bernoulli equation for split flow, fluid divided into 2 different branches

[caroline_d]

I have another question about the bernoulli equation. Let's says that I have a 200 gallons per minute flow on a 5 inch pipe and at 26 psia then the flow is divided equally into 2 pipes (each with a 2 inch diameter). And I wanna know what is the pressure on those 2 pipes. We can ignore any friction losses. All 3 pipes have the same elevation. I want to know if my bernoulli equation is correct

(P1/density)+((V1^2)/2G)=(P2/density)+((V2^2)/2G)+(P3/density)+((V3^2)/2G)

Where
P1 is pressure on the main pipe (the 5 inch pipe)
V1 is the velocity on the main pipe
P2 pressure on one of the 2 smaller pipes
V2 is the velocity on on the smaller pipes
P3 pressure on one of the 2 smaller pipes
V3 is the velocity on on the smaller pipes

Those pressure, flow and velocity numbers are made up. What I really wanna know is if the bernoulli equation above is correct fore this system, where a pipe with a given diameter is divided into 2 other pipes with a smaller diameter. And the flow is split in half between the 2 pipes.

Can someone tell me if I'm right?

[mr_homm]

Unfortunately, that is not correct. In fact, Bernoulli's equation is NOT sufficient to determine how the flow splits into the two pipes, because the flow in each pipe also depends on conditions downstream of the split. For instance, if one pipe is partially clogged later on, you will see less flow into it here at the junction.

Here is an analogy which may be useful in thinking about these problems (assuming you are familiar with electric circuits). The Bernoulli formula B gives the energy per unit mass carried by the flow, just as the voltage gives the energy per unit charge carried by electric current. The mass flow rate m' acts just like electric current I, in that both mass and charge are conserved. Power carried by the flow is B*m' and power carried by an electric current is V*I. Current splits when two wires are parallel, and mass flow splits when two pipes are parallel. The voltage drop across two parallel wires is equal, and the drop of B (head loss) across two parallel pipes (i.e. they split and then rejoin later) is also equal. Voltage changes across two resistors or batteries in series sum up, but the currents through them are equal, and head changes across two clogs or pumps in series sum up, but the mass flows through them are equal.

The analogy is quite exact, and all the same kind of reasoning can be applied. In the case of your branching pipes the analogy would say that since the voltages of three wires that are joined at a point are all equal, then B in the three pipes must all be equal at the junction (though they may change later due to clogs or pumps). This means that you DO NOT ADD the B values for the two smaller pipes. Instead, they are EACH equal to the B of the larger pipe. Therefore, your equation should read:

(P1/density)+((V1^2)/2G)=(P2/density)+((V2^2)/2G)=(P3/density)+((V3^2)/2G).

You will also be able to apply the equation of continuity, m'1 = m'2 +m'3, which in terms of velocity is v1*A1*density = v2*A2*density + v3*A3*density, where A is area.

There are 4 unknowns (two velocities and two pressures) but only 3 equations, which is why more information about conditions downstream is needed, just as with electric circuits.

By the way, many people make the mistake of thinking the pressure must be equal in all 3 pipes. This is not true. B must be equal in all 3 pipes, because it represents the energy per unit mass carried by the flow. If B were higher in one exit pipe than in the other, then it would also be higher than B in the inlet pipe (since the average value of B in the inlet and exit flows must be equal by conservation of energy). But that would mean that you could concentrate energy into one flow without doing any mechanical work. If that were true, you could build a system that would violate the second law of thermodynamics, by using heat o drive the flow in the first pipe, then concentrating the energy into one flow, and then converting it back to heat. Since there would be more energy per unit mass, the final temperature would be higher than the initial temperature of the heat source that was driving the flow, so you would have made heat flow from a cold temperature to a hotter one with no applied work. If that were possible, I would already have patented it.

Hope this helps!

--Stuart Anderson

[caroline_d]

Thank you so very much for your input Stuart, it has helped me a WHOLE LOT, thank you!!!

[caroline_d]

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Mr. Stuart Anderson or anybody that is willing to help me, let me make sure I got this right. I must say this system could not be considered as a closed loop, because the 2 flows come out of the branches and then are discharged on the same vessel, so the flows do not rejoin again. I dont know if this vessel could be considered as a point of ''reunion'' so to speak.

After stating that, which I dont know if it is important or not for this problem, I would like to explain to you what I think is correct for resolving this system. What I want to do is find the pressure on the main pipe (the one that's red, which has a bigger diameter than the other 2). I know the pressure and the flow on the 2 other smaller pipes. You told me that I could compare my problem to an electrical circuit. The circuit that is similar to this one is the one where resistances (pipes) are in parallel and the current (flow) is divided. In these types of circuits the voltage (total energy of the flow) is kept constant throughout the circuit and the resistances. Comparing that to the pipes and applying the Bernoulli equation to express the total energy of the fluid we have (ignoring friction losses):

(P2/density)+((V2^2)/2g)=(P3/density)+((V3^2)/2g)

2 is for the smaller pipe
3 is for the main pipe

If I know the pressure on one of the 2 smaller pipes (lets says P2), the velocity on pipe 2 (V2) and the velocity on the main pipe (V3), then I could find the pressure on the main pipe. Am I right?

