INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS
Come Join Us!
Are you an Engineering professional? Join EngTips now!
 Talk With Other Members
 Be Notified Of Responses
To Your Posts
 Keyword Search
 OneClick Access To Your
Favorite Forums
 Automated Signatures
On Your Posts
 Best Of All, It's Free!
*EngTips's functionality depends on members receiving email. By joining you are opting in to receive email.
Donate Today!
Do you enjoy these technical forums?
Posting Guidelines
Promoting, selling, recruiting, coursework and thesis posting is forbidden.

flow split in branch pipes from a main pipe

nupin (Chemical) (OP) 
15 Apr 07 9:32 
_______ : : : : : :
I have a main pipe that's divided into 2 branches. All 3 pipes have the same elevation. I have a manometer on each one of the smaller pipes, so I know the pressure there, plus I know the velocity of the fluid on the smaller pipes and on the main pipe. But I'd like to know the pressure on the main pipe just before the fluid splits in two.
I was comparing this system to an electrical circuit with 2 parallel resistors, where the voltage is kept constant throughout the system and the current is divided into the 2 resistors. I don't know if this system could be compared to an electrical circuit because the fluids do not rejoin, they're discharged on a vessel. But anyways, I considered the total energy of the fluid represented by the Bernoulli equation as the voltage on the electrical circuit and I did the following (ignoring friction losses):
(P3/density)+((V3^2)/2g)=(P2/density)+((V2^2)/2g)
3 for the main pipe 2 for the smaller pipe
The only thing I dont know in the equation is P3 so according to what I'm thinking I could calculate it. Am I right? Is it possible to use HardyCross here, since I know the flow on each pipe to figure out the pressure on the main pipe just before the fluid splits in 2?


Hi nupin
Your logic seems fine to me in terms of Bernoulli however should there be another set of figures on to small pipe side or at least the small pipe side should by multiplied by 2 because you have 2 branches.
regards desertfox 

nupin (Chemical) (OP) 
15 Apr 07 10:56 
I see your point desertfox, I think that would make sense if the 3 pipes were in series and they are not in series.
There is a main pipe and there are the other 2 parallel pipes, that is why I was comparing this system to a current (in this case, flow) divider circuit where the voltage (total energy of the fluid) is kept constant throughout the circuit. Can somebody tell me if I'm right? 

Hi nupin
If you have a node in an electrical circuit and you have 3 connections to it then I1+I2+I3=0
Therefore if you have 3 pipes connecting the same applies if you only have one V2 in your equation were is the flow accounted for in the other smaller branch?
regards
desertfox 

nupin (Chemical) (OP) 
15 Apr 07 12:03 
In a current divider electrical circuit with, let's say 2 resistor in parallel, we have 3 different currents:
I1=I2+I3
I1 is the total current I2 is the current that passes through resistor 2 I3 is the current that passes through resistor 3
In this type of electrical circuit, the voltage is kept constant throughout the circuit. Let's say I want to now the voltage on this circuit, I only have to do the following:
V=I1*R1=I2*R2
Notice that the voltage is the same troughout the circuit, so you dont have to sum the voltage throght every resistor because they're the same. I'm comparing the total energy of the fluid given by the Bernoulli Equation (P/density)+((V^2/2g) with the voltage on the elctrical circuit. Where
P pressure on the given pipe V velocity of the fluid
So, what I'm saying is if i know the pressure and the velocity of the fluid on one of the smaller pipes, the volcity and the pressure on the main pipe, I could apply this equation and find the pressure on the main pipe:
(P3/density)+((V3^2)/2g)=(P2/density)+((V2^2)/2g)
3 for the main pipe 2 for the smaller pipe
I think I dont have to sum the energy on the smaller pipes because I'm considering the total enregy of the fluid to be the same on the whole system, since the 2 branches are in parallel just like a current divider circuit where the resistors are in parallel. 

