acousticssujithnotes

8/3/2019 AcousticsSujithNotes

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Acoustic Instabilities in Aerospace

Propulsion

Dr. R. I. Sujith

Associate Professor, Department of Aerospace Engineering

Indian Institute of Technology Madras

Compiled and Collated by

Vipluv Aga


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Contents

1 Introduction 5

1.1 Motivation for studying acoustic instabilities in combustion . 51.1.1 Instabilities in Solid Motors . . . . . . . . . . . . . . . 5

1.1.2 Instabilities in Liquid Motors . . . . . . . . . . . . . . 8

2 Acoustics 9

2.1 The wave equation . . . . . . . . . . . . . . . . . . . . . . . . 92.1.1 Derivation of the 1-D wave equation . . . . . . . . . . 112.1.2 1-D Plane travelling wave . . . . . . . . . . . . . . . . 122.1.3 Plane propagating transverse wave . . . . . . . . . . . 14

2.2 Helmholtz equation . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Acoustic Energy, Intensity and Power . . . . . . . . . . . . . 15

3 Reflection and Transmission of plane waves 19

3.1 Reflection at rigid wall . . . . . . . . . . . . . . . . . . . . . . 203.1.1 Normal incidence on a rigid boundary . . . . . . . . . 203.1.2 Oblique incidence on a rigid boundary . . . . . . . . . 21

3.2 Acoustic Impedance . . . . . . . . . . . . . . . . . . . . . . . 21

3.3 Reflection at an impedance discontinuity . . . . . . . . . . . . 223.3.1 Normal incidence . . . . . . . . . . . . . . . . . . . . . 223.3.2 Oblique incidence . . . . . . . . . . . . . . . . . . . . . 23

3.4 Waves in an Impedance tube . . . . . . . . . . . . . . . . . . 25

4 Radial and cylindrical wave equations 27

4.1 Spherically symmetrical waves . . . . . . . . . . . . . . . . . . 274.2 Waves in cylindrical ducts . . . . . . . . . . . . . . . . . . . . 284.3 Rectangular Ducts . . . . . . . . . . . . . . . . . . . . . . . . 29

5 Combustion-acoustic Interaction 31

5.1 Coupling between acoustics and flames . . . . . . . . . . . . . 325.1.1 Effect of heat addition on oscillations . . . . . . . . . 325.1.2 Rayleigh Criterion . . . . . . . . . . . . . . . . . . . . 33

5.2 Solution of the wave equation with heat release . . . . . . . . 35

5.2.1 Greens Function Technique . . . . . . . . . . . . . . . 35

3


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4 CONTENTS

5.2.2 An Example . . . . . . . . . . . . . . . . . . . . . . . 38

5.3 Sensitive Time Lag Hypothesis . . . . . . . . . . . . . . . . . 395.4 Active Control . . . . . . . . . . . . . . . . . . . . . . . . . . 415.5 Various Sources of Instabilities . . . . . . . . . . . . . . . . . 43

6 Modal Analysis and Aerodynamic Noise 45

6.1 Galerkins Method . . . . . . . . . . . . . . . . . . . . . . . . 456.2 Aerodynamic Noise or Aeroacoustics . . . . . . . . . . . . . . 46


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Chapter 1

Introduction

For most aerospace applications, combustion occurs in a specially designatedzone, known as the combustion zone. For the case of a gas turbine this couldbe a combustor or for a liquid rocket it is the flame holder. It is importantin these cases for the flame to be held stably in a particular re-circulationzone, which may have some bluff bodies which shed vortices causing vortexshedding noise. An instability in the flame structure could lead to blow outor in some cases to higher N Ox emissions. Hence it is important to studythe effects of instabilities on combustion processes.

Combustion instabilities keep occurring when acoustic oscillations keepgrowing.

Acoustic oscillations are basically pressure variations caused due tovortex shedding noise (eg. whistle, flute, mouth organ) or synthetic noise(eg. speaker, synthesizer). Once an oscillation has been perpetrated it keepson occurring due to the feedback from the heat release.

Feedback is a process in which part of the output is returned to theinput. A common example is the loud shrill sound when one places a mi-crophone in front of a speaker.

1.1 Motivation for studying acoustic instabilities

in combustion

1.1.1 Instabilities in Solid Motors

A prevalence of pressure oscillations within solid rocket motors would leadto a strictly continued increase in the amplitude of oscillations due to pos-itive feed back from propellant combustion. This would lead to limit cycleoscillations due to presence of damping.

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6 CHAPTER 1. INTRODUCTION

Limit cycle oscillations are those in which the damping overtakes the

driving. Such oscillations may result in the resonation of the structuralmembers in the vehicle, causing catastrophic destruction. These could alsoharm other systems like guidance and control electronics in flight.

At this point a distinction should be made between combustor and com-bustion instability.

Combustor instability occurs due to pressure oscillations related tochamber geometry. These could be of acoustic/non acoustic origin.

Combustion instability, on the other hand is only a specific part of theproblem. It is the combustion response of propellant to oscillations in the

chamber.Combustion instabilities may manifest themselves as thrust oscillations.

For instance, a 0.1 bar oscillation in pressure amplitude over a 50 bar meanpressure results in 5 to 10 times the mean thrust in thrust oscillation.

The Decibel scale is used to measure the degree of oscillation withregards to a mean quantity. The Sound Power Level in dB is given by:

P W L = 10 log10Power

1012W(1.1)

The Sound Pressure Level in dB is based on the normal threshold forhuman hearing and is given by:

SP L = 20 logp

pref(1.2)

pref = 2 105P a in air and 106P a in other media. Correspondingly,150 dB 2000P a on a standard scale. The reference level pref corre-sponds to the normal threshold of human hearing at 1kH z. The minimumdetectable sound by a human ear is 105 P a.

Such oscillations can cause pre-ignition of fuel rich exhaust due to en-hanced mixing with ambient air. This would interfere with ground-based

navigation leading to dangerous situations. In fact just 15Hz thrust oscil-lations in the Space Shuttle SRBs (Solid Rocket Boosters) cause jitters tothe astronauts which is a major inconvenience.

DC Shift

This is the shift of the mean pressure, (known as mean pressure excur-sions) due to oscillations. Such a shift takes place when there are mass fluxoscillations from the solid propellant itself.

Oscillations are unforseen


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1.1. MOTIVATION FOR STUDYING ACOUSTIC INSTABILITIES IN COMBUSTION7

Usually oscillations like those described above are unforseen and hence

because they are too complicated to be detected a priori, they are not ac-counted for in the preliminary design stage. These are then detected quitelate during the motor development program when options on design changesare limited. In the industry, more often than not, tricks of the trade areemployed. Hence expertise to handle these problems is based on a holisticapproach, knowing full well that solutions for different mechanisms shouldntbe conflicting.

Incidences of Instabilities

Some facts that throw light on the importance of the studies of instabil-ities are as follows:

1. One estimate is that over 70% of solid rocket motors (SRMs) exhibitoscillatory behavior

2. More than 50% of SRM research funding in the United States is mo-tivated by instability problems

3. Most motors naturally whistle

Examples of oscillations in some known SRBs

The Ariane V MPS of the European Space Agency, exhibited the first

three modes of oscillations and the dominant frequency even shifted amongthe modes. The problem was tackled after enormous research efforts andfunding in Europe.

