## Extremely basic question about compressible flow

## Extremely basic question about compressible flow

(OP)

Hi Guys,

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

In

But in

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

2) Is

Thank you!

SR

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

## RE: Extremely basic question about compressible flow

kinetic energy (velocity) must be constant).Assuming basically, no change in elevation, steady and incompressible flow, no work/heat added or subtracted, no dissipation (due to viscous effect) in a given control volume we have:

Total pressure = static pressure + dynamic pressure (1/2 * rho * V^2)

For instance dynamic pressure decrease would be due to velocity decrease (e.g. diverging section). Static pressure will increase.

When flow is compressible, under same other assumptions as above, the equation above must be thought off in terms of total (stagnation) enthalpy conservation. In addition, if flow is choked further considerations apply.

Maybe you will find this interesting:

https://en.wikipedia.org/wiki/Fanno_flow

You need to be more precise on your working assumptions to obtain answers on 1/ and 2/.

If you plan an escape, you must succeed as if you fail, you will be punished for trying. Never say or write down your plan. Heart is the only place where secrecy is granted.

## RE: Extremely basic question about compressible flow

For

incompressibleflow in a no-elevation, constant cross sectional area conduit, KE (velocity) must be constant in order to preserve the continuity equation (conservation of mass). So all of the energy associated with lost static pressure (head loss) due to friction has to be converted to heat. Please correct me if I'm wrong but I think that's right.I'm looking at the known flow energy equation as I'm talking about this: http://hyperphysics.phy-astr.gsu.edu/hbase/pber.ht...

So for

compressibleflow in the same conduit, the way I'm imagining it... as static pressure drops, that will lower density and increase volume since it's a gas, and that higher volume within the cross sectional area (in order to adhere to conservation of mass) will cause velocity (KE) to increase to compensate for the lost energy associated with static pressure drop. And another portion of the lost static pressure energy will manifest as heat. Is that in the ballpark?## RE: Extremely basic question about compressible flow

Velocity of the single pipe (0-200') increases decreases over the length of the piping due to friction losses; once the flows converge it drops rapidly.

Temperature is the opposite, it increases slightly over the initial run then once converged rises much more quickly.

Pressure drop starts off as linear for the single pipe, then becomes non-linear after the flow is combined.

## RE: Extremely basic question about compressible flow

Thanks for your input.

Dry steam I'd think would behave as

compressiblefluid and so pressure drop for it is typically nonlinear as is the increase in velocity. The fact that pressure drop is being shown as linear and velocity as almost-constant tells me it's being treated as anincompressiblefluid here. Though velocity decreasing doesn't make sense unless the pipe diameter is changing...as velocity is typically constant forincompressiblefluids in a constant cross sectional area pipe. But if we pretend the velocity shown here is constant, then yes I'd say these graphs are in line with my understanding ofincompressibleflow. The temperature increasing is indicating that the pressure drop from friction is generating heat some of which is raising its temperature (some of it likely lost to surroundings).My biggest confusions are primarily about understanding

compressibleflow behavior (it's more complicated) in terms of my questions above.Do you have an input on that end?

Thank you.

## RE: Extremely basic question about compressible flow

For incompressible flow in a no-elevation, constant cross sectional area conduit, KE (velocity) must be constant in order to preserve the continuity equation (conservation of mass). So all of the energy associated with lost static pressure (head loss) due to friction has to be converted to heat. Please correct me if I'm wrong but I think that's right.Have you stated anywhere in your original post that you are considering "constant cross sectional area conduit" ? I suppose No. Therefore in absence of this assumption, reader may assume velocity can decrease or increase while satisfying continuity equation.

If the flow is compressible, it is more appropriate to use the concept of enthalpy (static and dynamic) instead of static pressure drop, this is why:

rho: density

p: static pressure

V: velocity

q: heat

w: work

q - work = [ (p / rho) + 1/2 * V^2 ]_2 - [ (p / rho) + 1/2 * V^2 ]_1

So it is not only about heat, it can be work which are two forms (degraded, resp. not degraded) of energy.

if we assume work = 0 and if we note hs (static enthalpy) = (p / rho) then

q - 1/2*(V_2^2 - V_1^2) = [hs2-hs1]

This shows that static enthalpy change corresponds to the changes of physical quantities you have mentioned, namely:

- Part is converted to heat

- Remainder corresponds to change of kinetics.

If you plan an escape, you must succeed as if you fail, you will be punished for trying. Never say or write down your plan. Heart is the only place where secrecy is granted.

