Undergrad General guidelines on the physics of flows at junctions?

[Freixas]

Let's start with a configuration in which one tube, A, splits into two, B and C. We can write the relationship
$$A_A v_A = A_B v_B + A_C v_C$$ When I first saw this formula (which was very recently), I wanted to solve for $v_B$ and $v_C$ given all other values. I can't of course. There are two variables and only one equation.

Let's say I closed off branch B, not at the junction, but somewhere down the tube. As I picture it, a compression wave would begin at the closed end, traveling at the speed of sound, to the junction. At that point, the velocity through branch B would be 0 and $v_C$ would be $\frac { A_A } {A_C} v_A$

My question is whether there are general guides for the effects branch B might have on C (or vice versa). Let's say the two branches have the same cross-section areas. With a well designed junction, I might expect the flow to divide evenly between the two branches. Some distance down branch B, however, the tube narrows. If I simply looked at the flow through B as though it were an independent tube, I could calculate the speed of the flow through the narrower section. I am not sure that what is true in a B-only configuration applies to the A-B-C branching situation. In other words, perhaps some part of the flow would find it easier to switch from B to C rather then continuing down the narrow part of B.

Again, I'm trying to get a sense of whether there are any broad principles that can be applied. I am not looking for a detailed derivation or a precise formula. Simplifying assumptions are fine. For example, if an abrupt switch from wide to narrow creates turbulence that complicates things but a gradual transition does not, it's fine to assume a gradual transition. I would certainly assume that all flows have reached a steady state.

 

[jrmichler]

Yes, this problem is solvable. The broad principle is as follows:
Calculate the pressure drop vs flow rate in B.
Calculate the pressure drop vs flow rate in C.

Find the flow rates in B and C such that:

Flow B plus flow C equals flow A
and
Flows B and C such that pressure drop B equals pressure drop C

If the tubes B and C are "long", then search Moody chart to find the procedure for calculating pressure drop vs flow rate. If the tubes B and C are "short", possibly with orifices or nozzles, then the Bernoulli equation might be sufficient.

 

[Freixas]

Since I'm new at this, let me work through it to see what I understand.

• I know $A_A$ and $v_A$, so I can calculate the mass flow.
• Let's say I also know $P_A$ (I haven't found a use for it yet).
• For the branches, I know $A_B$, and $A_C$.
• To simplify, let's add that $A_A = 2A_B = 2A_C$.
• Branch B has a constriction a short distance from the junction. Its cross-section narrows by 1/2.

"Flow B plus flow C equals flow A" translates to $A_A v_A = A_B v_B + A_C v_C$ (given the above, this means $v_A = \frac {(v_B + v_C)} {2}$).

I'm having trouble translating "Flows B and C such that pressure drop B equals pressure drop C" into anything I can use. In the first equation, the unknowns are the velocities $v_B$ and $v_C$, so I'm assuming the second statement sets up a second equation in which, again, the two velocities are the unknowns.

Using the simplified problem above, I tried applying Bernoulli's equation to branch B and was able to express the exit velocity in terms of the entry velocity (the velocity at the junction, which is the one I need). However, if I use the same approach for C, the pressure drop is 0. Trying the set the two pressure drops equal results in branch B having a velocity of 0 as well, so it's clear I don't understand exactly across which two points I calculate the pressure drop.

You don't need to solve the above problem for me, but could you describe the formulas you would apply and the steps you would take in a little more detail?

 

[gmax137]

Google
bernoulli pipe network or
bernoulli branched flow

 

[jrmichler]

If the pipe is short and has no other restrictions, the flow is controlled by entrance loss. Search pipe entrance loss to learn more. If the pipe is short and has a restriction (orifice or nozzle), then both the entrance loss and restriction need to be considered.

If the pipe is longer than short, then entrance loss, restriction, and pipe friction all need to be considered. In that case, I would use an iterative approach:

1) Assume a flow through B. Calculate the pressure drop using a Moody chart and equation. Add pressure drop from entrance loss and restriction. The sum of the pressure drops is the total pressure drop for that flow rate.
2) Calculate the flow through C from the assumed flow through B. Calculate the pressure drop using a Moody chart and equation. Add pressure drop from entrance loss and restriction.
3) Compare. If the pressure drops are not equal, adjust the flow through B, and repeat steps 1 and 2.
4) Iterate until the pressure drops through B & C are equal. You then know the two flows, and the two flows sum to the flow through A.

There are numerical analysis techniques to speed up the process by reducing the number of iterations, but this approach will get you to a solution. If you study the Moody chart, you will see that pressure drop from pipe friction is not simply proportional to velocity squared.

 

[Freixas]

Google
bernoulli pipe network or
bernoulli branched flow

Thanks.

It helps to know which keywords to use. Sometimes I use what I think is a good search phrase and find nothing. I change it a little and, all of a sudden, I get a ton of relevant results.

 

[Freixas]

If the pipe is short and has no other restrictions, the flow is controlled by entrance loss. Search pipe entrance loss to learn more. If the pipe is short and has a restriction (orifice or nozzle), then both the entrance loss and restriction need to be considered.

Thanks again for your help. I'll need some time to process this, but wanted to acknowledge your assistance.

Just so you know what you're helping with, I'm looking into the design of a melodica, a wind instrument with a piano-like keyboard. Press a key and a path opens up allowing air to flow past a reed, which vibrates to sound a note.

Reeds vary in size, with the lowest notes having the largest reeds. The reed gap, the hole through which the air flows is almost literally a hair's-width larger than the reed. Generally, all the reeds sit in one big air chamber. Pressing more than one key creates a complicated fluid dynamics branching problem. Ideally, you want each reed to have equal volume when played together. In practice, the lower notes seem to get more than their share.

There are some techniques used to even out the sound. The physics behind some of these seem questionable.

Rather than having one big air chamber, I've considered creating individual pathways to the reeds. Because of the different reed sizes, we come up with roughly the problem I've described above.

Flow within a big air chamber is particularly difficult to analyze as you can get all sorts of eddies and whirlpools—I've resorted to using SimScale to model the whole thing. But it's good to understand the general principles before getting too crazy with simulations.