Physics Puzzle

[Bob Scott]

Lately I've been independently trying to figure out how acoustics works. Thiele and Small applied electrical formulas to loudspeaker baffle design. I've been trying to extend their work if I can. For the last few days I've been trying to figure out the acoustic impedance or resistance of a hole, the hole being like a base reflex port in a cabinet. I think that I have almost found the answer.

Here is a problem I dreamed up:

There is a one cubic meter box set up like a piston and cylinder. The cylinder is a 1 meter cube with no fixed top. On top is a 1 square meter frictionless piston that slides to the bottom. The box bottom has a square hole in it 10cm X 10cm.
The top has a 1 Kg weight on top that forces it down, resulting in the air pressure in the box being 1Kg per square meter. If you let go of the 1 Kg mass, the 1 cubic meter of air in the box is forced out the bottom hole in a given time period.

The questions are:
1) How many seconds does it take the air to evacuate from the box?
2) What is the velocity of the air as it leaves the hole in the box?

You need to know:
The density of air is 1.2 Kg per cubic meter at 20 C (room temperature).

I thought that I would have to experiment to get the answer, but I figured it out. Wanna try?

Hint: If you have a complicated scientific problem, always try tracking the energy.

[jimmy101]

The velocity of the air is not constant so question #2 doesn't make any sense.

The pressure in the box as given is the gauge pressure and not the absolute pressure. The absolute pressure would be assumed to be 1 ATM + the 1kg/m^2. Since the air stays in the box the actual pressure in the box is 1 ATM until the piston is released. That's ok since 1kg/m^2 is only about 0.000097 ATM.

Simplest solution, and one that is probably in the ball park, is to ignore the box and the hole and the air and just ask how long does it take an object to fall 1 meter (the height of a 1m^3 volume with a 1m^2 top). That's how long it takes to pump the air out'a the box.

D = 1/2 A T^2
D = 1 meter
A= 9.6 m/sec^2

T= sqrt(2D/A) = 0.46 seconds

A more precise answer would still ignore the hole but takes into account the energy needed to accelerate the mass of air. The air doesn't "fall" like the 1kg lid does. The density of dry air at 1 ATM is 1.3Kg/m^3. So the mass of air is more than the mass of the weighted piston. Hence the energy needed to move the air is significant compared to the force from the falling weight.
But, the mass of the air decreases as the piston falls since once the air has passed the hole it no longer needs to be accelerated. I'm to lazy (and my physics and math to rusty :p) to work out the exact equations but I suspect the answer using this model is pretty darn close to the original answer. Early on in the falling process there is a lot of air to move but the piston is moving slowly so the acceleration of the air is small, which in turn mens little energy is being used. Later on the piston is moving faster but there is less air to move so less energy is required.

An even more precise answer has to take into account the friction of the air moving through the hole. This could get to be extremely complicated. The shape of the hole, the smoothness of the hole, the thickness of the container wall all affect the drag force on the air moving through the hole. At low speeds things are calculable. At high speeds turbulence may set in and an exact solution really isn't possible.

[Bob Scott]

Hi Jimmy.

For this experiment, ignore the acceleration, inertia, impulse, or momentum of the weight. Lets just say that the weight causes a downward force on the top of the piston equal to the weight of 1 Kg. f= m X a. "a" is the acceleleration due to gravity which is 9.8 m/sec^2, so the downward force on the volume of air is a contant 9.8 Newton.

Since the top of the piston is 1 m^2 (one square meter), the air pressure inside the box, as the weight drops, stays constant at 9.8 Newton/m^3.

Hint: Think of getting all of the air out of an air mattress. You squeeze the mattress to get the air out. The air comes out through the filler cap. The harder you squeeze, the faster the air comes out. Given the size of the filler hole of 10cm X 10cm, and the pressure of 9.8 N/m^2 squeezing the mattress, and the density of air at 1.2 Kg/m^3, and the volume of air in the mattress of 1m^3, how long will it take to empty the air?

This all started with me trying to figure out the acoustic resistance of a hole. There is air pressure, acoustic voltage in Kg/m^2 (analagous to electric voltage), a rate of volume change with time in m^3/s (analagous to electric current in (coulombs per second)). R = V/I.

