Fundamentals of Acoustically Coupled Combustions

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6-1

DYNAMICS OF COMBUSTION SYSTEMS:

FUNDAMENTALS, ACOUSTICS, AND

CONTROL

A Short Course of Lectures

VI. FUNDAMENTALS OF ACOUSTICS

F.E.C. CULICK

California Institute of Technology

September 2001

2001 by F.EC. Culick

All Rights Reserved


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FUNDAMENTALS OF ACOUSTICS

According to the experiences cited and the reasoning given in Section I,

combustion instabilities may be regarded as unsteady motions closely

approximated as classical acoustical motions with perturbations due to

combustion processes.

The framework for analyzing and interpreting combustion instabilities has

from the beginning been constructed with strong guidance by the principles

of classical acoustics. Moreover, as a general strategy in studying

combustion instabilities it is useful to concentrate first on gas dynamics as a

means of interpreting observed behavior. Deficiencies then require appeal

to other processes, notably those associated with combustion.

It is therefore essential to have a sound understanding of the behavior of

small amplitude motions in a compressible medium with no combustion

processes.

This section covers the parts of classical acoustics which in some degree

influence the subject of combustion instabilities.

The material included here covers only the essentials of small amplitude

motions in chambers and tubes, with no combustion and average flow.

Inhomogeneities in the medium are not included; interesting problems

therefore arise only if boundaries and sources are explicitly present.

In the narrow form considered here, acoustics comprises problems of wave

motion interior to regions partially or totally enclosed by prescribed

boundaries.

References: Morse, P.M., Vibration and Sound

Morse, P.M. and Ingard, K.U., Theoretical Acoustics

Landau, L.D. and Lifschitz, E.M., Fluid Mechanics

Temkin, R., Elements of Acoustics



6.1 LINEARIZATION OF THE EQUATIONS OF

MOTION; THE WAVE EQUATION AND THE

VELOCITY POTENTIAL

The influences of viscous effects are normally small within the

volume; they are examined later. Here the discussion is based

on the inviscid equations of motion for a homogeneous medium

free of inhomogeneities and sources:

Conservation of Mass

Conservation of Momentum

Conservation of Energy

Equation of State

0=⋅∇+∂

∂ )u(

t

r ρ

ρ

0=∇+∇⋅+∂

∂ puu

t

u rrr ρ ρ

02

1

2

1 22 =⋅∇+

+∇⋅+

+

∂

∂ )u(pueuue

t

rr ρ ρ

RT p ρ =

Remove the kinetic energy from the energy equation by

subtracting

0or 0 =⋅∇+=⋅∇+∇⋅+∂

∂u p

Dt

De u peu

t

e rrr ρ ρ ρ

( ) :equationmomentum⋅ur

6.1.1 Conservation Equations of Motion



The equation for the entropy of a fluid element is:

Substitution of the mass and energy equations gives:

Because all irreversible processes have been ignored, the motions

within the volume are necessarily isentropic.

∇⋅+∂

∂=−= u

t Dt

D

Dt

D

Dt

De

Dt

Ds r

p ;

ρ

ρ ρ ρ

0)( =⋅∇+⋅∇−= u p

u p Dt

Ds rr ρ

ρ ρ

Changes of density are given by:

will turn out to be the speed of propagation of small disturbances —

the “speed of sound”

dpadp p

dp p

ds s

d s s p

2=

∂

∂=

∂

∂+

∂

∂=

ρ ρ ρ ρ

Where we assume an isentropic process and

s

dpa

∂= ρ 2



VELOCITY POTENTIAL

6.1.1 Conservation Equations of Motion (cont’d)



The continuity equation can be written now for the pressure:

For more general applications it is useful to obtain this equation for

the pressure by adding the mass and energy equations with de =

C vdT and using the equation of state:

)( ∗=∇⋅+⋅∇+∂

∂ 02 puua

t

p rr ρ

( )

( ) (2) 0

(1) 0

=∇⋅+⋅∇+∂

∂

=∇⋅+⋅∇+∂

∂

T uuC

p

t

T

T uut

T

v

ρ ρ

ρ ρ ρ

rr

rr

( ) ( )

01

0

=∇⋅+⋅∇

++

∂

∂

=∇⋅+⋅∇

++

∂

∂

puu pC

R

t

p

T uuC

pT T

t

v

v

rr

rr ρ ρ ρ

But R = C p – C v, so R/C v = γ − 1 and equation (∗) is recovered

because γ p = ρa2 for the conditions assumed here,

ρ

γ pa =2

Add (1) and (2):



VELOCITY POTENTIAL




Note also the relation for the internal energy of a perfect gas,

because e = e(T), is a function of temperature only for a

perfect gas.

and the momentum equation can be written:

dT C dT T

edv

v

ede v

vT

=

∂

∂+

∂

∂=

For isentropic processes of a perfect gas,

00

γ

ρ

ρ

= p p

)( ∗∗=∇

+∇⋅+

∂

∂ 0

11/

0

0

p p

puu

t

u γ

ρ

rrr

Wave Equation for the Pressure

Differentiate (∗) with respect to time and substitute (∗∗) for

with a2 = γ p/ρ:t / u ∂∂r

( ) ( ) put

ut

puu p

p/p

p

p

pa

t

p∇⋅

∂

∂−⋅∇

∂

∂−∇⋅⋅∇=

∇⋅∇

−

∂

∂ rrrrγ γ

γ 1/00

202

2

)(



VELOCITY POTENTIAL




The boundary condition for this equation is set by taking the

component of (**) normal to the boundary:

6.1.2 Linearization of the Conservation Equations

Let ε be a small parameter characterizing the amplitude of

time-dependent motions superimposed upon a uniform state of

a gas at rest:

( )

∇⋅⋅+

∂

∂⋅

−=∇⋅ uun

t

un

p

p pn

rrr

ˆˆˆ 0

1/

0

ρ

γ

uu

p p p

′=

+=′+=

rrε

ερ ρ ρ

ε

'0

0

Nonlinear behavior arises in the present context mainly from

convection (i.e. etc.) and from the dependence

of the speed of sound on amplitude, .

pu ,uu ∇⋅∇⋅ rrr



VELOCITY POTENTIAL


ρ γ γ / p RT a ==



Substitute the assumed forms and expand to second order:

( ) ( )

( )

...11

...2

111

1

11

2

00

2

0

200

2

0

22001/

00

20

2

02

2

01/

01/

0

+

′∇−

′∇

′−+

′∇=

∇⋅∇

+

′

++

′−=

+=

p

p

p

p

p

pa p

p

pa p

p/p

p

p

pa

p

p

p

p

p'/p p/p

γ γ

γ ε

ε

γ

γ ε

γ ε

γ

γ γ

Substitute in the wave equation for p, collect terms according to

powers of ε and divide by ε :

The boundary condition becomes

Problems of linear acoustics are described by the equations obtained

in the limit ε = 0:

...11

)

2

00

2

0

200

022

02

2

+

′∇−

′∇

′−+

′∇⋅′

∂

∂−′⋅∇

∂

′∂−′∇⋅′⋅∇=′∇−

∂

′∂

p

p

p

p

p

pa p

pu( t

ut

p )uu( p pa

t

p

γ γ

γ

γ ε rrrr

( ) ...ˆˆ1

ˆˆ000

+

′∇⋅′⋅+⋅

∂

′∂′−⋅

∂

′∂−=′∇⋅ uunn

t

u

p

pn

t

u pn

rrrr

γ ερ ρ

nt

u pn

pat

p

ˆˆ

0

0

2202

2

⋅∂

′∂

−=′∇⋅

=′∇−∂

′∂

r

ρ



VELOCITY POTENTIAL

6.1.2 Linearization of the Conservation Equations (cont’d)



It is often convenient to introduce the scalar and vector potentials φand from which the velocity is found by differentiation: A

r

Aurr

×∇+∇−=′ φ

With this representation, the dilatation and curl (rotation) of the

velocity field are separated:

Auurrr

×∇×∇=′×∇−=⋅∇ ∇ ; 2φ

Here only the scalar potential is required for linear motions

because for small amplitudes, the pressure and momentum

equations become the equations for classical acoustics:

01

0

o

o

=′∇+∂

′∂

=′⋅∇+∂

′∂

pt u

ut

p

ρ

γρ

r

r



VELOCITY POTENTIAL

6.1.3 Velocity Potential



The pressure fluctuation is found from the relation:

022

2

2

=∇−∂

∂φ

φ oa

t

t p

∂

∂=′

φ ρ o

The second equation gives which means

Although it appears that non-zero may arise in

this limit, that is not the case in the present context.

,0=′×∇ ur

. A 0=×∇×∇ r

Ar

Combine the above two equations and substitute to fix the equation

for the velocity potential:



VELOCITY POTENTIAL

6.1.3 Velocity Potential



6.2 ELEMENTARY SOLUTIONS TO THE WAVE

EQUATION: PLANE, SPHERICAL, AND

CYLINDRICAL WAVES

6.2.1 Plane Waves

The basic property of linear problems is that the principle of

superposition applies. Solutions for more complicated problems can

often be constructed by superimposing elementary solutions.

Wave equation: 02

22

2

2

=∂

′−

∂

′∂ ∂ x

pa

t

po

can be factored 000 =′

∂

∂

−∂

∂

∂

∂

+∂

∂

p xat xat

and a general solution has the form

Pressure Waves t)a(x g t)a f(x p 00 −++=′

t):a(x g 0−

t):a f(x 0+ Wave to the left

Wave to the right

x: direction of propagation, so wave fronts or planes of

constant phase are normal to the x-axis.



Density Waves

For a wave traveling to the right:

[ ]t)a(x g t)a(x f a p

p002

00

0 1' −++=

′=

γ

ρ ρ

Velocity Waves

Integrate [ ]t)a(x g t)a(x f p

a

t

p

p x

u002

0

0

0

1−′−+′=

∂

′∂−=

∂

′∂

γ γ

[ ]t)a(xgt)a(xf p

au 00

0

0 −−+−=′γ

00a pu ρ

′=′

For a wave traveling to the left:00a

pu

ρ

′−=′



CYLINDRICAL WAVES

6.2.1 Plane Waves (cont’d)



Solution to the initial value problem

Initial conditions P(x) )(x, p =′ 0

Q(x) )(x,t

p=

∂

′∂0

( )0=t

The functions f and g are determined from

[ ](x) g (x) f aQ(x) )(x,dx

pd ′−′==

′00

g(x) f(x) P(x) )(x, p +==′ 0

Solution:

[ ] ∫

+

−

+−++=′t a x

t a x

d Qa

t)a P(xt)a P(xt)(x, p0

0

)(2

1

2

1

000 ξ ξ

Example: Rectangular pulse initially at rest



CYLINDRICAL WAVES




Example: Reflection of a pulse

Initial conditions:

00

0

=∂

′∂

=′

)(x,t

p

P(x) )(x, p( )00 ≥= x ,t

Boundary conditions:

00

00

=∂

′∂

=′

)(t,t

p

)(t, p

( )00 ≥= t , x

Solution:[ ]

[ ]

≤−−+

≥−++=

t a xt)a P(xt)a P(x

t a xt)a P(xt)a P(x

p' (x,t)

000

000

2

1

2

1



CYLINDRICAL WAVES




Sinusoidal plane

waves)i(

(

0

)i(

)

0

00

00

t kxt)a x

ai

t kx

t a(x

a

i

Be Bet)a g(x

Ae Aet)a f(x

ω

ω

ω

ω

−±−±

+±

+±

==−

==+

frequency

λ

π ω

λ π π ω

2

22

0

0

==

==

ak

a f

wavenumber

6.2.2 Spherical Waves

Assume motions symmetric about a point: wave will propagate

inward or outward.

