Pressure drop in a pipe (ie) Bernoulli 5

[mechengdude]

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

 

[25362]

BigInch thanks for the star. The subject was discussed with your participation not long ago in thread378-184261.

 

[BigInch]

Ummm. Yes now I can see my toothache painkiller has a few undesireable affects. Thanks for the reminder.

http://virtualpipeline.spaces.msn.com

 

[BRIS]

Upstream momentum exceeds downstream momentum then flow is from upstream to downstream - pressure may increase - water rise along a side spillway etc - .

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.

 

[BRIS]

Sorry - Pressure plus momentum upstram is greater than pressur plusmomentum dowstream (total force causing flow is less than total force resisting flow) then flow is from a lower to a higher pressure.

 

[elpepe2]

Hi, I don't know if the discussion is closed. Nevertheless it's an impressive subject..., after all these years I realized I got no solid basement in Hydraulics!. I was reading again my text books and want to remark something that maybe was wrong stated. Feric posted that velocities should be the same in the main and secondaries branches. I agree with everybody else that velocities in the two branches (downstream) must be a half from the upstream velocity.
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?

 

[BigInch]

Hola El Pepe2,

I think some of us wish it were closed, but here we go again....

Hi, I don't know if the discussion is closed. Nevertheless it's an impressive subject..., after all these years I realized I got no solid basement in Hydraulics!.

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,

Feric posted that velocities should be the same in the main and secondaries branches. I agree with everybody else that velocities in the two branches (downstream) must be a half from the upstream velocity.

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.

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.

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.

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?

Yes it is divided; however ratioed by flow area to each branch.

Let us know if you are still troubled.

http://virtualpipeline.spaces.msn.com

 

[dickon17]

I have just found this interesting thread. Finding out flow rates in different sections of pipe having different lengths and different diameters and fed from the same source is a common hydraulic problem. To read this thread would make you think the solution would be impossibly complex. However, calculating the individual flow rates is a quite simple, although a very tedious, procedure.

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.

 

[BigInch]

In the typical network problem, you're absolutely right dickon17. You would also be able to judicially specify enough boundary conditions to solve most network problem iteratively, using Hardy-Cross or a matrix formulation similar to the Kirchoff formulations used for electrical networks. Where solution of electrical network current flows generally requires only one matrix solution, the hydraulic solution requires an iterative matrix solution, since hydraulic flow is not linearly dependent on hydraulic resistance as is electrical current to electrical resistance.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir2.html#c1

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.

http://virtualpipeline.spaces.msn.com