The Subroc Motor showed wild instability for the first time in the32nd test firing. The Aluminum size was suggested to be reduced from 30microns to 15 microns. Following this the motor developers changed halfthe Aluminum contents to 5 micron for fear of processing problem and burnrate change but the instability still persisted. So finally a complete changein aluminum was effected.

The Minuteman Missile was induced into service with oscillations.Which led to random flight test failures although it had passed the statictests. The production log revealed supply of aluminum from different plants.Later the same supplier was asked to prepare the aluminium with the samespecifications.

The Space Shuttle SRB had slag accumulation and ejection in 1994,leading to pressure and thrust spikes. This was countered by a change inAmmonium Per chlorate supplier for want of better purity.


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8 CHAPTER 1. INTRODUCTION

The Retro Rockets of the Mars Pathfinder originally had only 2%

Al propellant in order not to contaminate Martian atmosphere but finally18% Al was used in order to suppress the oscillations.

1.1.2 Instabilities in Liquid Motors

The droplets dispersed into the combustion chamber of a liquid rocket motorare ignited and hence act like little flames. An acoustic oscillation couldmove these droplets back and forth, as a pressure oscillation would effectdrag on the droplets. The problem is compounded due to the variable heatrelease and burning rates due to the oscillations. These phenomena cause

local heat differences which when they try to convect cause expansion waveswithin the chamber, which is not designed to meet such calamities. Furthermore, these pulsations could also be propagated through to the pipelinesand manifolds delivering the liquid fuel.

In liquid motors too, the combustion instability is caused by heat re-lease fluctuations due to chemical reactions. The system instability on theother hand is caused due to oscillations in injectors, manifold pipe lines orpropellant tanks.

In general, in primary combustor, whenever we have a stagnant regionit gives rise to recirculation zones. In practice, recirculation zones are shearlayers. We have longitudinal, transverse and radial modes of vibration. The

worst part about tangential modes is that there is no fixed diameter alongwhich the nodes exist. They keep precessing.The basic mechanisms of sound generation are due to combustion sound

and vortex shedding sound. Heat release happening in tune with vortexshedding also produces very large sound. Particular case: LNGT (low N OXgas turbines). We have high temperatures in a combustor that gives rise toN OX pollution, so a solution came up i.e. we cool the combustor. But thatreduces efficiency so there is a trade off. Nowadays we suppress N OX levelswithout cooling, i.e. by operating at the lean limit.

Oscillations assume significance in the presence of Fuel inflow oscillationsand mixture fraction oscillations.


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Chapter 2

Acoustics

Sound should not be a misnomer for the frequencies in the range of hu-man perception. Sound is a disturbance of pressure (normal stress) whichpropagates at finite speed in a compressible medium (solid, liquid or gas).

There are Two basic types of waves

1. Longitudinal or Compression wave: Here particle motion is alongthe direction of wave propagation. The term particle does not meanindividual molecules but an infinitesimal volume of the fluid. A me-chanical analogy can be given by the model of a spring which is com-pressed and released thereby the compression wave travels along the

length of the spring.

2. Transverse waves or Shear wave: In this wave, the particle mo-tion is perpendicular to the direction of the wave propagation. Themechanical analogy is the wave caused by plucking a stretched spring.

Fluids do not support shear deformation. They support compression.But the above argument is strictly true for inviscid fluid. Transverse wavesencounter viscous damping mostly; we deal with longitudinal wave propa-gation.

2.1 The wave equationThe variables used in the equation are pressure p, particle velocity v, particledisplacement (used sometimes) and density . In order to derive the waveequation, two conservation equations and one equation of state is used.

Continuity equation

The continuity equation conserves the mass in a control volume. Assum-ing a 1-D control volume,

(tu) |x (tu) |x+x = t

(tx1)

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10 CHAPTER 2. ACOUSTICS

(tu)

|x+x = (tu)

|x +

x

(tu)x + . . .

(tu) |x (tu) |x x

(tu) =tt

x

This gives the general continuity equation

tt

+ (tv) = 0 (2.1)

For the sake of simplicity a 1-D flow is considered and hence Eq. 2.1reduces to

tt

+

x (tv) = 0 (2.2)

It must be noted that this equation describes a stationery medium wherethere are no particle velocities unless the acoustic wave passes through.

Momentum equation

Conservation of linear momentum implies, F = ma. Again a controlvolume is taken and using this Eulerian frame of reference the linear mo-mentum of the C.V. is conserved by

external f orces = x1 DuDt . This

implies (ext. forces)= Rate of change of momentum inside C.V.+changein mass flux across a control surface.

Now the external forces are considered to be made up of pressure forces,viscous forces and body forces. For simplicity it is assumed that the fluid isinvisid and there are no body forces. The pressure force across the controlvolume is obtained by using a Taylors expansion and neglecting higher orderterms.

pt |x pt |x+x=

ptx

x

The momentum equation then comes about in the following way

ptx

=

t(tu) +

x(tu

2) (2.3)

This equation can be represented in the form of primitive variables, i.e.quantities which can be directly measured. The above equation is expandedand the terms u tt +u

x (tu) evaluate to zero because of continuity thereby

givingt

utt

+ tututx

= ptx

(2.4)

The third equation comes from the state equation which is a relation betweenpressure, density and temperature/entropy.

pt = pt(t, so) (2.5)

These are non-linear coupled equations. Analytically, they have very limitedexact solutions. By some suitable assumptions and approximations, thesethree can be linearized to obtain the linearized 1-D wave equation.


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2.1. THE WAVE EQUATION 11

2.1.1 Derivation of the 1-D wave equation

The linearization process essentially entails, neglecting those terms whichare higher than the 1st order in the equation.

Order of variables: A quantity consists of a perturbation quantitysuperimposed over the mean quantity, for e.g. pressure can be written as

pt = p0 + p where p is the perturbation quantity1. Algebraically it is small

and so powers or orders greater than 1 can be ignored. 0 is a zero orderquantity, is first order while u is a second order quantity because it is aproduct of two perturbation variables.

Eq. 2.1 is expanded and terms containing the higher order like x (u)are neglected giving

t + 0u

x = 0 (2.6)

Similarly the momentum equation (Eq. 2.3) is linearized to give

0u

t= p

x(2.7)

Linearizing the equation of state could involve a graphical approach where ptand t values are read of the graph and in the near vicinity it is somewhata straight line. Mathematically, such an approximation is manifested byexpressing pt as a Taylor series and neglecting higher order terms therebygiving, pt = p0 +

ptt

|s0 (t

0). This reduces to

ptt

= c2 (2.8)

Here c2 is a constant by definition of the equation of state. It can also beargued that a very general equation of state can be written as

dp = c2d +

s| ds (2.9)

Here c2 is the isentropic speed of sound and is given by Eq. 2.8 (by def-inition). This equation of state when valid for the entire flow, refers to

a homogenous fluid. When density depends on pressure then it would becalled barotropic and when the fluid is homogeneous and the entropy uniform(ds=0) then it is called homentropic. For the given case of linearization, thelinearized equation of state is given by

p = c2 (2.10)

1Throughout the text, for all quantities the subscript 0 shall be used to denote mean

quantities and a non subscripted variable shall mean perturbation quantity, unless other-

wise specified. For eg. p0, 0, v0 are mean pressure, density and velocity, while p,,v are

perturbation quantities of the same.


