Yes, 100 mm H2O is a tiny fraction of atmospheric pressure. But it is enough to feel significant in your mouth.
For what it’s worth, you can find a draft of my modelling in this post, Blowing machine - #163 by Tunborough - Whistle - Chiff & Fipple In this model, K is a “flow loss coefficient”. In this thread, you’ll also see references to the discharge coefficient, Cd. You can work with either one, Cd = 1/sqrt(K).
I hope to post an updated draft this weekend, after I dig out the driveway.
Thanks Terry and Tunborough - it’s become a lot clearer now. The main thing missing now is what value rho has [in v = sqrt(2 * 9.807 * P / (rho * K))], so I’m closer to understanding how to do the calculations. The way the pressures are converted from mm to pascals makes it look as if the pressures aren’t absolute pressure readings, but just the additional amount above atmospheric pressure. Have I got that right? I always find this kind of thing hard to follow if there isn’t an example provided of one complete computation showing how you get from the measured values read off the gauges to the end result.
rho (Greek letter ρ) is the symbol for density, in this case the density of air. At 21 C, sea level (where Terry is), 60% humidity, the density of air is in the neighbourhood of 1.19 kg/m^3.
You are correct that the pressure readings are relative, usually relative to the external atmospheric pressure. Terry did do some measurements of other pressure differentials, such as the difference in pressure between one side and the other of one of the calibrators.
Here’s an example calculation. A while back, Terry did some measurements on the old Generation whistle playing notes. For the lowest note, D5, he measured a flow of 7.7 L/min and a “mouth” pressure of 16 mm H20. The windway exit of the Generation averages around 1.17 mm high and 7.85 mm wide. From the air flow, I estimate the mean air speed at the windway exit to be:
v = 7.7 / (0.060 * 1.17 * 7.85) = 14.0 m/s
For that whistle mounted in Terry’s blowing machine, I’ve calculated K = 1.22, so from the pressure, I estimate the mean air speed to be:
v = sqrt(2 * 9.807 * 16/(1.19 * 1.22)) = sqrt(216.2) = 14.7 m/s
The two estimates aren’t exactly the same, but they are within 5%, which is as close as I expect to get.
I’d be quite happy to do this, just to see what we learn. If nothing, then we have learned something!
Can you suggest specifications? Eg accept or alter the following:
Body OD, 13.5mm to fit Whistle Connector
Body ID, would 8mm be big enough? Or 10mm?
Body Length? 10mm, or more? Does that include the metal plate with the hole in it?
Metal plate thickness eg 0.8, 1.2, 1.6, 2mm?
Size of hole?
Hole left as a cylinder or chamfer lead in and lead out?
Any other important specs?
And turning to the actual testing, have the orifice at the end of the food chain? Or above the flow meters and use Differential Mode on the Manometer?
Which is just as well, as this period flageolet is so badly made, it would be very hard to get a detailed profile! You’ll see I’ve given you three window heights for entry and exit, the Left, Middle and Right as viewed from the blowing end. And an average for each. Best of luck!
English Flageolet
Length H(L) H(M) H(R) H(Ave) Width
0 2.52 2.42 2.76 2.57 9.85
35.3 0.86 0.96 1.29 1.04 8.5
36.1 Block protrudes into window
Those tubes should work nicely, as long as you don’t mind going high and loud with the short one. Let’s start with the Feadog mouthpiece, with the tube pushed all the way in. I’ll want to know what the length ends up between the splitting blade and the end of the tube. For this, remember, we’re measuring frequency as well as flow and pressure, going up as high as you dare, and back down again, to test the limits of the hysteresis in the register shifts. Because we want to get as close as we can to the register shifts, we can’t limit ourselves to fixed intervals like 4 L/min.
OK, I’ll get to that soon.
Now trill, here are the revised figures for the Old Gen head, now at the end of the food chain.
BUT SEE BELOW!
