Pressure drop in a pipe (ie) Bernoulli
Pressure drop in a pipe (ie) Bernoulli
(OP)
Forgive me if this is a bonehead question...I have someone telling me one thing that I thought knew what they were talking about but now not so sure...Anywho
If you have a pipe with liquid in it (incompressible) and you know the Pressure, temperature, flow rate, velocity etc. It branches into two pipes of equal size then the pressure does not change correct?! This is assuming no loss due to any fittings, pumps, valves etc.
thanks
If you have a pipe with liquid in it (incompressible) and you know the Pressure, temperature, flow rate, velocity etc. It branches into two pipes of equal size then the pressure does not change correct?! This is assuming no loss due to any fittings, pumps, valves etc.
thanks





RE: Pressure drop in a pipe (ie) Bernoulli
If the cross-sectional area is changed, doesn't the flow have to change? Wouldn't that be reflected in the pressure drop?
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RE: Pressure drop in a pipe (ie) Bernoulli
RE: Pressure drop in a pipe (ie) Bernoulli
RE: Pressure drop in a pipe (ie) Bernoulli
anyone else care to weigh in...
thanks
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
The velocity of the flow in the branch will be half the velocity upstream, so the kinetic energy upstream of the branch will be 4 times the kinetic energy downstream. Now, this energy has to go somewhere. mechengdude, stated there is no loss due to the fitting, so the energy has to be retained in the fluid, and it is retained by increasing the internal energy. This manifests itself as an increase in pressure, as the fluid is incompressible.
Therefore the pressure in the two branches is greater than the pressure in the common pipe upstream of the branch.
This pressure rise falls out of the Bernoulli Equation as (v1^2 - v2^2)*density/2
Where v1 is the uptream velocity and v2 is the downstream velocity.
RE: Pressure drop in a pipe (ie) Bernoulli
The upstream pipe has a pressure of 100 psig it splits in two ... the pressure in the downstream pipes is NOT 50 psig.
same assumptions as stated earlier apply... thanks
RE: Pressure drop in a pipe (ie) Bernoulli
What you have to do is pick some known distance downstream of the tee to base your calculations on. At the instant that the flow breaks, both streams are at 100 psig. If you assume equal flow rates (which never really happen by the way) then using a real fluid flow equation like Hazen Williams you can back into the downstream pressure from the known flow rate, known length, and the starting pressure.
David
RE: Pressure drop in a pipe (ie) Bernoulli
I have a suggestion for you.
Instead of asking all kinds of questions and covering what if cases, do a search over the Internet for free pipeline network software.
I am convinced that there are some free demos that you can download.
Once you get your hands on some free demo software, create a sample case that you would like to model and see what the software provides you with as output results for your input data set.
Keep changing the input data until you get all the answers to your questions.
Even eFunda.com -- http://www.efunda.com -- has a simple pipeline online calculator where in no time you can get some feedback in terms of the output data.
In my opinion, this is the best way to learn and get either right or wrong your answers and assumptions.
When I was a student, there was no software and no computer tools as they are available today. Most of the calculations were done by hand ...
Good luck!
Thanks,
Gordan Feric, PE
Engineering Software
http://members.aol.com/engware
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
Feric - not sure what you mean by all sorts of questions. I thought it was pretty specific as to what occurs when one pipe branches into two; relative to pressure.
I have a CFD program but I was less interested in a specific value which I could get by modeling the case and more interested in trying to visualize and understand the relationship and equations.
RE: Pressure drop in a pipe (ie) Bernoulli
School of engineering says that you need to put on a piece of paper:
- what you would like to model
- write down the basic governing laws, equations and assumptions
- for given set of input data define and determine the system steadt state conditions
- do basic steady state modeling and simulation
- plot the system parameters
- make observations and draw conclusions
- share the findings with people and engineers who know the subject area
In my opinion, there is no better way than this when trying to visulize an engineering problem.
That is what I meant.
Believe me, one thing is solve problems in your head, but in order to master the subject area you need to do what the school of engineering says. It is time consuming, requires effort, but if you do what you need to, you will get the feel for the subject area and become an expert and get ready to help others who are getting familiar with the subject area.
I am not trying to be smart, just trying to explain what works and what school of engineering says how it should be done ...
Thanks,
Gordan Feric, PE
Engineering Software
http://members.aol.com/engware
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
I have a suggestion.
Can mechengdude provide us with some kind of schematic layout so everybody can be on the same page?
This should not be difficult to do, for example, MS PowerPoint can be used ...
