Lecture 2-2 Part 1. Baffled Piston Problem 1

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En provenance du cours de Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

7 notes

Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

7 notes

Learners might have learned the basic concepts of the acoustics from the ‘Introduction to Acoustics (Part 1).’ Now it is time to apply to the real situation and develop their own acoustical application. Learners will analyze the radiation, scattering, and diffraction phenomenon with the Kirchhoff –Helmholtz Equation. Then learners will design their own reverberation room or ducts that fulfill the condition they have set up.

À partir de la leçon

K-H Equation & Baffled Piston Problem

First, you will learn Kirchhoff-Helmholtz equation, then you will adapt into the baffled circular piston problem.

Lecture 2-2 Part 1. Baffled Piston Problem 1 30:32

Lecture 2-2 Part 2. Baffled Piston Problem 212:03

Lecture 2-2 Part 3. Radiation from a Finite Vibrating Plate 19:32

Week 2 Summary Video4:37

Rencontrer les enseignants

Yang-Hann Kim
Professor
Mechanical Engineering

Let me continue to talk about the, Radiation

from A baffled,

Piston.

Okay, if I draw a baffled piston and it look like that,

and say It has a radius of a for simplicity,

let's see the baffle has a round shape.

And it is oscillating in minus.

Okay, [COUGH] and assuming that it has

some velocity profile, like that.

And I like to denote that is Un.

The reason why we are studying this,

as I explained in the last lecture,

if you know the radiation pattern of this,

then we could have some good idea.

How the radiation due to for

example, this kind of plates.

If the plates is vibrating with some different faces,

then we could have some good idea how

the radiation pattern of this kind of plate could be.

So, and then we would like to expand the idea of having

understanding about this kind of radiation, we could further.

[SOUND]

Expand what would happen if we have this

kind of opening and how the, this opening will radiate the sound.

And then we can go to the radiation of this kind of thing,

because this is the simply the half space of this kind of radiation,

so understanding the radiation of this kind of pattern

would lead us to understand this kind of diffraction problem, and so on, so on.

And we do have some theoretical theory

that can, they can allow us to predict

the sound radiation due to this kind of geometrical arrangement.

And that was pLv)=.

That simply says the sound

at any point Spatial distribution

sound at any point can be said

as the sum of the.

Sum of the sound

due to, for

example [SOUND]

add a p d g and

the at ds0.

And is the normal vector from

the surface of scatter for example.

And in this case, the unit vector would be this one.

In some literature or in the book we often

write this one as the p r

has to be integration of g d

p d s d phi d r zero vector and

minus PdG/ dr0 vector, okay?

And dS0 and saying that this is the object or boundary.

And n beta go that way.

And this is surface of S0 and

this is r0 vector and r vector is somewhere over there.

And then looking at this, we could say that this is on dS0,

therefore, the contribution of this one has to be normal direction,

otherwise because we assume there's no viscosity therefore all this has to be,

this can go to the dn

and this can go to dn because of no viscosity.

[SOUND] Okay,

so lets go

back to this

problem.

And lets apply the to predict the sound of radiation from this.

So let's denote this as R0.

And we would like to predict the sound at any point over here.

And you can write the sound at any point.

It will be sum of S0 I, I would say,

I will write S0 G B P D N,

minus P G G G N, dS0.

Okay, that is true.

As I mentioned, some of the books use the 1 over 4 pi over there.

Or some of the books, we can see this term comes here and that term goes away.

But that all depends on the way to select the green's function.

Green's function scale effect that can

somehow accommodate the one over four pi or or

even one over two pi or minus and plus sign over there.

Okay, then we can see that here, this is rigid wall.

Therefore fluid particle cannot penetrate in the wall in

other words the velocity in this direction has to be zero over there.

Okay?

And we can think that this problem can be

I mean there's some pressure due to

this buffled piston can be regarded

as the same as the sound field due to this kind of radiator.

