1

I've read a lot on J-T effect but it isn't still very clear...

What is the role of the plug or valve ? It seems to me a very strange "non physical" requirement... It is often said that because of the plug the pressure decreases. But why ?

Why the J-T expansion isn't the same as the adiabatic reversible expansion of a real gas ?

Anarchasis
  • 1,343
  • The process between the pistons is adiabatic and takes place so that the input and output enthalpies are the same; the heat generated by the friction in the throttle you are referring to shows up in the output enthalpy of the fluid. One is not allowed to say that the enthalpy is constant everywhere during the process because the fluid inside the throttle goes through an irreversible process, although far away from it is in effective equilibrium. – hyportnex Jul 17 '17 at 18:47
  • " the heat generated by the friction in the throttle you are referring to shows up in the output enthalpy". Does this mean that because of friction the fluid does not cool as much as it would have if there were no friction ? – Anarchasis Jul 18 '17 at 00:31
  • I would turn your question around and say that if there was no friction in the plug then you could not throttle, and the enthalpy would thus not be conserved. – hyportnex Jul 18 '17 at 01:25
  • OK. So the fluid heats while going through the plug ? – Anarchasis Jul 18 '17 at 10:45

1 Answers1

1

The plug allows gas to flow slowly enough from one chamber to the other for the pressures in the two chambers to be kept roughly constant (but different from each other) by pushing one piston slowly 'inwards' and pulling the other slowly outwards. Thus the volume of the gas in one chamber is reduced to zero, while the volume of gas in the other chamber rises from zero. But (in the classic porous plug experiment) the whole process is fast enough to be essentially adiabatic. Speed is, as they say, of the essence!

For what it's worth, I'd rate "because of the plug the pressure decreases" as thoroughly confusing.

Reversible adiabatic expansion of a gas involves the gas expanding quasi-staically doing work against a piston. The gas can be made to go through the whole sequence of states in reverse by reversing the sequence of forces on the piston (and adding an infinitesimal extra force). This cannot be done with the porous plug experiment.

Philip Wood
  • 35,641
  • Thanks. OK. If it is quasi static and irreversible, this means that there is friction. Right ? And if there's friction, heat must be generated. Where does this heat go ? – Anarchasis Jul 17 '17 at 07:23
  • I have an apology. I have removed the reference to the porous plug experiment being "roughly quasi-static", as it is misleading. The process is not quasi-static. But I'm not sure that 'friction' is a useful term when considering the flow of gas in this experiment. What do you mean by 'friction' here? – Philip Wood Jul 17 '17 at 07:57
  • To get a feel for what's going on in the porous plug experiment, and in particular why the gas heats up or cools down (it may do either) I think you really have to apply some thermodynamics to a model gas, for example a Van der Waals gas. There are some websites which do this, but they may take you further than you want to go. – Philip Wood Jul 17 '17 at 12:26
  • The gas has to be forced through the porous plug to overcome the force of resistance it experiences. This force of resistance must be friction. Right ? If not what else could it be ? – Anarchasis Jul 18 '17 at 00:46
  • @ Anarchasis I see where you're coming from: on a molecular level, collisions between molecules as they pass through the fine tubes of the plug transfer energy from ordered KE of flow to random molecular. I suspect that this effect is negligible, because the textbooks say that an ideal gas doesn't change its temperature in this experiment (and an ideal gas DOES experience viscous forces). Real gases can either heat up or cool down, depending on initial and final temperatures and pressures. This can be understood in terms of intermolecular forces and molecular volume. – Philip Wood Jul 18 '17 at 09:14
  • 1
    @ Philip Wood. Yes, correct ! So if this viscous forces are negligible, what's the purpose of the plug ? I suspect that because of the plug, the fluid cannot acquire a high velocity. I suspect that the reason of the plug is not to allow some part of the energy transform into KE. – Anarchasis Jul 18 '17 at 10:51
  • And because the fluid cannot acquire a high velocity, you can maintain a pressure difference between the two sides - which is what I'd give as the purpose of the plug! – Philip Wood Jul 18 '17 at 11:24
  • 1
    Yes it makes sense. Thanks ! And because the velocity is not high, heat generated through viscous friction is low, maybe be negligible. It is very surprising that no one in any book tells the purpose of the plug. Suddenly, out of nowhere, a porous plug appears without any explanation – Anarchasis Jul 19 '17 at 07:29