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Given the following definition of flow work:

enter image description here

Reading and following the derivations I understand that in theory, this is only accurate when the fluid accelerates from rest. However, if we consider that a fluid is already in a state of steady flow we are still expected to use enthalpy instead of instead of internal energy. Essentially, this flow work is compensation work (between the inlet and outlet) required to keep a continuous flow. But, if we have fluid in a frictionless pipe already moving steadily (at constant velocity), I understand that we should still expect flow work. Therefore, my question is, what is this flow work compensating for?

To explain the compensation aspect I would suggest an FBD of each particle in a steady flow. There would be no net force and equal and opposite Pressure/Area forces from every direction. Neglect Gravity. Now this flow work (Force*distance) is defined as maintaining flow, so what is the force that is resisting flow? We would assume surface friction but I believe that this still exists in frictionless pipes.

enter image description here

Basically, my question is what is causing this change in energy (IN THEORETICAL SENSE) and why can we represent it as Pv?

Intuitively, the only explanation I can come up with is that this has something to do with the friction within the fluid particle itself due to intermolecular bonds, or that in the frictionless theoretical example flow work does not exist and that flow work is defined to be an approximation to the friction in reality.

In the case of intermolecular bonds, would we not expect there to be specific constants due to the strength of bonds? Also couldn't this friction become negligible when the particles are at the same speed? It seems odd that it would make such a difference.

If it is the friction between the pipe, shouldn't the calculating be concerned with surface area and coefficient of friction?

Please let me know the faults in my logic and the intuition behind what this term actually is. In addition, why Pv is a valid representation. Thanks in advance

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  • $\begingroup$ Friction and impacts between particles / molecules. $\endgroup$
    – Solar Mike
    May 9 at 4:03
  • $\begingroup$ so yes intermolecular forces? what friction are you referring to? and in theory, we can represent the particles to not collide with each other. $\endgroup$
    – cav3
    May 9 at 4:23
  • $\begingroup$ In “theory” you can make any assumptions you want. Whether it helps with reality is another question. For a good explanation check out the publications of Feinmann. $\endgroup$
    – Solar Mike
    May 9 at 4:25
  • $\begingroup$ See feynmanlectures.caltech.edu $\endgroup$
    – Solar Mike
    May 9 at 5:41
  • $\begingroup$ Fluid in a pipe doesn't do work, it loses energy to friction. Your illustration shows the work being done on a piston. $\endgroup$
    – Tiger Guy
    May 9 at 18:58

1 Answer 1

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In many thermodynamics analyses, transformation of a system from one state into another can be simplified by assuming that the system is conceptually removed from the world, transformed, and inserted back into the world in its new form. This can avoid potentially complex analysis of multiple events that the system might undergo, which are irrelevant if we're interested only in state variables (such as energy, pressure, volume, temperature, and entropy, for example). These depend only on the beginning and end states.

Enthalpy is a useful quantity when we wish to consider the internal energy of a system plus the work needed to move a constant-pressure environment out of the way as we conceptually push the system into our world. Enthalpy is what we need to provide, and enthalpy is what we collect when we remove the system. Simple bookkeeping can be done with these values.

For a surrounding fluid, the work is specifically the pressure–volume work. The pressure is the surrounding pressure $P$, and the volume is the system volume $V$. This is consistent with the enthalpy definition $H\equiv U+PV$.

Friction is completely unrelated, except inasmuch as it might change the system state, which we'd need to accommodate when we conceptually reinsert the system into the environment.

For more complex problems where constant pressure can't be assumed (such as fluid filling a balloon, where the balloon skin stored strain energy), we would still need to incorporate flow work, but the relevant potential would no longer be the enthalpy. The reason enthalpy is so ubiquitous in mechanical and chemical thermodynamics texts is that our atmosphere is essentially a constant-pressure reservoir and that many of our fluid systems operate at controlled pressures.

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  • $\begingroup$ but why is work needed to push a constant-pressure environment out of the way if it's already at steady flow and in a frictionless pipe? We would assume that without any extra forces, it would continue with its velocity and "push itself out of the way" $\endgroup$
    – cav3
    May 10 at 21:40
  • $\begingroup$ In that very simple case, the input flow work on on one side exactly offsets the output flow work on the other side. Real scenarios don't always allow that simplification. $\endgroup$
    – Chemomechanics
    May 10 at 22:37
  • $\begingroup$ so if they exactly offset, then there should be no net flow work to keep the flow continuous? $\endgroup$
    – cav3
    May 11 at 13:37
  • $\begingroup$ Yes, unless I’m misunderstanding you $\endgroup$
    – Chemomechanics
    May 11 at 14:11
  • $\begingroup$ then flow work doesn't exist ie. is 0. So why do we use enthalpy over U $\endgroup$
    – cav3
    May 11 at 18:21

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