Acoustic Instabilities In Aerospace Propulsion

Course conducted by

Dr. S.R.Chakravarthy And Dr.R.I.Sujith

Department Of Aerospace Engineering

IIT MADRAS

This Typing Work Is Done By

A. Pandu Ranga Reddy

M.S Scholar

1

SYLLABUS

Motivation

Derivation of wave equation

Traveling wave solutions

Standing wave solutions

Impedance, Impedance tube techniques

Effect of in-homogenous media (High temperature, mean flow)

Oscillating heat release, response function

Rayleigh criteria

Combustion instability, combustion noise

Pulse combustion, different types of pulse combustors

Flame transfer functions

Vortex sound measurement

Transducers data acquisition processing

Two microphone technique

Instabilities in engines

Solid rocket, liquid rocket and gas turbines.

References

1). L. kinsler & Fray A.R., A. B. Coopers and L.V. sanders

“Fundamentals of acoustics”, John Wiley & sons, 1982.

2). P. M. Morse & Ingard U. K. “Theoretical acoustics”

Princeton university press, 1968

3). “Non-steady burning & combustion stability of solid propellants”, edited by

Deluca and E. W. Price, 1992.

AIAA progress in Aeronautics and Astronautics, Vol. 193.

4). A.A. Putnam, 1971, “Combustion driven oscillations in Industry”, Elsevier.

2

Rotameter – used to measure flow rates

M < 0.3 (incompressible)

Since +/- 5% deviation in density is acceptable

0.05 kg/m3

= 4500 Pa

=11 m/s !!!

SLIDE SHOW:

Importance of appreciating why acoustics is important with respect to gas turbine?

i.e. on vortex shedding noise, e. g. whistle, flute,

Reed noise e.g. mouth organ

Synthetic sounds e.g. Synthesizer, tape recorder

Combustion instabilities keep occurring when oscillations keep growing. The

feedback is provided by the heat release.

Common example of \feed back – place a mic in front of speaker. This results in a

shrill system. Flame is held stable in a re-circulation zone. Any bluff body sheds

vortices. Blow over a water bottle sound is produced due to shear layer oscillations

Motivation

Photograph of “blown off rocket nozzle due to combustion instabilities” during static

testing.

Solid propellant combustion instability

-Means prevalence of pressure oscillations in solid rocket motor.

-Strictly continued increase in the amplitude of oscillations

-Caused by positive feed back from propellant combustion

-Practically limit cycle oscillations due to presence of damping

-Any driving mechanism is bad enough

Combustor vs combustion instability

- Mostly we talk of combustor instability

Pressure oscillations related to chamber geometry

Acoustic/non acoustic

- Combustion instability is only a specific part of the problem

The combustion response of propellant to oscillations in the chamber

3

Natural sound

Thrust oscillations

- 0.1 bar oscillations ( pressure amplitude) over 50 bar mean pressure results 5 to 10

times thrust oscillations.

1 dB = 20 log (p / pref), pref = 2.0*10-06

150 dB 2000 Pa (Standard scale)

-Can cause pre-ignition of fuel rich exhaust due to enhanced mixing with ambient air

Interfere with ground-based navigation

-15 Hz oscillations in space shuttle (SRBs – Solid rocket boosters) makes astronauts

feel jitters- due to thrust oscillations

Waterfall plot for SRB

Pressure vs frequency at regular time intervals

DC shift-mean pressure excursions due to oscillations

-Affected through propellant mass flux oscillations

Structural oscillations

-Limit cycle oscillations may resonate with structural members in the vehicle causing

catastrophic destruction

-Other systems could include guidance and control electronics

Static test success, but flight test failure

Oscillations unforeseen !