Hint: (Yes, another one) At the beginning of the experiment, the 1KG weight has a height of 1 meter. At the end, it has dropped 1 meter. The energy expended during the experiment is 9.8 Newton-meters.

PE =force X distance. PE = m X g X h. PE is potential energy. Energy is also known as "Work". In science books, the symbol used for energy is "W".

[MrAl]

Is this going to be on the test?

Notes:
1. he he
2.
3. Repeat above 10 times.

[jimmy101]

As a starting point for calculating the impedance of a hole in air this particular example is actually trying to calculate the "resistance" of the hole. The impedance of the whole won't be the same.

The impedance of a hole can be darn near infinity even for cases where the resistance of the whole is fairly small. For example, the open end of a flute provides nearly zero resistance to air flow at normal flow rates (say under lung power). But the open end of the flute has very high impedance at the natural resonant frequency of the flute. The open end of the flute behaves almost as if it is closed. High impedance but nearly zero resistance.

[jimmy101]

Dang, multiple cases of confudling "whole" with "hole"!

[CeaSaR]

Ah, but since air is considered a type of fluid, shouldn't fluid dynamics be a part of the equation?

Since you are dealing with an opening smaller than the size of the "piston", that opening will only
allow X amount of mass through at a given pressure. And, since you are trying to force the same
amount of "fluid" through the opening as you would be pushing out of the "box" at 9.8 M/S, you can
only have equilibrium if the fluid being forced through the opening increases in rate of
flow. At some point, due to friction of the openings (both edge shape and wall roughness) there
will be oscillation. This would be the point of resonance, ie., the point at which turbulance consumes
the greatest amount of energy and makes the piston seem to stand still - the speaker equivalent
of bass resonance where the cone has very little movement and the port produces most of the sound.

I will have to brush up on my fluid dynamics (quite a lot!) in order to come up with equations based
on these assumptions, but I believe that some of the old storm water management programs I used to
use will help with things like coefficients, etc., even though they deal with water, not air.

CeaSaR

[Bob Scott]

The way I see it, the experiment starts out with the 1 Kg weight starting to push the 1 m^3 cube of air downward, releasing 9.8 Newton-meters of potential energy and giving the air in the box 9.8 Newton meters of kinetic energy. 1 Newton-meter = 1 Joule.

How fast? How long?
The air is forced out of the box through a 10cmX10cm hole, extruding it, just like squeezing toothpaste out of a tube. The extrusion is 10cm X 10cm X 100 meters long and still 1m^3 in volume. The equation for kinetic energy is KE = 1/2 X m X v^2. M is the mass of the air at 1.2 Kg. KE is 9.8 Joules.
V = SQRT(2 X KE / m)
or
V = ((2 X K) / m)^0.5
= 4.04 m /s

The extrusion is 100 m long, so it takes about 25 seconds (100 / 4.04) to finish. The 1 Kg weight takes 25 seconds to sqhish all the air out. What happens to the air after it leaves the box is not a concern, but I think that its velocity and energy is lost through turblence and through friction, all ends up as heat, like just about all energy we use.

I took some time yesterday to do multiple calculations at different speeds and hole sizes. I was puzzled because the resistance value calculated as pressure/volumetric velocity did not stay constant. The resistance changes with pressure, so a hole does not make a resistor equivalent.

Too bad. By the way, it takes 4 times the pressure to double the velocity of pushing air out of an air mattress. And since the velocity is doubled, 8 time the power is required to halve the time of emptying the air out of the mattress. No wonder it doesn't help much to squeeze harder.

[CeaSaR]

Bob,

Any particular size hole (dependent upon shape) will flow X amount of "fluid", be it air, water or other. As
you know, flow consists of rate, volume and density. The maximum flow can only be achieved through a
combination of these factors. In order to change one, you must change at least one other. To increase
rate, you either have to change volume or density. Increase volume, change rate or, most likely, density.
Increase density, you have to change (probably both) rate or volume. This, all assuming perfect openings
and wall shapes/surfaces, otherwise you need roughness coefficients for them.

This is why, as you say, it really doesn't deflate the mattress twice as fast when squeezing harder, because
the opening will only flow "so much".

