How Does Acoustic Resonance Affect Frequency in Open Tubes?

[faytmorgan]

Accoustic Resonance of an open tube -- help please

So I am trying to make sense of the different varying formulas that I have found. Some of the formulas are not fully explained hence my questions. This is a music related topic.

I am attempting to figure out what the formula is to determine the frequency of a stainless steel pipe (the one constant) with air pushed through it with these following variables.

(A) Length of pipe
(B) Wall thickness of pipe
(C) CFM going through the pipe
(D) Inner Diameter of pipe

Also I have read that overblowing a pipe will cause the frequency of the note to jump an octave. At what point is it considered to be overblowing a pipe? Is there a formula for the max cfm prior to the said overblowing?

 

[faytmorgan]

I am working with 57132.3958333 cfm btw

 

[faytmorgan]

I am working with a 336.6 CID sbc. At 7000rpm VE is at 83.8. Peak VE is at right around 5500rpm being at 92.1

RPM is a factor of cfm, it doesn't make sense to me that cfm is not a factor in the sound and the rpm is seeing that they are related. The higher the rpm the higher the cfm and vise versa.

The formula should technically be able to find the frequency giving the cfm coming out of the engine. valve sizing as far as I can see and all other factors of the engine cause the ve to be where it is at, so ve is the only number that should matter there. which again is a factor of the cfm.

Also one should be able to even out the pulses of the exhaust running first through an x pipe back into a single pipe. That single pipe is the one that matters in this formula.

As for the formula given I understand that "N" is an integer being "1,2,3" what I don't get is what it is supposed to represent. It says the resonant node, are they trying to say the frequency that we are attempting to achieve, or something else that I am missing here?

For example. if f=nv/2(L+.3d) and the req we are looking for is say middle c (261.626 hz) at 20c, which would make speed at 343 mps (meters per second) the length is say about 1 meter and the diameter is .0254 meters (1 inch) then f would equal 44546.0087328hz which makes no sense to me that appears to be very high. A 3 foot open pipe with 1 inch diameter openings. Unless that happened to be the limit of the pipes ability to make x amount of HZ with max amount of allowable cfm to travel through the pipe. Which this does not tell us when it is too much cfm for the pipe and then the octave raises.

Show me where I am wrong here, because I feel like I am missing something.

 

[Enthalpy]

The frequency relates essentially to the length of the pipe. You have an integer number of half-waves in one pipe length, and sound propoagates over one wavelength within one period, which is the reciprocal of frequency.

Overblowing can't be predicted easily. It doesn't depend just on throughput.

 

[ilhan8]

I am a musician. I am an expert of nay. ( a kind of end-blown reed flute)
an about an inner radius of 16mm and an about a length of 72cm pipe gives nearly A (440 hertz) when end blown.
however when side blown it is different.
different wind instruments have different mouth pieces and give different frequencies at the same pipe length.
therefore, the resonator has an important role on the frequency.

 

[Integral]

I am working with 57132.3958333 cfm btw

WOW! What kind of meter did you use to get that many digits?

Or is that just calculator vomit?

Please post measurements with a reasonable number of significant digits.

 

[sophiecentaur]

The frequencies of the overtones of a pipe will depend upon whether it can be regarded as open ended or closed ended.
Look at this link to see what I mean. The first overtone can be at around 2f or 3f, depending.
In a real situation, the openness or closedness will be a bit approximate and will depend on what is connected to the ends and the overtone may be a long way off from a harmonic frequency. Is this to do with a car exhaust?