## RE: Extremely basic question about compressible flow

Yes, I'm sorry I didn't specifically mention constant cross sectional area as an assumption in the original post but that's right, I want to hold that constant so I can better understand the dynamics. And yes, assume work is also zero.

I understand your point about enthalpy, it makes sense. What I don't understand is how any of it would be converted to kinetics for

incompressibleflow. I mean the velocity in constant cross section pipe is constant so shouldn't the static enthalpy change only correspond to conversion to heat?## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

Thank you for the input georgeverghese. You've gone to the exact root of my question. Let me get a bit more clarity if you don't mind...

So the pressure energy change due to friction, represented by that VdP term, is

COMPLETELYconverted to KE manifesting as increased velocity? None of it is converted to heat the way it is inincompressibleflow? That's how that sentence reads to me, though your next two sentences seem like they're suggesting otherwise. Please correct me if I'm wrong.If you

didn'tmean thatALLof it is converted to KE and some of it is actually converted to heat... then my question is, as my original post asks, what determines how much of it is converted to each form?I hope I'm making sense. Thanks!

## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

Sorry, you lost me a bit there due to a combination of a few things.. let me try to parse it out..

One of them is I'm rusty on my thermo with respect to interpreting what VdP is but anyway...

When you say friction loss, that is referring to pressure loss specifically correct? If so, that's one reason why I thought VdP represented pressure loss because it's got pressure change (dP) in it and not a volume change (expansion).

Let's go through the causal chain, hopefully that makes it clearer...

From my understanding, pressure/friction loss is converted to velocity

THROUGHthe mechanism of volume increase (expansion) and that that conversion can't happen any other way without that mechanism. In other words, the causal chain is the following: when pressure is lost, that increases lowers density and increases volume (expansion), and the fact that volume is increased and the cross sectional area hasn't means that velocity (KE) increases. So I don't see how the expansion is an independent causal phenomenon that is contributing to increase in velocity (KE) instead of being the effect of pressure/friction loss. I also don't see how pressure/friction loss can directly convert to increase in velocity (KE) -- through which mechanism? -- how does it cause an increase in velocity (KE) when cross sectional area hasn't changed? I mean, inincompressibleflow, pressure/friction loss doesn't convert to increase in velocity (KE).You're speaking as if pressure loss and expansion are two independent phenomena both contributing to increase in velocity (KE) rather than being in the same causal chain with pressure loss being the primary cause.

I know I must be misinterpreting something.. thanks for your help again!

## RE: Extremely basic question about compressible flow

a)Frictional loss occurs

b)Mechanical work of expansion VdP is expended to make up for frictional loss. This VdP may be iso thermal or adiabatic or in reality somewhere in between. Some of this work goes to KE gain.

c)Coincident with (a) and (b), frictional loss is converted to KE gain.

The VdP term does not mean molar volume is constant during this incremental change. It is the mathematical differential expression for work done. The V and P terms are directly related by PV=constant for isothermal expansion, and P.(V^k)= constant for adiabatic or partially adiabatic expansion, where the term k < ϒ. The reason for k lower than ϒ is due to the KE incremental contribution from (b) and (c) above.

I may not have my thermo grammar all in order here, but loosely speaking, this is how I see it.

## RE: Extremely basic question about compressible flow

Excellent! Now it makes perfect sense. Exactly what I was trying to understand. Thank you so much!!!

1) I haven't heard of the Y before (I know k is ratio of specific heats). Are you conceptualizing it here just as a way to explain the dynamics here or is it an actual entity? If it's the latter, can you tell me what it's called? I'd like to look it up and understand it better.

2) By the way, any thoughts on what determines where in between isothermal and adiabatic the expansion falls? Is it primarily determined by which gas we're talking about and what its k value is? (I know that k for a real gas decreases with temp).

3) I've heard from quite a few sources, including Crane TP410, that a recommended simplifying assumption for compressible flow is to assume relatively long pipes to be isothermal and relatively short ones to be adiabatic. Do you know why these assumptions are appropriate?

I know that even if the gas temperature and outside ambient temperature are the same (i.e., no temperature differential), the gas temperature will tend to fall along the length of the pipe due to the expansion. But I'm not sure how this relates which kind of flow it is estimated to be and why.

