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Velocity and Pressure Drop
8

Velocity and Pressure Drop

Velocity and Pressure Drop

(OP)
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!

RE: Velocity and Pressure Drop

Quote:

Bernoulli's equation says that a lower pressure (gauge pressure after 100 ft of piping) will correspond to a higher velocity.

Not really, there are other factors involved.  Bernoulli's equation says the change in potential energy (pressure and elevation) plus the change in kinetic energy (velocity) minus frictional losses equals zero (conservation of energy).

Good luck,
Latexman

RE: Velocity and Pressure Drop

(OP)
Such that an included head loss term in Bernoulli will be able to show me a velocity decrease with the pressure decrease?  

RE: Velocity and Pressure Drop

Liquid velocity is essentially constant in a pipe of constant cross-section. If this is school work you should ask your teacher. School work is not allowed here.

RE: Velocity and Pressure Drop

For an incompressible fluid, and no leakage along the way, flow in must equal flow out.  If areas are equal, velocities will be equal.  Q = V * A = constant.

Ted

RE: Velocity and Pressure Drop

2
Damn guys, doesn't anyone remember fluid mechanics?  mass flow rate must be constant for a flow with no additions or removals within the control volume.  To achieve that with a real fluid (who's density is a function of pressure and temperature), velocity must be ever changing.  If velocity keeps changing, then volume flow rate must also keep changing.  There is nothing in fluid mechanics that requires velocity to be constant and it isn't.  The only way for a fluid to move in a pipe is if the pressure on one end is lower than the pressure on the other end.  For a fluid like water over a short distance, this change in density and velocity can be very small, but it can't be zero.  The OP contention that velocity can remain constant with a pump is simply nonsense.

With regard to Bernoulli's equation.  His first assumption as he developed it from Euler's equation is that the flow is incompressible and inviscid.  That means zero change in density and zero friction drop.  Consequently it is only valid for very short distances.  Typically it is used across the length of an airplane wing, or to determine the pressure change across a diameter change.  It is not applicable to try to determine the pressure drop DUE TO FRICTION down 100 ft of pipe.

David

RE: Velocity and Pressure Drop

Gee David, I am sitting here laughing my head off.  After first reading premise put forward by the OP, if you read your post too fast it would almost sound like you are saying that with a long enough pipe line you would be pumping in at one at a given velocity and having none at the other end when in fact the opposite it true.

At least that is what the OP made it sound like was going to happen when referring to 0 velocity.  Composite Pro said "essentially" so he covered himself with respect to our post.

But, the OP also mentioned reducers, and once you change the cross sectional area of the flow path, with respect to velocity considerations everything changes.  Most of the subsequent posts seemed to assume constant flow area.

If the OP is a student, he/she needs to read the chapter before trying to work the problem.  If the OP is a practicing engineer, he/she needs to dig the old fluids text book out and re-read the applicable chapter.  If the OP is not an engineer, he/she needs to find one to explain it to him/her.

Bernoulli needs to roll back over in his grave because the laws of physics didn't change after all.

But, good job of nailing the essence of the matter.

rmw

RE: Velocity and Pressure Drop

ok, the OP's question and possible class room question, must be a compressible flow or some telescopic pipe, The annswer could be a .25 in pipe at first and a 200 mile diameter pipe at the end.

RE: Velocity and Pressure Drop

2
You have to use the "long" version of Bernoulli's equation, one that includes the head loss due to friction term.  
Z1 + P1/γ1 + V1^2/2/g - HL = Z2 + P2/γ2 + V2^2/2/g

HL, head loss between points 1 and 2, can be found using any pipe head loss equation for frictional flow, Fanning, Darcy, Colebrook, Churchill, etc.  Use one that is suitable for your fluid and flow conditions.  Now the Bernoulli equation applies to any length of pipe, 1 foot or 1000 miles.

The common friction equations assume constant fluid density and viscosity but if your density changes between point 1 and point 2 (its a compressible fluid), perhaps you can approximate by using the average density in a short segment.  If you have a long piece of pipeline and density changes significantly, you will have to break that piece of pipeline into shorter segments.  Same for viscosity; use the average viscosity, if it changes significantly in any segment of pipe.   If the pipe diameter changes, that's a great place to break the pipe into another segment.   

