i am trying to find the answer to a question which i hope someone here can help with
i have 3 stainless steel spheres, diameters of 15,20,and 32cm
if i were to fill them with water what would the resonant frequency of the sphere be???
i need to know so that i know exactly what frequency to hit them with.
this is for ultrasonic cavitation project.
upto now i have been using 42khz which has proved effective, but i would like to try with a resonant frequency for better results.
a formula would be good if there is one.
last part is what size sphere would be needed for 42khz resonant frequency???
User E M devices is a genius antenna man
I also know he has interest in this project He is a great member and has always answered my PM's
Also if you can't wait on the piezotransducer info user Tinsel Koala made some suggestions on this topic[achieving resonance in a fluid]] he MAY have a source for the piezo transducers
Chet
Hi,
Perhaps this paper even if not on all your questions but may give some hints, or the References at its ends:
http://www.wooster.edu/physics/jrIS/Files/Patton_web_article.pdf
I found this by some search, I am not an expert on this unfortunately.
rgds, Gyula
OK, this is from a guy that really does not know what he is talking about, but, here goes. the resonant frequency of the empty sphere is easy to determine. Filling it with water, which has a dampening effect, will change that, to a lower frequency. Someone else on another topic suggested to hit the item in question with a frequency generator. By doing this, you can determine at least one resonant frequency even though there will be several and also , harmonics of these frequencies.
It is kind of like tuning a guitar or a piano. The sphere (filled with water) will "ring" at a resonant freq. I don't know how to explain this in more detail and this info may not be very helpful. This might get you started tho.
Bill
hi
i know that when an object hits resonance it will "ring", but what i dont know is how to find it with a formula. i can do it with trial and error for instance, by setting the ultrasonics away at a known frequency and scan through the frequency until it starts to feed back into the oscilloscope, there will be a peak when this happens. this is resonance.
the reason i would like a formula is so that i can try different size spheres and instead of scanning each one by trial and error i would like to be in the range needed..
i would like to operate the ultrasonics in or around the 40 khz ish mark so it may involve using a harmonic with the bigger spheres
Quote from: Pirate88179 on December 11, 2008, 10:05:22 PM
It is kind of like tuning a guitar or a piano. The sphere (filled with water) will "ring" at a resonant freq.
What you need to do is plot a graph of frequency versus amplititude, getting the amplitude
of vibration from a strain gauge, possibly.
It should be the architypal upside down pudding basin, with resonance at the top.
Paul.
The problem is that there are several (many) resonance "modes" for a sphere.
You could try
f = v / (2*pi*r)
where f will be the characteristic resonance
v is the velocity of the wave in the medium (i.e. the speed of sound in water)
r is the radius of the sphere
Also f = v / 2*r is supposed to be the frequency of resonance of the "first quadrupolar mode" of the hollow sphere. I don't know what happened to the "pi" factor here.
One ambuguity, as I'm sure you have found by now, is whether you are talking about the container, which would be a hollow sphere, or the water in it, which would be a solid (liquid!) sphere.
Since the theoretical situation is somewhat murky, I'd proceed empirically. Use Piezos and a known sphere size to determine a resonance frequency. Then generalize that finding to spheres of arbitrary dimension.
hi tinsel koala
cheers for the info, however im not too sure how to use the formula, i worked it out for a 20cm diameter sphere and my answer was less than 1, it wa 0.0**** cant remember the exact figure.
could someone post an example for me please
@ Tinsel Koala:
You always have a lot of good information.
I do have one question....it seems to me that anyone attempting to calculate this would have to know more about the sphere, such as the wall thickness, density of the exact type of steel used in the construction, etc. Also, depending on the method used to create the sphere, the wall thickness and density may vary throughout the structure. It would seem to me that unless someone knew all of these factors, the results would be sort of a WAG. (wild ass guess) Of course, I believe you said as much anyway. I was just throwing this out there.
Bill
thanks
Yes, that's right, and it also depends on whether there's a hole and/or a neck in the sphere, and whether the fluid being resonated is inside, outside, or both.
But, as a general rule, it's easier to think (for me) about resonance wavelengths, rather than frequencies. (Wavelength is the inverse of frequency, right? But it also captures the characteristic velocity of propagation in the medium concerned.) So, just like with electrical resonances in a transmission line or on an antenna, there is a characteristic length that corresponds to a particular resonance in a mechanical system, and integral multiples of that length will relate to resonances.
I think.
Wavelength = v / f
where v is the characteristic velocity (or phase speed) in the medium. This will be c for EM waves, and it will be the speed of sound in water or air, and it could even be the speed of sloshing in a basin or tub, or the speed the breakers roll in at your favorite surf beach.
and f is frequency.