No, I don't have the math yet, just my years of work with underground pipes (civil design - all CAD, very little
hand work) and more closely related, intake and exhaust ports and associated tuning (mostly reading here).

CeaSaR

BTW, did you find the type of relationship in all your calcs? Obviously it wasn't linear, so is it close to LOG or EXP?

P.S. I know, it is pretty much a rehash of my first statement, but it still holds true. This is not a simple problem with
only a few variables. There are many to be considered.

[gerty]

Don't forget the fact that air is compressable, and fluid isn't.

[CeaSaR]

Applause! Yet another factor. Actually, there is an amount that normal fluids can be compressed, but
it is so minute as to be disregarded. Regular fluids have very little space between molecules where the
fluid "air" (gaseous form is a fluid) has very large (comparitively) amounts of space between molucules,
hence the "compressability". It all comes down to density, because acoustics behave differently in air
and other things such as water and even metals.

Ok, 'nuff for now.

CeaSaR

[Bob Scott]

Hi Gerty.

Don't forget the fact that air is compressable, and fluid isn't.

Yes, but at very low amplitudes, you can calculate "small signal characteristics". For example. 1 kilogram sitting on top of a 1 square meter of piston only has a pressure of 0.0014 psi. There's not much compression going on there considering that the ambient barometric air pressure is ~14.7 psi.

[Bob Scott]

Bob,

Any particular size hole (dependent upon shape) will flow X amount of "fluid", be it air, water or other. As you know, flow consists of rate, volume and density. The maximum flow can only be achieved through a combination of these factors. In order to change one, you must change at least one other.No, I don't have the math yet, just my years of work with underground pipes (civil design - all CAD, very little
hand work) and more closely related, intake and exhaust ports and associated tuning (mostly reading here).

I was thinking of a hole as a resistance. For example, a water tap is a variable hole, and so are the adjustments in those air distribution controls in the air ducts that fine tune the amount of heated air from your central heating furnace plenum to different parts of a your house.

BTW, did you find the type of relationship in all your calcs? Obviously it wasn't linear, so is it close to LOG or EXP?

It is exponential. Electric current has no mass. Current is proportional to voltage. But in acoustics, air has mass and velocity does not track air pressure. I'm finding that I can calculate the resistance of a hole quite well at 1 newton per m^2, but if I raise the pressure, the calculations don't produce the same result. The apparent resistance of the hole changes with pressure. Either that or I'm making calculation mistakes or I'm not competent enought to figure this out.

I wonder if my old buddy was right. He made himself a pair of big industrial grade "JBL 030" speakers, and he swore that the fidelity and transients sounded better the louder he played them. If their characteristics changed with pressure level, he could be right.

[CeaSaR]

So the hole is not analogous to resistance, rather it is impedance.
The hole has reactivity with relationship to both pressure and
frequency. Frequency determines the resonance peak of the hole
and pressure determine the range in which that peak falls. The
resonance peak should be analogous to the point at which the
hole is most efficient. Don't forget that a hole needs to have 3
dimensions in order to be modeled properly. Resonance changes
with length in respect to frequency and volumetric pressure (set
by the size of the driver and the box in which it is housed) for
a given opening size.

I suggest reading into how tuning intake runners and exhaust
pipes works. There should be some theory and math there that
could help further your research. Oh, also how they design mufflers.
They work on acoustics also, just in the cancellation of sound.

CeaSaR

OOPS - forgot to mention that phase needs to be studied too, since this
ultimately is leading to "AC" operation, not just the "DC" operation thus far
discussed. The phase at the work end (output) must be in phase with the
speaker cone in order to reinforce that particular movement. Therefore,
the output of the "hole" must be in time and 180 degrees to the cone
movement inside the enclosure. Now we are getting into Wave Mechanics...

'Lucy, I tole you it wass more complicated than you though...'

[MrAl]

"Oh Ricky, waaa waaaa waaaaaa..."


Bob,

I'd like to see your calculation thus far and see what's what.

There is a difference with pressure and dont forget the difference with hole size. If the hole is small enough even small pressures will effect the outcome. There's the venturi effect and also possibly later the choked effect, but im willing to ignore all this for the time being and see what you've come up with so far.

What could the resistance be for a hole in plywood that is large compared to the thickness of the plywood if there is no substantial flow.