4) Finally... least important: In Crane TP410 around pages 17-19 (depending on the version you have), within in "Principles of Compressible Flow in Pipe" section, they present the Complete Isothermal Equation applicable to

long gas pipelinesand then immediately present a simplified version of it saying that the "acceleration can be neglected because the pipeline is sufficiently long". This confuses me because, isn't it the case for compressible flow, and particularly for isothermal flow, that all of the lost pressure works purely to increase the velocity every step of the way? So isn't that all acceleration by definition?Feel free to answer those you're familiar with. 2) and 3) are of most interest.

## RE: Extremely basic question about compressible flow

2)There is a rather complicated expression in my copy of Coulson and Richardson text on Chem Engg Vol 1, that uses the ideal Y to account for the non ideal k in compressible ideal( ideal meaning z=1 or the same at both ends of the line) pipe flow. Pls refer to this text for this expression - I have not used it so far.

3) Havent heard or read about this. Isothermal compressible expressions are easy to use but tend to overpredict pressure drop somewhat, which is fine for inplant piping and flare lines. Adiabatic expression is more complicated but is closer to reality. When pressure drop is high in a line, temperature will drop since this is approximately adiabatic.

4)Yes, there are 2 versions of the isothermal compressible expression. The simpler one, with one term less, is perhaps more amenable to manual calcs, but the longer expression is easily used these days with everyone having access to spreadsheets. It may this 2nd term that is called "acceleration" in Crane. I am referring to the isothermal compressible flow expression in Perry Chem Engg Handbook. I would tend to retain the 2nd term for long lines, as suggested in Perry also.

## RE: Extremely basic question about compressible flow

In most cases the differences will be far less. I have shown a typical comparison for nitrogen flowing at about 270 ft/s in a line with L/D = 300 where I compared the adiabatic and isothermal models with each other, and also with the incompressible case. The two compressible cases differ by less than 0.1% while the incompressible assumption introduces an 8% difference. See https://www.katmarsoftware.com/examples/aioflo-exa...

3) I agree that most texts distinguish between long and short pipes, but I believe the important parameter is the velocity rather than the length. It just so happens that long pipes tend to have much lower velocities than short pipes. If you are working at less than Mach 0.2 just use the isothermal assumption and be aware that you might be slightly over-estimating the pressure drop. For greater than Mach 0.2 it is probably worth doing both calculations.

4) I would disagree with the statement "for compressible flow all of the lost pressure works purely to increase the velocity every step of the way". In the Bernoulli model for incompressible flow in a horizontal pipe, which ignores friction, it is true that if the diameter changes (and therefore the fluid velocity changes) the pressure energy is all converted to kinetic energy. In compressible flow the velocity change is caused by the loss in pressure rather than a change in diameter, but you cannot ignore friction. The friction term is proportional to Velocity squared so if you change the velocity you also change the friction loss. So some of the lost pressure works to increase the velocity and some of it works to overcome the increased friction.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Extremely basic question about compressible flow

I know I'm bringing this thread back from the recent dead. After taking in your explanations, I had to go and study and look into a bunch of things to enlighten myself in a lot of details before I could ask any further questions. Thank you again for your responses. In the meanwhile, I got my hands on that Coulson & Richardson book and I'm going to go through it.

I did want to ask you guys about something you both mentioned...

Why is it the case fundamentally that adiabatic flow gives a higher flow rate for a certain pressure differential? Put a different way... why is it that isothermal flow gives a bigger pressure drop for a certain flow? What's the fundamental mechanism that causes this?

Thank you gentlemen.

## RE: Extremely basic question about compressible flow

The isothermal model does not predict any cooling (by definition!) so it will tend to predict higher volumetric flows towards the end of the line and therefore higher pressure drops.

## RE: Extremely basic question about compressible flow

Sorry, I thought you said in the previous post, where you quoted Coulson & Richardson's, that adiabatic conditions give higher flows? Here you're saying isothermal conditions give higher flows?

## RE: Extremely basic question about compressible flow

The two statements have different criteria. It's apples and oranges.

With the same initial conditions and the same pressure drop, flow

_{adiabatic}> flow_{isothermal}.Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

## RE: Extremely basic question about compressible flow

why is it that isothermal flow gives a bigger pressure drop for a certain flow?" and my comments were based on the assumptions implicit in that question.Let me try again. Under the assumption of

the samethe isothermal model will predict a higher rate of expansion for the gas than the adiabatic model would. The reason for this is the temperature behavior I explained earlier. This higher rate of expansion leads to higher velocities and therefore higher pressure drops for the isothermal model.massflow rate, with the same compressible fluid and the same pipe dimensionsConversely, under assumptions of the same pressure drop the isothermal model will predict lower mass flow rates.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Extremely basic question about compressible flow

Ah! My fault guys.. I overlooked that detail. Thanks for being patient with me. Now there's no confusion on that end. I now understand these macro patterns.