-Usually oscillations are unforeseen

-Too complicated to predict priori

-Generally encountered late in the motor development program

-Too few options on design changes

-“Tricks of the trade” employed rather than scientific approach

-Solutions for different mechanisms should be implicating

- Need expertise based on holistic approach

Incidents of instabilities

-One estimate is that over 70% of SRMs exhibit oscillatory behavior (R. S. Brown)

-More than 50% of SRM research funding in the U.S motivated by instability

problems (E. W. Price)

-Most motors naturally whistle - F. Vuillot

Case study: subroc motor

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-While instability witnessed for the first time in the 32nd test firing

-Aluminum size was suggested to be reduced from 30 microns to 15 microns

Motor developers changed half the Aluminum contents to 5 micron for fear of

processing problem and burn rate change

-Instability persists

-Complete change in aluminum effected

Case study: minuteman missile

- System induced into service with oscillations

- Random flight test failures

Static tests success

- Production log reveals supply of aluminum from different plants

Same supplier, same specifications

Case study: other motors

- Space shuttle slag accumulation and ejection (1994)

- Led to pressure and thrust spikes

-Due to change in Ammonium Per chlorate supplier (better purity)

-Retro rockets of mars path finder (1997)

Originally 2% Al is propellant in order not to contaminate Martian atmosphere

Finally 18% Al in order to suppress oscillations

-Ariane V MPS (Motor Propergol solid)

Exhibits first three modes

-Shift in dominant frequency among modes

-Enormous research efforts and funding in Europe to tackle the problem

-PSLV first stage motor

Shift in dominant frequency during burn

Instabilities in liquid rockets

In this combustion chamber each and every droplets have a flame. And acoustics

could move the droplet back and forth. /this is like drag acting on the droplets. There

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The pulsations could propagate to the pipelines

is also a pressure oscillation, which means variable burning rates. And in turn we

could have variable heat release and since local heat difference try to convect, we

could have an expansion wave.

Combustion instability

-Primarily caused by heat release fluctuations due to chemical reactions

System instability

- Oscillations in injectors

- Oscillations in manifold pipe lines

- Oscillations in propellant tanks

Gas turbines

In general in primary combustor, whenever we have a stagnant region it gives

rise to recirculation zones. In practice, recirculation zones are shear layers.

We have longitudinal, transverse and radial modes of vibration.

Radial oscillations Tangential modes

We have flow going some length along the radial path and then coming back and

forth and so on. The worst part about Tangential modes is that there is no fixed

diameter along which the nodes exist. They keep précising

In a duct like this

For larger lengths lower frequencies

If we look at the water fall plot

A A

Longitudinal mode Transverse mode

After burner/Ram jet combustor

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We know where nodes exist, so we can accordingly adjust pressure taps

Fuel injectors

The liquid comes out and then vaporizes. We see turbulent flames but these are time-

averaged picture. Heat release happening n tune with vortex shedding produces very

large sound

Particular case: LNGT (low NOX gas turbines)

We have high temperatures in a combustor that gives rise to NOX pollution, so a

solution came up i.e. we cool the combustor. But that reduces efficiency so there is a

trade off. Nowadays we suppress NOx levels without cooling, i.e. by operate at the

lean limit

Oscillations assume significance

Fuel inflow oscillations

Mixture fraction oscillations

Basic mechanisms o interest in

Combustion sound

Sound generation

Vortex sound

Coupling between driver and driven

A A

f f

[In a flute we keep altering the length so we keep changing the vortex shedding

frequency].

Feed back from the driven to the driver in the coupled system

Acoustics generation, propagation and effects of sound

We will generally be dealing with the first two

Sound should not be a misnomer for the frequencies in the range of human perception

it can fall into any category

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Lock on

Sound-a disturbance of pressure (normal stress) which propagates at finite

speed in a compressible medium (solid., liquid or gas)

Two basic types of waves

(a) Longitudinal (compression wave) here particle motion is along the

direction of wave propagation

-By particle we do not mean individual molecules but infinitesimal

volume element of fluid around a part

- Mechanical analogy

(b) Transverse waves (shear) – particle motion perpendicular to the direction

of the wave propagation

- Mechanical analogy (a stretched string)

Fluids do not support shear deformation. They support compression.

But the above argument is strictly true for inviscid fluid.

Transverse waves encounter viscous damping mostly; we deal with longitudinal wave

propagation.

Amplitude: e.g. Air as a medium Po

P-minimum detectable sound by

Human ear -10-5 Pa t

Total pressure Pt = Pt (t, r) = Po + P (t, r) ------------- acoustic pressure

Typically | p | << Po

Acoustic pressure in dB = 20 log (P/Pref)

Logarithmic scale depends upon Pref, in air Pref = 20 Pa-the lowest that a human ear

can hear.(in water Pref = 1 Pa )

typical speech p(dB) = 20 log 10(10-2 / 20*10-6)

DERIVATION OF HE WAVE EQUATION:

In 1- dimension, in fluids , methodology space time – problem, variables are

P- pressure,V- particle velocity, - particle displacement, - density.

