Extremely basic question about compressible flow 9

[SilverRule]

Hi Guys,

I've searched the forum and haven't found this exact question asked which is very basic and fundamental.

In incompressible flow, I know that static pressure energy is converted to heat (or vibration/noise) exclusively because kinetic energy (velocity) must be constant in this type of flow due to law of continuity.

But in compressible flow, is it true that static pressure energy is not exclusively converted to heat because the pressure drop in this case is converted to increase in volume and therefore shows up as increase in kinetic energy (velocity) for the most part and only some of it is converted to heat?

Is that right?

What determines how much is converted to volume/KE/velocity increase and how much is converted to heat?

Please let me know if any of my premises is wrong.

Following up on the above...

1) Is Compressible Isothermal flow (P1V1 = P2V2) a case where all of the static pressure energy (decrease in pressure) is converted to KE (increase in volume) and none of it is converted to heat?
2) Is Compressible Adiabatic flow a case where only some of it is converted to KE and the rest of it is converted to heat and shows up as rise in temperature?

Thank you!

SR

 

[TiCl4]

I don’t think that is correct. Both P and V are singularly defined in thermo books, meaning they have the same meaning throughout.

If taking an expansion while flowing and both P and V are changing, then the expression is d(PV), or Wec = d(PV)*mdot. Note that V is an intensive property here, not extensive.

I’m pretty sure that V in the polytropic equation PV^k=constant is extensive, and this means something entirely else. Please correct me if that is wrong.

 

[SilverRule]

Hey, thanks for the input.

I don't believe V has to necessarily be an intensive property here. It is in h=u+Pv (small case, intensive) which seems to be just as valid/applicable as H=U+PV (capital case, extensive) where it is an extensive property. In the wiki link here, they show it in capitals to represent extensive:

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

If you scroll down, they do a proof that h=u+Pv remains constant in terms of intensive. So again, I think both are valid. Correct me if I'm wrong.

And yes, I believe V in PV^k=constant is extensive.

I didn't mean that P and V have separate meanings/definitions between these two contexts. I didn't even necessarily mean that the PV quantity has separate meanings/definitions between these two contexts (unclear wording on my part). But more so that the change in the PV quantity is different between them. If the way you'd calculate PV expansion work was with d(PV) like you're saying (which was how I thought it would be calculated as well as that was intuitive), then there wouldn't be a difference. But actually, it's not calculated that way. See this short webpage:

https://www.chemteam.info/Thermochem/PV-Work.html

Or look at Khan academy video here where they calculate it manually through integration of the PV curve. Even though both P and V change, how you calculate the PV expansion is by assuming P is constant for an infinitesimal increase in V and repeat that till you cover the whole area:

https://www.khanacademy.org/science...l-energy-sal/v/pv-diagrams-and-expansion-work

The wiki page says that the change in PV represents the net work done by the fluid. Since the fluid does net work here, the magnitude of that PV increase is the amount of net work done.

Does that make sense? Let me know if you disagree.

There could be a more accurate explanation that someone else could extract from that wiki page as I'm not able to grasp some of the info on there.

 

[TiCl4]

d(PV) is the correct expression. P is NOT invariant with V, and is a function of it. From my thermo textbook (Elliot and Lira), the rate of work from a flowing fluid, Wflow, is

Wflow = (PV)in*mdotin - (PV)out * Mdotout. P is not assumed to be a constant here.

Also, the classic expression is Wec = integral (P) dV. Here, P is a function of V, and in most intro thermo classes the teacher will have the students solve by using the ideal gas law:

Wec = integral (nrt/V) dV from V1 to V2.

Trying to say that since P is constant over an infinitely small volume change, it is okay to take P out of the integral. That's just bad calculus.

If I had Y = integral (Z) dx, and Z = ln(x) (from X1 to X2), would you just take Z out of the equation and say the solution is Z*deltaX?

From your chemteam link:
This last equation has two more points about it: (1) we make an assumption that P remains constant (a fairly defensible one, I think).
NO NO NO. This assumption is patently false. He is assuming that a certain PV that expands in a piston against atmospheric pressure will have NO change in P. That (false) assumption is how he arrives at PdeltaV as his work term.

That is how I understand it, anyways. I'm always open to correction.

 

[SilverRule]

I'm slightly out of my element here as my thermo was always weak haha

I think we're conflating two different types of work but unfortunately they're both called PV work which is confusing. Though these might be related in some way I don't well understand. You've been talking about flow work which can be applicable to any fluid while I've been talking about expansion work which is only applicable to gases. But both are called PV work in a lot of places. The expansion work should be correctly called PdV work to avoid confusion. But you can see the two screenshots of the same wiki page on Enthalpy that I have below which are pulled from different parts of the page.

A mistake on my part I just realized that should clarify the confusion with that classic integral expression is that the Joule Thomson is an irreversible process. The way that you're doing that integral is only applicable to reversible expansion processes. Watch this short video and it'll be very clear.

https://www.youtube.com/watch?v=sYe0ulf428w

So my mistake on saying that P is held constant in this case. It's not held constant but simply the final pressure (P2) is multiplied with the DeltaV. P1 is ignored. P2 is technically called P[sub]external[/sub]. So that Khan academy video I posted in my previous post was showing only how to calculate the expansion work for a reversible process, which is not applicable to Joule Thomson.