I tend to try to understand all the fundamental underlying theory, so I was studying the wikipedia page on Joule-Thomson expansion effect closely in detail, because as I understand it, that is essentially the fundamental expansion that we see in compressible flow, which we've been discussing in this thread (yes, I understand that gas flowing in

insulatedpipe isn't as clean of an example of this effect as flowing across a throttled valve as this flowing in uninsulated pipe isn't quite adiabatic/isenthalpic but the effect is still in play at a milder level). I was confused by something in this section of the wiki page -> https://en.wikipedia.org/wiki/Joule%E2%80%93Thomso..."In a Joule–Thomson expansion the enthalpy remains constant. The enthalpy, H, is defined as H = U + PV where U is internal energy, P is pressure, and V is volume.

Under the conditions of a Joule–Thomson expansion, the change in PV represents the work done by the fluid. If PV increases, with H constant, then U must decrease as a result of the fluid doing work on its surroundings. This produces a decrease in temperature and results in a positive Joule–Thomson coefficient.Conversely, a decrease in PV means that work is done on the fluid and the internal energy increases. If the increase in kinetic energy exceeds the increase in potential energy, there will be an increase in the temperature of the fluid and the Joule–Thomson coefficient will be negative."That PV increase being referred to in

bold.. this is how I'm interpreting it (please correct me if this is wrong): If P1 and V1 are 100 and 75, respectively, and P2 at some downstream location is 50, then V2 will have to be higher than 150... in which case, PV hasincreaseddue to the PV expansion work having done by the system. If PV is said to be constant in isothermal flow, meaning it could only expand up to 150, how could the PV increase in adiabatic flow? Shouldn't there be smaller expansion in adiabatic flow? So I would expect PV to decrease instead.Any idea what I'm misunderstanding here? I'm almost sure I'm comparing to apples to oranges here again but can't pick out what exactly.

Sorry if this is a bit too esoteric/theoretical. You guys have been very helpful so far.

## RE: Extremely basic question about compressible flow

^{1}= constant) to question adiabatic behavior (PV^{k}= constant)?Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

Oh no.. I'm not asking this question -> "Since PV must be constant for isothermal flow, how is it that they're saying PV increases in adiabatic flow?" That would be definitely be an absurd question blatantly trying to compare apples to oranges.

It's more like this -> "We recognize that PV = constant for isothermal behavior. And we recognize that PV

^{k}= constant for adiabatic behavior. So for adiabatic behavior, if PV^{k}= constant, then it would be correct to say that PV itself decreases for adiabatic ... in other words V2 wouldn't be as high as it would have been in the isothermal case. So based on this assumption, I'm questioning why that wiki page says that PV increases, rather than decrease, suggesting that V2 would actually be higher than it would have been in the isothermal case."Hope my question is clear now

## RE: Extremely basic question about compressible flow

^{k}= constant and PV = constant on a graph, but don't go past the critical pressure ratio like on the previous example.Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

boldin my earlier post) that PV increases?## RE: Extremely basic question about compressible flow

Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

Well no, I'm not saying wiki got it wrong and needs fixing necessarily (unless you're saying that) ... I don't know if it's right or wrong, I'm just seeing a dissonance between what you and katmar have said (volume not increasing sufficiently enough to keep PV constant for adiabatic) and what wiki is saying about PV increasing when there's a cooling. The logic in wiki seems to make a lot of sense when you look at the equation H=U+PV ... because temperature is captured within U, when it goes down, PV needs to go up in order to keep H constant (to adhere to the isenthalpic requirement).

I'm just not sure why there seems to be a dissonance. Do you think the wiki has it wrong?

The current focus is not really on what makes JT coefficient positive or negative, it's just that wiki's explanation of how the value of PV changes is nested inside its explanation of what makes JT coefficient positive or negative. We're really only interested in the cooling case (positive JT coefficient) where it says PV increases as this is the most common case you're dealing with.

## RE: Extremely basic question about compressible flow

The

IF, in front makes me think they are talking hypothetically and not about a specific case. This is supported further down before their "proof" with the discussion on a gas changing sign of the JT coefficient before/after it's inversion point:They are covering all the cases, not just the usual expansion with a drop in temperature we see in compressible flow of most gases and vapors. I think from that respect what they say is correct.

Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

"Conversely, a decrease in PV means that work is done on the fluid and the internal energy increases. If the increase in kinetic energy exceeds the increase in potential energy, there will be an increase in the temperature of the fluid and the Joule–Thomson coefficient will be negative."