Need to have three equations, three conservation equations and equation state.

Continuity equation

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Conservation of linear momentum equation:

In Eulerian form

(External forces) = rate of change of momentum inside the control volume +

change in momentum flux across the control surface.

(External forces) = pressure forces +viscous forces + body forces

Assuming no body forces and in viscid.

Pressure forces are predominant in acoustics propagation in fluids.

Net pressure force in the x-direction:

(Ptx - Ptx+x)

Pt = Po + P (x, t) assume Po = constant (uniform fluid assumption)

(Ptx - Ptx+x) = - .X (neglecting higher order terms in tailor expansion of

Ptx+x about x)

In mass conservation statement,

The other form of expressing an equation would be what is called in terms of

primitive variables, because primitive variables can directly be measured.

Pressure density relationship

Thermodynamic variables are t, Pt, Tt, st etc.

Postulate of classical thermodynamics: A thermodynamic state Is defined by two

variables in an instantaneous single phase uniform fluid.

For example Pt = Pt (t, st)

For sound propagation, it is assumed to be an adiabatic phenomenon, no heat flow in

and out of the fluid particle, hence st = so + s(x, t) or Ds/Dt = 0.

Pt = Pt (t, so) implies unique relationship between Pt,t, so.

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Momentum equation for 1-D:

DERIVATION OF THE ONE DIMENSIONAL WAVE EQUATON:

1). Continuity Equation:

2). Momentum equation:

3).equation of state: Pt = Pt (t, so)

Non-linear coupled equations, have very limited exact solutions,

Line arise, because p << po etc.

Linearization process: the second term on the L.H.S of the first two equations is the

non-linear terms.Perturbations in p,, T etc are fairly small when compared to the

mean values. By postubating the above, we are confining ourselves to the realm of

LINEAR ACOUSTICS. We assume a homogeneous, fluid; po,o, To etc are

independent of the medium.

Quiescent medium implies po,o, To are independent of the time and uo = 0.

We have pt = po+ p & p / po << 1.

Linearised equations:

1). Continuity Equation:

-------------(1)

2). Momentum equation:

-------------------(2)

3).equation of state:

P = c2 --------------------------(3)

Perform: to get

From equation (3) we have

Therefore from the above equations we get

-------- This is the one-dimensional wave equation.

The pressure is the dependent variable here. when we examine it mathematically, we

look at it as

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- The wave equation is linear. Advantage of linearity is that we can superpose

Solutions.

- The wave equation is also homogeneous.(no dependent term on R.H.S)

- Also no forcing terms, nothing drives or dampens the pressure disturbance.

- The equation governs phenomenon in a domain with no sources of sound

- Hence we are concerned with the phenomenon of a non attenuated wave

- Later, see terms added terms to R.H.S.

A similar wave equation can be derived for propagation of electromagnetic

waves in free space, but different GE’s similarly, Schrodinger wave equation.

Historically, 1-D wave equation derived by D’Alembert’s in 1747 for oscillation

of strings.euler derived PDE equation specifically for sound propagation in fluids

We consider elasticity as of the medium of propagation instead of kinetic

approach as indicated by the spatial derivative.

The wave equation is of second order equation, therefore infinite solutions.

Specific solutions can be obtained, given initial &boundary conditions

1-D Plane traveling wave:

We seek solutions for p(x, t), (x, t),etc.,

Where x is the direction of the wave propagation.

Acoustic properties are constant in a plane at any x.

“Plane “ waves are encountered in ducts at low frequencies and away from sources.

Solution procedures we will adopt.

Easy to remember way.

We note cross-Derivatives are equal for smoothly varying quantities.

The wave equation then can be written as

The solution to (1) is p=f(x, t), where f is an arbitrary function of (x-ct), mark of

ingeniousness of d’Alemberts

Check if p=f(x-ct) satisfies (1).

Let = x- ct; f=f () only.