Okay, that is also oscillating exponential minus [INAUDIBLE] and

it has distribution of lower velocity like that.

But also has a same distribution of lower velocity.

Okay.

What we can anticipate from this kind of oscillation because it is symmetry.

The velocity over this line has to be zero because of the symmetry.

So this problem and this problem is the same problem.

Okay, that's the idea.

Okay now, let's say this is the surface,

S0 minus and this is surface S0 plus.

And we have to find out all this

four times on this surface.

Okay, the pressure, r0 on this surface and

the pressure r0 on the other surface.

Would be same because there's no,

Pressure on our S0 minus in this [INAUDIBLE] process the same surface.

And the pressure is a scalar of the vector.

And the G, the [INAUDIBLE] function

on S0 minus, S0 plus, sorry,

Grange function on S0 plus,

would be the same?

And then what about the DG/Dn?

DG/dn on S0 minus,

DG/dn on S0 plus.

Its magnitude is the same, but it has different sign.

So if I say this is plus, this has to be minus.

Because the direction of the normal direction.

Therefore, we can immediately see the contribution of this term,

contribution of this term.

Is okay, there is a contribution of this one because dP/dn and

the G of S0 and S minus is not necessarily cancel out.

But this term, dG/dn will be cancelled out.

S0 minus and S0 plus, okay?

Therefore, this is the only remaining term.

Now Therefore, I can write

The pressure at any point

is equal to integration,

surface integral, S0 plus

S0 minus G dP/dn, dS0.

It's a rather simplified form.

Okay, and the G we can select.

What is the G?

The G is the Grange function that satisfy the in homogeneous wave equation.

So let's see and then the physical meaning of G is it

propagates the velocity which related to the dP/dn from the surface.

So let's say Okay,

let's say this is r vector and

that is simply r minus r generator.

That it is simply, its magnitude is simply

the distance from any point on the surface of that is

vibrating to the the observation point, okay?

Then I can say G,

what a candidate of G would be monopole.

1 over R exponential jkR,

okay, but I selected this.

And what about this term, dP/dn?

I think some student at Carnegie Institute of Technology

asked me a similar question that is related with this, okay.

DP/dn, that is dP/dn

on S0, right?

That has to be related with this.

In such a way that the information is propagating using this function, okay?

Then what is dP/dn on surface S?

S0 plus or S0 minus.

What is this?

That is related with the velocity because of the Euler equation.

So Euler equation says,

the gradient of pressure in space is related with

This, okay.

And noting that, And

noting that, I'll erase this.

Okay, we have velocity,

its magnitude.

I can say, so Un in this case,

exponential [INAUDIBLE] t Therefore,

this one I can write minus gradient of

pressure is equal to rho 0 Un minus j omega, okay?

If I take out the time dependency,

I can say this is gradient of p,

where p is the p exponential minus j omega t.

Okay, that p is this p, because we are only handling

spatially dependent pressure magnitude, okay?

So now what I can say is in normal direction,

I can write minus dpdn is equal to rho 0 Un minus j omega.

Therefore, I can write dpdn is equal to plus,

because I have minus and minus over there, rho 0 Un j omega.

And I use this Sperger relation that is k equal to omega over c over here.

And I can write that is rho 0 Un.

I have j over there, and omega is kc.

Therefore, I can write instead of omega kc.

And I will put c over there because that is the characteristic impedance of medium.

So rho 0 ck Un.

That is the dpdn.

Okay, the strangest things I did.

Instead of solving integral equation,

I took advantage that dpdn is velocity on the surface,

so I don't have to solve integral equation.

Ha!

Okay, so

having the magical mathematical property

associated with this Kirchhoff Elovich equation,

I certainly have succeeded to predict some pressure over there with everything known.

So let me rewrite what we've done.

Then what I have is,

Using this blackboard.

What I have now is the pressure

at any point is surface integral, okay.