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(It shall be conclusively proved in subsequent sections that c is indeed the

speed of sound.)The following operation is peformed: t (Eq. 2.6) -

x (Eq. 2.7). Also it

is noted from Eq. 2.10 that 2p

t2= c2

2t2

. This gives

2

t2

2p

x2= 0 (2.11)

2p

t2= c2

2p

x2(2.12)

This is the 1-D wave equation.

The wave equation is linear which allows superposition of solutions. The

wave equation is also homogeneous.(i.e. no dependent term on R.H.S).Again, there is no forcing term which means that nothing drives or dampensthe pressure disturbance. The equation governs phenomenon in a domainwith no sources of sound, hence we are concerned with the phenomenon ofa non attenuated wave. A similar wave equation can be derived for propa-gation of electromagnetic waves in free space, but governing equations willbe different. For eg., Schrodingers wave equation.

Historically, 1-D wave equation was derived by DAlembert in 1747 foroscillation of strings. Euler derived a PDE specifically for sound propagationin fluids. Elasticity is considered as the medium of propagation instead ofkinetic approach as indicated by the spatial derivative. The wave equation

is of second order, therefore it has infinite solutions. A particular solutioncan be obtained, given initial boundary conditions for a 1-D Plane travellingwave.

2.1.2 1-D Plane travelling wave

Cross-derivatives are equal for smoothly varying quantities.

2p

xt=

2p

tx(2.13)

The wave equation then can be rewritten as

x

+1

c

t

x

1c

t

p = 0

px

+1

c

p

t= 0

px

1c

p

t= 0

By experience and intuition the solution to the above two equations wasfound to be p = f(x ct) and p = g(x + ct) by dAlembert, where f andg are arbitrary functions. It can be checked that the above solutions do


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2.1. THE WAVE EQUATION 13

indeed satisfy the equations. Therefore the general solution of a plane wave

propagating in the x-direction is p(x, t) = f(x ct) + g(x + ct). It can berearranged to give:

p(x, t) = f1(t xc

) + g1(t +x

c) (2.14)

Physically f1(t xc ) could imply a transient signal of some shape. (t xc )can be regarded as retarded time, . At any point is preserved. So ifwe have two points then,

(t1 x1c

) = = (t2 x2c

)

x2 x1t2 t1 = c

This also gives the speed of sound, as c. It can be clearly seen that forf1(t xc ), since x2 > x1 as t2 > t1 it indicates a forward propagating wave.Similarly g1(t +

xc ) indicates a backward propagating wave.

Relation between p and u for a plane wave

Using the linearized Euler equation: 0ut = px . This is integrated in

the positive x-direction.

u+ =

1

0 p

x

dt

Now change variables to = x ct. This gives

u+ = 10

f

(

c)

=f

0c

u+ = p+

0c

The constant of integration vanishes due to initially quiescent medium. Herethe plane wave has velocity in the positive x-direction for positive p+ andmagnitude proportional to pressure. Similarly for negative x-direction, the

variable is changed to = x + ct, giving u = p0c .In general p

+

|u+|= p

|u|= 0c. This quantity is called the characteristic

impedance of the medium.

Different media have different characteristic impedances. This is howthe acoustic wave is able to distinguish boundaries. The amount of soundtransmitted/reflected depends on the extent of impedance mismatch at theboundary.


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2.1.3 Plane propagating transverse wave

A forward propagating wave can be expressed as

p(x, t) = A cos

2f(t x

c) + 0

(2.15)

This is a plane wave because it only depends on x and t and harmonicbecause it is of the form of a cosine function which is oscillatory. After atime period T the wave repeats. For sines and cosines this happens whenthe argument changes by 2.

2f(t + T xc

) 0

2f(t xc

) 0

= 2

2f T = 2 T = 1

f

The spatial extent of one cycle is called its wavelength, often denoted by .The time taken to cover one wavelength is T and hence they are related asT = c i.e. c = f . The wave number is denoted as k =

2fc =

c where is

called the angular frequency. p can then be written as,

p(x, t) = A cos[(kx t) + 0] (2.16)k can be interpreted as a spatial analogue of

2.2 Helmholtz equation

The Euler relation ei = cos + i sin is used to obtain the solution of thewave equation in order to separate the time and space quantities. As such,the solution for a wave travelling in the positive x-direction can be writtenas,

p(x, t) = Re{Aei(kxt)+0} (2.17)Eq. 2.17 gives the real part of the complex pressure. The complex pressureis of more use mathematically as it leads to a factorization of space and timevariables.

p(x, t) =

Aei0eikx

eit

= p(x)eit

where p(x) is known as the complex amplitude. It can be easily observed thatthe complex pressure satisfies the wave equation. The Helmholtz equation isobtained when p(x)eit is substituted in Eq. 2.11. Owing to the separationof the variables, the Helmholtz equation is obtained as

d2p(x)

dx2+ k2p(x) = 0 (2.18)


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2.3. ACOUSTIC ENERGY, INTENSITY AND POWER 15

In fact the density t can also be considered as having a perturbation

component, 0 + , which can then be expressed as = eit

, then thecontinuity equation can be written as

i + 0uu

x+

xu = 0 (2.19)

The energy equation is given as

p

t+ p0

u

x= 0 (2.20)

which can be converted to the harmonic domain as

i p + p0 ux

= 0 (2.21)

From Eq. 2.19 and Eq. 2.21 the expression for is derived as

=p

c2+

iu

0x

(2.22)

But this is only used when density is not considered constant.

Acoustic velocity and wave velocity

Since p(x) is a a complex number in the Helmholtz equation, the realpart of p(x)eit is given by

Rep(x)eit

=

p2r(x) +p2im(x) [cos(t + )] , tan =

pim(x)

pr(x)(2.23)

From the momentum equation, Eq. 2.7

eit0iu(x) = eit dp(x)dx

(2.24)

u(x) = 1i0

dp(x)

dx(2.25)

Here U(x) is the acoustic velocity, which is different from the wave velocity,

as it is the actual velocity of the individual particles which are oscillatingwhile wave velocity is the speed at which the wave propagates.

2.3 Acoustic Energy, Intensity and Power

The starting points to derive the expression for acoustic energy are thelinearized mass and momentum conservation laws for the acoustic quantities(from Eqs. 2.6 and 2.7), which are described for a quiescent, inviscid andnon-conducting fluid. The expressions for the acoustic intensity and energy,will take into account dissipation due to m and f which are the perturbation


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quantities for mass source and body forces respectively, the mean quantities

being zero.

t+ 0 v = m (2.26)

0v

t+ p = f (2.27)

Using p = c20 and suitable algebraic manipulation of Eqs. 2.26 and 2.27 theconservation law for acoustic energy in 3-D is obtained. This means thatv is the velocity perturbation vector having all three components for thefollowing analysis. The 1-D equation is simply obtained by considering allthe vectors to have only a single component.

1

20c20

p2

t+

02

v2

t+ (pv) = pm

0+ vf (2.28)

This can be interpreted as

E

t+ I = D (2.29)

where D is the dissipation which is mostly considered zero, while E theenergy comprises the two terms as given in the equation below, as the kineticand potential energy. I is known as the acoustic intensity

E = p220c20

+ 0v22

(2.30)

I = pv (2.31)

D = pm

0 vf (2.32)

d

dt

V

E dV +

S

I n d =

VD dx (2.33)

Physically the acoustic intensity vector I is the acoustic power flow per unitarea.