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=

Substituting we get

Similarly solution to (2) is of the form

P= g (x +ct)

Again use transformation = (x + ct); to show that the above solution works.

The general solution of plane wave propagation in the x-direction is

P(x ,t) = f(x - ct)+ g(x + ct)

Where f & g are two arbitrary functions.

We ended up with arbitrary functions as we are trying to solve for 2nd order PDE’s

with out specifying the boundary conditions. Rearranging,

P (x, t) = f1(t-x/c) + g1(t + x/c)

Physically f1(t-x/c) could imply a transient signal of some shape.

Fig:

f1

t=t1

(t - x/c) – retarded time

At any time point A, is preserved.

I.e., (t2 – x2/c) = A = (t1-x1/c)

(x2 - x1)/(t2-t1) = c. “speed of sound”.

We will have ever- expanding or ever- collapsing medium, and f1(t-x/c), indicate

wave propagation in positive x- direction.i.e., forward propagating wave.

(t-x/c) – retarded time for forward plane wave.

For p (x, t) = g (t+ x/c)

(t + x/c)= = retarded time – preserved for any part of the signal.

(t2 + x2/c) = A = (t1+ x1/c)

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AA

XX

x2 = x1 – (t2- t1)c we know that c> 0.

So for t2 >t1, we have x2 < x1 – g1(t + x/c) is a backward propagating wave.

Features of 1-D plane wave.

Amplitude remains constant (no atténuation)

The wave shape does not change during propagation.

Relation between p & u for a plane wave:

Linearized Euler:

U = particle velocity along the wave direction.

We know p(x, t) = f(x- c t) + g(x+ ct). in the positive x-Direction.,

change variables = x- ct; =( x, t), dt=-1/c.d;

u+=p+/0c The constant of integration of vanishes due to initially quiesecent medium.

Plane wave has velocity in the positive x- direction, for positive p+ and magnitude

proportional to pressure.

In the negative x-direction p- = g(x - ct)

Change variables : = x+ c t. dt = 1/c.d and

Therefore u- = -p-/0c

u- = p-/0c( -i )

Particle velocity is in the direction of wave propagation. For p- > 0.

Important concept:

Have applications in sound transmission and reflection at boundaries. Different media

has different impedances; this is how the acoustic wave is able to distinguish

boundaries.

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u- = c/gdgc/1)dnc/1(

dxdg/1 000

The amount of sound transmitted or reflected depends on yhe extent of impedance

mismatch at the boundaries.

Plane wave p/u=oc characteristic impedance of the medium.

Units of impedance: kg/m2s = 1 rayleigh.

Typical values:

- air (at STP) = 25o c 105Pa.

- o= 1.2 kg/m3 c = 1500 m/s. o c =400 Rayl.

Similarly for water(at STP)

o= 1000 kg/m3 , c = 1500 m/s. , . o c = 1.5e6 rayl.

The enormous mismatch in o c of air and water causes most of the incident plane

wave from air at the water surface to be reflected, and only a small quantity to be

transmitted.

For p= 10e-2. Pa.

u= p/o c = 2.5e-5 m/s for air,7e-9 m/s for water.

The impact in the water is much is very less.

Waves of constant frequency:- harmonic waves.

Physically this requires a sound source that is constantly vibrating at a single

frequency. There is a simple and ideal but fundamental case of wave propagation.

Fundamental transient signals can be expressed as a superposition of harmonic waves.

Plane propagating transverse wave:

Prompts us to use sines and cosines for a forward propagating wave, we can write

P(x, t) = A cos [2f(t-x/c) - o]

This is of the form f (t-x/c)

This a is plane because f1= f1 (t-x/c) & not (x, y& t)

Progressive: f1= f1 (t-x/c) & (t + x/c)

Harmonic: cosine oscillating at single frequency.

o = initial phase

Harmonic waves can be visualized as a function of time and space at a given location

and vice-versa.

X = 0

P (0, t) T

A ------- At x = 0 & t = 0; p(x, t) = Acos(-o)

Acos(-o)

14A --------------- -

This picture defines A, o, etc., o is

Arbitrarily depending

upon the

Choice of coordinate axes

&

the origin.

T time period for one cycle.

The idea of repeats itself after a time period T.

