## flow split in branch pipes from a main pipe

## flow split in branch pipes from a main pipe

(OP)

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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 Hardy-Cross 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?

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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 Hardy-Cross 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?

## RE: flow split in branch pipes from a main pipe

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

## RE: flow split in branch pipes from a main pipe

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?

## RE: flow split in branch pipes from a main pipe

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

## RE: flow split in branch pipes from a main pipe

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.

## RE: flow split in branch pipes from a main pipe

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

## RE: flow split in branch pipes from a main pipe

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## RE: flow split in branch pipes from a main pipe

All 33 pipes have the same elevation, so I'm ignoring the z term. Can somebody tell me if I'm right?

## RE: flow split in branch pipes from a main pipe

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.

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

• 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 mass-and-energy balance would show that:

a. the

sumof the flow rates in the two branches equals that in the header, andb. the

sumof the mechanical energies in the two branches equals the mechanical energy in the header.## RE: flow split in branch pipes from a main pipe

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

## RE: flow split in branch pipes from a main pipe

Take a look at this site it shows 3 parallel pipes coming from a water tank with no losses.

It states the incoming pressure is the same in each leg.

http://www.electronicstheory.com/html/e101-12.htm

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

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 Hardy-Cross unless you have loops. From the system I think you described, you would have to make 2 virtual loops. Not worth the trouble.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

The op was neglecting friction losses, I pointed out like

you that their was no pressure difference in that case.

regards

desertfox

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

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

## RE: flow split in branch pipes from a main pipe

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

http://

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

"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

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

Regards

## RE: flow split in branch pipes from a main pipe

In practice there are energy "losses" in 90

^{o}T-junctions as a result of flow separation.## RE: flow split in branch pipes from a main pipe

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.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

Hardy-Cross 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?

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

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

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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?

## RE: flow split in branch pipes from a main pipe

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 H-C 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 "B-C" 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 "B-A" loop to get distributed there. The same would happen when there was any unbalanced flows at "A", which would also redistribute over to "B-C". So you can see that whenever the flow in branch C was not 0, the redistribution to B-A would automatically handle that little problem for us, and likewise C to B-A, 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 tie-overs 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 tie-overs B-C. Likewise B-A.

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 H-C 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.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

The subject of dividing and combining manifolds has been discussed in past threads, and is treated in textbooks on fluid mechanics.

## RE: flow split in branch pipes from a main pipe

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

## RE: flow split in branch pipes from a main pipe

## RE: flow split in branch pipes from a main pipe

http://www.eng-software.com/demo/download/

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Christy Bermensolo

Engineered Software

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