This doesn't seem to be the usual case because in the usual case the fluid would be doing the work (rather than not getting work done on) and thermal kinetic energy (temp) decreases rather than increase.

Something is not adding up here on a semantic level..

Let me ask you this in case we've got the semantics flipped: When a fluid is doing (expansion) work as opposed to getting work done on, would the PV value in H=U+PV increase or decrease? And during the usual expansion in compressible flow, is the fluid doing the work or getting work done on?

## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

But the gentlemen here have done an awesome job clearing everything up and I really believe this is the last outstanding unanswered question in this "thread" so to speak. I find it hard to believe that either wikipedia or the guys here are wrong, so I'm trying to find the missing piece that would reconcile the two.

## RE: Extremely basic question about compressible flow

I was studying the wikipedia page on Joule-Thomson expansion effect closely in detail, because as I understand it, that is essentially the fundamental expansion that we see in compressible flow, which we've been discussing in this thread (yes, I understand that gas flowing in insulated pipe isn't as clean of an example of this effect as flowing across a throttled valve as this flowing in uninsulated pipe isn't quite adiabatic/isenthalpic but the effect is still in play at a milder level). I was confused by something in this section of the wiki page -> https://en.wikipedia.org/wiki/Joule%E2%80%93Thomso...

"In a Joule–Thomson expansion the enthalpy remains constant. The enthalpy, H, is defined as H = U + PV where U is internal energy, P is pressure, and V is volume.

Under the conditions of a Joule–Thomson expansion, the change in PV represents the work done by the fluid. If PV increases, with H constant, then U must decrease as a result of the fluid doing work on its surroundings. This produces a decrease in temperature and results in a positive Joule–Thomson coefficient.Conversely, a decrease in PV means that work is done on the fluid and the internal energy increases. If the increase in kinetic energy exceeds the increase in potential energy, there will be an increase in the temperature of the fluid and the Joule–Thomson coefficient will be negative."That PV increase being referred to in

bold.. this is how I'm interpreting it (please correct me if this is wrong): If P1 and V1 are 100 and 75, respectively, and P2 at some downstream location is 50, then V2 will have to be higher than 150...which represents the PV increase.BUTif PV = constant in isothermal flow meaning it could only expand up to 150 (V2 = 150), and if PV^{k}= constant for adiabatic flow meaning PV itself would decrease (i.e., P2*V2 < P1*V1) and so V2 wouldn't be as high as it's in the isothermal case...HOW COMEthe wiki quote says that PV increases suggesting that V2 would actually be higher than it's in the isothermal case? How could the PV increase in adiabatic flow? Shouldn't there be smaller expansion in adiabatic flow? So I would expect PV to decrease instead.## RE: Extremely basic question about compressible flow

Maybe this thread has gotten too deep and messy and maybe I should just make a new clean independent thread with the question?

Thank you!

## RE: Extremely basic question about compressible flow

The Joule-Thomson effect is approached in an insulated valve, so minimal heat is transferred. A valve is extremely short, comparatively. Friction is not accounted, because it is negligible to the process. L/d approaches 1.

Apples and oranges.

Good Luck,

Latexman

Pats' Pub's Proprietor

## RE: Extremely basic question about compressible flow

I'll first lay out all the expansion scenarios. I've got this from a wide mix of sources:

Joule-Thomson (JT) expansion effect is isenthalpic, so H=U+PV must be constant. U consists of thermal KE (temperature) and thermal PE (attractive Van der Waals forces between molecules).

- If a real gas is expanding into a vacuum across a throttle, they call it Free Expansion instead of JT expansion because it's not expanding against a pressure. Since it's not going against a pressure, it doesn't have to do any PV expansion work. But still, as it has attractive molecular forces that need to be overcome for it to expand, the temperature drops and is converted to thermal PE. So overall, PV doesn't change since it's Free Expansion and U doesn't change on a macro level. Just the magnitudes of the constituents of U change. So H is constant overall.

- If an

idealgas 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.- If a

realgas is expanding against a pressure, AKA regular flow, across a throttle, this is the JT expansion. There are attractive forces to overcome, and since it's going against a pressure, PV expansion work needs to be done to overcome these forces. So here temperature is sacrificed/dropped and converted to do PV expansion work.Here is where wikipedia says PV increases. It doesn't mention if the magnitude of thermal PE also increases. I'm assuming it does since molecules have more separation now.So here, PV changes, and thermal KE and thermal PE that make up U also change. Again, the changes in U and PV would balance each other out to keep H constant.These are the fundamental mechanisms. I thought laying this out would help understanding of what's below.