Wave equation: 01 2

2

20

2

2

=

∂

′∂

∂

∂−

∂

′∂

r

pr

r r a

t

p

Try a solution: )(1

t r,r

t)(r, p ψ =′

02

2202

2

=∂

∂−

∂

∂

r a

t

ψ ψ

and a solution for the pressure has the form:

Pressure Waves [ ]r)t G(ar)t F(ar

t)(r, p −++=′00

1

r):t F(a +0

r):t G(a −0

inward traveling wave

outward traveling wave



CYLINDRICAL WAVES




An initial value problem for spherical waves

Initial conditions

Assume p′ finite at r = 0 :

To satisfy initial conditions:

Pressure Field:

W(r) )(r,t

p

V(r) )(r, p

=∂

′∂

=′

0

0)0( =t

[ ] [ ]t)G(at) F(ar

(r,t) pr

r 000

0

1+

=′

→→

)()( ξ ξ F G −=⇒

Hence [ ]r)t F(ar)t F(ar

(r,t) p −−+=′00

1

)(1

)()(

)()()(

0

ξ ξ ξ ξ

ξ ξ ξ

W a

F F

V F F

=−′−′

=−−

Assume medium initially at rest, W(r) = 0

>−−+++

<−−−++=

t)a(r t)at)V(r a(r r)t r)V(at (a

t)a(r r)t r)V(at (ar)t r)V(at (a

r p'(r,t)

00000

00000

2

1



CYLINDRICAL WAVES

6.2.2 Spherical Waves (cont’d)



Example

>

≤<=

0

00

0

0)(

r

r pV

ξ

ξ δ ξ

>−

>

≤−<

−−=

>−

<

≤−<

−−=

00

0

000

0

00

0

000

0

0

0

2

1

0

0

2

1

r t)a(r

t)a(r

r t)a(r p

t)a(r r

r r)t (a

t)a(r

r r)t (a p

r)t (ar

p'

δ

δ



CYLINDRICAL WAVES




Wave motion subsequent to an initial stationary pulse

Unlike the case in one dimension, the outward propagating

compression wave is followed by an outward propagating

rarefaction produced by reflection, at the origin, of a wave

propagating inward from the edge of the initially pressurized

region



CYLINDRICAL WAVES




Velocity in an outward propagating spherical wave

+′+−=

∂

′∂−=

∂

′∂r)t (a F

r r)t F(a

r r

p

t

u002

00

1111

ρ ρ

∫∞−

′−′+−=′t

t r)d t F(ar

r)t F(ar a

(r,t)u 020

000

1111

ρ ρ

∫∞−

′′+′

=t

t d p

r a

pu(r,t)

000

1

ρ ρ

00a

pu(r,t)

ρ

′=

Spherical wave

Plane wave

In a spherical wave, the velocity disturbance may be zero where

the pressure disturbance is nonzero unless

∫ ∫∞−

∞

=′⇒=′′t

rdr pt d p0

00

for the wave to be confined to a finite radial extent



CYLINDRICAL WAVES




Unlike planar and spherical waves, a cylindrical wave necessarily

possess a wake. To see this examine how planar and cylindrical

waves are produced by infinite planar and linear arrays of point

sources:

For an infinite linear array of point sources, the outward propagating

pressure wave is

( ) ( )

z d r

r t aG p ∫

∞

∞−

−=′ 0t, ρ

dr d r t ar

rdr dz z r −=−=

−=+= ξ ξ

ρ ρ ;;; 022

222

Hence the observed signal for a cylindrical wave can never

be discrete: there is always a “wake”.

( ) ( )( )

( )1

2,

0

220

0

0

0

ξ ξ

ρ ξ ξ ξ ρ

δξ ξ

ξ

ρ

d Gt a

t ad Gt p

t a

∫

∫

+

−

∞−

→

+−=′

( )

δξ ξ ξ ξ

ξ

+<<

≠

00in

0 and largefor Gt


EQUATION: PLANE, SPHERICAL, AND CYLINDRICAL

WAVES

6.2.3 Cylindrical Waves



6.3 AN ESTIMATE OF THE INFLUENCE OF

HEAT CONDUCTION

Linearized

2

2c

x

T

C x

u

C

p

x

T u

t

T

vv ∂

∂=

∂

∂+

∂

∂+

∂

∂ λ ρ ρ

One-dimensional energy equation with heat conduction

2

2

00

0

x

T

C x

u

C

p

t

T

v

c

v ∂

′∂=

∂

′∂+

∂

′∂

ρ

λ

ρ

Eliminate by using the continuity equation

)( x

T

Cx

p

C

p

t

T2

2

v0

c

v20

0 ∗∂

′∂=∂

′∂−∂

′∂ ρ

λ

ρ

Linearized momentum equation

01

0

=∂

′∂+

∂

′∂

x

p

t

u

ρ

Motions are non-isentropic, and assume p = p(ρ,T )

dT T

pd

pdp

∂

∂+

∂

∂= ρ

ρ T

xu ∂′∂

Reference: Morse and Feshback (1952) Methods of Theoretical

Physics, Vol. 1, Chapter , Problem . .



Combine with momentum and continuity equations:

)( ∗∗=∂

′∂

∂

∂−

∂

∂

∂

∂−

∂

∂ 0

''2

2

2

2

T2

2

00 x

T

T

p

x

p

t ρ

ρ

ρ

ρ

( ) ( )t kxit kxi eT T e ω ω ρ ρ −− =′= ˆ;ˆ'

( )

( ) 0ˆˆ

0ˆˆ

0

22

0

2

20

02

0

c

=

∂

∂−−

∂

∂⇒∗∗

=

−

+⇒∗

ρ ρ

ω

ρ ρ

ω ρ

λ ω

ρ T

vv

pk T

T

pk

C

piT k

C i

Assume sinusoidal travelling waves:

Substitute in (∗) and (∗∗):

Non-trivial solutions for and only if: ρ ˆ T

( )∗∗∗=

−

∂

∂

+

−

∂∂+

∂∂

0 222

0

c

220

0

T

2

0

00

ω ρ ω ρ

λ

ω ρ ρ ρ

T v

v

pk

k

C

T p

C p pk i


HEAT CONDUCTION



Under this condition, the imaginary part of (***) must be

nearly zero:

If heat conduction is negligible, from (*), with δ x~λ, δt ~τ=1/ f :

1ˆˆ

20

c2

0

c <<⇒<<

λ ρ

τ λ

τ λ ρ

λ

vv C

T T

C

( )

( )1

k

C

k

k 2

/2 2

v0

c2

22 <<

⇒==

ω ρ

λ

ω π

ω π

λ

τ

→

∂

∂

+

∂

∂

==

∂

∂+

∂

∂=

0

2

0

c20

02

22

0

20

022

00

00

ω ρ

λ

ρ ρ

ω

ρ ρ ω

ρ

ρ

k

C T

p

C

p p

k a

T

p

C

p pk

vvT

vT

or


HEAT CONDUCTION



Thermodynamics for a perfect gas gives the result:

→

=

∂

∂=⇒

∂

∂+

∂

∂=

∂

∂

∂

∂+

∂

∂=

∂

∂

02

0

c20

20

0

000000

ω ρ

λ γ

ρ

ρ ρ ρ ρ ρ ρ ρ

k

C

RT p

a

T

p

C

p pT

T

p p p

vS

vT S T S

In the inverse limit:

→= 0

k

CRTa

2c

v020

ω

λ

ρ

(Rapid heat transfer, isothermal sound propagation)

Numbers for air:

K -gcal0.17

scm103

cmg101.2

K -cmcal100.73

40

33

4

°=

×=

×=

°×=

−

vC

a

0

c

ρ

λ

ω ω ρ

λ

ω ρ

λ 10

200

c2

0

c 104~ −×=

⇒

aC

k

C vv

⇒ Isentropic propagation in the

audible range, 10–20,000 s-1

(Negligible heat transfer, isentropic sound propagation)


HEAT CONDUCTION



6.4 ENERGY AND INTENSITY ASSOCIATED WITH

ACOUSTIC WAVES

This relation must be written to second order in the isentropic

fluctuations, e.g.:

( )

∫∫∫∫

∫∫ ∫

⋅−⋅

+−=

⋅∇−

+⋅∇−=

+

∂

∂

S d uS d uu

e

dV u pdV u

eudV u

et

rrrr

rr

ρ ρ

ρ ρ

2

22

2

22

( ) ( )

200

2

000

2

22

00

p

2

1'

'

2

1'

00

ak e

eeee

S S

ρ ρ ρ

ρ

ρ

ρ ρ

ρ

ρ ρ ρ

′++=

∂

∂+

∂

∂+=

Integrate the energy equation over a closed stationary volume:

Eventually the result is:

∫∫ ∫∫∫ ⋅′′−⋅′−=∂∂ S d u pS d udV t

rrrrEE



E

( ) 0=′′⋅∇+∂

∂u p

t

rE

is the acoustic energy per unit volume and is the

intensity, the flux of acoustic energy (energy/areas-s).