Assume an inlet pressure and a flowrate.  Calculate the velocity in each segment of pipe and in each fitting.  Now  calculate the head loss for each segment of pipe using whatever friction equation you have chosen, add the head losses from the fittings using a table of equivalent lengths for fittings, or by using fitting "K" factors.  The sum of all those individual head losses is HL.  Calculate the outlet pressure.  If the outlet pressure is less than zero (absolute), increase the inlet pressure, or reduce the flowrate, or make some combination of those two and recalculate again.    

If you keep track of any fluid density changes, the velocities will change between points 1 and 2 and everywhere else, even if you have a constant diameter pipe.  It is much easier if you approximate a constant density in each segment and in each fitting by using the average density in each segment.

If you have fluid streams entering or leaving the pipeline, then simply recalculate the velocity in each segment to account for the increase or decrease in mass flow according to how much fluid has entered or left the pipeline.

If the temperature changes along the pipeline; no problem.  you may need to recalculate the density and perhaps viscosity too.  If there is a big difference in density, recalculate the velocity.  You may need to keep track of viscosity changes too.  Calculate viscosity at each point and if there is a large difference, then just as always, try to use the average.  If at any place using the average value of any of those variables is not accurate enough, make the pipe segment length shorter, recalculate the average value for the variable giving problems, then use a new average value of the shorter pipe segment and recalculate.

OK.  I think Mssr. Bernoulli can now go back to sleep ...  

As you can see, it is much easier if you can assume constant diameter, density, viscosity, velocity, mass flow.  You can even assume no friction, if your pipeline is very short, but if you use the proper form of the Bernoulli equation, you can accomodate any changes you may need to consider for your particular conditions.  That increases the complexity of the calculations quite a lot, so if you're going to do a lot of these calcs, do a simple one by hand to see how it works, then buy some pipe hydraulic analysis program.  Have a look at www.aft.com.
  

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/

RE: Velocity and Pressure Drop

BI - a star for someone that knows fluid mechanics.

Good luck,
Latexman

RE: Velocity and Pressure Drop

"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?"


Bernoulli equation does not apply here;it is the frictionless form of the adiabatic energy equation. And BTW, you can have zero gauge with velocity; happens every time you use a hose.

Your problem in analyzing friction is you haven't considered the effect of pressure in the momentum equation.

Consider this example for clarification:
If you take a section of straight  pipe say 1 foot long and invoke conservation of momentum you get  that the differential pressure drop times the area should be equal to friction force + the mass change of momentum. Since the mass change of momentum for this section is zero (constant velocity) you get that the pressure drop= friction force*area. Your model forgot to include pressure.

 

RE: Velocity and Pressure Drop

zekeman,

I explained how the Bernoulli equation can be applied to cases where friction is significant by including HL.  Please don't say that "Bernoulli's equation doesn't apply", even if you only mean that in its basic form.  That conclusion is already obvious and further discussion of unapplicability only serves to confuse the issue.

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 unsteady.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/

RE: Velocity and Pressure Drop

BigInch,
In spite of Latexman's unqualified support of the equation you presented, zekeman is right.  There is no friction or HL term in the Bernoulli Equation.  Sorry, but a "long" version of the equation that you presented is some sort of after-the-fact construct that ignores the fact that in the derivation of this useful equation from Navier-Stokes (through the Euler simplification) discards the friction terms and converts the derivatives of density with regard to time and space to a constant.  If friction is not close enough to zero to be safely ignored then the derivation is invalid and all airplanes could be expected to fall from the sky.

The zero psig point on the pressure continuum does not have a physical significance in most fluid mechanics, and certainly not in the Bernoulli Equation.  With the exception of some poorly constructed empirical equations, arithmetic with pressure must always use absolute pressure.

The OP's "problem" is a nearly total lack of understanding of the subject.