For sines and cosines, this happens when the argument changes by 2

[2f (t + T - x/c) - o] – [2f (t - x/c) - o] = 2

2f T = 2 f T = 1. or f = 1/T ; at t = 0 , plot p (x, 0)

P (x, 0)

A -------

A cos(-o)

x

-A ----------------

l=Wavelength spatial extent of one cycle. Obviously T or f and l are related.

T is the interval between one compression peak and next.

During this time, the peak was traveled a distance l at a speed c.

l/T = c. c = f l

Wave number k= 2/l = 2f/c = 2f is called angular frequency.

p (x, t) = Acos[2f (t - x/c) - o]

= Acos [ (t- x/c) - o]

= Acos [( t- kx) - o]

= Acos [(k x- t) + o]

k is interpolated as spatial analogue to .

Complex notation :

Euler relation : ei = 1 + i + ……….., Where i = -1 by definition.

ei = cos + isin

p (x, t) = Acos[[(k x- t) + o]]

= Re {Aei[(k x- t) + o]}

actual (physical) pressure

complex pressure : p(x, t) = Aei[(k x- t) + o] }

Advantage: factorization of time and space quantities.

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l

i.e,. p (x, t) = A (eio eik x) e-i t

=

Where = A (eio eik x) complex amplitude.

To show that complex pressure staisfies the wave equation :

For plane harmonic wave in – ve x direction is p(x, t) = Acos[(k x- t) - o].

HELMOLTZ EQUATION :

Wave equation :

Substitute p (x, t) with .

this is the HELMOLTZ EQUATION.

Solve this for spatial dependence of complex amplitude with B.C’s in .

Acoustic energy corollary

Recall, Euler equation:

u times (1) gives ,

but

Linearized continuity equation:

but p = c2

=

…………………Use this in EQN in 3, to get

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Acoustic Energy corollary

The above equation is for 1-D case.

In 3 –D vector form, this can be written as

Where w =

And pointing vector or acoustic intensity vector.

It is an instantaneous local quantity. In integral form equation 5 can be written as

Time rate of change of acoustic energy inside the control volume

= Intensity flux (in – out) across the control surface

Law of conservation of energy for non-distributive medium in 1-D

or I = I = pu

Remarks:

1. These energy quantities (W,I) are of second order, this implies very small

numerically.

2. Acoustic intensity is a vector physically implies acoustic power flow/unit area.

Acoustic intensity

In 1-D = I i, I = pu Instantaneous local quantity.

= (x, t),

Specific case: Plane wave

- Unit vector along the direction of wave propagation.

Then Intensity can be deduced with measurement of pressure.

Harmonic plane wave: P (x, t) = A cos (k x –wt)

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Call w =

.

So we are getting rid of time dependence and retaining only the spatial variation. For

periodic waves o= Best example = harmonic waves for non – periodic signals. is

“long enough”. Such that the average ( <....>) does not depend on o.

(<......>) is simply called acoustic intensity.(This is what sensed by the human ear).

= p2rms = root mean square pressure.

For a harmonic plane wave, p(x, t) = A cos( kx-t)

p2rms =

=

= .T/2 = , there fore prms = A/2.

Acoustic power: Pav = ds. Just a number independent of space and time.

Units are [I] = W/m2 [P] = W.

P is a very powerful quantity. Pav is usually independent of sigma in 3-D

This is because the 1/r dependence is offset by the 4r2 surface term and a property of

the source. Hence useful quantity.

Note: This is true even when background noise is present.

This is because if the control volume contains only the source term the net flux of the

background noise across the control surface is zero.

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t

T/2

A

-A

t

T=2/

In period is useless The averaged intensity is useful. Iav(x) = < I(x, t) > =

Specific cases:

Plane harmonic wave p(x,t) = Acos (kx – wt)

let us suppose we consider a duct.

K = 2 / l l > > D plane wave

P = ds

= ds

No intensity flux across solid walls.

P does not vary across S.

Pav = ds = < I >S = P2rms S / oc

Given a certain power level for the source

Prms decreases as S increases.

Velocity and particle displacement for harmonic plane wave.

Momentum: = mean density,

, = Acoustic velocity and pressure.