To aid the math, wavelength has the unit: length
v has the units: length/time
and f has the units: counts/time or cycles/time, with Hz (Hertz) meaning specifically cycles per second.
So let's say you have a sphere, no hole, diameter 10 cm. So you could start with using 10 cm as the wavelength, and with water inside use 1500 m/sec or 150000 cm/sec.
wavelength = v / f
So f = v / wavelength
So f = 150000 / 10
So f = 15 kHz
Check the units: F is in cycles/time or 1/time, v is in length/time, wavelength is a length. SO units check.
And I would look at 1/4 wavelength so f=(150000 / 2.5)
and then integral multiples of the above
Now that's assuming what might be called a "transverse mode" of oscillation: the diameter of the sphere "houses" a sinusoidal pressure oscillation. But it is possible to imagine many other modes of oscillation, so the diameter might not be the appropriate length to try for those modes. I can imagine a circumferential oscillation that would include a "pi" term in calculating a characteristic wavelength from the sphere's radius.
I don't really know how a sphere of water can oscillate, but I'll bet it isn't simple and there will be multiple resonances.
So, yeah, pretty much WAG, but with a little guidance from the sphere's dimensions.
Resonating spheres with holes in them are called Helmholtz resonators and are extensively analyzed.
Someone in one of the threads asked if I had a source for high-power piezo transducers. Sorry, I don't really. I get most of my experimental materials from several local surplus outlets. The small transducer in the video cost 50 cents at Active Surplus. I also got some heavier duty ones there for a couple bucks each. And every once in a while I see some really high-power ex-military stuff at another local junker. SO I would recommend looking in places like that first.
and if f = v / (2 x pi x r) and the sphere is 20 cm diameter, r = 10 cm and in water v = 150000 cm/sec
then
f = (150000)/(20 x 3.14)
f = 2389 Hz
And to find the sphere size for 42 kHz, you work backwards (ASSUMING the formula is correct for the resonance you wish to excite):
f = (v / wavelength) generally, and if your resonance equation is f = v / (2 x pi x r) then the resonant wavelength is of course (2 x pi x r) and we are interested in finding r.
So rearranging we have
r = v / (2 x pi x f)
substituting knowns, we have
r = (150000) / (42000 x 2 x pi)
SO
r = a little over half a centimeter.
Which actually doesn't seem unreasonable for an ultrasonic resonator sphere filled with water.
Theoretical Resonant Frequency Of A Sphere Filled With Water...
F = V / (2*PI*R)
5 cm Diameter = 95 khz
9 cm Diameter = 53 khz
12cm Diameter = 39.8 khz
15cm Diameter = 31.8 khz
20cm Diameter = 23.9 khz 1st harmonic = 47.8khz Most appropriate size/frequency for testing cavitation
32cm Diameter = 14.9 khz 1st harmonic = 29.8khz
48cm Diameter = 10 khz 1st harmonic = 20 khz
The idea i have is to focus the frequency to the centre of the sphere. if i then place a steel ball bearing in the centre of the sphere cavitation will take place on the all bearing surface. the bearing and surrounding water will then heat up.
well thats the theory anyway.
any ideas / comments welcome
Hmmm-one of us is making a math error.
It's probably me.
What value are you using for v ?
EDIT: well, I've checked my math and I'm still coming up with numbers a tenth of yours. Damn those decimals anyway.
I have a feeling its all about resonance of the water, from possibly 2 directions inside the sphere...itself serving to conduct and reflect the waves created be the 2 sources, so that the resonant waves clash.
I am no expert in this field...this is just my take...hopefully it may lead to something.
Regards...
Maybe if you lightly strike the whole unit when full of water and record it with a good mic on your PC then you could use a WAV editor with a spectrum analyser function (like goldwave) to give an idea about the systems mechanical freq. response. Only good up to 22kHz though.
edit: idea is similar to sonoluminescense experiments aka star in a jar.
edit2: wow this forum´s getting shakey, just clicked on "modify" in cap-z-ros post by mistake and was taken to the editor :o
Yucca.
Quote from: Hydro-Cell on December 12, 2008, 05:43:04 PM
Theoretical Resonant Frequency Of A Sphere Filled With Water...
F = V / (2*PI*R)
Can you adapt this formula to apply to hemispherical bells open
to the water they are submerged in? I am thinking of the Peter
Daysh Davey heater (thread to be found here:
http://www.overunity.com/index.php?topic=4083.430)
Description:
http://www.free-energy-info.co.uk/D15.pdf
on page 20
Paul.
I think in that case the Helmholtz resonator formulae would probably be appropriate--although I don't know about the hemisphere part--
HydroCell
This resonator could speed things along
http://www.overunity.com/index.php?topic=6225.50;topicseen
Thanks Fellows
Chet