David

RE: Velocity and Pressure Drop

(OP)
Thanks everyone - you have all been very helpful :)

RE: Velocity and Pressure Drop

I agree with zds04.
Bernouilli's equation comes from Navier-Stokes equations with the hypothesis of dealing with an ideal fluid, where ideal means:

1)    Incompressible
2)    Inviscid
3)    Steady state (this means partial derivative with respect to time equal  to zero and flow-in equal flow-out)

Bernouilli's equation is a particular case of the most general mechanical energy balance, where the resistance to the fluid motion due to its viscosity is equal to zero. Entering the head loss in the Bernouilli's equation accounts for the effect of the fluid viscosity, but assuming "steady state hypothesis" stands still, implies heavy simplifications of the problem.
 

RE: Velocity and Pressure Drop

desertfox,
One of the long-time posters on eng-tips.com used to have a signature line that said something like "anything your read on the Internet is suspect until proved otherwise".  I looked at your last link and I have to say that "wayne.edu" got it wrong and could use a bit more "edu" himself.  His original statement of Bernoulli's equation was lazy and incomplete.  Saying you could extend it to viscous flow is just wrong.

There are many empirical equations in the world that describe real flows pretty effectively.  The one I use most often for gas is called the "AGA Equation", but there are others as comprehensive.  My point is that every flow equation does not have to be labeled "Bernoulli".  Daniel Bernoulli did an amazing piece of work deriving the equation that bears his name, but he never claimed it was a closed-form solution to the Navier Stokes Equation, and it isn't.  It has MANY places where it does a good job of describing fluid flows and it is invaluable in those places.  Pipeline flow is not one of those places.  Adding a friction (or head loss) term to his elegant equation and calling it meaningful is just wrong.
 

David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips Fora.

"It is always a poor idea to ask your Bridge Club for medical advice or a collection of geek engineers for legal advice"

RE: Velocity and Pressure Drop

David,

I admire your devotion to keeping the Bernoulli equation pure and unadulterated and in it's original form with it's classical assumptions and boundary conditions.  However there have been many engineers for many years that have been taught the principles of the Bernoulli equation mainly using a "mechanical energy balance" between two points in incompressible pipe flow that includes a frictional head loss term.  I know the textbook for my first fluid mechanics course, McCabe and Smith's Unit Operatioins of Chemical Engineering, used it.  Crane Technical Paper 410 uses this mechanical energy balance with a frictional head loss term too.  Crane TP410 does not call it the Bernoulli equation, but do say it was derived from the Bernoulli equation.  In fact TP410 says, "All practical formulas for the flow of fluids are derived from Bernoulli's theorem, with modifications to account for losses due to friction."  I don't have access to my McCabe and Smith now, so I can't say what they called it.  I'll look later.  Given all this, it is not surprising to me that a lot of folks identify this mechanical energy balance by the label "Bernoulli equation".  This mis-labelling is wrong, and I agree with you there.  Having a mechanical energy balance with a frictional head loss term for incompressible pipe flow is fine though.  It's worked for me for many years.  You know this could be mainly a Petroleum and Chemical Engineering thing.  After all, Petroleum and Chemical Engineering were originally together (over time they became separate disciplines), and McCabe and Smith has probably been used in both curriculums for well over 40 years (my textbook was the 3rd Edition in 1977).  Yep, I gotta check my McCabe and Smith when I get back to the office.

Good luck,
Latexman

RE: Velocity and Pressure Drop

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

RE: Velocity and Pressure Drop

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.
 

RE: Velocity and Pressure Drop

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.

RE: Velocity and Pressure Drop

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

RE: Velocity and Pressure Drop

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

RE: Velocity and Pressure Drop

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/

RE: Velocity and Pressure Drop

"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.

RE: Velocity and Pressure Drop

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.

RE: Velocity and Pressure Drop

"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.

RE: Velocity and Pressure Drop

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/

RE: Velocity and Pressure Drop

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.
 

RE: Velocity and Pressure Drop

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 Δg, 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 Σ E.   

 

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
http://virtualpipeline.spaces.live.com/

RE: Velocity and Pressure Drop

Didn't I say ΣE in post #2?

Good luck,
Latexman

RE: Velocity and Pressure Drop

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/

RE: Velocity and Pressure Drop

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

RE: Velocity and Pressure Drop

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/

RE: Velocity and Pressure Drop

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/

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