In practise the time-averaged quantity I is the quantity that is actuallysensed by the human ear. If a signal is periodic then E is zero (as ithas fluctuating quantities in the numerator with mean quantities in thedenominator) and so the time-averaged equation for Eq. 2.33 becomes

P=

SI n d =

VD dx (2.34)

where P is the acoustic power flow across the surface S. The left handside simply corresponds to the mechanical work performed by the volumeinjection m/0 and the external force field f

on the acoustic field. P is a


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2.3. ACOUSTIC ENERGY, INTENSITY AND POWER 17

useful quantity because it is independent of in 3-D and is a property of the

source. Even if background noise is present, the net flux of the backgroundnoise across the control volume is zero and that is what the D term in Eq.2.34 represents.

For e.g. if a plane harmonic wave in a duct is considered, then there willbe no intensity flux across solid walls, as p does not vary across S.

P =

I n d

=p2rms0c

S

It is also observed that given a certain power level for the source, prmsdecreases as S increases.


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Chapter 3

Reflection and Transmission

of plane waves

Before looking at the reflection of waves, it would be beneficial to under-stand how standing waves are produced by reflection at the boundaries. Astanding wave consists of superposition of two progressive waves travellingin opposite directions.

A quarter wave tube is considered (so called because the no. of harmonicsformed within it is in terms of quarter of the wave tube length) with oneend closed and the other open. The boundary conditions are x = 0, u = 0and x = L, p = 0 with L being the open end. The velocity in the harmonic

domain can be written as

u = 1i

p

x= 1

i

ikAeikx ikBeikx

(3.1)

From boundary conditions, A = B and the Eq. 3.1 will give the velocity inthe tube. The pressure is then given by

p = 2A

eikx + eikx

2

= 2A cos kx (3.2)

substituting for p(L) = 0 the natural modes are obtained as

f =(2n + 1)

4

c

L(3.3)

Similarly using the same analysis for a closed-closed tube, only the boundaryconditions change to x = 0, u = 0 and x = L, u = 0. Thus the pressure andvelocity expressions are

p = 2A

eikx+eikx

2

= 2A cos kx (3.4)

u = 1i iAc

eikx+eikx

2

2 = 2A

csin kx (3.5)

19


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20CHAPTER 3. REFLECTION AND TRANSMISSION OF PLANE WAVES

This gives the fundamental frequencies as

f = nc2L

(3.6)

after applying the boundary condition at x = L.

3.1 Reflection at rigid wall

3.1.1 Normal incidence on a rigid boundary

The boundary condition at a solid surface is required. The general case tostart off would be from a vibrating surface, which itself generates waves inthe surroundings. On this surface, Xs(t) is the surface displacement and (t)

is the fluid particle displacement for a particle in contact with the surface.The continuity of displacement warrants that Xs(t) = (t) and so Vs(t) =

v(t), the two being surface and fluid particle velocities respectively. For ageneral 3-D case these can be given as

Vs(t) ns = v(t) ns (3.7)When the linear Euler relation 0

vt = p is applied at the fluid at the

rigid boundary the equation obtained is

0

tv ns = p ns (3.8)

Therefore if the boundary is stationary than the LHS will become 0 and thecondition derived will be

p ns = pns

= 0 (3.9)

When a wave i is incident normally on a plane rigid boundary a reflectedwave r is sent back in the opposite direction of the incident wave. Thereforelet pi = f(t x/c) and pr = g(t + x/c). Pressure at any point can be givenby p = pi +pr as we are taking the linearized forms and thus the quantitiescan be superimposed on each other. The rigid boundary condition is

pix

x=0 +

prx

x=0 = 0 (3.10)

1c

f +1

cg = 0 (3.11)

f = g (3.12)where f = f , = t x/c and g = g , = t + x/c.

It can be inferred from Eq. 3.12 that rigid boundaries simply redirectthe wave in the opposite direction, without altering the shape or phase ofthe wave. This also means that at the wall, since p = pi +pr, f+ g = 2pi.There is hence a pressure doubling at the wall.


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3.2. ACOUSTIC IMPEDANCE 21

3.1.2 Oblique incidence on a rigid boundary

Consider a plane wave which is incident on a rigid boundary at angle ofincidence i and reflects at angle of reflection r, both measured from thenormal at the point of incidence. The pressure must be expressed in termsof the distance along the direction of propagation si and so is given as

pi = fi(t si/c). From geometry it is easily seen that si = x sin i y cos i.The expression of pressure is

pi = fi

t

x sin i y cos i

c

(3.13)

Similarly if sr is the distance measured along the reflected wave then, sr =n

r x,

s

r= x sin

r+ y cos

r. The expression for pressure thus becomes

pr = fr

t

x sin r y cos r

c

(3.14)

To obtain a relation between the two waves the B.C. vn = 0, vni +vnr = 0is used.

vni =pi

0cni ey

= pi0c

cos i (3.15)

vnr =pr

0c

nr

ey

=pr

0ccos r (3.16)

So at y = 0

cos i0c

fi

t x sin i

c

+

cos r0c

fr

t x sin r

c

= 0 (3.17)

t, x at y = 0, i = r and fi = fr this equation will be satisfied. It shouldbe noted that the net tangential component vti + vtr = 0 at y = 0. At anypoint then pressure is the summation of pi and pr and at y = 0, p = 2pi.

3.2 Acoustic Impedance

Acoustic impedance is the measure of the amount by which the motioninduced by a pressure applied to a surface is impeded. Or in other words: ameasure of the lumpiness of the surface. Since frictional forces are, by andlarge, proportional to velocity, a natural choice for this measure is the ratiobetween pressure and velocity. It is given as

Z(x, ) =p(x, )

v(x, ) ns(x) (3.18)


22/47


at a point x on a surface S with unit normal vector ns pointing into the

surface. Being a complex number the real part of this impedance is calledresistance, the imaginary part is called reactance and the reciprocal 1/Z iscalled admittance. For plane waves a fluid may be said to have the impedance0c0 and hence this value is known as the fluid impedance. This inherentimpedance of a fluid is used to non-dimensionalize the Z leading to Z/0c0which is known as the specific acoustic impedance.

If this is compared to electrical circuits where p, the acoustic pressure islikened to voltage and v ns equivalent to electric current, then the termsresistance, impedance etc. can be easily understood. This is useful becausea lot of mathematical tools developed for electrical circuits can be exploitedfor these calculations too.

3.3 Reflection at an impedance discontinuity

When a wave is travelling from a medium, having a characteristic impedancedefined as 1c1 to another having 2c2, this jump in characteristic impedanceacross the discontinuity causes part of the wave to transmit into the nextmedium and part of it to be reflected back.

