Accoustic Resonance of an open tube HELP

[vega]

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 "musical exhaust" 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? I am working with 57132.3958333 cfm btw

 

this is the closest thing i could find.

http://en.wikipedia.org/wiki/Acoustic_resonance

 

it does not say what the value of the integer is supposed to be.

Edited September 20, 2012 by vega

[Leon]

What are you trying to accomplish?

 

In the given formula, n (the integer) represents the "resonance node" as they called it. A lower number means a lower frequency, i.e. harmonic. n=1 is the "base" resonant frequency, aka the first harmonic.

 

Also, temperature has a fairly large effect on the speed of sound through air, but there are tables and formulas that are easily available to calculate that.

Edited September 20, 2012 by Leon

[zero]

There seem to be a few things here that, while similar, are not really talking about the same thing.

 

Open tube resonance has little to do with the material that the pipe is made of. It is really just talking about the interaction of the air with the pipe and when there will be a complete wave at the ends of the pipe, but is not dependent on the pipe's composition. There will be a resonant frequency at every integer value because a 'complete' wave will be at both ends of the pipe when that frequency is introduced to the pipe.

 

A separate question is the natural frequency of a pipe, but here we are talking about the pipe itself. All objects will vibrate at a particular frequency when struck. There are ways to sort of predict this for common shapes, but there are two ways to measure it directly. One is to strike the tube and measure it with a speaker/software (pretty easy to do with a computer nowadays) and another is to use a more advanced program to play frequencies at the tube and measure the volume of what comes back. There will be a spike in amplitude at the resonant frequency.

 

Now for musical instruments the resonant frequency of the air is of main importance. That's why most instruments use openings in the vibrating column of air to manipulate the effective length of the pipe.

 

The natural frequency of the pipe itself is often important for getting a certain tone. Instruments that play a single note (like an organ pipe for example) will exploit this.

 

 

What is the end goal of what you're trying to do here?

[Leon]

There seem to be a few things here that, while similar, are not really talking about the same thing.

 

Open tube resonance has little to do with the material that the pipe is made of. It is really just talking about the interaction of the air with the pipe and when there will be a complete wave at the ends of the pipe, but is not dependent on the pipe's composition. There will be a resonant frequency at every integer value because a 'complete' wave will be at both ends of the pipe when that frequency is introduced to the pipe.

 

A separate question is the natural frequency of a pipe, but here we are talking about the pipe itself. All objects will vibrate at a particular frequency when struck. There are ways to sort of predict this for common shapes, but there are two ways to measure it directly. One is to strike the tube and measure it with a speaker/software (pretty easy to do with a computer nowadays) and another is to use a more advanced program to play frequencies at the tube and measure the volume of what comes back. There will be a spike in amplitude at the resonant frequency.

 

Now for musical instruments the resonant frequency of the air is of main importance. That's why most instruments use openings in the vibrating column of air to manipulate the effective length of the pipe.

 

The natural frequency of the pipe itself is often important for getting a certain tone. Instruments that play a single note (like an organ pipe for example) will exploit this.

 

 

What is the end goal of what you're trying to do here?

Nicely laid out, zero.

[vega]

What is the end goal of what you're trying to do here?

 

I have a theory in that based on what you just said about the pipe organ (there are two kinds of pipe organs open and closed tube. the closed tube has a varied opening). The theory is taking an air pump (engine) that produces a max amount of the given cfm (57132.3958333)when let into the tube. I want to know what size tube will give certain musical notes. That is what I am trying to figure out.

 

I calculated cfm using this formula CFM = (CID * RPM * VE)/3456

 

As an example (not the note I am looking for in particular btw) Middle c on a piano plays at 261.626 hz. What dimensions of a tube would be needed to create a 261.626 hz frequency when forcing 57132.3958333 cfm through the tube? There must be some equation for what I am looking for.

Edited September 21, 2012 by vega

[SUNNY Z]

This thread

 

 

------------

 

 

My head

[Swervey McZCar]

Me thinks your cfm calc may be incorrect.

 

cfm = ( (RPM * CID) / 3456 ) * VE

 

Using ( 7000rpm x 170ci / 3456 ) * .85ve I get 292 cfm for an NA engine.

 

another:

 

cfm = (RPM*Disp in L * (VE*10) ) * PressureRatio) / 5660

 

Using ( (7000 * 2.8 * 85) * 1) / 5660 = 294cfm used this one for Turbo

 

The differences here are attributable to the metric to CI conversion.