[mr_homm]

Yes, that is exactly right. Even if the flows never rejoin, the "voltage" must still be the same in all 3 branches of the pipe. You would need to know the values of enough variables to solve for the others. Basically, you have continuity m1' + m2' = m3' (1 equation), Bernoulli B1 = B2 = B3 (2 equations), and the flow relations m1' = rho*v1*A1, m2' = rho*v2*A2, m13' = rho*v3*A3, (3 equations). This is 6 equations in 12 variables ( 3 "P"s, 3 "m' "s, 3 "v"s, 3 "A"s), so if you are given any 6 you can find the other 6.

If all you want to know about are the pressures and velocities, then you have Bernoulli (2 equations) and 6 unknowns (3 "P"s and 3 "v"s), so if you know any 4 of them, you can find the other 2. Even more simply, if you have P and v for one pipe and v for one of the others, you can find P for that other pipe (1 Bernoulli equation and 4 variables, 3 of which you know). This last case is exactly the one you described.

Hope this helps!

--Stuart Anderson

[Long]

Hi there,

your discussions are interesting. I am wondering what if the headloss cannot be neglected in Caroline's case. Will the headlosses in the two parallel pipe be the same? and Why?

Any opinion would be appreciated. Thanks in advance

Cheers

[mr_homm]

Hi Long,

Yes, as a matter of fact, the head losses must be the same on both paths. This is very similar to the voltage drop in an electric circuit, where the voltage changes along two parallel paths are equal.

One way to see that this must be true is to think about what happens when the flows rejoin. If one branch has a lower head than the other, then after they mix, the combined exit stream must have a head that is the average of the two. But then the stream with the least head will be one of the branches, not the exit stream, which means that flow will want to back up into that branch (because fluid flows from higher to lower head when there are losses). If the fluid does that, it will reduce the flow rate in the lower head branch, and since head loss depends on flow rate, this will reduce the head loss. So you can see that the fluid automatically compensates, and the flow divides itself in just the right way to make the two head losses equal.

Hope that helps!

--Stuart Anderson

[Long]

Hi Stuart,

Thanks for the detailed explanation of how the head-losses achieve the same on both pipes. I have thought further about it and then got an inherent question which I look forward to being advised.

My question is: Will the pressure in both parallel pipes are the same?

As you suggested that (P1/density)+((V1^2)/2G)=(P2/density)+((V2^2)/2G)=(P3/density)+((V3^2)/2G), then the following should be right

((P2/density)+((V2^2)/2G)+HL2=(P3/density)+((V3^2)/2G)+HL3.
where HL is head loss on the pipes.

I guess the pressure at the end of both pipes have to be the same. Otherwise the water would flow from the pipe with lower pressure to the high pressure pipe. If that is correct, and given the headlosses on both pipe are also the same, then the velocities through both pipe have to be equal in order to make the above equation to be right!

I have a gut feeling that the velocities are not the same in these pipes considering that they are in different diameters.

So, ??? - please advise. Thank you.

[mr_homm]

Hi Long,

The pressure in both pipes does not need to be equal where they meet, which is very surprising at first. However, the "stagnation pressure" IS equal in both pipes. Stagnation pressure is defined to be what you would get if you took the pressure and kinetic terms in the Bernoulli formula and calculated the pressure that would give the same total value all by itself (i.e., if velocity were zero). This is equivalent to saying that if you placed a small cup in the flow, so that within the cup some fluid was caught and stopped, while the rest of it passed around the cup, then the pressure inside the cup would be the stagnation pressure. This is higher than the fluid's actual pressure, due to the effect of fluid ramming into the cup and coming to a stop.

One way to see that the pressures need not be equal is to consider that a high velocity flow must have a lot of forward momentum. If a high velocity jet of water is injected into a tank full of water, then it can force its way in even if the pressure in the tank is higher than in the pipe carrying the jet, because the stronger pressure may only slow the forward motion, not stop it altogether. There are two possible cases: If the stagnation pressure of the jet is greater than the pressure in the tank, that is the same as saying that the pressure that would need to be developed in order to halt the fluid in the jet is more than what the tank has, therefore the jet will proceed forward. On the other hand, if the pressure in the tank exceeds the stagnation pressure in the jet, then there is more than sufficient pressure to stop the jet, and once it is stopped, the excess pressure will drive the fluid back up the pipe that the jet tried to come out of.

Therefore, since pipes must be at the same height to join, it follows that P_stagnation + d·g·z is the same for both pipes. But since P_stagnation = P + 0.5·d·v^2, this is just the same as saying that the value of the Bernoulli formula is the same in the pipes where they meet. Pressure alone is not a consideration, only the total Bernoulli value (total head) of the flow.

You are right that the velocities will not be the same in both parallel pipes. Basically, as I mentioned in an earlier post, the velocities will adjust so that the heads match at the inlets of both pipes and again at the outlets. This ensures that flow will not back up into either pipe and that head loss will be the same on both paths.

Hope this helps!

--Stuart Anderson

[leonhenry]

Hi, Sturt,
Your replies are very detailed and brilliant. I don't know if you are still in the forum.
Can you or somebody else help me about the question?
If the two parallel pipe have elevation change, let's say, 10 feet?
how can we address the head loss, pressure loss? It head loss still same? If so, that means elevation won't affect head? Thanks.