The first term on the right-hand side is of third order and

must be dropped; the result implies the differential equation:

u p ′′r

202

00

2

2

1

2

1u

a

p′+

′= ρ

ρ E

E


ACOUSTIC WAVES



6.4.1 Results for Sinusoidal Plane Waves

Wave to right

( )

( )

00

2

200

200

ˆˆ

ˆ

ˆ

a

pu p I

a

pa

pu

euu

e p p

kxt i

kxt i

ρ

ρ

ρ

ω

ω

++++

++

++

−−++

−−++

′=′′=

′=

=

=′

=′

E

Wave to left

( )

( )

200

2

200

200

ˆˆ

ˆ

ˆ

a

pu p I

a

pa

pu

euu

e p p

kxt i

kxt i

ρ

ρ

ρ

ω

ω

−−−−

−−

−−

+−−−

+−−−

′−=′′=

′=

−=

=′

=′

E

( ) ( )∫+

′=τ

τ

t

t

t d 1

00

2

200

2

22

a2

ˆ

2

ˆ

ˆ2

1

ρ

ρ

++

++

++

=

=

=′

p I

a

p

p p

E

00

2

200

2

22

2

2

ˆ

ˆ2

1

a

p I

a

p

p p

ρ

ρ

−−

−−

−−

=

=

=′

E

More generally: ( ) ( )

( )

( ) ( )*u pu* pu p I

uu p pua

p

euue p p t it i

′′+′′=−=

′⋅′+′′=

+=

=′=′ +−+−

rrr

rr

rv

41

21

0412

0200

2

41

cosˆˆ

**ˆˆ

ˆ;ˆ

ψ ϕ

ρ ρ ρ

ϕ ω ϕ ω

E

where ( )* denotes complex conjugate.


ACOUSTIC WAVES



6.4.2 The Decay or Growth Constant

α ω ω i−→

and the variables of the motion have the behavior in time

t t -i ee α ω

where for this definition, α>0 means waves grow.

Normally in practice, |α|/ω << 1, which means that the fractional

change of amplitude is small in one cycle of the oscillation. Thus

when time averaging is carried out over one or a few cycles, e-αt isapproximately constant and the average energy is

In practice, if there is no external source of energy, waves in a

chamber will decay; or in case there are internal sources of

energy, waves may be unstable, having amplitudes growing in

time.

In ‘standing’ or ‘stationary’ waves in a chamber, the frequencyis complex

+=

2

0200

2

t ˆˆ

4

1u

a

pe ρ

ρ

α E


ACOUSTIC WAVES



6.4.2 The Decay or Growth Constant (cont’d)

Hence we have the important interpretations of α which serve

as the basis for measuring α:

dt

d

dt

pd

p

E

E

12

ˆ

ˆ

1

=

=

α

α

Note: the sign of α is a matter of definition.


ACOUSTIC WAVES



6.4.3 The Decibel and the Response of the Human Ear

rms4rms log2074102

log20dB P P

+=×

=−

=

1

210log10dB

I

I

Conventional reference level:

( )

102

00

24

0 ⇒×

=−

a I

ρ

(difference in sound

levels)

where P rms is expressed in dynes/cm2

Definition of the decibel

Relation between dB and pressure in psia (rms)Copied from Morse, Vibration and Sound


ACOUSTIC WAVES



RELATION BETWEEN THE RMS AMPLITUDE

OF PRESSURE AND THE DECIBEL


ACOUSTIC WAVES

6.4.3 The Decibel and the Response of the Human Ear (cont’d)



6.5 BOUNDARY CONDITIONS; REFLECTIONS

FROM A SURFACE

6.5.1 Impedance and Admittance at a Surface

Linearized boundary

conditions on the pressure:

Acoustic impedance, z:

Acoustic admittance:

For harmonic motions:

z, y complex functions of ω

nt

u pn ˆˆ 0 ⋅

∂

′∂=′∇⋅

r

ρ

p z

nu ′=⋅′1

ˆr

p yi p z

i pn ′=′=′∇⋅ ω ρ ω ρ

00ˆ

z y

1=

Characteristic acoustic impedance, ρ0,a0:

s-gm/cm42 200 =a ρ

Acoustic impedance ratio:

00a

z

ρ ζ =

Acoustic admittance ratio:ζ

η 1

=

reactanceacoustic :Im(z)

resistanceacoustic :Re(z)

For air at atmospheric

pressure and temp.

(standard, sea level)



6.5.2 Reflections of Plane Waves from a Surface

Plane waves propagating in direction k k ˆ2

λ

π =

r

)1

)

00

t r k g( a

,t)r ( u

t r k g( ,t)r ( p

ω ρ

ω

−⋅=′

−⋅=′rrrr

rrr

t yk(x

t r k

g k

k

au

g p

i

ii

ii

i

ii

iii

ω θ θ

ω ξ

ξ ρ

ξ

−−=

−⋅=

=′

=′

)cossin

)(1

)(

i

00

rr

r

rr

r i k a

k rr

==0

ω

Incident Wave Reflected Wave

t yk(x

t r k

g k

k

au

g p

r r

r r

r

r r

r r r

ω θ θ

ω ξ

ξ ρ

ξ

−+=

−⋅=

=′

=′

)cossin

)(1

)(

r r

00

rr

r

rr


FROM A SURFACE



6.5.2 Reflections of Plane Waves from a Surface (cont’d)

Surface Impedance

)sin(cos)sin(cos

t)sin()sin(00

0t kx g t kx g

kx g t kx g a

u

p z

r r r iii

r r ii

y y ω θ θ ω θ θ

ω θ ω θ ρ

−+−−

−+−=

′−

′=−

=

In general: z(x) z =

Assume z independent of x: True if ( ) ( )

=

==

ξ β ξ

θ θ θ

ir

ir

g g

( ) θ β

β ρ

cos1

100

+−

+=−⇒ a z

Reflection coefficient:00

00

cos

cos

a z

a z

ρ θ

ρ θ β

+

−=

E.g. θ = 0: no reflection if z = ρ0a0

(perfect impedance matching)

θ ≠ 0: non-zero reflection (!)

Nonsense: because the transmitted wave,

required to conserve energy, has not been

accounted for

⇒ the result is not valid for z = ρ0a0


FROM A SURFACE



6.6 WAVE PROPAGATION IN TUBES

0)

0

0)()'(

00

0

0

=′∂

∂+′

∂

∂

=∂

′∂+

∂

′∂

=′∂

∂+

∂

∂

S u( x

pS)T ( t

C

x

p

t

u

S u x

S t

V ρ

ρ

ρ ρ

Linearized equations for flow in a non-uniform tube having

cross-sectional area S(x):

t ie ω -

Wave equation for the pressure:

6.6.1 Waves in Uniform Tubes; Normal Modes

Normal modes: time dependence,

real wave number, k

0

11

2

2

20

=∂

′∂−

∂

′∂

∂

∂

t

p

a x

p

S xS

complex bemayˆˆ; ˆ (x) p pe p p t i ==′ ± ω

0ˆˆ 2

2

2

=+∂

∂ pk

x

p



( )

0ˆ

0ˆ0ˆ0

ˆˆ

:equation Momentum

=⇒=⇒=′=

±−=

dx

pd uu , L x

uidx

pd ω

Example (1): tube, length L, closed at both ends

A solution:

6.6.1 Waves in Uniform Tubes; Normal Modes (cont’d)

=== 00

ˆcosˆ @ x

dx

pd kx A p

0cosˆ

== kLkAdx

pd

⇒ L

k π l

=




Example (2): Normal modes for a tube open at both ends:

Example (3): Normal modes for a table closed at x = 0, open

at x = L:


L )l ( kkx A p

212cosˆ

π +==

Lk kx A p

π l sinˆ ==

0ˆ

0ˆ

0

==

==

p L x

dx

pd : x

β

Example (4): discontinuity in cross-section

At discontinuity, match solutions in uniform tubes

Boundary conditions:




( ) x Lk B p L x L

kx A p L: x

−=≤<

=<≤

β β sinˆ

cosˆ0

Example (5): discontinuity in cross-section (cont’d)

(ii) Continuity of acoustic mass flow: integrate wave equation

over L – ε ≤ x ≤ L + ε :

Possible solutions:

Matching conditions:

( ) ( )

( ) ( )∗−=→

+=−

:

1sincos :0

ˆˆ pressureof Continuity (i)

kL BkL A

L p L p

β ε

ε ε

0ˆˆ

2 =

+

∫

+

−

dx pS k dx

pd S

dx

d L

L

ε

ε

( ) ( )ˆî.e.ˆˆ

ε ε ε ε

ρ ρ −+−+

=

=

L L

L L

uS uS dx

pd S

dx

pd S 00

Substitute wave forms:

( ) ( )∗∗−−=− Lk BS kL AS 1cossin 21 β

Combine (∗) and (∗∗):

( ) Lk kLS

S 1cottan

2

1 −= β A transcendental

equation for k

Mode shapes are accurate except in a transition region

at the discontinuity





Experimental results

(ref.: Mathis, Derr, and Culick)





6.7 THE IMPEDANCE TUBE

Ref.: Morse, Vibration and Sound

• A forced system used to obtain the impedance function for

surfaces, or objects such as nozzles or injectors

Steady waves are sustained by the driver at a chosen fixed

frequency. The power supplied by the driver precisely

balances the power absorbed by the test sample and dissipated

by viscous effects at the walls.

• Assume all variables

• If there are no distributed losses long the tube, then in the

absence of mean flow, the waves are represented by

t -ie~ ω

ikx

ikx

Ae p

Be p

−−

+

=

=

ˆ

ˆ

ikx

ikx

ea

Au

ea

Bu

−−

+

=

=

00

00

ˆ

ˆ

ρ

ρ real k



Impedance at z = 0:

002 i 0 πβ πα ψ +=−== Ae: B z

( )( )00

002

2

2

000 /1

/1

1

1

ˆ

ˆ

a z

a z e

e

ea

u

p z

x ρ

ρ ρ ψ

ψ

ψ

+

−=

+

−=

−=

=

[ ]ψ

ψ

ρ

2

00

2

ˆ

ˆ

+−

+−

+−=

−=

ikxikx

ikxikx

eea

Au

ee A p

Values of ψ = π α0 + i π β0 are inferred frommeasurements of the envelope of the modal structure

along the impedance tube

( ) ( )ikxk Aeee Ae x p ikxikx +−=−−= −−+ψ ψ ψ ψ ψ sin2ˆ

( )

+=

=

+=+λ

β β

α α

β α π ψ xiikx 20

0

( ) πβ πα β α π πα ψ coscosh2sin2ˆ 22 −=+= Aeik e A p




Maxima: L,2

3 ,

2

12 @ cosh2ˆ 0max

±±=+=λ

β πα πα x Ae p

Minima: L2,1,2 @ 1cosh2ˆ 02

min ±±=+−=

λ β πα πα x

Ae p

πα α α coth p

p :from found

min

max0 ==

022

0

00

coshcosh2Ae p

wheresample, at the :from found

πβ πα

β β β

πα

−=

=

The first minimum occurs at:

min0min0

21 1

2 x x

λ β

λ β −=⇒=+




6.8 VISCOUS LOSSES AT AN INERT SURFACE

• Dissipation of acoustic energy at inert surfaces is often asignificant contribution to losses in combustion chambers

• The problem of computing the losses will be analyzed in

three different ways

Consider two-dimensional flow in the vicinity of a rigid

impermeable wall, both velocity and temperature oscillating far

from the wall; the pressure is uniform but time-varying:

6.8.1 The Acoustic Boundary Layer



0'

0 =

∂

′∂+

∂

′∂+

∂

∂

x

u

y

v

t ρ

ρ

6.8.1 The Acoustic Boundary Layer (cont’d)

The equations for this incompressible boundary layer flow are:

Continuity

Momentum

Energy

t

u

y

u

t

u

∂

′∂+

∂

′∂=

∂

′∂ 002

2

0 ρ µ ρ

t

p

y

T k

t

T c p

∂

∂+

∂

′∂=

∂

′∂ ˆ2

2

0 ρ

Note: Both and can be taken as functions of x without

affecting the equations and hence their solution.0u 0T

Boundary conditions0

0

=′

=′

T

u0= y

t -i

t -i

eT T T

euuu

ω

ω

00

00

ˆ

ˆ

=′→′

=′→′ ∞→ y

t

T

ct

p

T cT

T p p y

p

p

∂

∂

=∂

∂

⇒

=−

=∞→

ˆˆ

ˆˆ

1ˆ ,For

0

000

00

ρ

ρ γ

γ




6.8.1 The Acoustic Boundary Layer (cont’d)

Appropriate solutions:

t -it i

t -it i

eT eT T

eueuu

ω ω

ω ω

1-

0

1-

0

ˆˆ

ˆˆ

+=′

+=′

Equations for and :u T

0=ˆPr 2ˆ

0=ˆ2ˆ

1221

2

1221

2

T i

dy

T d

ui

dy

ud

δ

δ

+

+

Solutions:

k

c µ

ρ

µ ν

ω

ν δ

p

0

Pr ;Pr ;2

===

[ ] y

y

eT T

euuPr 2

0

20

1ˆˆ

1ˆˆ−

−

−=−=

( )i−= 11

δ λ




6.8.2 Energy Losses Associated with the Acoustic

Boundary Layer

One way to compute the losses is to find the time-averaged total

energy dissipation, defined as

∫∞

+

0

2

0

2

ˆˆ dydyT d

T k

dyud µ

Substitution of either the real or the imaginary parts of the

solutions for û and leads to the important result:T

−+

=

2

0

2

00

ˆ

Pr

1ˆ

22

1

p

p

a

u γ ων

γρ

time-averaged energy

loss per unit area

Standing waves in a closed tube, length L, radius R l,

(kz)a

Au(kz) A p sinˆcosˆ00 ρ

==

( ) L Ra

Al 2

200

2

4π

ρ =E






6.8.3 Another Way of Computing the Decay Constant

Treat the viscous stresses and heat conduction as processes

distributed in space. The linearized equations are

(3)1

(2)'

(1)0'

00

0

0

qC

uC

p

t

T

F pt

u

u

t

′⋅∇=′⋅∇+∂

′∂

′=⋅∇=′∇+∂

′∂

=′⋅∇+

∂

∂

rr

rtr

r

ρ

τ ρ

ρ ρ

Form the equation for the pressure:

(4)'

0 qC

Ru

t ′⋅∇=′⋅∇+

∂

∂ rrγρ

ρ

Wave equation for the pressure: (4) and substitute (2):t ∂

∂

( ) F aqt C

R pa

t

p′⋅∇−⋅∇

∂

∂=′∇−

∂

′∂ rr 22202

2





(cont’d)

Boundary condition, (2):⋅n

F nnt

u pn ′⋅+⋅

∂

′∂−=′∇⋅

rrˆˆˆ 0 ρ

Harmonic motions, , etc. (k complex):kt -iae p p 0ˆ=′

f pn

h pk p

ˆˆˆ

ˆˆ22

−=∇⋅

=+′∇

( )α ω

ωρ

ω

ia

k

F nnui f

qc

R

a

i F h

−=

⋅−⋅−=

⋅∇−⋅∇=

0

0

20

1

ˆˆˆˆˆ

ˆˆˆ

rr

rr







(cont’d)

{ }∫ ∫∫++= dS f dV h E

k k nn

n

n ψ ψ ˆˆ12

22

Because is ‘close’ to , set in the denominator

and later in numerator as well:

p nψ

∫= dV E nn22 ψ

For viscous losses in the boundary layer:

2

2

2

2

ˆˆˆ

ˆ'

ˆ

y

T

y

qq

y

u F

c y

||

∂

∂=

∂

∂=⋅∇

∂

∂=⋅∇=

λ

µ τ

r

rtr

∂

∂−∇⋅

∂

∂−+=

+∇⋅−=+

∫∫

∫∫ ∫∫

dydS y

T

C

R

a

idydS

y

u

E k k

dV F dS f dV h

ncn

n

n

nnn

|| ψ λ ω

ψ µ

ψ ψ ψ

2

2

20

2

2

2

22ˆ1

ˆ

r

r


n p ψ =ˆ





6.9 PROPAGATION OF HIGHER-ORDER MODES IN

TUBES

The behavior discussed in this section is included partly

as an example of departures from the purely one-dimensional

wave propagation assumed in most of the preceding analyses;

and partly because the results are applicable to slender

passages in combustion; i.e. regions having relative large

values of length/transverse dimension.

6.9.1 Traveling Plane Waves and Reflections in a Duct

If the frequency of a wave traveling along the axis of a rectangular

tube is larger than the speed of sound divided by twice the largest

lateral dimension of the tube, then the motion cannot be treated as

perfectly parallel to the axis. That frequency is called the “cut-off”

frequency, f c.

For f > f c, modes other than the simple one-dimensional

mode will propagate in the duct, i.e. purely axial waves lie in the

region of lower frequencies.

Consider a two-dimensional channel having width w in the

y-direction and extending indefinitely in the x-direction:

Reference: Morse and Ingard



and the corresponding wavelength is 2w, the wavelength for the fundamental standing wave for motions normal to the axis

For f > f c, purely one-dimensional waves will propagate in

principle; for the higher frequencies interesting things are

possible, namely wave structure in transverse planes.

Consider a plane wave incident at angle to the axis onthe entrance to the tube.

6.9.1 Traveling Plane Waves and Reflections in a Duct (cont’d)

For the case illustrated, the cut-off frequency is exactly

w

a f c

2

0=

θ π −2 /

What gets through the tube?

Answering the question requires treating reflections from the

walls of the tube.

First treat reflection from a rigid surface: 0ˆ =⋅′ nur


TUBES



The surface impedance is infinitely large and the reflection

coefficient β = 1 (§5.5). Hence

g g g r i ==

and the velocity component normal to the wall is:

( ) ( )[ ]

( ){ } ( ){ }[ ]t y z k g t y z k g a

g g a

v r r ii

ω θ θ ω θ θ ρ

ξ ξ ρ

−+−−−−=

+−=

cossincossincos

cos

00

00

For sinusoidal waves, ( ) : Ae g iξ ξ =

[ ]

( ) t)i(kz

kyikyit)i(kz

ekya

iA

eeea

Av

ω θ

θ θ ω θ

θ ρ

θ

ρ

θ

−

−−−

=

−−=

sin

00

coscossin

00

cossincos

2

cos


TUBES




:t)i(kz

e ω θ −sin

wave having wavenumber k sin θ traveling in

z-direction parallel to the wall; solid lines

represent pressure maxima

sin (ky cos θ): standing wave structure normal to the wall,

unchanging in time

• The dashed line identifies a surface where all pressure

maxima always intersect. Hence the velocity vanishes on

that surface


TUBES




only for particular values of ωW.

This condition defines “higher order modes” for

propagation in a tube or duct.

A Plane Wave Reflecting from a Rigid Surface

The velocity vanishes on all planes such that

( ) 0cossin =θ ky

In particular, on the surface y = W , the normal velocity vanishes

if the frequency has the values given byk a0=ω

0cossin0

=

θ

ω

a

W

Wave propagates parallel to the wall, the velocity

is parallel to the wall and for all

W and ω.

0cossin0

=

θ

ω

a

W :2/π θ =

:2/π θ < 0cossin0

=

θ ω

aW


TUBES




( for waves incident from the left)

Cut-off Frequency

Fix W ; then requires0cossin0

=

θ

ω

a

W

π θ

ω

l a

W

=

cos

0

or

W l

W

al

2cos 0 λ

ω π θ ==

20 / π θ ≤≤

Special cases:

l = 0:

l = 1:

2

π θ = and motions are one-dimensional

parallel to the surface,

0

sina

k k ω

θ ==

W

a

W

a 00 1cos π ω ω

π θ ≥⇒≤=

Critical or ‘cut-off’ value is

W a

W

a

c

c

22 0

0

==

=

ω π λ

π ω


TUBES




For and θ is not real.1cos 00 >===<ω

ω

ω π θ π ω ω c

cW

a ,

W

a

Hence, a wave having frequency and the form defined

above (p. 5-55) cannot exist if the normal velocity must vanish on y =

W .

At the critical frequency, the wavelength is that for thefundamental standing wave in For , only

purely one-dimensional waves propagate: the axial velocity

fluctuation is uniform between the surfaces.

cω ω >

.0 W y ≤≤

Conclusion: For a plane wave reflecting from a rigid

surface, there are planes parallel to that surface

on which the normal velocity vanishes. Such a plane may be replaced by a rigid surface,

forming a channel in which a wave travels in

the x-direction.

Propagation of a Wave in a Tube, with Internal Reflections

Note: Solid lines represent surfaces of constant

phase (not rays! — this is not ray acoustics)

cω ω <


TUBES




Imagine that for the situation sketched, the planes are maintained a

fixed distance apart, but θ is increased from θ = 0

Plane waves reflect between the plates, always

propagating in the y-direction.

The boundary condition is satisfiedsimultaneously in the two surfaces only at one

wavelength,

0ˆ =⋅′ nur

The boundary conditions are satisfied

if λ is decreased as θ is increased so

W.2=λ

θ λ cos2W =

Now let θ be non-zero:

0ˆ =⋅′ nur


TUBES




When W is fixed, waves with will not propagate in

the first mode.

If a wave for which is incident at angle

the motions penetrate into the tube but reflections from the

surfaces cause destructive interference.

cλ λ >

2 / π θ ≠

Conclusion: For , the amplitude decays

with distance into the tube and the

wave will be totally reflected

CUT-OFF FREQUENCY, :cω ω =

If only purely one-dimensional waves propagate; i.e. For

only planar axial waves propagate.

Waves having higher frequencies (i.e. shorter wavelengths) purely

plane waves cannot propagate.

Above ‘cut-off’ waves having higher frequencies will propagate but

they must possess structure in transverse planes. These are called

‘higher order modes’.


