flow split in branch pipes from a main pipe

[nupin]

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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?

 

[desertfox]

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]

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?

 

[desertfox]

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]

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.

 

[desertfox]

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]

Nupin, I believe the units in the quoted Bernoulli equation aren't dimensionally correct. They should be:

[•] P/density + v²/2 + z[×]g,
gives the components of mechanical energy per unit mass.

[•] P/(density[×]g) + v²/2g + z,
has the dimensions of length, or height, or head: metre or feet.

[•] P + (density)v²/2 + z(density)g,
has dimensions of force per unit area: Pa, or N/m², or psi

 

[nupin]

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?

 

[desertfox]

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]

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]

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 mass-and-energy 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.

 

[desertfox]

nupin

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

 

[desertfox]

nupin

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

 

[nupin]

Thanks for your input and the webpage desertfox.

 

[BigInch]

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 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

 

[desertfox]

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]

Sounds pretty logical BigInch. That's what I'm going to do, thanks.

 

[sailoday28]

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]

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.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[BigInch]

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.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com