Directly to your comment: I also thought initially that the Joule-Thomson (JT) expansion effect only applied tovalvesonly as the heat transfer is negligible in that scenario. But after diving deep, I learned that JT is actually in effect in any conduit that is adiabatic. So if a pipe is perfectly insulated the JT effect is working EXACTLY the same way there throughout the whole pipe. In fact that's not even the full picture...in reality, even in an UN-INSULATED pipe the JT effect is happening...it doesn't have to be adiabatic...only difference is that the effect is milder (you don't see as big a temp change). So adiabatic will show the purest, cleanest and maximal JT effect and non-adiabatic (polytropic) will show a lesser JT effect and isothermal willeffectivelyshow none.I say isothermal will

effectivelyshow none because just like polytropic and adiabatic, it will also cool, as it flows over a unit length of pipe. Why will it cool? Because as the pressure drops it has to cool in order to expand -- since this is a real gas going against pressure, there is no expansion possible without JT effect of cooling (as per the mechanism described above). So the reason this isending upas isothermaleffectivelyis because of the heat transfer which raises the temperature back up. So my point is that JT expansion is notfundamentalonly to adiabatic flow. It's just that...effectivelythat is the flow that sees its maximal impact.Again, for a real gas, JT expansion is inherently built-in to the entire flow, not just across the valve. There is no flow without JT expansion, whether that cooling is being "reversed" by the heat transfer/input that may also happening at the same time (isothermal) or not (adiabatic). This was a big revelation to me. Hope that makes sense.

The sources that made this particularly explicit/clear to me:

1) My flow simulation program: It automatically does the JT calculation across a valve. And if you select the adiabatic model for the pipe as if it's perfectly insulated, it again automatically includes the JT calculation. It won't even let you uncheck that calculation. If you select a non-adiabatic (heat transfer) model for the pipe, then it lets you check or uncheck the JT calculation. See image below.

2) https://www.cambridge.org/th/files/1613/6690/0251/...

Look at the bottom of page 13 in the

section. It says the following:ISENTHALPIC PV WORK- The Irreversible Case"Pippard (1966, pp. 68–72) points out that an isenthalpic expansion need not have a throttle, but could take place for example in a tube in which the entropy-producing effect of the throttle is replaced by viscous drag along the walls, or possibly by other entropy-producing processes such as turbulence, chemical reactions, and so on. In other words, the essential element of a Joule-Thompson expansion is not the presence of a throttle; it can be any adiabatic irreversible change between two equilibrium states having the same enthalpy."

Please let me know if anything I said is confusing or needs any clarification. I would really appreciate it!!

## RE: Extremely basic question about compressible flow

Has anyone been able to have a chance to digest my last post?

SR

## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

I think I may have figured out the answer to my own question (possibly) if anyone is interested..

As I expected, I think it's confusion arising from semantics..

I think the PV quantity in the enthalpy equation H=U+PV has a slightly different meaning than the PV in PV

^{k}=constant.The change in PV in the former context is at the "micro" infinitesimal level strictly as a means of calculating the expansion work which would take the generic form of W=P*DeltaV. So at the micro level, P is basically held constant as the volume changes/increases (recall back to how you manually would integrate the area under a PV curve). So of course PV would increase in that micro expansion causing U to decrease leaving H constant. This is how PV is used here even though the overall quantity of PV decreases at macro level for all practical purposes because P actually decreases in reality instead of holding constant.

This was really bothering me so I feel much better now. I guess my whole confusion came back from completely forgetting the theoretical basics.

If anyone sees anything off with this, let me know.

Thanks a lot to EVERYONE who's responded in this thread. I really appreciate y'alls help!

## RE: Extremely basic question about compressible flow

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.

## RE: Extremely basic question about compressible flow

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%E2%80%93Thomso...

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/chemistry/ther...

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.

## RE: Extremely basic question about compressible flow

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. :)

## RE: Extremely basic question about compressible flow

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

_{external}. 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 :)

## RE: Extremely basic question about compressible flow

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.

## RE: Extremely basic question about compressible flow

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/epgpdata/uploads/epgp_...

## RE: Extremely basic question about compressible flow

- 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?

## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

## RE: Extremely basic question about compressible flow

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.

## RE: Extremely basic question about compressible flow

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.

## RE: Extremely basic question about compressible flow

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)?