Note that although I had balanced the two flow meters at 20 L/Min using the higher one’s needle valve, by the time I got back to that end of the scale, it had shifted slightly.
Old Gen head after Flow Meters in parallel, no stop valve
LH RH L+R Press. Res.
20 19.8 39.8 286 0.425
18 17.6 35.6 220 0.417
16 15.7 31.7 179 0.422
14 14 28 144 0.429
12 12.2 24.2 105 0.423
10 10.3 20.3 77.5 0.434
8 8.1 16.1 45 0.417
6 5.6 11.6 23 0.413
4 4 8 16 0.500
2 2.5 4.5 12.5 0.786
Average Resistance 0.47
Average Resistance above 4L/Min 0.42
Hold the presses! I noticed that the tape holding Old Gen’s beak into the tubing now looks like it’s shifted. So regard the figures above as dodgy, I’ll leave them there to prove my point!
When you look at one of these whistles, you see what we are up against. We have round tubing, but the whistle beak is nothing like round. And not even smooth. The poly tubing I’m using is flexible, but not soft, so it doesn’t conform well to awkward shapes. And the beak of the whistle is tapered, and the compressor is doing it’s best to spit the whistle out. And the beak goes a fair way in, leaving very little room for taping before you come to the window. And I don’t think we’d want to tape over the window!
I cast around again for solutions, but am not finding much. So I tried this desperation stunt. I put a lot of turns of tape around the blowing end of the beak, reckoning that this will make the beak seem rounder, smooth the awkward bumps, and prevent the beak going so far in, leaving me more room to tape it onto the end of the tube. It seemed to work, so I ran the test again:
Old Gen head after Flow Meters in parallel, no stop valve, taped better.
LH RH L+R Press. Res.
20 20.2 40.2 490 0.551
18 18.3 36.3 360 0.523
16 16 32 280 0.523
14 14.1 28.1 214.5 0.521
12 12.2 24.2 151.5 0.509
10 10.6 20.6 116 0.523
8 8.3 16.3 68 0.506
6 6.1 12.1 35 0.489
4 4.3 8.3 25 0.602
2 3 5 17 0.825
Average Resistance 0.56
Average Resistance above 4L/Min 0.52
Note the much higher pressures and resistances.
Ideally we’d come up with a more reliable interface between system and whistle beak, but its not jumping out at me yet. There are quite a range of shapes and sizes to accommodate. But maybe the approach of taping the beak before inserting it will help with others too.
I also made one little change to the system. You’ll remember I used to have a flow regulator after the pressure regulator, but got rid of that as it didn’t seem to be necessary. It did make setting the Pressure Regulator a little finicky, so I reasoned that including a bit of resistance in the line between the Pressure Regulator and the Flow Meters might help by making the setting more progressive. I made up a “resistor” - some delrin rod about 35mm long, reduced diameter at each end to plug into the 1/4" tubing, and with a 2mm or thereabouts hole through it. This seems to work nicely - I now need a few turns of the Pressure Regulator knob to get from 0 to 40L/Min rather than less than one. You’ll remember I squished this tube with a G clamp to halve the flow rate and then compensated by advancing the pressure, and got the same results. So I can’t see any downsides from this development.
Eagle-Eye Terry !
Malua’s own McGee
spotted shifted tape
and looky-looky see !
The Cd now descends
like I hoped it should
a simple little leak
almost turned my head to wood ! 
(Octave markers included for reference)
Well,
First off, thank you Terry for spotting the tape and re-running the case. And also, of course, for indulging my wondering about the 2 meters in parallel for the whole range.
Looking now,
So, since the shut-off valve has been shown to be a non-issue, how about another case as follows:
You know what they say:
Old age and Treachery
will always overcome youth + skill !
Absolute Genius !
I’ve never heard a player admit to leaky lips.
So, now, how about “artificial lips” ?