For harmonic wave = p (x) ei t = Re( P (x) ei t)

= v (x) ei t

p(x) is a complex number

P (x) ei t = [Pr (x) + i Pins (x)][cost + i sint ]

= [Pr (x) cost - Pins (x) sint] + i [Pr (x) sint + Pins (x) cost]

= Re(P (x) ei t) = pr(x) cost - Pins (x) sint

= [Pr (x) cost / -

Pins (x) sint / ]

= [ cos (t + )]

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S

microphone

D x

Sound source

People do not pay much attention to phase because a lot of times sound is absorbed.

ei t (i ) v(x) = - ei t

V(x) = - 1 Acoustic velocity is not the speed of the wave.

It is just the velocity of the particle. e.g. Standing in line in OAT, one pushes, then we

see that the push has traveled forward rapidly. This is the wave velocity. And an

individual’s speed is like the acoustic velocity.

Derive algebraic expressions for relations between acoustic pressure and acoustic

velocity for plane waves moving to

(a) Left

(b) Right

-----------Particle displacement

= ei t

V ei t = - i ei t

= u / i

Show that when the mean density is not a constant i.e., = (x), this wave equation is

And derive the corresponding Helmoltz equation.

, p = P ei t

p(x) = A sin (kx) + B cos (kx)

= A eikx + B e –ikx, where A & B are complex.

Let A = a + ib & B =c + id

Hence,

P = Re[(a + ib)(cos(kx) + isin(kx)) + (c + id)(cos(kx) - isin(kx))] cos(kx) + isin(kx)

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= a cos kx cos t – b sin kx cost + c cos kx cos t + d sin kx cost – b sin kx sint

- a cos kx sin t –c sin kx sint + d cos kx sint

= (a + c) cos kx cos t + (b-d) sin kx cost – (b + c) sin kx sint + (d-a) cos kx sin t

= cos kx cos t + sin kx cost + sin kx sint + cos kx sin t

= C1 cos(t – kx) + C2 sin(t-kx)

Standing waves in ducts

We get standing waves when there is a superposition of two or more

progressive waves travelling in opposite directions.

Boundary conditions:

p = 0 infinite reservoir OPEN END

u = 0 RIGID BOUNDARY

Linear combination p +

Consider 1-D waves in a duct with variable area

(a) Show that the energy equation is

Momentum:

Derive the wave equations for p and u. Do not assume to be constant.

Can you get solutions for these equations for atleast some ?

Complex amplitude denotes the phase difference.

--------------Helmoltz equation

P = A eikx + B e –ikx

What is interesting about this closed and open boundary condition?

= 0. Then the acoustic energy of the system remains constant.

Perfect conditions: where acoustic pressure or velocity = 0.

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Standing wave – superposition of progressive waves travelling in opposite directions

u = V ei t = 0 u = 0.

At x = 0, V = 0.

V = A = B.

To get acoustic velocity m acoustic pressure, we use the momentum equation.

P = 2A

P(L) = 0 coskL = 0 kL =(2n+1)/2

As 2fL/c = (2n+1) /2 f = (2n+1)c / 4L

n = 0 f = c / 4L ; n = 2 f = 5c / 4L.

n = 1 f = 3c / 4L ; n = 3 f = 7c / 4L 

if x = 0, p = po.

P(x) = po cos(kx).

p(x) = po cos(kx) ei t

= Re(po cos( t ) + isin( t ) cos(kx))

Physically particles move towards each other or away from each other.

Across pressure minima, phase changes by 180o.

CLOSED – CLOSED:

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OpenClosed

x = 0 x = Lu = 0 p = 0

x = 0 x = L

p

negative cos(kx)Positive cos(kx)

Whatever is on the left is 180 o out of phase with the other side

At x = 0, u = 0. u =

A= B, P = 2A

u =

=

at x = L, u = 0. sinkL = 0. kL = n

f = nc / 2L c/2L, c/L, 3c/2L

OPEN – OPEN:

At x = 0, p = 0 A + B = 0. A = - B.

P(x) = Aeikx – Ae-ikx = 2A sin (kx).

P(L) = 0, sin(kL) = 0.

2f / L = n

f = nc / L c/L, 2c/L, 3c/L.

Derive the wave equation in the presence of a mean flow . Assume = constant.

Derive the corresponding Helmoltz equation (i.e., in frequency domain). Solve this

equation.