[Swervey McZCar]

<P>double post, sorry.</P>

Edited September 21, 2012 by Swervey McZCar

[zero]

I have a theory in that based on what you just said about the pipe organ (there are two kinds of pipe organs open and closed tube. the closed tube has a varied opening). The theory is taking an air pump (engine) that produces a max amount of the given cfm (57132.3958333)when let into the tube. I want to know what size tube will give certain musical notes. That is what I am trying to figure out.

 

I calculated cfm using this formula CFM = (CID * RPM * VE)/3456

 

As an example (not the note I am looking for in particular btw) Middle c on a piano plays at 261.626 hz. What dimensions of a tube would be needed to create a 261.626 hz frequency when forcing 57132.3958333 cfm through the tube? There must be some equation for what I am looking for.

 

I think the issue here is that just pushing air through a tube does not create a frequency. You have to introduce vibration either by a reed valve, the block and fipple of a recorder, or the similar thing on an organ(though I don't know what that's called) And the frequency of the vibration isn't really related to the cfm through the tube, but the length of the air column. The CFM can have an effect on the amplitude, but not the frequency (until you get to weird things that happen with reed valves, like overblowing).

 

With cars, the pitch of an exhaust note is related to the RPM, not the CFM pushing through. Even in exhaust systems where there is funky harmonic tuning going on(the Infiniti cars come to mind) it's independent of CFM because it occurs at all throttle openings(all VE values). Don't forget that you are introducing a frequency to the pipe with the pulses of the various cylinders into the exhaust. This is what most 'tuned' exhausts tune to.

 

Another issue is that there is an 'effective' frequency because those pulses, in many engines, aren't evenly spaced, and even if they are they don't necessarily travel the same distance in the header before merging in the collector. This can make tuning the exhaust particularly difficult, because you're not tuning to a 'clean' frequency. The effective frequency is sort of an average that your mind perceives when you hear it because some pulses are closer than others, which is why flat crank v8's sound higher than most.

[vega]

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 comming 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 again 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.

Edited September 21, 2012 by vega

[Leon]

Me thinks your cfm calc may be incorrect.

 

cfm = ( (RPM * CID) / 3456 ) * VE

 

Using ( 7000rpm x 170ci / 3456 ) * .85ve I get 292 cfm for an NA engine.

 

another:

 

cfm = (RPM*Disp in L * (VE*10) ) * PressureRatio) / 5660

 

Using ( (7000 * 2.8 * 85) * 1) / 5660 = 294cfm used this one for Turbo

 

The differences here are attributable to the metric to CI conversion.

Nice, I was just about to post that. Fifty-seven thousand CFM would be a pretty good flowing engine...

 

I bet vega used VE% instead of actual VE in decimals since the calculated number is two orders of magnitude too big, i.e. VE=90 instead of VE=0.90.

 

Vega, I'm not sure if you're approaching this problem from the correct angle. CFM and pipe composition are not directly tied into the natural frequency of the air passing through the pipe, meaning that changing the CFM through the pipe is going to change the volume of the sound but not the frequency at which it resonates. As zero mentioned, the composition of the pipe is tied into the natural frequency of the pipe itself, e.g. a tuning fork.

 

What matters more, in terms of exhaust design at least, is pipe length, valve timing, RPM, and exhaust temperature. This is the same thing as discussed in the sticky: exhaust resonance. It looks like you're looking for a specific sound instead of a performance gain. The concepts are the same, except now you're going from known specs to a required frequency rather than the other way around.

 

[EDIT: Damn! Took too long to post, zero covered some of this already.]

Edited September 21, 2012 by Leon

[Leon]

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 higher the rpm, the higher the CFM, the LOUDER the sound.

 

 

The formula should technically be able to find the frequency giving the cfm comming 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.

There is no formula relating CFM to frequency, as we've said, that doesn't make sense. The amount of air flowing through the pipe will change its volume (or "amplitude" as zero already mentioned), not frequency.

 

 

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.

You make it sound easier than it really is. There are a bunch of waves interacting with each other, especially in a multi-cylinder engine. How are you going to deduct a given frequency from before the X-pipe to after? In other words, how are you going to relate what the engine is doing to what's coming out of the end of the pipe? As zero already mentioned, differing lengths of primaries will cause the waves to reach each other at different times thus resulting in a different (average) frequency.

 

 

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?

I covered this in my first reply. N represents the harmonic that you're calculating, e.g. N=2 to calculate the 2nd harmonic.

 

 

For example. if f=nv/2(L+.3d) and the req we are looking for is say again 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.

I don't really get what you're doing here. Aren't you trying to solve for frequency (f=261Hz)? In that case, to get the middle-C frequency of 261Hz, the length of pipe needed in your example would be 25.5 inches (.65m).