The thought had crossed my mind. I wondered how I could phrase a grant application to the Australia Council to buy something appropriate from an on-line Adult outlet. Hmmm, perhaps “appropriate” isn’t quite the right word…
How about something involving silicone? Either a flexible sealant (one that rubs off plastic cleanly) or something intended to make casts. You could probably make a soft oval ring to push the beak into - maybe with cork grease applied.
For the orifice plate-like calibrator as near as possible to a zero length calibrator. So that the entry side geometry is the same as the calibrators (same tube diameter and distance from anything upstream, the hole is 3.94 mm, and the same geometry on the outlet side. Assuming the calibrator is in the end of a tube sticking out of it then as little tube beyond the plate as possible. As well as giving another look at the meter scales it would give an L=zero case for Tunborough’s K equation which, as he notes, assumes some things either side of the calibrator/whistle can be ignored - so we need to keep them the same.
Still looking forward to playing with whistles but I haven’t forgotten that some of this begin with a windway that was made tapered to reduce backpressure. Also that some on this forum find ‘forming an embouchure’ with the lips effects tone. So what goes on when the mouth is not much larger than the windway entry may matter.
Edit to add. Sorry didn’t answer all the questions about the plate. Plate as thin as you can work with (? 0.8mm), the sides of the hole however it they are on the calibrator. Although a proper orifice plate has the tubes either side of the plate what goes on there are things not included in Tunborough’s draft model, for example we are not worrying about the speed of the air in the tube that the calibrator is plugged into. Fluid dynamics seems especially to be a world of treating things as not significant to keep it manageable - the test is does the model replicate what is observed well enough for the purpose?
It shall be done, m’Lord…
I’ll have a think. (That can take some time…)
For the orifice plate-like calibrator as near as possible to a zero length calibrator. So that the entry side geometry is the same as the calibrators (same tube diameter and distance from anything upstream, the hole is 3.9 mm, and the same geometry on the outlet side. Assuming the calibrator is in the end of a tube sticking out of it then as little tube beyond the plate as possible.
So it would be OK to have nothing beyond the end of the whistle connector tube apart from the metal plate with the hole in it?
As well as giving another look at the meter scales it would give an L=zero case for Tunborough’s K equation which, as he notes, assumes some things either side of the calibrator/whistle can be ignored - so we need to keep them the same.
Still looking forward to playing with whistles but I haven’t forgotten that some of this begin with a windway that was made tapered to reduce backpressure. Also that some on this forum find ‘forming an embouchure’ with the lips effects tone. So what goes on when the mouth is not much larger than the windway entry may matter.
Yeah, I’m not altogether convinced that changing the mouth geometry changes the tone. I wonder if it changes the player’s perception of the tone. It would be a good exercise to practice “changing the embouchure to change the tone” until you reckon you can do it on demand, then record yourself doing it and listen back, and do it in front of the domestic rent-a-crowd and see if they hear anything. Oh, and watch it on a spectrum analyser live, and while listening to the playback. What’s actually happening?
Plenty of downloadable spectrum analysers for phones or PC.
To use with the data from ther calibrators yes, assuming there is no tube on the end of the calibrator. If there is then the same length as the calibrator.
Neither am I, though for some reason I find myself making faces for some notes. At one point Tunborough raised the possibility of a change in flow regime in the windway. If that is a possibility I wonder if making an embouchure could push the flow from laminar to turbulent or vice versa and somehow influence the jet. Once you are happy that pressure and flow measurements can be “either or” could you put a clamp close up the the entry side of the beak and just log flow? It may not be a useful avenue to go down - thought reported results might be referred to next time it comes up.
Getting back to flow meters in series.
If the density of dry air at 20C and atmospheric pressure(100 kPa) is 1.19 kg/m3 then at atmospheric pressure plus 1300 mm of H2O (112.7 kPa total) it would be 1.34. So its volume would go up by about 13%. As mass is conserved with flow meters in series that is the difference in flow rate that we would expect from expansion with a clamp between the two so that the first was operating at 1300mm H2O more than the second. There would also be a difference due to the drag on the float changing.