3.3.1 Normal incidence

If F1, F2 stand for the right moving waves and G1, G2 represent the right

moving ones, and x = y is the point of discontinuity, then at x = y, u2 = u1and p2 = p1, and t

is the time after the wave has reached y thus

F1(y c1t) + G1(y + c1t) = F2(y c2t) + G2(y + c2t) (3.19)1

1c1

F1(y c1t) + G1(y + c1t)

=

1

2c2

F2(y c2t) + G2(y + c2t)

(3.20)

The shape of the wave could be anything, depending on the initial conditions.Consider the case when the sound source is in region 1 and region 2

being an infinite duct does not have any reflected wave. Thus G2 = 0. G1becomes the reflected wave and F2 is the transmitted wave for the incident

wave F1.To get a relation between the reflected and incident wave write G1 in

terms of F1, from Eqs. 3.19 and 3.20 and putting G2 = 0

F1(y c1t)

1 2c21c1

+ G1(y + c1t

)

1 +

2c21c1

= 0

F1(y c1t)

2c2 1c12c2 + 1c1

= G1(y + c1t

)(3.21)

All these relations are in terms of the discontinuity location y. In orderto convert them to the general spatial coordinate x, recollect the fact that


23/47

3.3. REFLECTION AT AN IMPEDANCE DISCONTINUITY 23

the retarded time remains constant. x + c1t = y + c1t and hence, t =

(x y)/c1 + t. ThusG1(x + c1t) = F1(2y x c1t)

2c2 1c12c2 + 1c1

(3.22)

The ratio between the reflected wave function and the incident wave is calledthe reflection coefficient, R, hence

G1(x + c1t) = RF1(2y x c1t) (3.23)It should be noted that if the impedances between the two mediums areequal, i.e., 1c1 = 2c2 then there will be no reflection.

Similarly the transmittance coefficient, T is obtained by writing F2 in

terms of F1, from Eqs. 3.19 and 3.20 and putting G2 = 0

2F1(y c1t) = F2(y c2t)

1 +1c12c2

F2(y c2t) = 22c21c1 + 2c2

F1(y c1t)

F2(x c2t) = T F1

y

1 c1

c2

+

c1c2

(x c2t)

(3.24)

3.3.2 Oblique incidence

Similar to the previous section the pressures due to incident and reflected

waves are represented as

pi = fi

t

x sin i y cos i

c

pr = fr

t

x sin r y cos r

c

The pressure and velocity expressions at the point of incidence, y = 0 are

p = fi

t

x sin i

c

+ fr

t

x sin r

c

(3.25)

v =

cos i

1c1

fi t x sin i

c +

cos r

1c1

fr t x sin r

c (3.26)

The impedance Z is constant hence

Z =p

v= constant (3.27)

For similar waves the algebra becomes simpler, so the simplification, fr =Rfi is carried out. Thus,

t x sin ic1

= t x sin rc1

i = r (3.28)


24/47


Also

Z =fi

t x sinic + fr t x sin rc cos i1c1 fi

t

x sin i

c

+ cos r1c1 fr

t

x sin r

c

(3.29)Apply the simplification and obtain

Z =fi[1 + R]

fi1c1

[1 + R] =1 + R

cos 1c1

[R 1] (3.30)

Now consider solutions in the harmonic domain

pi = Re f ei(t nixc ) (3.31)pr = Re

gei(t

nr xc

)

(3.32)

Define, kx =c1

sin = k sin and ky =c1

sin = k cos . Also amplitude

reflection coefficients can be defined as g = Rf. This simplifies pressureexpression to

p = pi + pr = f eikx

eikyy + Reikyy

(3.33)

Normal velocity for the incident wave can be expressed as

vi n = pi1c1 cos (3.34)

Normal velocity for the reflected wave can be expressed as

vr n = pr1c1

cos (3.35)

The negative sign shows that normal velocity is away from the surface.Normal velocity of both the waves together can then be given by

vn =cos

1c1

f eikx eikyy + Reikyy (3.36)

The impedance in the simplified form using the reflection coefficient Rcan be given as

Z =1 + R

cos 1c1

(1 R) (3.37)

Thus the expression for R is

R =Z 1c1cos Z+ 1c1cos

(3.38)


25/47

3.4. WAVES IN AN IMPEDANCE TUBE 25

3.4 Waves in an Impedance tube

An impedance tube is a device used to study standing waves and determinequantities like admittance etc. It is provided with its own sound generationsources like speakers and it behaves like a tube whose one side is rigidlyclosed but the other side can be changed depending on the specific acousticimpedance of the fluid or conditions maintained at that end.

Just like the specific acoustic impedance Z/0c0, a non-dimensional ad-mittance Y = (u/p)0c0 can be defined. A standing wave is formed insidethe tube and the pressure can be given by

p = Aeikx + Beikx (3.39)

here A and B are complex and given by, A = aei1 and B = bei2 . Thusthe pressure expression becomes

p = aei1+ikx + bei2ikx (3.40)

p = a cos(kx + 1) + b cos(kx 2) + i sin(kx + 1) ib sin(kx 2) (3.41)The magnitude of p is

|p|2 = pp = a2 + b2 + 2ab cos(2kx + 1 2) (3.42)

In order to determine, 1 2, first find the places where there are max-ima and minima. These are |p|

2max = (a + b)

2

and |p|2min = (a b)

2

. Soat these extremums, a = (|p|max + |p|min) /2 and b = (|p|max |p|min) /2.Substituting these values in the minima point we get

cos(2kxmin + 2 1) = cos (3.43)

2kxmin + 2 1 = (3.44)The minima is chosen because the amplitude is sharp and not distributed

as in the case of maxima. Also, at the minima the particles are oscillatingin different directions on either side of the minima due to which there is thephase difference of c. IfR = ba e

i(21) is the reflection coefficient than the

pressure magnitude can be written as

|p|2 = a2

1 + |R|2 + 2|R| cos(2kx )

(3.45)

where = 2 1The acoustic velocity is similarly obtained from, the linearized momen-

tum relation, i.e. the Euler relation in Eq. 2.7 and substituting the expres-sions for u and p in the harmonic domain as

u = 1i0

p

x(3.46)


26/47


Substituting the value of p the expression finally becomes

u = k0

[a cos(kx + 1) + b cos(kx 2) + i sin(kx + 1) ib sin(kx 2)](3.47)

Noting that k/ = 1/c, the amplitude is obtained as

uu =1

c220

a2 + b2 2ab cos(kx )

(3.48)

and in terms of the reflection coefficient R the above equation takes on theform

|u|2 =

a

0c

2 1 + |R|2 2|R| cos(2kx )

(3.49)

The maximum and minimum values of u can be obtained as

|u|max = a0c

1 +

b

a

(3.50)

|u|min = a0c

1 b

a

(3.51)

R = b/a the reflection coefficient can then also be obtained by an algebraicmanipulation of the above relations to get

R =b

a=

|u|max |u|min|u|max + umin (3.52)

By substituting R = 1 it can be verified that the pressure and velocity

expressions turn out to be identical to those for a closed-closed tube. R = 0,will give expressions for a progressive wave with no impedance, while R = 1gives expressions for a tube in closed-open configuration, with the reflectionscausing a 180 phase change.

The non-dimensional admittance is given according by Y = (u/p)0c0

Y =

aeikx bei(kx)aeikx + bei(kx)

Y =

eikx Reikxeikx + Reikx

(3.53)

At x = 0, we can calculate Y = YRe + YIm thus

Y =

1 |R|ei1 + |R|ei

(3.54)

expanding and separating the real and imaginary parts gives,

YRe =1 + |R|2

1 + 2|R| cos + |R|2 (3.55)

YIm =2|R| sin

1 + 2|R| cos + |R|2 (3.56)

For any material that absorbs sound, YRe < 0.