Hi nupin
Well in that case you don't need to calculate the pressure because according to what you have written the pressure is the same as in either of the small pipe branches and you already said you know the pressure there in your first post and your neglecting losses.
regards
desertfox 

25362 (Chemical) 
15 Apr 07 13:07 
Nupin, I believe the units in the quoted Bernoulli equation aren't dimensionally correct. They should be:
• P/density + v^{2}/2 + z×g, gives the components of mechanical energy per unit mass.
• P/(density×g) + v^{2}/2g + z, has the dimensions of length, or height, or head: metre or feet.
• P + (density)v^{2}/2 + z(density)g, has dimensions of force per unit area: Pa, or N/m^{2}, or psi


nupin (Chemical) (OP) 
15 Apr 07 13:27 
Desertfox I really dont think the pressures on all 3 pipes are the same, because the bigger pipe has a bigger diameter than the other 2, so according to the Bernoulli equation, there should be a bigger pressure and a smaller velocity on bigger pipe than on the other 2.
All 33 pipes have the same elevation, so I'm ignoring the z term. Can somebody tell me if I'm right? 

Hi nupin
According to what you have written: V=I1*R1=I2*R2
V=your pressure across your system and you clearly state that I1*R1=I2*R2=V therefore the pressure in pipe 3 must be equal to either the pressure in one of the smaller pipes.
Further a current divider as you rightly pointed out as constant voltage, voltage is seen as pressure in a electrical circuit do you agree or not? You will get a difference in pressure in your system due to losses which you are not accounting for.
Otherwise your theory of a parallel circuit is correct.


nupin (Chemical) (OP) 
15 Apr 07 14:45 
I'm not comparing voltage to pressure, I'm comparing voltage to total energy of the fluid, which is given by the equation (P/density)+((V^2)/2g 

25362 (Chemical) 
15 Apr 07 14:46 
Nupin,
• You didn't refer to the error in dimensions. • To answer your question: since we are assuming an incompressible fluid, no accumulation of mass or energy, and no friction, a massandenergy balance would show that: a. the sum of the flow rates in the two branches equals that in the header, and
b. the sum of the mechanical energies in the two branches equals the mechanical energy in the header.


nupin
sadly voltage cannot be compared to the total energy of a fluid in a system 

nupin (Chemical) (OP) 
15 Apr 07 16:24 
Thanks for your input and the webpage desertfox. 

BigInch (Petroleum) 
15 Apr 07 16:37 
You seem to be forgetting or ignoring pipe friction losses. If frictional losses are very very small, that's OK since the pressure at all points of equal elevation would be effectively equal and you have no problem left to solve. If frictional losses are not small enough to ignore, do it like this, Take the pressure at the manometers on each of the branches. For each branch, take the pressures at the branch manometers and add the head lost to friction (using the Darcy equation for example) in the segment between the manometer back to its junction. If both branches leave from the same junction point, you should get the same pressure for each branch calculation. From that pressure at the junction, add the head lost in the main between its manomater to the junction. Once again you should get the same pressure at the junction as you got for each branch calculation. In other words, the pressure will be the same at the connection of any number of pipes to any given node. You have only to determine a flow loss due to friction between a point where pressure is known to any point where it is unknown. P.S. You don't need to use HardyCross unless you have loops. From the system I think you described, you would have to make 2 virtual loops. Not worth the trouble. BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

hi big inch
The op was neglecting friction losses, I pointed out like you that their was no pressure difference in that case.
regards
desertfox 

nupin (Chemical) (OP) 
15 Apr 07 17:13 
Sounds pretty logical BigInch. That's what I'm going to do, thanks. 

25362 (Chemical)Correctly describes two general equations, one for energy balance and one for mass balance. Consider the upstream stagnation pressues as known, there are then 3 equations with 4 unknowns. (With p2, p3,U2 and U3, representing the downstream "T" conditions. Even if losses are known there would still be 3 equations and 4 unknowns. If one estimates the minor losses for the "T", Then the stagnation pressures on the downstream side of the "T" are known and standard hydraulic calculations may be made.
Regards 