Thanks,
Gordan Feric, PE
Engineering Software
http://members.aol.com/engware
RE: Pressure drop in a pipe (ie) Bernoulli
Bernoulli
z1 + P1/γ1 + v1^2/2g -HL = z2 + P2/γ2 + v2^2/2g
incompressibility => γ1 = γ2
short branch fitting => z1 = z2 and => frictional Hl ~= 0
outlet areas same as inlet area =>
Voutlet velocity = 1/2 inlet velocity
z1 + P1/γ1 + (V)^2/2g = z1 + P2/γ1 + (1/2 V)^2/2g
P1/γ1 + (V)^2/2g = P2/γ1 + (1/2 V)^2/2g
for convenience γ=1
P1 + (V)^2/2g = P2 + (1/2 V)^2/2g
P1 - P2 = 1/4 V/2g - 1 V/2g
P1 - P2 = -3/4 V/2g
P2 = P1 + 3/4 V/2g
P2 > P1 => flow is reversed
====================== 2nd ==============================
Upstream Pressure = 100 psi
Downstream pressure P2 = 100/γ1 + 3/4 V1/2g
P2 > P1 => Flow is Reversed
Since velocities are known, as per the OP, you can solve for P2.
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RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
Please correct me if I am wrong, but why do the flows reverse? The question is based on a system in which there are no losses. The total energy through the system stays constant so there is not need to reverse the flow. Fluids can and do flow from a low pressure region to a high pressure region so long as the TOTAL ENERGY reduces. Given a very low friction environment, the water flowing through a canal into a lake has a lower top water level than the water in a lake. The water moves up hill as the kinetic energy is converted to potential energy.
The equations presented are based on a system with the fluid flowing in a particular direction. If the flow reverses then a new set of equations would have to be set up to represent this alternative situation.
RE: Pressure drop in a pipe (ie) Bernoulli
I don't know of any flow that goes from a low pressure point to high pressure point. That's one of the few times I can use "ALWAYS" flow is from high pressure point to low pressure point.
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RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
Your example, "Given a very low friction environment, the water flowing through a canal into a lake has a lower top water level than the water in a lake. The water moves up hill as the kinetic energy is converted to potential energy."
.. . is not correct.
If such a thing occurred (and it can and does), it would be due to the velocity head being converted to static head, which technically is known as a hydraulic jump. If an open channel led into a reservoir, one of three things can happen,
1. the velocity of the channel is subcritical and the liquid level from channel to reservoir would fall,
2. the velocity of the channel is at critical velocity, in which case there would be no change in liquid level from the channel to the lake
3. the velocity of the channel is supercritical, in which case the level of the liquid would increase going from channel to reservoir
In ALL CASES the TOTAL ENERGY UPSTREAM OF ANY FLOW is greater than TOTAL ENERGY DOWNSTREAM OF ANY FLOW.. OTHERWISE THERE CAN BE NO FLOW!
Bernoulli describes total energy Z potential due to gravity, P potential due to pressure, V kinetic energy, HL friction. If you don't want to consider friction for some reason, make HL zero as I did above.
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RE: Pressure drop in a pipe (ie) Bernoulli
An analogous situation is that in a vertical column of water the pressure is lower at the top than at the bottom, but the water has no problem in flowing from the zone of low pressure to that of high pressure. The potential energy term in Bernoulli takes care of this effect.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Pressure drop in a pipe (ie) Bernoulli
Niagra falls, pressure is 0 at the top and the Lake Eire level and at Lake Ontario. Agreed, but IMO trivial and irrelavent to the OP.
Katmar. Not a good analogy. There is no flow in a barrel of water without some component of total energy being converted to drive it, yet you don't explain what it might be.
In every case of the OP branch problem, pressure increases in the branch outlets and flow is reversed OR there is NO flow, hence no pressure change, assuming potential z1 = z2.
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RE: Pressure drop in a pipe (ie) Bernoulli
Let me try to go back to the very beginning of the problem.
If I got it right, initially we do know inlet pressure, temperature, cross sectional area of the pipeline and velocity. Friction losses can be ignored.
Therefore, the volumetric flow rate is:
Q = v1*A1
At all the time, at any point of the pipeline, the volumetric flow rate is constant -- Q = constant.
The pipeline cross sectional area is fully filled with the flow resulting in A = A1 = A1wetted.
At a certain point, the pipeline splits in two branches and the cross sectional area stays the same.
A = A2 and A = A3 and A2 = A3
If I am not mistaken, v1 = v2 and v2 = v3 and v2 = v3
As much volumetric flow rate goes in, as much volumetric flow rate goes out ...
The continuity equation needs to hold resulting in
Q = v1*A1 = v2*A2wetted + v3*A3wetted
Assuming that A2wetted = A3wetted = Awetted
v1*A1 = v1*Awetted + v1*Awetted
v1*A1 = 2*v1*Awetted
Awetted = A1/2 = A2/2 = A3/2
In my opinion this is what is happening, the wetted cross sectional area in branches is half size the cross sectional area of the pipeline.