27/47

Chapter 4

Radial and cylindrical wave

equations

4.1 Spherically symmetrical waves

In 3-D we have a general solution for spherically symmetric waves (i.e. de-pending only on radial distance r ). They are rather similar to the 1-Dsolution, because the combination rp(r, t) happens to satisfy the 1-D waveequation. Since the outward radiated wave energy spreads out over thesurface of a sphere, the inherent 1/r-decay is necessary from energy con-servation arguments. It should be noted, however, that unlike in the 1-Dcase, the corresponding radial velocity, vr is rather more complicated. Thevelocity should be determined from the pressure by time-integration of themomentum equation (Eq. 2.3), written in radial coordinates. The radialwave equation can be written as

2p

r2+

1

r

p

r=

1

c22p

t2(4.1)

If a linearization process similar to the one in 1-D case is carried out thenthe pressure can be expressed in terms of f(r ct)/r and g(r + ct)/r basedupon the 1-D solution similar to Eq. 2.14

p = 1r

f(r ct) + 1r

g(r + ct) (4.2)

The velocity is obtained as

vr =1

0c

1

rf(r ct) 1

r2F(r ct)

1

0c

1

rg(r + ct) 1

r2G(r + ct)

(4.3)

where F(z) =

f dz and G =

g dz.Normally a separation of variables method is used to get the exact ex-

pression for p, i.e. first consider p = R(r)T(t) and then substitute in the

27


28/47

28 CHAPTER 4. RADIAL AND CYLINDRICAL WAVE EQUATIONS

equation to separate the r and t dependant derivatives. The separation will

change Eq. 4.1 to1

R

2R

r2+

1

Rr

R

r=

1

c2T

2T

t2= k2 (say) (4.4)

The equation in terms of r alone has the form

2R

r2+

1

r

R

r+ k2R = 0 (4.5)

The solution isR(r) = AJ0(kr) + B

Y0(kr) (4.6)

where

J0(kr) = 2kr cos(kr 4 ) (4.7)Y0(kr) =

2

krsin(kr

4) (4.8)

The harmonic domain solution is given by

H10 = J0 + iY0 (4.9)

H20 = J0 iY0 (4.10)Here the H10 =

2

kr ei(kr/4) indicates backward propagation and H20 =

2kr e

i(kr/4) indicates forward propagation.

4.2 Waves in cylindrical ducts

The cylindrical wave equation without cylindrical symmetry can be derivedas

2p

r2+

1

r

p

r+

1

r22p

2+

2p

z2=

1

c22p

t2(4.11)

This is again to be solved by separation of variables by substituting p =R(r)Z(z)T(t)() in the above equation reducing it to

1

R

2R

r2+

1

rR

R

r+

1

r2

2

2

f(r,)

+

1

Z

2z

z2

f(z)

=

1

c2T

2T

t2

f(t)

(4.12)

The separated equations are equated to k2 terms for solving. It should benoted that kc =

r2

R

2R

r2+

1

r

R

r

+ k2r r

2 =1

2

2= m2 (4.13)

2T

t2+ 2T = 0 (4.14)

2Z

z2+ k2z Z = 0 (4.15)


29/47

4.3. RECTANGULAR DUCTS 29

The compatibility condition between these equations is

k2r + k2z = k

2 (4.16)

The solutions can be given as

R(r) = AJm(krr) + BYm(krr) (4.17)

T(t) = Aeit + Beit (4.18)

Z(z) = Ceikzz + Deikzz (4.19)

() = Eeim + F eim (4.20)

4.3 Rectangular Ducts

The acoustics in rectangular modes are characterized by the presence oftransverse modes. The wave equation is given by

2p

x2+

2p

y2+

2p

z2=

1

c22p

t2(4.21)

Similar to the previous analysis the individual differential equations aresolved for p

1

F

2F

x2

=k2x+

1

G

2G

y2

=k2y+

1

H

2H

z2

=k2z=

1

c2T

2T

t2

=k2(4.22)

p(x,y ,z ,t) = (c1eikzz + c2e

ikzz)(eikxx + c3eikxx)(eikyy + c4e

ikyy)eit

(4.23)The compatibility condition is

k2x + k2y + k

2z = k

2 (4.24)

The boundary condition for rigid walls would imply

p

x= 0 x = 0, b (4.25)

p

y = 0 y = 0, h (4.26)

p

z= 0 z = 0 (4.27)

resulting in the conditions, kx =m

b and ky =nh . The solution can be

expressed as

p(x,y ,z ,t) =

m=0

n=0

cos(mx

b) cos(

ny

h)

c1m,neikzm,nz + c2m,ne

ikzm,nz

eit

(4.28)


30/47

30 CHAPTER 4. RADIAL AND CYLINDRICAL WAVE EQUATIONS

kzm,n = k2 (m

b)2

(

n

h)2

1

2

(4.29)

If this kzm,n is imaginary the mode may decay along z. It propagates unat-tenuated only if this is real. Thus, any particular mode will propagateunattenuated when

k2 ( mb

)2 ( nh

)2 > 0

( 2

)2 > (m

b)2 + (

n

h)2

< 2

( mb )

2 + ( nh )2

1

2

(4.30)

This gives rise to the transverse modes.


31/47

Chapter 5

Combustion-acoustic

Interaction

The acoustic effects of combustion are in general unfavorable for operationof the apparatus in which they are observed. The sound from combustionprocess is of two types:

Combustion noise or roar

Combustion driven oscillations

The roar or sound from a flame, in many cases is like a fingerprint ofthe turbulence in the flame, to the effect that turbulence in a flame can bemeasured from the noise measurements. The combustion roar in combina-tion with resonance is a broad spectrum noise which causes environmentalsound pollution.

In industrial parlance, combustion instability deals with the control andcombustion dynamics deals with management of combustion processes.

Combustion driven oscillations

When combustion drives the oscillations, it is likened to a feedback cyclethat converts chemical energy to sound. In this case the spectra is verycrisp, with pressure oscillations limited to almost just one frequency. Theamplitude itself is very high - almost 50% of the mean pressure. This candecay or grow depending on the individual phase differences and frequenciesbetween the combustion driven oscillations and the heat addition.

When a flame adds more energy to a resonator than the amount whichis dissipated, acoustic energy grows in time, thereby causing instability, onthe other hand limit cycle oscillations are caused when the acoustic drivingis balanced by energy dissipation.

31


32/47

32 CHAPTER 5. COMBUSTION-ACOUSTIC INTERACTION

5.1 Coupling between acoustics and flames

The causes for the coupling between the oscillations and flames are as fol-lows:

Fluctuating mass flow Acoustic velocity fluctuations induce density andpressure fluctuations, thereby changing the rate of mass flow (fuel andair) through the burner. The fuel fluctuations will then cause heatrelease fluctuations. Thus if flow instability occurs the flame areachanges.

Vortex shedding If the dominant flow frequency due to vortex shedding

coincides with the natural frequency of the burner a strong couplingwill be observed.

Fluctuating equivalence ratio A fluctuating fuel/air ratio leads to anoscillating heat addition. This could also lead to droplet formationand breakup, again leading to oscillatory heat release.

Such a coupling if not controlled, can lead to the following problems:

1. Component melting due to excessive heat transfer

2. Burning rate modification due to changes in combustion processes

3. Excessive vibrations leading to mechanical failure

4. Interference with control systems and electronics

5. Excessive noise, flame excitation and costly system/operation failure

5.1.1 Effect of heat addition on oscillations

Any change in the oscillating state of a system is due to the heat addition toit by the flame. The oscillations of a system, will grow or decay in amplitude

depending on which part of the cycle the heat is added.From Fig. it is observed that when heat is added at a maxima (i.e. in

phase with oscillations), driving occurs resulting in the growth in amplitudeof oscillations, while the frequency remains the same. Similarly if the heataddition is carried out at a minima, then there will be damping, i.e. a decayin amplitude of oscillations.

