Velocity and Pressure Drop 9

[KC12]

Hi All,

Lets say you have a pipeline of 100 ft with various fittings, reducers etc, and a velocity of about 5 ft/s entering the pipe line intially. I want to find how much allowable pressure drop there can be so that 5 ft/s does not reach 0 ft/s after its journey through the pipe. Which calculation methods are good for this? The velocity will remain constant if there are no forces there to decelerate or accelerate it. Frictional losses would decelerate the velocity. Pump design would take into effect the frictional losses in the pipe required to push through the constant velocity of 5 ft/s.

Now for my clarification:

Bernoulli's equation says that a lower pressure (gauge pressure after 100 ft of piping) will correspond to a higher velocity. Im having a hard time wrapping my head around this and using bernoulli in my case above would give bizarre results! Can someone clarify? Reality wise, a gauge pressure of zero on the line would mean no flow?

TY!

 

[zdas04]

ione,
That paper is a pretty good review of the waterfront of pipeline empirical equations, and most of what I saw there looked useful. Thanks for posting it. The equation I use is more properly referred to as the "AGA Fully Turbulent Gas Equation", I had forgotten about the Partially turbulent version.

I read the paper really quickly, but I did catch one point where the authors said that Reynolds Numbers under 2100 represented Laminar flow which is fine, but then they went on to say that RN over 2100 is turbulent which is not fine. The lower limit for turbulent flow is usually stated at around 4,000. The area between Laminar and turbulent is usually called "transition" and none of the friction correlations are valid in that range.

Again, stuff on the Internet can be very useful, but it is all presented as caveat emptor and it is difficult to know how well a document or calculation has been checked.

David

 

[ione]

David,

Just posted the link to emphasize that there around many correlations developed over the years, applicable to those real world applications, where Bernoulli’s equation fails. I considered this paper interesting for those people who need to touch with their hands to believe.

Further to the distinction between laminar and turbulent regime, I considered there was a typo in the paper, since it is reported:

“As known, for Re smaller than 2100 the flow is laminar, whereas for Re above 2100 the flow is considered turbulent. Between laminar and turbulent flows there is a transition region, for which there are no available pressure drop correlations.”

If for Re smaller than 2100 the flow is laminar and above 2100 is turbulent, what in between? I mean no range available for the transition region, though the authors mention it.

 

[kpg452]

don't mean to beat a dead horse here...but as a young engineer (3 months out of school) now I am getting confused reading these posts. I think my confusion can be explained best with an example. Below I am using the Darcy Weisbach equation.

If I have water flowing through the inlet of a 10" ID pipe at 70F and 2500 GPM and a length of 5000', my Moody Friction factor is about .015. The pressure drop is about 64 psig by using Darcy-Weisbach. But if my pressure becomes less and less as you move down the pipe, the velocity will decrease due to friction. So that means the mass flow rate decreases if the density stays relatively constant? Would that mean I have to iterate the pressure drop in sections instead of doing it all at once?

I am thinking no, since my analogy for an incompressible flow is if I push a solid bar at one end with some velocity, the other end of the bar will move at the same speed. If I don't apply enough pressure (less than 64 psig), then the bar won't move from ground friction sort of like fluid through a pipe.

I think I am confusing myself here between incompressible and compressible flow, but never really had it explained to me very well.

 

[katmar]

While we are splitting Bernoulli's hairs, can I split one too? The zone between the Reynolds numbers of 2100 and 4000 is often referred to as the "transition" zone but this is confusing because the zone between the "smooth pipe flow line" (von Karman and Prandtl) and the "rough pipe flow line" (Nikuradse) is also called the transition zone. I prefer the term "Critical Zone" for the interval between Re=2100 and Re=4000 because I think Critical describes the behaviour better than Transition does. There is no smooth transition here.

The terminology I favour is used on the Moody Diagram in the Schroeder article referenced by ione. This is the same Moody Diagram as presented in Crane TP410. I do believe it is worth the effort of making this distinction and not having two "transition zones" on the same diagram.

End of hair splitting rant.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

kpg452, your analogy of the solid bar is exactly right. Water is incompressible (to a first approximation) and the density and mass flow remain constant. Even the velocity is constant if the pipe diameter is constant. Water behaves like a very flexible steel bar.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

Had to double check I wasn't at yahoo.answer.com

What's important about the Bernoulli equation is not Z + p/[γ]+ v^2/2/g. Who cares about the equation, original or modified. Maybe we need to modify it for the Lorentz transformation for plasma near black holes too. Read the concept Bernoulli expressed in words, then tell me its not valid. Is the original body of the US Constitution invalid because it didn't foresee 27 amendments? Sounds like I'm talking rubbish now doesn't it.

Sorry, but if you guys insist on splitting hairs?, no, more like nano-machining ameba flagella, I'm not hanging around for it.

P.S. 1
kpg452, Steel bar is good. Run with it.

P.S. 2 Critical Reynolds numbers in pipe flow have been reported up to 10,000.



**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[zekeman]

"If I have water flowing through the inlet of a 10" ID pipe at 70F and 2500 GPM and a length of 5000', my Moody Friction factor is about .015. The pressure drop is about 64 psig by using Darcy-Weisbach. But if my pressure becomes less and less as you move down the pipe, the velocity will decrease due to friction. So that means the mass flow rate decreases if the density stays relatively constant? Would that mean I have to iterate the pressure drop in sections instead of doing it all at once?"

You can't spec a flow rate And a pressure drop for an existing piping system .

You simply write;
fL/D*V^2/2g=pressue drop
You iterate f to get V.
Only f is not known apriori.

