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Extremely basic question about compressible flow
4

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 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

I don't know if this is really correct (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

(OP)
rotw,

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.

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 compressible flow 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

Does this image help at all? These graphs are the result of two steam pipes converging into one pipe at length = 200 ft.

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

(OP)
RVAMeche,

Thanks for your input.

Dry steam I'd think would behave as compressible fluid 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 an incompressible fluid here. Though velocity decreasing doesn't make sense unless the pipe diameter is changing...as velocity is typically constant for incompressible fluids 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 of incompressible flow. 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 compressible flow 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

(OP)
rotw,

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 incompressible flow. 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

The expansion work done by the gas is incrementally given by the VdP term, while frictional losses are all converted to KE increment. In isothermal compressible flow, this VdP term is based on constant temp, while adiabatic behaviour applies in adiabatic flow.

RE: Extremely basic question about compressible flow

(OP)

Quote (georgeverghese)

The expansion work done by the gas is incrementally given by the VdP term, while frictional losses are all converted to KE increment. In isothermal compressible flow, this VdP term is based on constant temp, while adiabatic behaviour applies in adiabatic 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 COMPLETELY converted to KE manifesting as increased velocity? None of it is converted to heat the way it is in incompressible flow? 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't mean that ALL of 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

The VdP term is not the friction loss term, it is the work of expansion. The friction loss term is ALL converted to KE, but is not the entire KE change. Some of the KE change comes from VdP.

RE: Extremely basic question about compressible flow

(OP)

Quote (georgeverghese)

The VdP term is not the friction loss term, it is the work of expansion. The friction loss term is ALL converted to KE, but is not the entire KE change. Some of the KE change comes from VdP.

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 THROUGH the 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, in incompressible flow, 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

The chain of events may be pictured as follows:
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

(OP)

Quote (georgeverghese)

The chain of events may be pictured as follows:
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.

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 pipelines and 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

1)Nomenclature used in my chem engg text uses Y as the ideal adiabatic constant, and k as the process specific adiabatic constant
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

2) It is easy to get lost in the weeds when trying to decide whether to use the isothermal or adiabatic model. For the overwhelming majority of cases it is irrelevant. Quoting from Coulson & Richardson's Chemical Engineering (Vol 1, 6th Ed): "The rate of flow of gas under adiabatic conditions is never more than 20 per cent greater than that obtained for the same pressure difference with isothermal conditions. For pipes of length at least 1000 diameters, the difference does not exceed about 5 per cent."

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"

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