G, that is 1 over R exponential jkR.

Make sure.

And I have dpdn,

that is over there.

j rho 0 c and

kUn, right?

And I have surface integral dS 0.

So rearrange what we have and

I can write, Okay, one thing I missed.

This surface integral has to be performed on both side.

Therefore, I have scale factor of 2.

In fact, it doesn't really matter, but If we stick to

the derivation then I have to put the scale factor of 2 over there.

And so that gives me

j2 and rho 0 c.

k can come out and Un.

And then I have surface integral

e to the jkR over R dS 0.

Okay, this Un has to go into over here.

And so what we have is minus

jk rho 0 C over 2 pi integration

R exponential jkR UndS 0.

The difference is, Okay,

this is and these are same.

Okay, and the difference is this.

Rho 0, c, k, rho 0, c, k, 2, j.

So difference is if I multiply 1 over 4 pi and

minus, then two equations are same.

So that's the scale factor difference.

If I use the Green's function as exponential jkR over R,

1 over 4 pi, and minus,

then I will get the same result.

So we can predict the sound using

either this formulation or

that formulation.

Now let's do this.

Let's do this.

Using that formulation, we can have non-irradiation

due to the baffled piston due to non-uniform Un, okay?

But assuming, for simplicity, that Un is constant.

In other words, baffled piston is oscillating this way.

Then the formulation just depends on what?

Integration R,

e to jkR dS 0.

So what we have to do is mathematically,

we have to calculate this mathematical expression.

Okay?

What physically means.

Physically means,

The radiation, Due to baffled piston,

when piston is oscillating with constant on Un.

Okay, now Un is constant.

It is simply the radiation due to

distributed on the surface.

With the velocity of Un.

This is rather surprising result.

And everything besides that is a scale factor, okay?

So what does it mean, what does it mean?

And it propagates 1 over R exponential jkR, and

capital R is the distance from here to there.

So to see the physics associated with this expression,

let's see what's going on on this line first.

And then what kind of radiation could happen

besides this line we will see later on.

Okay, what you can guess the radiation on this line?

Okay?

If wavelength is very long, then what I

can see on this line would be the radiation

coming from mainly from the center, but

a small contribution from the edge, okay?

I don't know whether that's going to be true.

In fact, that's not true if you get the result.

Actually what we will have is the radiation from the center and

then radiation from the edge.

And then nothing really significant will contribute in between.

Okay, so in terms of sound radiation having effective sound radiation,

and if you want to have some pressure maximum over here,

you don't have to derive all the disk at the same time.

You only have to derive the ring and the center.

Why that happen physically?

Because it radiates a sound as if there are many monopoles.

Those monopoles' contribution somehow cancel out, and

the contribution from the edge and the contribution from the center only exists.

That's one thing that we learned from this theory, okay?

And the other one is when the wavelength is small compared with the size of a,

then the contribution of each for this edge and

the center are not following what I just described.

But obviously one is the contribution from the center would be very strong.

Getting stronger and stronger as ka getting smaller and smaller.

In other words, wavelengths is small compared, wavelengths is?

Can ka getting larger and larger,

it means that wavelengths is small compared with the disk size.

We will see the directivity, because if

the wavelength is small compared with the size of a,

for example, we have three waves over here,

then this contribution will be very strong and

the contribution away from the center will be

diminishing, having some directivity factor.

So for ka is large, then we have very large directivity.

The other way to understand the physics.

Suppose you have fluid particle over here.

When ka is small, which means that wavelengths is large

compared with the disk size, and when it is oscillating,

then all the fluid particles would tend to oscillate

at the same time in the same manner, equal phase.

So it radiates like that.

But as ka is getting larger and larger,

the wavelength is small compared with the disk size,

all these fluid particles tend to oscillate in different manner,

in other words, different phases, okay?

So let's see what really the theory

exhibits what we just imagine.

Okay, let's see the theory.

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