BigInch (Petroleum) 
15 Apr 07 19:04 
Since he knows the pressures at 2 points on the outlets of the pipe branches, and the velocities in all segments, it is a determinate system. No Loops. With velocities he can calculate the flow rates in each segment and the frictional losses in each segment using a frictional loss equation of his choice. From the manometers in the branches he adds the frictional losses going upstream to get the pressure at the branch junction. Both branch calculations should result in the same pressure at the junction. Once he has the pressure at the end of the main line (at the junction point of the branches), he already knows the mainline flow velocity and hence flowrate, so calculation of the pressure at any point anywhere in the mainline (by using the frictional loss equation of his choice) becomes a trivial matter, as is the case for any point along the length of either of the branches as well. If there are fittings or valve, strainers, elbows, reducers, etc. with additional losses of significance, those losses can be estimated from standard tables and included in the head loss calculations of the corresponding segments in which they are contained. As such this is a totally determinate system. No iteration required. BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

BTW Not only is it determinate, its overlydetermined, since you can calculate the pressure at the inlet of the branches using the friction loss formula for EACH branch, then use one value to check the other. BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

BinginchYou are correct, I did not pay attention to the original statement "I have a manometer on each one of the smaller pipes, so I know the pressure there, plus I know the velocity of the fluid on the smaller pipes and on the main pipe. But I'd like to know the pressure on the main pipe just before the fluid splits in two." You are also correct in that the losses for the branch and run may also be determined BUT they are for "only those flow conditions"
Regards


rcooper (Mechanical) 
16 Apr 07 8:35 
Would there be any merit in using the force/momentum equation at the branch to get some handle on the loses through the branch? 

rcooper (Mechanical)Very good question, but how is it to be done?
Regards 

25362 (Chemical) 
16 Apr 07 8:52 
In practice there are energy "losses" in 90^{o} Tjunctions as a result of flow separation.


BigInch (Petroleum) 
16 Apr 07 11:10 
Most standard fittings and valve types have generally well known loss characteristics which can usually be found in the pipe and fitting reference texts. Sailo, Frictional flow losses for pipe with standard flow conditons, as I'm sure you know, are well correlated to a number of equations which are not limited to one flow condition. In fact, having actual data with which to work, as the OP has above, should allow him to make further corrections if necessary and apply the general equations to virtually any hydraulically similar flow condition without suffering significant loss of accuracy. Generally large hydraulic systems are designed relying entirely on standard frictional loss equations, without any benefit of having the tiniest bit of actual data with which to work and the results are normally quite acceptable, so I don't think there will be much problem for the OP to extend the range of his equations to any practical limits he chooses. BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

nupin (Chemical) (OP) 
16 Apr 07 23:01 
Biginch, thanks for the website. You said in a previous post that if I were yo use the Hardy Cross method to solve this problem, I would have to create a virtual or pseudo loop. It was my understanding that you create pseudo loops to join two fixed grade nodes. I dont know if that would apply to my case because even though I know the pressure on the branches, they are supposed to be the same. 

No, you can include elevation differences. I'm not too keen on discussing the HardyCross method where its not needed as in your case, but the typical textbook HC methods do not usually include elevation changes from one node to another, since all they are really concerned about is balancing flows and frictional head losses around the loops (Σ = 0) and the elevation changes are typically known values anyway, they prefer to discuss flat systems of loops subject only to differential heads created by friction, but you can include elevation heads if you want to and you're willing to do the extra math. And they also do not usually include any entry or exiting pipes with flows into or out of the looped circuits either, since those are determinate "free bodies", once the node pressures in the loops and the loop flows have been solved. HardyCross is only usefull for indeterminate system analysis with a looop or with flows across loops, so if you don't have any loop(s) (or if you have loops with flows in all segments are known), you must create some in order to have flows "to balance". What you have described is a "Y" pattern, so in order to make some loops, you would have to create 2 pipe connections, each beginning at an arm at the top of the Y and looping down to join at the bottom. But since you already know the flows at each arm, and also the leg Q0= Q1+Q2, you have no flows that remain for you to balance anyway. HC becomes only a method for distributing any errors you may have that could have resulted from the Darcy based analysis, but the frictional flow losses in HC are also based on similar Darcy equations, so what are you checking? Nothing, just doing the problem twice and ... including the hard way. In any pipe segment, if you have any two of the 3 items, P1 inlet pressure (or head), P2 outlet pressure (or h) or the flow Q, and know the pipe parameters, roughness, diameter, length etc., such that friction loss can be calculated using a Darcy equation, the missing variable is solvable. That's why radials to/from the loops are ignored in the textbooks... the radial analysis is trival after loop analysis has determined the flow and pressure at the connection points to the loops are solved against the problem's boundary conditions, the total inlet and outlet flows =0 and at least one given reference head somewhere in the system. Did I explain? BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