TUBES


cλ λ >

cλ λ >

cω ω <

cλ λ >



6.9.2 Higher Order Modes as Solutions to the Wave Equation

in Three Dimensions

Solve for fixed:0ˆˆ 22 =+∇ pk p 0 /ak ω =

0ˆˆˆˆ 2

2

2

2

2

2

2

=+∂

∂+

∂

∂+

∂

∂ pk

z

p

y

p

x

p

Assume

( ) t it) z (k ie pe x,y P p z ω ω −−

==′ ˆ

for motion in a rectangular duct. The acoustic velocity is

∂

∂+

∂

∂+

∂

∂−=∇−= k

z

p j

y

pi

x

p pu ˆˆˆˆˆˆ1ˆ

1ˆ ρ ρ

r

Hence the boundary conditions on P(x,y) are

W , y y

P

V , x x

P

00

00

==∂

∂

==∂

∂


TUBES



Solution:

( ) ( ) ( )t z k i y x

z e yk xk A p ω −= coscosˆ

where

and

W l k

V l k

y y

x x

π

π

=

=K , , , ,l l y x 210=

2222

y x z k k k k +−=

The wavenumber for motions in the z-direction along the

axis is

+

−

±=

222

0 W l

V l

ak y x z

π π ω

Note: For the assumed behavior

if we require then is positive or

negative, representing wave propagation to

the right or left respectively.

( ) ,e P p t z k i z ω −=′

,0>ω z k


TUBES

6.9.2 Higher Order Modes as Solutions to the Wave Equation in

Three Dimensions (cont’d)



For a two-dimensional duct, and (correctly)

From the above formula, is real only if z k ,cω ω >

22

0

+

=

W l

V l a y xc

π π ω

is the critical frequency or cutoff frequency. For

waves of the form specified above, for given values of l x and l y,cannot propagate in the duct.

∞→V

W l a yc

π ω

0

=

+

−

±=

222

0 W l

V l

ak y x z

π π ω

,cω ω <


TUBES

6.9.2 Higher Order Modes as Solutions to the Wave

Equation in Three Dimensions (cont’d)



The formula for can be written

and if is imaginary:

t i z k z e P p ω −=′ m

,cω ω >

z k

z k

Then the pressure field is represented by

2

0

2

0

−

±=

aak c z

ω ω

z c

z k iaa

ik ±=

−

±=

2

0

2

0

ω ω

For the pressure to remain finite as in the tube, the

+ sign must be taken:

( ) t i z k ee x,y P p z ω −−=′

∞→z

so the amplitude | p′| of the pressure decays as the wave

propagate into the tube.


TUBES





Phase Velocity

The phase velocity is the velocity with which surfaces

of constant phase (e.g. nodal surfaces on which )

propagate, i.e.

pv

constantt z k z =−±= ω ϕ

Hence (drop the ):±

2

0

2

0

2

0

1

−

=

−

==

=

=

ω

ω ω ω

ω ω

ϕ cc

z const p

a

aa

k dt

dz v

and the wavelength is

2

0

1

22

−

===

c

p p av

f

v

ω

ω

ω π

ω π λ

Because the phase speed depends on frequency, waveshaving different frequencies will tend to ‘separate’ during

propagation---the medium is said to be dispersive.

In particular, for the phase velocity and the

wavelength become infinitely large(!)

,cω ω →

0ˆ = p


TUBES





Thus as and neither information nor

energy is carried down the duct for the mode in question.

Although the phase velocity is infinite, signals, information

and energy travel with finite speed, the group velocity.

Information (and hence net transfer of energy) occurs only

if the waves have nonuniformities, for example pulses.

The envelope of a pulse, or a ‘group’ of waves, travels with

speed , g v

,cω ω →

dk

d v g

ω =

[See good books on electromagnetic theory, or acoustics]

Here and

2

0

2

0

2

−

=

aak cω ω

2

0

2

0

2

0

202

0 1

−=

−

==

ω

ω ω ω

ω ω cc

g aaa

ak av

0→ g v


TUBES





6.9.3 The Cutoff Frequency for a Cylindrical Tube

Represent the motions in circular cylindrical coordinates:

The wave equation is

where

02

=+∂

∂+

∂

∂+

∂

∂

∂

∂ pk

z

p p

r r

pr

r r ˆ

ˆˆ1ˆ1 2

2

22

2 φ

( ) ( ) t ie z r p z;t r, p ω φ φ −=′ ,,, ˆ

QUESTION: Under what conditions will a wave

propagate along the tube?


TUBES



( ) 011 22

2 =−+

∂

∂+

∂

∂

∂

∂ F k k

F

r r

F r

r r z 2

2

φ

Propagation along the tube is represented by the form

( ) ( ) ( )φ φ Φ= r Rr F ,

Substitution in the wave equation gives the equation

for F :

This equation is soluble by separation of variables:

Assume

( ) ( )t z k i z er, F p ω φ

−=′

and satisfyΦ R,

0md

d 2

2

2

=Φ+Φ

=

−−+

φ

01

2

222 R

r

mk k

dr

dRr

dr

d

r z


TUBES

6.9.3 The Cutoff Frequency for a Cylindrical Tube (cont’d)



( ) ( ) areand r Rφ ΦThe solutions for

K , , ,mm

m321

cos

sin=

=Φϕ

ϕ

( ) ( )r J r R mnm α =

with 222 z mn k k −=α

The values of are set by satisfying the boundarycondition on the pressure at the rigid lateral boundary:

mnα

0ˆ

=Φ=∂

∂ z ik z edr

dR

r

p

which ensures that the radial velocity vanishes on thewall. This condition requires

( )0=

= Rr

mnm

dr

r dJ α


TUBES




For each value of m there are an infinite number of values

of of which the lowest few are given in

the table:

( )K , , ,n Rmn 210=α

0 1 20 0 3.83 7.02

1 1.84 5.33 8.54

2 3.05 6.71 9.97

m

n

The wavenumber for propagation along the tube is

2

20

2

mn z a

k α ω −±=

which is real only if

cmna ω α ω => 0

Waves of the specified form and having frequency lessthan will not propagate.cω

Because the critical or cutoff frequency is lower

for larger diameter tubes

, R

~mn

1α


TUBES




Example:

( )

( )( )

Hz , ,

f

,

. )(

)( R

.aa

R

.;" R

c

mnc

695342

000218

000218

58610161

7400

58610

58610161

00

10

==

=

=

==

==

π

π

π α ω

π α

Thus the cutoff frequency for a 1/8” tube is about 35 kHz,

well above the bandwidth of the human ear.

Why, then, is the audio reproduction so poor in those systems

used in airlines to bring you music and the latest films?


TUBES




6.10 NORMAL MODES: RESONANT FREQUENCIES

AND MODE SHAPES

DEFINITION: A normal mode of an oscillating system is amotion in which all parts of the system oscillate

sinusoidally at the same frequency, with fixed

relative phases and with amplitudes constant in

time.

6.10.1 An Elementary Example of Normal Modes: Two

Degrees of Freedom

A system of two masses connected by springs:

The coordinates are measured from the equilibrium

positions of the masses at rest. Assume small amplitude

motions along the x-axis; the equations of motion are:

21, x x

21222

2

12121

2

kx ) xk(xdt

xd m

) xk(xkxdt

xd m

−−−=

−+−=

(1) Formulation in State Space; Normal Coordinates





Non-trivial solutions for exist if:21 ândˆ x x

( )0

2

220

20

20

20 =

−−

−−

λ ω ω

ω λ ω

In matrix form; with 2ω λ =

( )( )

0ˆ

ˆ

2

2

2

1

20

20

20

20 =

−−

−−

x

x

λ ω ω

ω λ ω

which gives the characteristic equation for λ

03420

220

2

=+− ω λ ω λ

The solutions for the two characteristic values or

eigenvalues are21 λ λ ,

0201 3ω ω ω ω == ;

Only at these frequencies will the entire systemoscillate in simple harmonic motion.

Corresponding to each characteristic value or

eigenvalue is a motion called a normal mode, an

eigenmode, or simply a mode defined by the relative

values of .21 ândˆ x x


Degrees of Freedom (cont’d)


AND MODE SHAPES



The mode shapes, i.e. the corresponding sets of values of

are found by substituting successively the values

for ω in the equations of motion:

( )0

ˆ

ˆ

2

2

2

1

2020

20

20 =

−−

−−

x

x

λ ω ω

ω λ ω

21 andω ω

)( i

)(

)( )(

x x

x x

112

11

20

11

20

01

ˆˆ

ˆˆ

=

−

=

ω ω

ω ω

)ˆ,ˆ( 21 x x

First Mode Second Mode

)( )(

)( )(

x x

x x

21

22

22

20

21

20

02

ˆˆ

0ˆˆ

3

=

=−−

=

ω ω

ω ω

Imagine an ‘abstract vector space’ having two orthogonal

axes defined by unit vectors along which thedisplacements are marked off

21 ândˆ ee21 ândˆ x x

1

0

2211

21

=⋅=⋅

=⋅

eeee

ee




AND MODE SHAPES



A motion of mass 1 in physical space is represented by

in this vector space and the normal modes can

be written11ˆ e

t ie x ω

( ) t i

t i )( )(

eee x

ee xe x X

1

1

211

21

211

11

ˆ

ˆˆ

ω

ω

+=

+=

( ) t i )(

t i )( )(

eee x

ee xe x X

2

2

212

1

22

212

12

ˆ

ˆˆ

ω

ω

m±=

+=

and the mode shapes are

( )2111ˆ ee += C X

where are normalization constant whose values

can be assigned according to some chosen rule; for example,

21, C C

e.g. 1)1( 21 == C C

211ˆ ee += X 212

ˆ ee +−= X

1ˆ2 =i X )(

( )2112

1eeE += ( )

2122

1eeE +−=

( ) 2

2211

222 because iii C C X =⋅+⋅= eeee




AND MODE SHAPES

( )2122ˆ ee += C X



In the second case, the objects and can

be interpreted as a new pair of unit vectors in the abstract

space introduced above:

11ˆ E= x 22ˆ E= x

The motions of the two masses are described equivalently

by specifying their physical coordinates or by specifying

the contributions of the normal mode coordinates

to their motions.

According to the figure, the unit vectors are related by the

transformation

( )21 ˆ,ˆ x x

21, X X

4wherecossin

sincos

2

1

2

1π θ

θ θ

θ θ =

−=

e

e

E

E




AND MODE SHAPES



and the physical coordinates are transformed to the

normal mode coordinates by the same transformation

( )21 ˆ,ˆ x x

−=

2

1

2

1

cossin

sincos

x

x

X

X

θ θ

θ θ

The square matrix represents an orthogonal transformation

[ ]

⋅⋅

⋅⋅=

=

−=

2212

2111

2221

1211

cossin

sincos

eEeE

eEeE

QQ

QQQ

θ θ

θ θ

having the properties

[ ] [ ]

1222

221

212

211

1

==+=+

=−

KQQQQ

QQ t

Hence, in shorthand notation the transformations from

‘old’ to ‘new’ configurations are the same for both thecoordinates and the unit coordinate vectors:

{ } [ ]{ } xQ X = [ ] [ ]{ }eE Q=

Note: the matrix [Q] is not symmetrical




AND MODE SHAPES



The equations for the normal mode coordinates

are uncoupled.