As an aside, but on the main challenge. No need for a blowing machine, just need a supercomputer. https://phys.org/news/2015-02-supercomputer-simulations-explore-air-reed-instrument.html Some interesting goings on in the windway in that first animation.
Thanks for the clear examples - it all makes sense to me now that I can see all the parts coming together.
Is the plan to go on from here to generate comparative lists for different types of whistle of the flow and resistance range for each note across two and a half octaves? (That’s the data I’d most want to see if I made a blowing machine.)
Have we tried this arrangement, to test how the flow meters react to downstream pressure?
stop valve → first flowmeter → manometer takeoff → second valve → second flowmeter → open air,
where the manometer is measuring relative to open air. Suppose we adjust the balance on the two valves to keep the flow on the second meter around 10 L/min and see what happens to the flow on the first meter at different pressures. (The needle valve on the second flowmeter would probably do as the second valve.)
When Terry squeezed the tubing between the flowmeters, it made a big difference to the reading on the upstream flowmeter. The only thing I can think of that would be making that difference is the downstream pressure. This would give us a chance to quantify this. Maybe density changes are enough to explain it, maybe not.
Here’s my model so far …
Let h and h1 be the height of the windway exit and entrance in mm,
w and w1 be the width of the windway exit and entrance in mm,
L be the length of the windway (the windway, not the whole whistle) in mm,
rho (ρ) be the density of air in kg/m^3,
Q be the air flow in L/min,
P be the mouth pressure in mm H2O,
v be the air speed leaving the windway.
If we know the flow rate, we can calculate the air speed in m/s as:
v = Q / (0.060 * h * w)
(The 0.060 is to convert from L/min to mL/sec.)
If we know the mouth pressure, we can calculate the air speed in m/s as:
v = sqrt(2 * 9.807 * P / (rho * K))
(The 9.807 is to convert from mm H20 to Pascals.)
Here, K is the flow loss coefficient, a dimensionless constant that is specific to a whistle. It is the sum of three parts:
K = Kb + Kdw + Ke
Kb comes from Bernoulli’s principle. If the cross-sectional area of the mouth is A0 mm^2, and the area of the windway entrance is A1 = h1 * w1,
Kb = 1 - (A1/A0)^2
Kdw comes from the Darcy-Weisbach equation:
Kdw = fd * T * L/Dh
Here, fd is the Darcy friction factor, which is close enough to 0.04 for pretty much everything we’re interested in.
Dh is the hydraulic diameter of the windway entrance, Dh = 2h1w1/(h1+w1).
T is a taper factor. Assuming most of the taper happens in the height, and the taper is more-or-less linear:
T = (1 + h/h1)/2 * (1 + h^2/h1^2)/2
Ke is an empirical adjustment. As yet, I don’t have a handle on this value. For the calibrators and the Feadog it is around 0.35; for the old Gen, Killarney, and Mellow D, it is around zero. When Terry put the flowmeter after a calibrator instead of before, Ke was around zero, so Ke may be nothing more than compensation for some idiosyncrasy of the flowmeter.
For the discharge coefficient, Cd = 1/sqrt(K)
For the “resistance”, resistance = sqrt(P)/Q = 0.2657 * h * w / sqrt(rho * K) = 0.2657 * h * w * Cd / sqrt(rho)
[ Crossing - I was typing this as you posted ! - it refers to the previous draft ]
As you are around Tunborough and the equations have come up, please can you confirm I have this right.
If p1 and v1 are the pressure and velocity on the inlet side
and p2 and v2 the same on the outlet side then the case of Bernoulli’s equation in use is:
p1 + rho/2v1v1= p2 + rho/2v2v2
and we are treating v1 as negligible and p2 as the same as the low pressure end of the manometer
so we would have p1= rho/2v2v2 for a zero length tube (i.e. K=1 so not used)