 

[desertfox]

hi zdas04

I can see what your saying with reference to the Bernoulli equation and I agree it didn't have the losses for friction etc in its original form.
However there are numerous references which extend the formula to include losses, the first reference I gave is from the University of Leeds, here is another reference from "edexcel" which is linked to the engineering council and various engineering institutes (see uploaded file).
If this theory is wrong or invalid why are so many institutions educating our engineering fraternity with it.
I suppose the final proof might be to provide an numerical example calculated by your method and one by the extended Bernoulli equation and compare the difference in answers.

 

[zekeman]

"zekeman,
...........Velocity at zero gage certainly can't be stated as a generalized rule for flow within any conduit, be it a rigid pipe or a hose, except for the one particular case where a conduit discharges into a volume that happens to be at reference pressure.

IMO the OP's problem is not that he hasn't considered momentum, as that is ignored in steady state flow. Nothing causes me to assume that the flows in this case are unstead"

1- I didn't know that Bernoulli wrote two equations, one with and one without friction.

2- who said that 0 gauge pressure was a "general" rule?

3- where does The OP accknowledge pressure as a force in the momentum equation.His confusion starts with the statement that the velocity should drop towards zero since the ONLY force stopping the stream is friction

4 your last paragraph-"IMO the OP's problem is not that he hasn't considered momentum, as that is ignored in steady state flow. Nothing causes me to assume that the flows in this case are unstead" -- is totally unintelligible to me.

 

[BigInch]

Velocity does not decrese in noncompressible flow inside a constant pipe diameter. The inlet pressure - outlet pressure supplies any force necessary to move the fluid column. Friction loss is the result of the movement.

Steel bar
sliding on concrete

F1 -----> [======================] <--- F2
<---friction

F1 is a bulldozer₁ pushing on the bar
friction is the result of movement of the bar.
if bulldozer F2 pushes with force = F1-friction, the bar continues at the same velocity to the right.
If F2 is less than F1-friction, the bar accelerates to the right and velocity increases (if the bar is fluid, or we consider air drag, friction increases too, so adjust F2 a little to compensate for that and maintain the faster velocity)
If F2 is greater than F1-friction, the bar acceleration becomes more negative and slows down (if the bar is a fluid column, friction decreases so adjust F2 a little to compensate for that) and maintain the slower velocity.
Now can you imagine that F1 is your inlet pressure x pipe inside area and F2 is your outlet pressure x pipe inside area.

If you want to consider compressible, steady state flow, then then the reduction in pressure due to friction would permit an expansion of fluid and velocity and volumetric flowrate would increase, but in steady state mass flow remains constant at all point, thus you must artificially set the fluid bulk modulus (inverse of compressibility) such that the exact volumetric expansion and resultant decrease in density needed to exactly maintain constant mass flow results from the decrease in pressure caused by friction between the points. In other words the heat capacity of the fluid, heat capacity at constant pressure (CP) to heat capacity at constant volume (CV). It is sometimes also known as the isentropic expansion factor, needs to be just right. If the heat capacity you need to do that job is close to the actual heat capacity of the fluid, you're home free. If there's a lot of difference, you need to reconsider your assumptions of steady state compressible flow, or let things move on to transient flow, use the actual value and recalculate everything.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[ione]

The principle at the basis of the formulation reported by BigInch in his first post, as I can understand it, is that in a the real world it is not possible to transfer the whole hydraulic energy a fluid posses in a point A to a point B of a conduit, without losing part of this energy, just because to the fact that the fluid moves from A to B.
Now I do not intend to candidate myself as zds04’s lawyer (I don’t think he needs one, and this is not a trial), anyway the pillars on which Bernoulli’s equation rely, represent heavy simplifications.
In many cases (i.e. when dealing with liquids) those assumptions which led to Bernoulli’s formulation are acceptable and the equation works well (very well).
There are anyway applications which invalidate Bernouilli’s equation. For those cases (compressibility, unsteady state, rotational) Bernoulli’s equation is misleading, not applicable and leads to wrong conclusions.

 

[BigInch]

Can we just concentrate about doing an energy balance for a fluid system. The ONLY thing that's important about the Bernoulli equation today is the concept. We know the original equation doesn't even work for an elbow. The terms represent the sumation of energy at and between points. If they don't sum up, find the reason it didn't and add another term to account for that. If its friction, add HL, if Haley's commet came too close, include [&Delta;]g, if its plasma flow... We're engineers here, we're supposed to be able to take scientific concepts, adapt them to suit our intended application and go build somthing.

The Bernoulli take-home lesson is not [p/&gamma;]
its [&Sigma;] E.



**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[Latexman]

Didn't I say [&Sigma;]E in post #2?

Good luck,
Latexman

 

[BigInch]

If the thread ended there, I could have lived happy everafter as a clam under the bottom of the sea.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[zdas04]

If you start with the Navier Stokes equations and assume that friction is zero, not constant, but zero, then a whole bunch of the most complex arithmetic falls out. With that mess gone, then you can further assume that density is constant and a bunch more of the equation moves a step closer to a closed-form solution. The result of these assumptions was the very elegant derivation of Bernoulli's Equation. To add friction or head loss back into this equation invalidates step 1 and the equation becomes meaningless. A "long" version of Bernoulli's equation is simply garbage science.

There are a lot of ways to solve the OP's original question. Invalidating the underlying assumptions in one of the few closed-form solutions (as opposed to empirical approximations) to the basic fluid mechanics equations is just sloppy science, sloppy engineering, and sloppy mathematics. If universities teach that garbage then shame on them.

David

 

[BigInch]

I guess you just have to be a pipeline specialist to understand the value, simplicity and use of Bernoulli's concept. Nobody I know uses anything else.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[BigInch]

Its like trying to understand why pump discharge pressure means nothing, but pump discharge head means everything.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/