Suppose that the system is set in motion following initial

values of displacements and velocities .

The subsequent motion of the system can be described

equivalently in the physical coordinates or in

normal mode coordinates .

)0(),0( 21 x x )0(),0( 21 x x &&

)(),( 21 t xt x

)(),(21

t X t X

The advantages of using normal mode coordinates is:

For the two-mass system:

1202

202

2201

201

2

2

x x x

x x x

ω ω

ω ω

=+

=+

&&

&&

Transform to normal coordinates:

{ } [ ]{ } { } [ ] { } [ ] { } X Q X Q x xQ X t ==⇒= −1

Here

−=

2

1

2

1

cossin

sincos

X

X

x

x

θ θ

θ θ




AND MODE SHAPES



Substitute:

( )

( )

( )

( ) θ

θ θ ω

θ θ ω θ θ

θ θ θ ω

θ θ ω θ θ

sinXsinXcos

XcosXsin2XcosXsin

cosXcosXsin

XsinXcos2XsinXcos

2120

212021

2120

212021

∗

−

−+−

∗

−=

−+−

&&&&

&&&&

Add:

:4π θ = 01201 =+ X X ω &&

02202 =+ X X ω &&and

2

22

1

2

01

2

01 sincossincos22 X X X X θ θ θ θ ω ω −+=+&&

For




AND MODE SHAPES



Solutions for :

(2) General Motion Expressed in Terms of Normal Modes

‘General Motion’ means the motion following imposition

of arbitrary initial conditions:

Initial conditions

202202

101101

00

00

x )( x; x )( x

x )( x; x )( x

&&&&

====

From the orthogonal transformations:

{ } [ ]{ } [ ] [ ]

{ } [ ] { } [ ] [ ] { } X M M x M x

M M x M X

t t

t

&&==⇒

==−

−

1

1

and

;

Hence the initial conditions on can be calculated:21, X X

{ } [ ]{ } } [ ]{ }0000 xQ X ; xQ X && ==

)(and)( 21 t X t X

202

210222222

101

110111111

1sincos

1sincos

X B; X A:t Bt A(t) X

X B; X A:t Bt A(t) X

&

&

ω ω ω

ω ω ω

==+=

==+=




AND MODE SHAPES



Transform to motions in physical space:

−=

−=

(t) X

(t) X

(t) X

(t) X

(t) x

(t) x

2

1

2

1

2

1

11

11

2

1

cossin

sincos

θ θ

θ θ

For example:

+

+

+=

t X t X

t X t X (t) x

220

2

220

1101

1101

sin1

cos

2

1

sin1

cos2

1

ω

ω

ω

ω ω

ω

&

&

No Beats Beats due to Coupling

(Normal Mode X 1) (Physical Motion x1)

= (contain 2 frequencies) e.g.02010

2010

==

=

X X

X X

&&




AND MODE SHAPES



6.10.2 Normal Modes for a Finite Line of Discrete Masses

Now consider the case of an arbitrary number N of masses

in a line:

Assuming that the only forces acting on the mass are

due to the springs connecting to the r – 1 and r + 1 mass

the equation of motion for the mass is

thr

thr

( ) ( )r r r r r

x xk x xk dt

xd m −+−−= +− 112

2

or

( )1120

202

2

2 +− +== r r r r x x x

dt

xd ω ω

with

mk =2

0ω

Seek normal modes by assuming all motions :t iω e~

0ˆ2ˆ 120

2201

20 =−−+− −− r r r x x x ω ω ω ω (a)


AND MODE SHAPES




The set of equations can be written

[ ] [ ] { } 0ˆ2 =− x I V ω

(1) Non-trivial solutions exist only if ω2 takes on

characteristic or eigen values, the roots of the secular

or characteristic equation

02 =− ijijV δ ω

N masses, N equations, N positive roots

N ω ω ω K,, 21

In matrix form:

( )

( )

( )

0

ˆ

ˆ

ˆ

ˆ

200

0

0020

02

0002

3

2

1

220

20

20

20

20

220

20

20

220

20

20

220

=

−−

−

−−−−

−−−

−−

N x

x

x

x

M

LL

M

OMM

ML

L

ω ω ω

ω

ω ω ω ω ω

ω ω ω ω

ω ω ω


AND MODE SHAPES




(cont’d)

(2) For every value of there is a corresponding

set of found as solutions to equations (a):

2ω

i x

S iS x2 ↔ω

The are the components of the normal mode

vector in physical coordinates, i.e. in the normalmode s,

si x th s

t iS i

S i

S e x x ω ˆ=

and is the amplitude of motion of the mass

in the mode.

S i x

th s

thi

(3) For each modal vector there is a multiplicative constant whose value can be set by some

normalization rule, e.g.

S i x

( ) ( ) ( ) 1ˆˆˆ22

2

2

1 =+++ S N

S S x x x K

Hence the can be interpreted as the

components of a unit eigenvector in the abstract

vector space spanned by the Cartesian unit vectors

lying along the axes

S N

S S x x x ˆˆ,ˆ 21 K sE

{ } N eee K,, 21 { } N x x x ˆ,ˆ,ˆ 21 K


AND MODE SHAPES



(4) Transformation to normal coordinates is accomplished

with the ‘modal matrix’ of the transformation. Let the

be unit vectors obtained by rotating the

system according to the rule N X X X ˆ,ˆ,ˆ

21 K

{ } N x x x ˆ,ˆ,ˆ 21 K

=

N N N N N

N

N

N x

x x

x x x

x x x x x x

X

X X

ˆ

ˆˆ

ˆˆˆ

ˆˆˆˆˆˆ

ˆ

ˆ

ˆ

2

1

21

222

22

11

2

1

1

2

1

M

L

LM

LL

M

The elements of the columns are the components of the

eigenvectors; i.e.

=

N N N N N

N

N x

x

x

X

X

X

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

2

1

21

22

12

112

11

2

1

M

L

LM

ML

L

M

EEE

EE

EEE

or and{ } [ ] X M x ˆˆ = { } [ ] { } [ ] { } x M x M X t

ˆˆˆ 1 == −

[ M ] = modal matrix

[ ] [ ]t M M =−1so [ M ] is orthogonal (not proved here)


AND MODE SHAPES


(cont’d)



(5) In the normal coordinates, transformed is

diagonal

[ ]V

[ ]{ } { } [ ][ ] [ ]

[ ] [ ][ ]

=

=⇒=

−

O

O

O

O

21

22

S

S S

S S

M V M

M M V E E V

ω

ω ω

[ ]

=

|

| N

|

|

|

|

E E E M K21

(6) The equations for the normal coordinates X (t )

are uncoupled

Equations for physical coordinates { } [ ]{ } xV x =&&

[ ]{ } [ ][ ] { } x M V x M =&&Transform

{ } [ ] [ ][ ] { }

{ } { } X X

X M V M X

S

=

= −

O

O

&&

&&

2

1

ω

Result: Transformation to normal or modal coordinates

diagonalizes the potential energy matrix and the

equations of motion for the modal coordinates

are uncoupled.


AND MODE SHAPES


(cont’d)



6.10.3 Normal Modes for Mass Continuously Distributed

on a Line

Finite discrete masses in a line:

The force exerted on mass m at position r depends onthe relative displacements of adjacent masses:

( ){ } ( ){ }11 −+ −+−−+= r r r r x xa F x xa F Net force

Expand in Taylor’s series about ra:

( ) ( ) ( ) ( )

[ ]

( )r r r

r r r r ra

r r ra

r r ra

x x x F

x xa x xadx

dF

x xadx

dF ra F x xa

dx

dF ra F

211

11

11

−+′=

−+−−+

=

−+

+−

−+

+=

−+

−+

−+


AND MODE SHAPES




on a Line (cont’d)

Equation of motion for r th mass:

( ) ( )[ ]

0

),(;

=

−−

−−

=≈

=′

=′=−−−′−

−+

=

=−+

a

x x

a

x x

aT x

a

m

pT F a

F

dx

dF F

dx

dF F x x x x F xm

r r r r r

x

xr r r r r

11

0

0

00110

1

ilitycompressibor tension

0

&&

&&

γ δ δ

Take limit finite linear density ρ =→→a

mma but0,0

2

2

11

1

11

z

x

z

x

z

x

a

a

x

a

x

aa

x x

a

x x

a

r r

r r

r r r r

∂

∂→

∂

∂−

∂

∂→

−

→

−−

−

−+

−+

−+ δ δ

In the limit:

0)( 2

2

2

2

=∂∂−∂∂ z xT

t x z ρ

02

22

2

2

=∂

∂−

∂

∂

z

xa

t

x,2

ρ

T a = wave propagation

speed

z : coordinate along the line (string or column of a fluid)


AND MODE SHAPES




on a Line (cont’d)

Normal modes: t ie z t z x ω ψ )(),( =

02

2

2

=+ ψ ψ

k

dz

d

a

k ω =

Spatial mode shape

( )φ ψ += kz A z sin)(

For a finite string (or column) k can assume only discrete

values set by the boundary conditions.