In general, there is no pressure los due to friction, and both continuity and Bernoulli equations are satisfied.
The flow is from the left to the right ...
Once we can agree that v1 = v2 = v3, then everybody is right.
I just do not know how it turned out to be that v2 = v3 = v1/2 ... -- this is against physics and common sense
I would appreciate if somebody can check my understanding of the problem and my outcome ...
Thanks,
Gordan Feric, PE
Engineering Software
http://members.aol.com/engware
RE: Pressure drop in a pipe (ie) Bernoulli
It is a common error to assume that an incompressible fluid fluid moves in a horizontal pipe (without friction) only from higher to lower pressure heads.
It does so from higher to lower "total" heads. This is what the Bernoulli equation tells us.
It is quite common to read "higher gage pressures" downstream when a pipe diameter is enlarged, as a result of a drop in "velocity head".
RE: Pressure drop in a pipe (ie) Bernoulli
I think you've got it very well indeed.
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25362
I don't think its a common "error"; its simply the typical beginning of the construction of a common generic pipe flow problem. In a single horizontal pipe, most people's initial thought of the problem is, "OK I'll assume constant diameter until somebody says its not, as well as, "flowing full" until somebody says its not, as well as no holes in the pipe wall between inlet and outlet, no partially open valves, the outlet end of the pipe is open, etc. Most people know that water will flow in a vertical pipe, if you put it in the top and the bottom of the pipe is not closed, so I think a lot of time was wasted above discussing things that (I thought) were extraneous to the OP.
IMO, in solving for velocities in the tee of the OP, its pretty much a waste of time to discuss the zero velocity solution, as well as ask for all the confirming information to arrive at what should be the more or less direct solution of interest, immediately discounting all the possible trivial variations.
---------------------------------------------
mechengdude,
Can you confirm that the tee is flowing full in all branches?
Can you confirm that the inlet is at the same elevation as the outlets?
Can you confirm that the velocities are nonzero?
Can you confirm that the inlets and outlets are all fully open?
We know that the fluid is incompressible and there is no friction, but can you confirm that the fluid does not expand?
Since you know all the velocities, why not list them (or just one of them) and I'll fill in the rest.
.......
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RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
- no change in elevation
- no change in pipe diameter
- all liquid flow, pipe full
- no losses due to friction
Out of curiosity, I'm sticking a quick model into COSMOSFloWorks to see what really happens and will post the results.
thanks
RE: Pressure drop in a pipe (ie) Bernoulli
Guys, you are good -- your credentials indicate so.
This is a good engineering discussion.
It is my pleasure to be able to make my contribution and be a part of it.
Engineering should be practiced and discussed like this.
As it has already been pointed out, it is tricky to quickly figure out what is in and what is out of the problem.
I am looking forward to having more discussions and analysis with you guys.
Thanks,
Gordan Feric, PE
Engineering Software
http://members.aol.com/engware
RE: Pressure drop in a pipe (ie) Bernoulli
Voltage is analogous to pressure (Pa)
Current is analogous to flow rate (m3/s)
Current is a measure of the flow of electrons i.e. a flow rate.
Voltage is what pushes the electrons through the wires. Pressure pushes the fluid through the pipe.
So ask your electrical engineering colleague if the voltage at a junction varies just in the branches assuming there is low resistance in the conductors.
RE: Pressure drop in a pipe (ie) Bernoulli
If anyone has a easy way to post pics here let me know.
RE: Pressure drop in a pipe (ie) Bernoulli
We can solve that and see what happend with pressure, velocity and maybe other things in our flow.
I know that there are other effects here but we have almost neglected all of them.
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
If "downstream" pressure is higher than upstream pressure, what usually happens?
I don't know of any flow that goes from a low pressure point to high pressure point. That's one of the few times I can use "ALWAYS" flow is from high pressure point to low pressure point.
Consider flow into a venturi?
What happens in a venturi, from the throat to the outlet?
Does the flow reverse?
RE: Pressure drop in a pipe (ie) Bernoulli
Not sure I understand your analogy. Pressure on the "downstream" side of a venturi is the delta-P between "upstream" pressure and the pressure at the venturi. That would mean that pressure on the "downstream" side would be less, if it wasn't it would be the upstream side.
Flow would only reverse if the pressures were reversed. The point holds true, flow will always go from high pressure to low pressure.
Greg Lamberson, BS, MBA
Consultant - Upstream Energy
Website: www.oil-gas-consulting.com
RE: Pressure drop in a pipe (ie) Bernoulli
As I should have remembered, in a frictionless environment, Bernoulli says nothing at all about flow. In fact in the Bernoulli equation, it is only the friction term that has anything at all to say about flow, so if friction is neglected, there's not much about flow that anybody can infer at all. Without friction, its just an energy balance equation.
