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I'm trying to understand the air flow within a melodica.

A melodica is a wind instrument that has a piano-like keyboard. Pressing on a key opens an airway so that air entering the melodica's air chamber can flow past a brass reed and exit the system. Low note reeds are bigger than high note reeds and the opening through which the air flows is just slightly bigger than the reed.

Here is a simplified diagram in which two keys are pressed simultaneously: simplified diagram of a melodica air flow with two reeds open At the moment, I am just looking at the mass flow, the total mass of air flowing past any point in a tube- or pipe-like system per given unit of time (at melodica air speeds, air can be treated as incompressible, so the mass flow can be specified in terms of volume per unit of time).

I am simplifying my analysis by considering only horizontal systems and only constant air flows.

In a system open at two ends, the mass flow at any point in the system is a constant. Therefore, if I play a single note, the mass flow through any reed is equal to the mass flow entering the system. This is not true if I play two notes.

I could apply the rule for parallel pipes; in the diagram above, the mass flow exiting A + B is equal to the mass flow entering the system. The mass flow for A would be twice that of B or 2/3 of the entry flow. The mass flow for B would be 1/2 that of A or 1/3 of the entry flow.

However, note that the path from the entry point to A is shorter (much shorter in real life than in this diagram) than the path to B.

My question is whether this introduces some additional factor that I have to account for; in other words, will A's mass flow be greater than 2/3 of the entry flow?

UPDATE 9/7/2020

I don't have an answer, but I have done further research and realize my question needs work.

First, the "rule" for parallel pipes is $A_1v_1 = A_2v_2 + A_3v_3$, which does not mean the mass flow rates would be 2/3 and 1/3. There are multiple solutions for $v_2$ and $v_3$ even when we know all the areas and the initial velocity. For instance, if I block one branch, its velocity becomes 0 and the other branch's velocity will increase to compensate.

Originally, I thought I could apply rules for pipes in parallel and pipes in series. In other words, once a flow branched, I thought each branch could be treated independently. Blocking one branch shows that the flows are not independent. This makes me think that, for example, if one branch is constricted even after it splits, it might still have an effect on the other branch.

In the case of the melodica, there is a large air chamber with the input flow closer to one outlet than the other. At this time, I don't know if there are any general rules that can be applied; perhaps a simulation (or experiment) is the only way to answer the question. I suspect that reed A gets more than it's share of the air.

As a test, I connected two melodicas using equal diameter, equal length tubes. If I play a low note on one and a high note on the other, the high note does not drop in volume the way it does when I play both notes on the same melodica. Topologically, it looks like the same situation, but it's clearly not.

Some potential changes to a melodica's design might be to direct the airflow first to the higher, smaller reeds. Given their smaller area, they might balance the mass flow distribution more equally. Another thought would be to have a more even distance to each reed. This could be done by having the airflow enter in the middle, separating the reeds into individual chambers and ensuring that the distances from the entry to each reed is the same.

I'm leaving the question open, but I'm going down the simulation path. If I get an answer, I will post it. If anyone knows a way to answer the question (even in a general way) without simulation, please do.

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This is more like a comment than an answer, but it's too long for the comment box.

I think that your model for how air flows past the reeds --- and in particular the assumption that the air flow is proportional the reed's width --- may be unrealistically simple. Airflow past the reeds has at least three distinct modes:

  1. a low-flow mode, where the reed does not sound
  2. a moderate-flow mode where, the reed oscillates with a well-defined frequency
  3. a high-flow mode, where the reed's frequency shift downwards (flat)

I would guess, as a physicist, that the transition between the silent and well-behaved modes corresponds to the onset of turbulence in the air passing over the reed, and the pressure difference / flow rate of this transition has less to do with the width of the reed than with the height of the gap between the vibrating end of the reed and the fixed part of the air channel. (Replacement harmonic reeds, which I think are the same technology, seem to have varying length but all the same width; it's also possible to adjust the frequency by changing the brass's thickness.)

Something you might try: finger a big chord while not blowing, and then slowly transition from "blowing very gently" to "blowing hard enough to play the entire chord." When I do this on my melodica (aside: I love my melodica) I can hear the $1\to2$ transitions for each reed separately, like a spelled barbershop chord. The order that the notes come in is mostly low-to-high, but not always. There's also a little bit of hysteresis: if I hold the lowest two Fs on mine (which bracket middle C), the F4 starts to sound before the F3, but if I ease off the F3 continues after the F4 has stopped. This particular observation seems germane to your question about two reeds whose widths (and naively, frequencies) are different by a factor of two.

If I were designing a melodica, I would try to choose or adjust reeds so that all of the notes in a "typical" chord started to sound at the same time; that might or might not mean that they all have comparable airflow.

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  • $\begingroup$ I am actually trying to understand the physics of the melodica by analyzing small bits of it rather than trying to tackle the entire complex system. For my question, I would like to remove the reeds from the picture entirely and just look at the mass flow at the points where the reeds would be. I was hoping this situation I drew above would be a simple analysis for a fluid dynamics expert. The alternative would be to create a test rig to measure the mass volume, which would be tricky given the tools I have at hand. $\endgroup$
    – freixas
    Sep 4 '20 at 23:13
  • $\begingroup$ I reproduced your experiment and results with chords. I think, however, the behavior as a reed begins to sound may may be due to magnifying manufacturing differences in the reeds—thus, for example, it's not always the lowest note that sounds first. What I'm more interested in is what happens when all reeds are in their "normal" operating state—or, even better, when the reeds are removed from the equation entirely (hopefully, there is a consistent mapping from the latter to the former). $\endgroup$
    – freixas
    Sep 4 '20 at 23:31

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