( )φ ψ

ψ

+=

==

kz Ak dz

d

L z dz

d

cos

,0,0

( ) 0cos:2

,2

5,2

3,2

0cos:0

=+=

±±±±===

φ

π π π π φ φ

Lk L z

l z K

02

=± π l Lk

L

l k

2

π ±= K,3,1=l

Example:


AND MODE SHAPES



6.11 NORMAL ACOUSTIC MODES FOR A

CHAMBER

Computation of acoustic modes, proceeds in four steps:

(1) Linearize the inviscid equations of motion without sources

(2) Form the linear wave equation for fluctuations of the flow

variables

(3) Seek solutions of the form

(4) Satisfy appropriate boundary conditions

( ) t ier ω ψ ±r

6.11.1 Linear Acoustic Equations (§ 5.1)

Conservation of Mass:

Equation for Pressure:

Momentum Equation:

Linearization forms, no average flow speed and

uniform:

pT ,, ρ

( ) 0=⋅∇+∂

∂u

t

r ρ

ρ

0=∇⋅+⋅∇+∂∂ puu p

t rrγ ρ

puut

−∇=

∇⋅+

∂

∂ rr ρ ρ

0

0'

0'

=′∇+∂

′∂

=′⋅∇+∂

∂

=′⋅∇+∂

∂

pt

u

u pt

ut

r

r

r

ρ

γ ρ

ρ ρ



6.11.2 Linear Wave Equations

02 =′⋅∇+∂

′∂ua p

t

p r

02

2

=

∂

′∂

⋅∇+∂

′∂

t

u pt

p r

γ

Assume isentropic unsteady processes:

pa

p p

p s

s p

p pS

′=′=′

∂

∂+′

∂

∂=

2

1'''

γ

ρ ρ ρ

Substitute in continuity equation:

(same as equation

for the pressure)

Differentiate with respect to time:

Substitute momentum equations:

01

2

2

=

′∇−⋅∇+

∂

′∂ p p

t

p

ρ γ

022

2

2

=′∇−∂

′∂ pa

t

p Wave equation for

pressure fluctuations

Differentiate momentum equation with respect to time

02

2

=

∂

′∂∇+

∂

′∂

t

p

t

ur

ρ


CHAMBER





6.11.3 Solutions for Normal Modes (cont’d)

02

=

∇−

∂

∂∇ φ

φ 22

2 a

t

Assumed form: t ieu ω φ ±−∇=′r

cat

=∇−∂

∂φ

φ 22

2

20=∇c

6.11.4 Satisfying Boundary Conditions

Rigid walls:

Apply momentum equation:

Computational problem for the normal modes of pressure:

0ˆ =⋅′ nur

( ) 0ˆˆ

0ˆ2

=′⋅∂∂=′∇⋅

=

′∇+

∂

′∂⋅

unt

pn

pt

un

r

r

ρ

ρ

0ˆ

022

=∇⋅

=+∇

n

nnn

n

k

ψ

ψ ψ


CHAMBER



6.11.5 Some Examples of Common Normal Modes

( ) ( ) ( )φ α ψ mr J z k mnmn coscos l=


CHAMBER



heat release

presence of mean flow

condensed material suspended in the gas

conversion of condensed material to gas

motion of the boundary

6.12 A METHOD OF SPATIAL AVERAGING FOR

SOLVING INTERNAL ACOUSTIC PROBLEMS WITH

SOURCES

Many possible kinds of sources of acoustic waves exist; e.g.:

M

Here assume time-varying heat sources distributed in the

volume, and distributed motions of the boundary

The equation for the pressure with heat sources (§ 4.5 )

QC

R puu p

t

p

V

&rr=∇⋅+⋅∇+

∂

∂γ

Boundary condition for a moving boundary

∇⋅+

∂

∂⋅−=∇⋅ uu

t

un pn

rrr

r ρ ρ ˆˆ

Linearized forms:

t

un pn

pt

u

QC

Ru p

t

p

V

∂

′∂

⋅−=′∇⋅

=′∇+∂

′∂

′=′⋅∇+∂

′∂

r

r

&r

ˆˆ

0

ρ

ρ

γ



t un pn

t

Q

C

R pa

t

p

V

∂ ′∂⋅−=′∇⋅

∂

′∂=′∇−

∂

′∂

r

&

ˆˆ

22

2

2

ρ

Form the linear wave equation (§ 5.11)

Define the functions h and f :

t

un f

t

Q

C

R

ah

V

∂′∂⋅−=

∂

′∂−=

r

&

ˆ

12

ρ

f pn

ht

p

a p

−=′∇⋅

=∂

′∂

−′∇

ˆ

12

2

2

2

Problem to be solved:



SOURCES



(1) If h, f are small, then should not differ

greatly from its form given by a solution to the

homogeneous problem

(2) Hence ‘subtract’ the two problems and devise

a scheme for calculating the difference, due to

non-zero values of h, f .

(3) Any solution to the homogeneous problem can

be expressed as a superposition of normal

acoustic modes:

6.12.1 Oscillator Equations for Unsteady Motions in a

Chamber

0ˆ

01

2

2

2

2

=′∇⋅

=∂

′∂−′∇

h

hh

pn

t

p

a p

Assume h, f ‘small’ perturbations and given

The idea:

(i) is in general complex but independent of t .

(ii) Each term in the series, and therefore the

sum, i.e. is an exact solution to the

homogeneous problem.

nC

( ) ,t ,r ph

r

( ) ( ) t innh

ner C t ,r p ω ψ rr

∑=′

Notes:

p′



SOURCES




Chamber (cont’d)

(4) The solution when sources are present will not differ greatly

from a normal mode or a combination of normal modes.

Hence, compare the perturbed (actual) solution p′ with one

of the normal modes .

(5) Formal Procedure:

(i) Multiply (1) by , (2) by p′, and subtract

(ii) Integrate over the volume of the chamber

(iii) Rewrite the first integral by using Green’s theorem and

substitute the boundary conditions (3) and (4):

nψ

(1) (2)

(3) (4)

ht

p

a p =

∂

′∂−′∇

2

2

2

2 1022 =+∇ nnn k ψ ψ

nψ

( ) h pk t

p

a p p nnnnnn ψ ψ ψ ψ ψ =

′+

∂

′∂−∇′−′∇ 2

2

2

2

22 1

[ ]∫ ∫∫ =

′+

∂

′∂−∇′−′∇ dV hdV pk

t

p

adV p p nnn

nnn ψ ψ

ψ ψ ψ 2

2

2

2

22

f pn −=′∇⋅ 2ˆ 0ˆ =∇⋅ nn ψ


SOLVING INTERNAL ACOUSTIC PROBLEMS

WITH SOURCES

[ ]

∫∫

∫∫∫

−=

⋅∇−⋅∇=∇−∇

dS f

dS n pn pdV p p

n

nnnn

ψ

ψ ψ ψ ψ 321

0

22 ˆ'ˆ'''



6.12.1 Oscillator Equations for Unsteady Motions in a Chamber (cont’d)

{ }∫ ∫∫∫∫ +−=′+

∂

′∂dS f dV hadV pdV

t

p

nnnnn ψ ψ ψ ω ψ 22

2

2

The equation in (ii) becomes

Note: This equation is the spatial average of the

difference between the perturbed and unperturbed

problems with an acoustics mode

as weighting function.

(iv) Now expand as a synthesis of basis functions,

selected here to be the unperturbed acoustic modes with

time-dependent amplitudes:

( )t r p ,r

′

( ) ( ) ( )r t pt r p mm

m

rrψ η ∑

∞

=

=′1

,

(∗)

(∗∗)



WITH SOURCES




Chamber (cont’d)

Substitute (∗∗) in (∗):

( ) ( ) ( ) ( )

{ }

( ) ( ) ( ) ( )

{ }∫∫

∑∑

∫∫

∑∑

+∫−=

∫+∫

+∫−=

∫+∫

∞∞

∞∞

dS f dV ha

dV r r (t) pdV r r p

dS f dV ha

dV r r (t) pdV r r p

mm

mmmnmmm

mm

mmmnmmm

ψ ψ

ψ ψ η ω ψ ψ η

ψ ψ

ψ ψ η ω ψ ψ η

2

1

2

1

2

1

2

1

rrrr&&

rrrr&&

Orthogonality: ( ) ( ) mnmmm E dV r r δ ψ ψ 2=∫ rr

{ }∫∫+∫−=+ dS f dV h E p

a

dt

d nn

n

nnn ψ ψ η ω

η 2

22

2

2

Oscillator Equations



SOURCES



Problem: Find ω and α (NOTE: α > 0 for instability)

6.12.2 An Example of Linear Stability

Assume (see above)

t

un f ;

t

Q

C

R

ah

V ∂

′∂⋅=

∂

′∂−=

r&ˆ

12

ρ

Examine the stability of the nth longitudinal acousticmode supplied or losing energy by distributed heat addition and

motion of the boundary at one end.

Assume slowly growing or decaying harmonic motion;

all variables have the form

wavenumber complexˆ

=−==

−

:k ik ae g g kt ai

α ω

Measure phases relative to the pressure oscillation:

kt aii

nkt ai

nn

kt aiikt ai

kt ai

nn

eeueuuun

eeQeQQe

Q

Q

−−

−−

−

==′=′⋅

==′

=

φ

φ η η

ˆˆˆ

ˆˆ

ˆ

r

&&&

Q&

nu

: distributed combustion, non-uniform

: non-zero on the end face at z = 0



SOURCES



6.12.2 An Example of Linear Stability (cont’d)

Longitudinal modes:

2cos

cos

0

22 LS dz k S E

Lnk

z k

C L

nC n

n

nn

==

=

=

∫

π

ψ

( )

( )

−=′⋅∂

∂

−=∂

′∂

−−

−−

kt aiin

kt aii

eeuk aiunt

ee z Qkait

Q

u

Q

φ

φ

ρ ρ ˆˆ

ˆ

r

&&

( ) ( ) ( )

( ) ( )

( ) n

n

kt ain

kt ai

V V

k adt

d

euk aiunt

f

e z Qk aiC

R

at

Q

C

R

ah

U

Q

η

η

ρ ρ φ

φ

ˆ

ˆˆ

ˆ11

2

2

2

22

−=

−=′⋅∂

∂=

+=∂

′∂−=

+−

+−

r

&&



SOURCES




Substitute in oscillator equations and cancel ,kt aie−

Remark: unQ ′⋅′ r& ˆ, are perturbations;

0≠≠ α ω ω ,n because

To first order in small quantities:

( )( )

u

Q

i

n

nn

Li

nn

V nn

eu

L

i

dz e z Q

z k C

R

L pii

φ

φ

η

γ ω

η ω ω αω ω

−

−

−

+=− ∫

ˆ

ˆ2

ˆ

ˆ

cos2

20

22

&

1ˆ

,ˆ

ˆ

<<n

n

n

uQ

η η

( ) ( )

( ) }

( ) ( )( )

( ) u

Q

u

Q

i

n

n

n

C

Li

nn

V n

C n

C C

i

nn

i

nV n

nn

eu

i

E p

S ai

dz e z Q

z k C

R

E p

S iii

RS ;dz S dV

dS ek uk ai

dV e z Q zk k aiC

R

a E p

ak a

φ

φ

φ

φ

η

α ω ρ

η α ω ω α ω

π

ρ

η ω

−

−

−

−

−+

−+=−

==

⋅−

∫

+−=+−

∫

∫∫

ˆ

ˆ

ˆ

ˆ

cos

0cosˆ

ˆcos1

ˆ

2

2

02

22

2

22

2222

&

&



WITH SOURCES



Internal feedback : either or both of and


Take real and imaginary parts:

( ) z Q

( )