nupin (Chemical) (OP) 
17 Apr 07 21:58 
Hi, I understand what you are saying I was just curious as to how I would solve this if I did not have the data I have.
My system is not actually Y shaped, there is a main pipe that disctributes flow, to, lets say, an n number of pipes, which in turn discharge into a tank:
_______< main pipe : : : : : :
It's more like a several squares. So if I had only the outlet pressures on all the branches, I could use the Hardy Cross method to solve this by creating a number of ficticious pipes that would join the branches, and then get on from there. right? 

Yes, as long as you had enough known variables to sufficiently specify the boundary conditions for the system, 1 reference pressure node somewhere in there and either a known flowrate or a known pressure at any other node that could allow fluid to enter or exit the system. (its assumed all pipe diameters, roughnesses and lengths are "known"). In your example, Let's say your reference pressure is at the "<" point of the main_line, a known inlet flowrate of 5 at the "e" point of the main_line, with branches (outlets?) "A", "B", and "C", with known outflow rates at A=2 and B=3 and C=0, you could tie B to both A and C to make virtual loops. Typically it would go like this, Start HC by assuming flows in branch "A" of 1, "B"=3 and "C" = 1, looking at node "C", you would have an unbalanced flow of 1, which would be distributed around the "BC" loop, and each time you sum the flows in that loop, any unbalance against "C" = 0 would always get pushed over towards "B" and enter the "BA" loop to get distributed there. The same would happen when there was any unbalanced flows at "A", which would also redistribute over to "BC". So you can see that whenever the flow in branch C was not 0, the redistribution to BA would automatically handle that little problem for us, and likewise C to BA, so they continuously fight eachother until the nodes balance. When flowrates at A, B & C finally balance, you would be left with 0 flows in the tieovers B>A and B>C. If you didn't have 0 flows, you stopped the iterations too soon. You know you must have 0 flow in the virtual loops, because they obviously can't transport any fluid if they really don't exist. You also know that if there was 0 flow from the outlet at C, you would have zero flow in branch "C", and a balanced inflow on B = outflow at B must result in ZERO flow in the tieovers BC. Likewise BA. You might also see that if you just happened to make very good guesses, knowing that B>A and B>C really cannot carry any flow and in the end must = 0, and used that knowledge to assume a flow of 2 in branch "A", 3 in branch "B", 0 in branch "C" to start the HC solution method, you would have arrived at the solution to all flowrates in the branches ... before you really ever started working on it. With known flowrates in each pipe and one reference pressure somewhere in the system, you can solve each for the differential pressure between their inlet's and outlets one by one with Darcy equations until all dPs for each pipe are known, then just start at the reference pressure node, chain the pipes together and algebraically add dP to get the pressure at the other end of each pipe. BigInchborn in the trenches. http://virtualpipeline.spaces.msn.com 

25362 (Chemical) 
18 Apr 07 1:56 
The subject of dividing and combining manifolds has been discussed in past threads, and is treated in textbooks on fluid mechanics.


nupin (Chemical) (OP) 
18 Apr 07 19:05 
thanks for the advice BigInch 

It sounds like your system is less than 5 pipelines. PIPEFLO has a free downloadable demo which does fluid calculations of up to 5 pipelines for free. You could use it to easily check your calculations. http://www.engsoftware.com/demo/download/ Full Disclosure: I work for Engineered Software developing this tool. Christy Bermensolo Engineered Software www.engsoftware.com 