Now if the heat addition were to happen at some intermediate point, thenthere is a change in the frequency and time period. Time period increaseswhen heat addition is carried out at a point on an up-going curve of thewave and decreases for a down-going part.


33/47

5.1. COUPLING BETWEEN ACOUSTICS AND FLAMES 33

Table 5.1: Heat addition effects on oscillationPhase of heat supply relative to pressure Effect on amplitude Effect on frequency

In phase Increases NoneOut of phase Decreases None

Quarter period before None IncreasesQuarter period after None Decreases

5.1.2 Rayleigh Criterion

The qualitative effect of heat addition on oscillations has been described inTable 5.1. An analytical consideration to describe these effects is presentedbelow.

Consider an acoustically closed-closed duct (so that an acoustic wavewould be reflected). The assumptions for this analysis are as follows:

1. Uniform steady state gas properties along a tube as Q = 0. In a rocketcombustor, p0 = const is not a bad assumption, but T0 = const is.

2. Gas is inviscid and non-heat conducting

3. Oscillations are 1-D (good if L >> D)

4. Cp, Cv & are constants

5. Fluctuations of u ,p,T, & Q are small. Q is the oscillatory heataddition per unit mass.

The continuity, monentum and energy equations are given respectivelyas:

t+

x(0u) = 0 (5.1)

0u

t+

p

x= 0 (5.2)

p

t+ p0

u

x= ( 1)0Q (5.3)

Multiply Eq. 5.2 with u and Eq. 5.3 with p0 and add the two to obtain

t

p2

20c2+

u2

2

+

x(pu) =

1

c2

pQ (5.4)

On spatially integrating the above equation, using the notation to meana temporally averaged quantity:

t

V p

2

20c2+

u2

2 dV +

S

(pu) dS = 1

c2

VpQ dV (5.5)


34/47


The Rayleigh criterion dictates that when the term on the R.H.S is

positive the energy will rise leading to an acoustic driving and if it is negativethen there will be a damping.

This can be simplified for p = p cos t and Q = Q cos(t + ). Thequantity, pQ computes to

1

T

T0

cos t cos(t + )dt =cos

2(5.6)

Thus if = 180, the combustion process will absorb all of the energy asthe RHS of Eq. 5.5 becomes negative. Correspondingly, for = 0, there ismaximum acoustic driving and at = 90 there is no driving or damping.

Flame transfer Function

The u, p and Q can be written in the harmonic domain as, p = peit, u =ueit and Q = Qeit.

It is desired to obtain a flame transfer function across a flame to relatethe pressure and velocity, just behind the flame front, (p1, u1)with that justin front of the flame front (p2, u2).

From the momentum equation, by substituting above quantities in Eq.5.2. (The derivative wrt time, leads to cancellation of the eit term fromboth sides of the equation)

limV0

i0

Vu dV + ( p2 p1)A = 0 (5.7)

p2 p1 = 0 (5.8)

Thus, acoustic pressure does not jump across a flame if it is thin.

Now considering the energy equation, from Eq. 5.3:

limV0

V

i p + p0

u

x

dV =

V

( 1)0Q dV (5.9)

p0 [u2 u1] A =

V( 1)0Q dV (5.10)

u2 u1 =

1

Ac2

VQ dV (5.11)

A relation can be derived across the flame and a matrix known as theflame transfer function can be made from Eqs. 5.8 and 5.11

p2u2

=

T11 T12T21 T22

p1u1

(5.12)

For the case discussed above there is no pressure jump but there is avelocity jump across the flame.


35/47

5.2. SOLUTION OF THE WAVE EQUATION WITH HEAT RELEASE35

5.2 Solution of the wave equation with heat re-

lease

The wave equation corresponding to the oscillatory heat release is to beobtained from Eqs. 5.1, 5.2 and 5.3. On eliminating u from Eq. 5.2 and 5.3by p00

x (Eq. 5.2) t (Eq. 5.3), the equation obtained is

2p

t2 c2

2p

x2= 0( 1) Q

t(5.13)

Substituting, c2 = p00 and using the complex quantities, p = peit, Q =

Qeit etc. Eq. 5.13 becomes

2p

x2+ k2p = i0 ( 1)kQ

c= f(x) (5.14)

Eq. 5.14 can be described as an Inhomogeneous Helmholtz Equation owingto the f(x) due to the heat release.

5.2.1 Greens Function Technique

A solution for such differential equations (Eq. 5.14) is of the form S = CF +P I. Obtaining the particular integral for the solution of an inhomogeneousdifferential equation is difficult. For Eq. 5.14, the complementary function

can be obtained as the solutions of the homogenous equation

2pxx2

+ 2xpx = 0 (5.15)

The solution for this equation should satisfy the solid wall B.C. namely,pnx = 0 at x = 0, L. Such a solution is given by pn = B cos nx, after

taking into account the boundary conditions. Now, n =nL , n = 0, 1, 2...

are eigenvalues for natural acoustic modes of the combustor.

The solution has to be constructed from these natural acoustic modes,by obtaining the Greens function. The Greens function f(x, ) satisfiesthe same equations and boundary conditions as pn. By enforcing a jumpand continuity condition at the flame location, f(x, ) can be obtained fromwhich the solution for p follows.

The momentum equation in the harmonic domain is given by:

i0u +dp

dx= 0 (5.16)

The energy equation is given as:

i p + p0du

dx= ( 1)0Q (5.17)


36/47


Assume Q = (), a delta function. In effect an infinitesimal heat source at

a point is considered. An integration is carried out from a point just behindthe heat source to a point just in front.

limx0

x+x

dp

dxdx = 0, (5.18)

p+ = p

limx0

x+x

i p dx + p0

du

dxdx

= lim

x0

x+x

( 1)0() dx (5.19)

p0(u+ u) = ( 1)0 (5.20)

(u+ u) = ( 1)RT0

(5.21)

From the Euler relation:

dp+

dx dp

dx=

( 1)p0iR2T20

(5.22)

Keeping these results at hand, first the Helmholtz equation with heat isderived. Eliminate u terms from the momentum and energy equations toobtain

d2p

dx2+

1

T0

dT0dx

dxdp

dx+

2

RT0p =

( 1)i

p0R2T2

0

Q (5.23)

This is the Helmholtz equation with heat release.Now considering the heat quantity as a Dirac delta function () and f

as the Greens function for that corresponding heat release, a technique toobtain the function is given below. The corresponding form of Eq. 5.23 isgiven as

d2f

dx2+

1

T0

dT0dx

dxdf

dx+

2

RT0f = () (5.24)

A fundamental solution for f is sought. It is assumed that T0 is constant andphysically, is the point of the heat source location. The equation reducesto

d2

fdx2

+ 2

RT0f = () (5.25)

The operator used is L = d2dx2 + 2

RT0, and the analysis is carried out for

x = . f satisfies the same B.C. as p so the complementary functions for L,before and after the heat release point are obtained as

f or < x, f = A sin kx + B cos kx (5.26)

f or > x, f + = Csin kx+ + D cos kx+ (5.27)