By the way, though it's not relevant to this issue anymore, I have an explanation below for why/how P is "held constant" to calculate the expansion work for a reversible process if you're interested in bold orange:

I think you may be misinterpreting how the calculus is being done. If you haven't, I really think you should look at that Khan academy video which is as clearly as it can be fleshed out. Summary is that they're not simply taking P as a constant for the overall process and multiplying it with the overall change in V. They're only doing it for a tiny slice of rectangle under the PV curve (recall that the area under the PV work IS the total expansion work). Then they repeat the process by taking A DECREASED P (note that THIS IS WHERE where P decreases) as constant and multiply it again with the next tiny change in V (next tiny slice of rectangle). Then repeat by taking the next decreased P as constant and so on... So effectively, P is not really being kept constant. It keeps decreasing. And at the end, each multiplication (slice of rectangle) they did is added together to get the total integrated area under the PV curve which represents the reversible PV expansion work done.

Let me know what you think

 

[TiCl4]

I watched that video. His explanation of piston work takes work done on the surroundings as the expression. Work done by the system = work done on the surroundings. His main assumption here is that the surroundings are a constant pressure against which the piston expands. For his PdeltaV expansion, he took that expansion step in one go. As number of steps increases, the number of integral boxes increases and the expression approaches integral PdV. The question remains: does this one-step change assumption apply to expansion across a throttling valve?

Also, I don’t think your statement - that an ideal gas expansion across a throttling valve (dH=0) has NO work - is correct. I took your statement to mean dPV=0 and dU = 0. In this case, I believe dPV = dU. The components of U are more than just Van der Waals, they include the spring motion of the atoms themselves that depend on monatomic/diatomic nature and temperature. To imply that the interior components of U just shift around with no PV work would imply that gas temperature could drop with no change in pressure or temperature.

 

[SilverRule]

Yes, one of the fundamental aspects of the Joule Thomson expansion is that it's an irreversible expansion and expansion through the throttle is the most irreversible situation because it happens the fastest (in one step if you will). I mean in fact, all real processes are irreversible and this includes gas expansion that's happening everywhere else in the pipe too (whereas the most extreme effect is through a throttle).

The components of U don't shift around for ideal gas expansion, that was for free expansion of a real gas which is to say the real gas is expanding into a vacuum and not against an external pressure (so no PdV work is needed)...the temp drops but not without a change in pressure...the pressure does drop which is what causes the internal components of U to shift around. NOTE though I didn't say the components of U were just Van der Waals (which is called thermal PE)...I said the other component of U was temperature (which is called thermal KE). Temperature lowers and thermal PE increases...this is the shifting I was talking about.

As for the ideal gas, see the screenshot below from the Joule Thomson wiki page. There is no PdV work done here either. I think it's because there are no attractive forces to overcome so the gas can simply expand enough to match the drop in pressure leaving temperature constant.

Most of the info I'm getting from that wiki page and this page:

http://epgp.inflibnet.ac.in/epgpdat...665/M019127/ET/1515661337CHE_P10_M5_etext.pdf

 

[TiCl4]

Read your post above
- If an ideal gas is expanding against a pressure, AKA regular flow, across a throttle, this isn't a JT expansion. This is because...since there are no attractive molecular forces to overcome, there is no PV expansion work that needs to be done, and the gas expands without the temperature having to drop. So, PV is constant. Since temperature is constant and thermal PE is constant, U is constant. Overall H is constant.

Here you say no PV work for an ideal gas expansion against a pressure. Maybe I’m misunderstanding this statement.

Do we agree that dPV = dU for the isenthalpic case for an ideal gas?

 

[SilverRule]

Yep agree that dPV = dU for isenthalpic ideal gas expansion. None of the quantities change. Even though there is an expansion. Is your question that how can I say there is no PV work being done at the same time saying there's an expansion?

 

[TiCl4]

Yes. For an ideal gas expanding against a pressure, there is dPV work being done. The statement I quoted contradicts that.

 

[SilverRule]

The PdV work we're talking about here is strictly expansion work, not flow work. So could it be that work only needs to be done to expand only in circumstances where there's attractive forces whereas ideal gas has no attractive that are needed to overcome (in a case where it's only a fluid channel ... not going against an external force from a piston or anything)?

I'm not entirely sure honestly. I do feel like there's a hole or two that needs to be filled conceptually that I'm missing. I was hoping someone here knew.

 

[TiCl4]

So could it be that work only needs to be done to expand only in circumstances where there's attractive forces...

No.
You’ve got the idea of work with an ideal gas all wrong. Work is not done because it has to expand vs internal attractive forces. If your statement were true, ideal gases could never do any work because the definition of an ideal gas involves no intermolecular forces. For the throttle case of dPV = dU, each term is non-zero, meaning expansion flow work is being done. For an ideal gas, the dU term will change due to a change in temperature of the gas.

 

[SilverRule]

"For an ideal gas, the dU term will change due to a change in temperature of the gas."

I'm not saying you're saying. But the question is how come the wiki page says that there's no change in temperature for an ideal gas through an expansion (see the screenshot I posted in my previous post)?