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RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
A constant area short dividing manifold could be another interesting example. In some geometries as flow is bled off the (static) pressure raises towards the end of the manifold resulting in a larger flow rate out of the end branches.
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
BigInch thanks for the star. The subject was discussed with your participation not long ago in thread378-184261: flow split in branch pipes from a main pipe.
RE: Pressure drop in a pipe (ie) Bernoulli
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RE: Pressure drop in a pipe (ie) Bernoulli
Bernoulli's equation gives the solution except in reality the velocity difference between upstream and downstream is lost. i.e it is practically impossible to design a fitting to recover velocity (kinetic energy) and convert it to pressure (potential energy) so in reality the pressure downstream will be less than upstream because the velocity head is not recovered.
RE: Pressure drop in a pipe (ie) Bernoulli
RE: Pressure drop in a pipe (ie) Bernoulli
Thinking aloud: volumetric rate in the upstream section must be equally divided between the two downstream branches. They can not be equal to the upstream branch because there's no additional source of fluid volume except the upstream branch itself. Am I right?
As the volume rate is a half of the upstream branch, and the section is equal to the last, velocity in each downstream branch must be the half of the upstream branch.
But I cannot get the clue about the mechanical energy involved yet. Think about having a enlargement to the double of section of the same upstream pipe. All the equations and assumtions are the same but one can clearly make a direct correspondance about energy at a point in the upstream part, and another point in the enlarged part. The energy must be conserved. But what happens in the two branch problem?, the energy of the upstream section should be divided between the two downstream branches? Maybe I am lost enough that I missed some basic notion. What do you think about it?
RE: Pressure drop in a pipe (ie) Bernoulli
I think some of us wish it were closed, but here we go again....
Well I don't think there is much solid about hydraulics and fluid mechanics
Jokes aside, and after putting in several example problems in my superduper BigInch simulator, I (think I) can say the following,
That is only true, if there are 2 downstream branches of equal cross-sectional flow area.
If pressures remain the same upstream, inside and downstream of the junction, no compressibility and there is no friction, all as specified in the original post (that means total head upstream = total head downstream, which also given that there is no elevation difference across the junction, it follows that velocity change must account for any differences between the energy from the upstream side to all downstream sides of the junction), flowrate (and mass rate in an incompressible system) must proportion itself according to area ratio and thus also by velocity in each branch, since V = Q/A. If it did not, the fluid would have to be compressible/expandable. Going back to basic physics, it should also be such that ingoing momentum must = outgoing momentum (sum of components in relavent coord directions assuming axial and shear forces in all pipe are balanced... nothing is moving around space), would imply that downstream flowrate in each branch must be proportional to the branch flow area ratio to the total downstream area of all pipes. What other mechanism is there to proportion flow otherwise? None.
I think you mean they must be equal to the upstream branch. So anyway, Volumetric rate in must be equal volumetric rate out. Are you right? Well, flow is divided, however NOT necessarily equally between the two downstream branches. As I said above, that division must be according to area ratio of branch area to total downstream area, so it would only be equal, if the downstream branches had equal areas.
Yes it is divided; however ratioed by flow area to each branch.
Let us know if you are still troubled.
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RE: Pressure drop in a pipe (ie) Bernoulli
If the system is fed by a positive displacement pump, then the total flow rate is pretty much fixed to begin with and all you have to do is find out how this specific flow is divided into the different outlets. At each point where the flow divides you need to make an educated guess about how much flow will go which way. When you have finished guessing how the total flow rate will probably be distributed you need to work backwards from each outlet and calculate the pressure at each point where the flow divides. These pressures should be easy to calculate because you will have all the data you need (flow, distance, pipe size etc). The pressure drops working backwards must be the same for each leg. If they are not you make some small correction as to how the flow is divided and do the pressure drop calculations again. You carry on this way for each point where the flow divides until you arrive back at the pump outlet. The total flow for each outlet must equal the pump flow and the pressures at each branch must be correct for the flow down that leg.
If the pump is a centrifugal, there is the added complication of matching the flow/pressure combination to the pump performance curve.
This kind of iterative calculation is tailor made for a computer program. There has to be something out there available for download.
RE: Pressure drop in a pipe (ie) Bernoulli
http://
In the branch problem of the OP no iteration was necessary, because the pressures are given upstream and downstream of the junction as being equal (OP) and friction does not exist, so no head loss anywhere. With equal pressures at each outlet, the flow has to be proportioned by area.
Yes download EPANET (developed by US EPA specifically for water pipeline network analysis). It won't do transients, but does do a lot of things, including this and ... its free.
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