37/47

5.2. SOLUTION OF THE WAVE EQUATION WITH HEAT RELEASE37

A,B,C and D are arbitrary and the relation amongst them is determined

from the jump and continuity conditions. Owing to the continuity of thepressure across the flame front according to Eq. 5.19:

(C A)sin k + (D B)cos k = 0 (5.28)Now the jump in velocity across the flame front/heat source is written. Tomaintain the usage off or p as the case may be, the velocity jump is relatedto the jump in dpdx or

dfdx , as given in Eq. 5.21.

df

dx= (A cos kx B sin kx)k (5.29)

df+

dx= (A cos kx+

B sin kx+)k (5.30)

Depending on the type of oscillating heat release function used, the RHSwill be like that obtained in Eq. 5.22, for generality here it is taken as afunction 1 . The jump across x = is given by

(D B)sin k (C A)cos k = 1

(5.31)

Let (D B) = v and (C A) = u. The system of equations from Eqs. 5.28and 5.31 can be written as

u sin k + v cos k = 0 (5.32)

v sin k u cos k = 1

(5.33)

This gives a solution matrix assin k cos k

cos k sin k

uv

=

01

(5.34)

The left hand coefficient matrix is inverted to give the solution matrix asuv

=

sin k cos kcos k sin k

01

(5.35)

u = cos k

, v = sin k

(5.36)

C = A + u, D = B + v (5.37)Thus the relation between the coefficients is determined. The values of Aand B will be determined from the boundary conditions.

From Eq. 5.23 and forcing the appropriate assumptions the equation tobe solved becomes

Lp = i

( 1)R2T20

p0Q (5.38)


38/47


As f is the Greens function it should satisfy the relation Lf = (). Thisproperty is used to facilitate integration by parts in the following integration.

L0

Lpf dx =L0

(d2p

dx+ k2p) dxL

0Lpf dx =

f

dp

dx

L0

L0

dp

dx

df

dxdx

+

L0

k2pf dx

=

f

dp

dxL

0

pdf

dxL

0+

L

0p

d2f

dx2dx

+L0

k2pf dx

=

f

dp

dx

L0

pdf

dx

L0

+ p()

As dpdx = 0 and p = 0 at boundary conditions, the first and second terms

in the above equation will vanish. The relations,L0 p() dx = p() and

Lf = () are used to obtain,

p() = L

0 L(p)f dx (5.39)

Substitute Eq. 5.38 in the RHS integral to get the solution as

p(x) =

L0

i

( 1)p0R2T20

Q()f(, x) d (5.40)

where

f(, x) = A sin k + B cos k, 0 < x (5.41)

f(, x) = (A +cos kx

sin k + (B sin kx

)cos k , x L (5.42)

A and B are obtained from the boundary conditions to complete the solu-tion. Thus the Greens function technique can be used to obtain the pressuregiven the heat release function. The real advantage of this technique lies inthe fact that whatever be the function of Q, the solution p can always becomputed, if not analytically then by numerical integration of the integrand.

5.2.2 An Example

Consider a 1-D problem of a thin, flame front at a distance a in a duct oflength L in an open-closed configuration, represented by two zones, 1 (before


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5.3. SENSITIVE TIME LAG HYPOTHESIS 39

the flame front) and 2 (after the flame front). The axial acoustic field can

be represented as

p1(x, t) = Aeik1(xa)+it + Beik1(xa)+it (5.43)

p2(x, t) = Ceik2(xa)+it + Deik2(xa)+it (5.44)

Here B and C denote the right running waves, while A and D denote theleft running waves. Q(t) = Qeit and based on this relation between theconstants can be determined as follows. The boundary conditions are x =l, p = 0 and x = 0, u = 0. Considering the right running wave given by,u = p/0c the velocities in the two zones are given as

u1 = 11c1

Aeik1(xa)+it + 11c1

Beik1(xa)+it (5.45)

u2 =1

2c2Ceik2(xa)+it +

1

2c2Deik2(xa)+it (5.46)

The boundary conditions lead to

Aeik1a Beik1a = 0 (5.47)Ceik2(la) + Deik2(la) = 0 (5.48)

The jump condition as given in Eq. 5.19 for this case is

u+ u = ( 1)Q 1

0c2(5.49)

0c2 = 1c

22 = 2c

22 (5.50)

Thus, the condition translates to

[A B] 1c12c2

[C D] = ( 1)Qc1

(5.51)

Now A,B,C,D can be obtained from Eq. 5.47 and the solution can be

obtained as shown in the previous section.

5.3 Sensitive Time Lag Hypothesis

The heat release Q is actually created by velocity perturbations existingat the flame holder (i.e. u(x = a, t)) but there is a time delay . Let = 1c1/2c2. The coupling is given by

Q( 1)0c2

= nu1[a, t ] (5.52)


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n is the interaction constant. It is typically 0-10, 10 being intense coupling.

At x = a this becomes

( 1)Qeit0c2

=nei

1c1[A B]eit (5.53)

( 1)Qc1

= n[A B]ei (5.54)

Eq.5.51 then is written as

[A B] [C D] = n[A B]ei (5.55)

[A B] 1 + nei [C D] = 0 (5.56)A,B,C,D can now be solved for as mentioned before by the following matrixequation.

1 1 1 10 0 eik2b eik2b

eik1a eik1a 0 01 + nei 1 + nei

ABCD

=

0000

(5.57)

Here l a = b The determinant of the coefficient matrix should be zero fornon-trivial solutions. This gives rise to the characteristic equation

cos(k1a) cos(k2b) sin(k1a)sin(k2b)

1 + nei

= 0 (5.58)

If has complex roots then there is exponential growth or decay of theoscillations by means of the instabilities. In order to better understand thetrends of growth or decay os oscillations some simplifying assumptions tothe above can be made, i.e. a = b (the flame is in the centre of the duct),c1 = c2 = c (the speed of sound is the same before and after the flame front)and so = 1 and finally, k1 = k2 = k. This reduces Eq. 5.58 to

cos2(ka) sin2(ka) 1 + nei

= 0 (5.59)cos(2ka)

1 +

nei

2

=

nei

2(5.60)

If n


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5.4. ACTIVE CONTROL 41

The wave number k can be written as k = k0 + k, the first and second

being the mean and perturbation components respectively. Substituting thisvalue in Eq. 5.60

cos(2k0a + 2ka) = cos(2k0a) cos(2k

a) sin(2k0a) sin(2ka) (5.62) = sin(2k0a) sin(2ka) (5.63)

The first term on the RHS is always zero because the k0 will always cor-respond to some fundamental frequency, and so 2k0a = n/2. This valuewhen substituted in Eq. 5.61 gives

sin(2ka) = 0.5nei (5.64)

2ka can be assumed to be small enough to assume, sin(2ka) = 2ka, giving

kRe + kIm =

n

4a[cos() i sin()] (5.65)

Comparing real and imaginary parts of the equation

kRe = n

4a[cos()] (5.66)

kIm =n

4a[sin()] (5.67)

There will be a growth in oscillations when kIm < 0. This implies that thesystem is linearly unstable when Im < 0 and linearly stable when Im >

0. The condition for an exponential increase in oscillations is therefore,sin() < 0. The condition for is therefore obtained as,

< 0 < 2

+ 2j < 0 < 2 + 2j

0+

2j

0

acousticssujithnotes

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