation for either process - condensation will of course change all this)

 

[MortenA]

Compositepro - a JT process is just another word for isenthalpic expansion. Thus an expansion process where the work assiciated to the expansion "stays" in the fluid as opposed to the isentropic expansion where the work "leaves" the system e.g. a pistion where the work is turned into mechanical energy.

 

[Latexman]

Thanks MortenA. I don't know how that slipped through. Must have been distracted. I fixed it.

Good luck,
Latexman

To a ChE, the glass is always full - 1/2 air and 1/2 water.

 

[runsaep]

No one has yet commented upon the statements from Campbell and Perry handbooks (recognized bibles for process engineers) Campbell in chapter "basic thermodynamic concepts" writes that work is zero unless some mechanisms is provided for its transfer; pumps, compressors, turbines and expanders connected to some driving or driven rotating shaft, piston in a cylinder, electrical wires crossing the system boundary,use of a magnetic field to transfer energy, atomic transformation"


Her is my interpretation of J/T effect based on Wikipedia that I now have gained a little bit more understanding of :
In the production separator vessel of about 100 m3 the gas molecules are moving around with a kinetic energy corresponding to a high pressure of 39 bar and 65 deg C (normal op. conditions). The molecules are forced together due to the high pressure, creating intermolecular forces. The internal energy of the gas inside is the sum of the kinetic and the potential energy.
When the blowdown valve opens outside the vessel and the gas expands to a low pressure of 2 bar, the molecules come farther away from each other. For this to happen, the intermolecular forces must be overcome, which requires energy that is taken/shifted from the kinetic term. In this way, the transformation from kinetic to potential creates the cold temperatures downstream the valve and is called the Joule-Thompson effect.

In fact if one in an old-fashioned way use a Moliere diagram and follow a isentropic path one can find a temperature somewhat lower than a computer calculates, but that is due to the fact that heat- and massexchanges are accounted for in the calculation program and makes it less conservative. In one article on internet dealing with this expansion phenomena in the vessel has ascribed the term "pseudo-isentropic" to it, acknowledging that it is not ideal.

MortenA; "the expansion (due to mass leaving the vessel) is very close to isentropic - " - how can the fact that molecules leaving the outlet nozzle on its way to the blowdown arrangement be close to isentropic?