 

Secondly, I think you made an error in the calc you did. With speed=343m/s, L=1m, and d=0.0254m, f=170Hz.

 

Since N=1, this is all based on the first harmonic.

[Leon]

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

FWIW... Flow=571.3CFM @ 7000RPM

 

As I thought, you were using the percent value for VE instead of actual VE (decimal).

[vega]

The higher the rpm, the higher the CFM, the LOUDER the sound.

 

 

 

There is no formula relating CFM to frequency, as we've said, that doesn't make sense. The amount of air flowing through the pipe will change its volume (or "amplitude" as zero already mentioned), not frequency.

 

 

 

You make it sound easier than it really is. There are a bunch of waves interacting with each other, especially in a multi-cylinder engine. How are you going to deduct a given frequency from before the X-pipe to after? In other words, how are you going to relate what the engine is doing to what's coming out of the end of the pipe? As zero already mentioned, differing lengths of primaries will cause the waves to reach each other at different times thus resulting in a different (average) frequency.

 

 

 

I covered this in my first reply. N represents the harmonic that you're calculating, e.g. N=2 to calculate the 2nd harmonic.

 

 

 

I don't really get what you're doing here. Aren't you trying to solve for frequency (f=261Hz)? In that case, to get the middle-C frequency of 261Hz, the length of pipe needed in your example would be 25.5 inches (.65m).

 

Secondly, I think you made an error in the calc you did. With speed=343m/s, L=1m, and d=0.0254m, f=170Hz.

 

Since N=1, this is all based on the first harmonic.

 

could you show me the math how you reverse engineered that? also i assume n2 would be the second harmonic?

 

thanks for clearing this up guys.

[Leon]

could you show me the math how you reverse engineered that? also i assume n2 would be the second harmonic?

 

thanks for clearing this up guys.

Just a little bit of algebra...

 

You're using:

f=nv/2(L+.3d)

 

You have a known frequency (f), speed of sound (v), pipe diameter (d), and the harmonic you're shooting for (n).

 

You want a length of pipe that accomplishes this. This means solve for L:

 

Multiply both sides by (L+.3d):

f(L+.3d)=nv/2

 

Divide both sides by f:

L+.3d=nv/2f

 

Subtract .3d from both sides, and voila:

L=(nv/2f) - .3d

 

Then as they say, "plug and chug".

 

n=1 is 1st harmonic

n=2 is 2nd harmonic

n=3 is 3rd harmonic

...

 

[edited typo]

Edited September 21, 2012 by Leon

[vega]

And solving for d?

 

Also what is the difference in the first second and third harmonic?

 

I was always bad at algebra.

Edited September 22, 2012 by vega

[zero]

That link you posted has a good explanation of the harmonics.

[vega]

Well I finally found the frequency I need to emulate that being 493.883hz tuned in 440hz which is N1 according to wiki.

 

It looks like using your equation to find length also varies depending on the diameter.

 

so i did some math and the diameter doesn't really seam to make that big of a difference. whether or not its as small as 1 inch or 4 inches.

 

maybe i am figuring for L wrong

 

440hz(343.2mps)/2(493.883 the freq. goal)

1510083/987.7663=152.878266853

taking that and subtracting the .3(d) doesn't vary that much from 1 inch to 4 inches

 

at 1 inch (.0254 meters) .3(.0254)=.00762

at 4 inches (.1016 meters) .3(.1016)=.03048

 

So there is only a range of 152.87064 meters to 152.847786853.

that is 501.544 to 501.469 feet. first off that is very long and I know that there is no way that a car is ever 500+ feet. So there is something either I am missing or there is something else that is bringing the exhaust sound up in hz.

 

I am going to go ahead and say that for the most part this theory must be wrong (or I did my math wrong) and the piping matters less than the muffler design. I am betting that the muffler systems on say a corsa vs a magnaflow are very different in where the baffles are placed. In the muffler one has baffles placed in a way to bring the lower grumble sound out of the tubing. The other has baffles placed so that the baffles are placed in a way to bring a crisp exotic sound.

 

I know exhaust pulses and the like are a huge factor etc etc etc. Though I think this shows (correct me if I am wrong) how different mufflers can change the sound extremely differently.

 

examples on a gt500

Corsa

 

vs

 

magnaflow and stock

Edited September 22, 2012 by vega

[Leon]

You did the math wrong, again.

[Swervey McZCar]

Looks like "n" should equal 1, 2 or 3 depending on which harmonic you are trying to match. Not 440... I get L = .347m at the first harmonic...