( )

n

un L

n

Q

nV n

n

n

unn L

n

Q

nV

nn

u

Ldz

z Q

z k C

Ru

L p

u

Ldz

z Q z k

C

R

L p

η

φ γ

η

φ

η α

η

φ γω

η

φ ω ω ω

ˆ

cosˆ

ˆ

cosˆ

cosˆ

ˆ1

ˆ

sinˆ2

ˆ

sinˆ

cos2

0

0

22

+

−

=

+

+=

∫

∫

&

&

nu

depend on pressure, i.e. nη ˆ

For example, ( )

nn

nnnnnn

uu

z k qq z Q

η

η ψ η

ˆˆ

cosˆˆˆ

===& ( )0≥uq,

( )

quantitysmall

coscos

sinsin222

+=

=−=

=−+=

n

nunQn

V

nunQnnn

L

B Bu Aq

C

R

p A Bu Aq

ω ω

γω φ φ α

ω φ ϕ ω ω ω

unnQnn

nnnqnnn

u Bq A

u Bq A

φ φ α

φ φ ω ω

coscos

sinsin

+−=

++=



SOURCES





6.12.3 A General Result for Linear Stability (cont’d)

Special Averaging ⇒ Oscillator Equations

Linear Stability ( linear in and )

{ }∫∫+∫−=

=+

dS f dV h E p

a F

F dt

d

nn

n

n

nnnn

ψ ψ

η ω η

2

2

2

2

2

n F nη nη &

kt ainn

kt ainn e F F e −− == ˆη η

n F ˆ has the general form

( ) ( )

( )nnnnn

i

n

r

nn F i F i F F inˆˆˆˆˆˆ

η ω η η =−=

Substitute:

) ( ) ( )

( ) ( )( )

( ) ( ) ( )

( )n

i

n

r

nn

in

r nn

ni

nr

nnn

F i F i

k a F i F k a

F i F k a

αω αω ω α

ω αω α ω

α ω ω

η η ω

≈<<

−−=−−

−=−−=

−=−

;

ˆˆ2

1;ˆˆ

ˆˆˆˆ

22

222

222

222

n

in

n

n

r n

nn

F

F

η ω α

η ω ω ω

ˆ

ˆ

2

1

ˆ

ˆ

2

1

)(

)(

−=

−=

( )nω ω ≈



SOURCES



6.12.3 A General Result for Linear Stability (cont’d)

( )

0

ˆ

ˆ>

n

in F

η

Small amplitude motions are unstable if α > 0

Instability:

This is essentially a generalized form of

Rayleigh’s Criterion for linear behavior

Remarks:

(i) accommodates all linear processes

(ii) The main problem is modeling

(iii) The frequency shift is normally

not important in linear behavior, but is

significant in nonlinear behavior.

n F

nω ω −



WITH SOURCES



6.13 APPLICATION OF TIME AVERAGING

• Commonly (not always) unstable oscillations in combustors haveslowly varying amplitude A and phase φ; that is, during one cycle,

or

12

11 <<<<dt d T ;

dt d φ

π τ A

A

where is the period of the oscillation.ω

π τ

21==

f

• The primary intention of the method is to change the

dependent variable from the rapidly varying oscillatoryamplitude ηn (governed by a second order equation) to theslowly varying amplitude A and phase φ (two first-order

equations)

• Two main advantages of the method:

(i) Because the oscillatory motions are removed,

numerical solutions to the equations are much cheaper

and quicker (perhaps several cycles for each interval

of integration)

(ii) The first order equations are convenient for certain

theoretical considerations and for understanding

qualitative behavior

12

;1 <<<<π

δφ δ

A

A



(i) Given the oscillations with forcing function ‘small’

of order, µ<< 1:

Procedure:

nnnnn G F µ η ω η ==+ 2&&

Remark: In combustors, µ is of the order of the Mach number of the mean

flow and )1(0~nG

(ii) Write in the equivalent form defining the pairs of

functions and :

( )t nη

( ) ( )( )t t nn φ , A

( ) ( ) ( )( ) ( ) ( ) t t Bt t At t t At nnnnnnnn ω ω φ ω η sincossin +=+=

( ) ( )

( )t A

t Bt

B A

B; A

n

nn

nnn

nnnnnn

1

22

tan

sincos

−=

+=

==

φ

φ φ

A

A A

(iii) One way to proceed follows a physical argument based

on the energy of the oscillator and its change in time.

The mechanical energy of the nth oscillator is the sum

of potential and kinetic energies:

222

2

1

2

1nnnn η η ω &+=E

( ) ( )( )t Bt nn , A




(iv) The time average values of energy and power input

to the oscillator due to action of the force are

( ) ( ) ( )nnnnnnnnnnnn t t t φ ω φ φ ω φ ω ω η +++++= cossincos A A A &&&

(vi) From (ii), the velocity of the oscillator is

t d F F t d t

t nnnn

t

t nn

′=′= ∫∫++ τ τ

η τ

η τ

&&11

E E

n F

where τ is an interval of time-averaging.

(v) Conservation of energy requires that the time-

averaged rate of change of energy equal the time-

averaged rate of work done by on the oscillator:

nnn F dt

d η &=E

Apply the ‘strong’ condition (Krylov and Bogoliubov)

that the velocity is always given by the formula for an

oscillator in force-free notion

( ) ( )( )

( ) ( ) ( ) 0cossin

cos

=+++∗

+=

nnnnnnn

nnnnn

t t

t t t

φ ω φ φ ω

φ ω ω η

A A

A

&&

&


n F



( ) ( )[ ]

dt

d

dt

d

t t

nnn

n

nnn

nnnnnnnn

nnnnn

A A

E

A E

A A E

22

22

222222

2222

2

1

2

1cossin

22

1

2

1

ω

ω

ω φ ω φ ω ω

η ω

=

=

=+++=+= &

(vii) Compute the time averages in (v.)

( )( ) ( )

( ) ( ) ( ) ( ) nnnnnnnnnnnnnnn

nnnnnnnnn

nnnn

F t t t

F t t dt

d

F

=++++−+

=+++

==

φ ω ω φ ω φ ω ω φ ω ω

φ ω ω φ ω ω

η ω η

sinsincos

sincos

2

2

2

A A A

A A

&&

&&

(a)

(b)

( ) t d F t A F F nnnn

t

t

nnn ′+′= ∫

+

φ ω τ

η τ

cos1

&

( ) t d t F dt

d nn

t

t n

n

n ′+′= ∫+

φ ω τ ω

τ

cos1 A

(viii) To find the equation for substitute in the oscillator

equation:

,n

φ &




( )nnnn

nn t F dt

d φ ω

φ ω +=− sin A

( ) t d t F dt

d nn

t

t n

n

nn

′+′−= ∫+

φ ω τ ω

φ τ

sin1

A

Substitute condition (∗), part (vi):

(ix) Time-average over ( ) :, τ +t t

(x) From definitions of and , part (ii):n A n B

( )

( ) t d t F dt

dB

t d t F dt

dA

n

t

t n

n

n

n

t

nn

n

′′−

=

′′=

∫

∫+

+

ω τ ω

ω τ ω

τ

τ

sin1

cos

1

t

6.13.1 Example: Linear Stability

First need to construct representation of the processes,

i.e., n F

nnnn F =+ η ω η 2&&

General form of for linear processes onlyn F

nnnnn E D F η η −−= &






6.13.1 Example: Linear Stability (cont’d)

( ) ( ) t A E B Dt B E A D

E D F

nnnnnnnnnnnn

nnnnn

ω ω ω ω

η η

sincos −−+−−=−−= &

( ) ( )t it nn

nnn eet θ ω α

η −±= 0 A

Method of

Averaging:

t d t F T dt

dB

t d t F T dt

dA

n

t

t n

n

n

n

t

t n

n

n

′′−

=

′′=

∫

∫

+

+

ω ω

ω ω

τ

τ

sin1

cos1

(i)

(ii) Assume nn ω π τ τ 2==

−=+−=

−=+=

n

nnnnnn

n

nnnnnnn

E A B

dt

dB

D B A

dt

dA

ω θ θ α

α θ α

2

12

1Results:

−=

=

nn

nnn

t

d dt

d

θ φ

α

d

A A

Combine these to form equations for and :n

A nφ

Solution:

i.e.:

022

2

=−+− nnnnnnn η θ ω ω η α η &&&




[ ]

[ ] t d t t Bt AT

K

dt

dB

t d t t Bt AT

K

dt

dA

n

t

t

t

t

′′′+′−

=

′′′+′=

∫

∫

+

+

ω ω ω ω

ω ω ω ω

τ

τ

sincossin

coscossin

22221

1

122221

1

Also:

6.13.2 Example: Coupled Linear Oscillators

Two coupled oscillators:

( ) ( )t Bt A 11 ,

( ) ( ) ( ) ( )21cossin

12222

21211

,it t Bt t At

K

K

iiiii =+=

=+

=+

ω ω η

η η ω η

η η ω η

&&

&&

are solutions to:

Assume and set12 ω ω = :11 2 ω π τ τ ==

21

12

1

1

22 A

K

dt

dB; B

K

dt

dA

ω ω −==

11

21

1

2

22 A

K

dt

dB; B

K

dt

dA

ω ω −==

All slowly varying functions satisfy the same equation:

( )212121

2

2

2

4 B ,;B A , A

K

dt

d =−= ψ ψ

ω

ψ

Initial Condition

t t

K

C tt

K

C ii

ii ω ω η ω ω η cos2cossin2sin 2211

=

=




6.14 REMARKS ON THE GENERAL RESULT OF

SPATIAL AVERAGING

Quite generally, the wave equation for the pressure

fluctuation takes the form

f f f pn

hhh

t

p

a

p

nl l

nl l

−=−−=′∇⋅

=+=

∂

′∂−′∇

ˆ

12

2

2

2

where ( )l denotes ‘linear’ and ( )nl ‘nonlinear’

The oscillator equations are

nnl

nl

nnnn F F F

dt

d =+=+ η ω η 2

2

2

where

{ }∫ ∫∫+−= dS f dV h E p

a F nn

n

n ψ ψ 2

2

and the are found as solutions to the problemnψ

0ˆ

022

=∇⋅

=+∇

n

nnn

n

k

ψ

ψ ψ



Thus splits into a sum of linear and nonlinear parts

{ } { }∫ ∫∫∫ ∫∫ +−+−= dS f dV h E p

adS f dV h

E p

a F n

nl n

nl

n

nl

nl

n

n ψ ψ ψ ψ 2

2

2

2

n F

If there is no linear coupling between modes

{ }∫ ∫∫+−+= dS f dV h E p

a F n

nl n

nl

n

nnnnnn ψ ψ η θ ω η α 2

2

22 &

and

{ }∫ ∫∫+−=+ dS f dV h E p

an

l n

l

n

nnnnn ψ ψ η θ ω η α 2

2

2&

is the formula for computing and .nα nθ

Note: Linear coupling between modes can be removed by formal transformation (see § 6.10.2)

6.14 REMARKS ON THE GENERAL RESULT OF

SPATIAL AVERAGING



End of Section VI

Documents

Fundamentals of Acoustically Coupled Combustions