Reflection and transmission of plane waves.

Boundary condition at a solid surface.

Simplest case: vibrating surface – generates waves in surroundings

(a) 1 – D case:

For a particle in contact with the surface, boundary condition continuity of

displacement: s(t) = (t). Vs (t)(surface velocity) = V(t)(fluid velocity)

General 3- D case:

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s(t)

(t)

So So = mean position of vibrating surface s(t) = s(t) = fluid particle displacement surface displacement

We are dealing with in viscid flow. Hence we do

not deal with no slip condition. The acoustic boundary layer is very small. We look at

the linear Euler equation:

Apply this to fluid at the boundary.

L.H.S = 0 at boundary.

At a stationary boundary

p / = c2 is it true when the temperature is changing

; p = p(x) ei t ; = (x) ei t

Where e is the total energy.

What are the different terms we have to consider here:

- Pressure

- Energy flux

- Unsteady term

Energy equation for a non-uniform duct.

For uniform area:

Cv =R / (-1) and

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A

A + dA

eAx)( .).()( dxeAux

eAx

Constant area:

ip + (Energy)

(Momentum)

after linearising:

Substitute for each acoustic term as that of comprising with amplitude to get

Substitute for to get

Plane wave incident on plane rigid boundary.

Normal incidence:

p = pe + pr

Boundary condition rigid boundary.

= at all times on the boundary

f= g at rigid boundary simply redirect the wave in the opposite direction , without

altering the shape or the phase of the wave.

At the wall p = pi + pr = f +g = 2p

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Pi = f (t – x / c)Pr = g (t + x/c)

incident

reflected

Pressure doubling at the wall.

Plane wave oblique incidence on a rigid boundary.

On a rigid boundary

Position vector

= distance measured in the direction of the incident wave.

Pi = fi {t –(x sini- y cosi ) / c}

Sr is the distance measure din the direction of the reflected wave.

)

to obtain the relation between the incident wave use the boundary condition

vn = 0 vni+ vnr = 0 at y = 0.

vin = =

=

vnr = =

=

For all t, x at y=0,

i = r & fi = fr this equation will be satisfied.

Pi = f{t –(x sini- y cosi ) / c};

Pr = f {t –(x sinr + y cosr ) / c}

Note: net tangential component

Vti + Vtr is not equal to zero.

At any point p = pi + pr , @ y = 0, p = 2Pi.

Reflection at discontinuities:

26

sryey

ii sn ,

Jump in characteristic impedance c

Laminar flame speed = 1 m/s.

Very small compared to wave speed.

Flame (impedance discontinuities)

x = y; u2 = u1; 2 = 1.

F(y – c1t) + G(y + c1t) = F2(y –c2t) + G2(y + c2t)

F1(y – c1t) / 1c1 - G1(y + c1t) / 1c1 = F2(y –c2t) / 2c2 + G2(y + c2t) / 2c2

The shape of the wave could be nay thing. It depends on the initial conditions.

Sound source in region 1, region 2 is infinite duct.

G2 = 0, G1 is the reflected wave

F2 is the transmitted wave.

Write G1, F2 in terms of F1

F1(y – c1t) / 1c1 – G1(y + c1t) / 1c1 =F2(y –c2t) / 2c2

F1(y – c1t)[1 - 2c2 / 1c1] + G1(y + c1t)[1 + 2c2 / 1c1] = 0

G1(y + c1t) = F1(y – c1t) [2c2 - 1c1] / [2c2 + 1c1]

Retarded time = x + c1t = y + c1t

G1(x + c1t) = F1(2y – x- c1t) [2c2 - 1c1] / [2c2 - 1c1]

t = (x - y) / c1 + t

y2 – c1t = y – c1 ( x – y)/ c1 – t c1 = 2y – x - c1t

G1(x + c1t) =RF1(2y – x- c1t), where R is the reflection coefficient.

2 F1(y – c1t) = F2(y –c2t)[1 + 1c1 / 2c2 ]

F2(y –c2t) = 22c2 F1(y – c1t)/ [2c2 + 1c1]

F2(x – c2t) = T F1[y(1 – c1/ c2) + c1/ c2 (x – c2t)]

y – c2t´ = = retarded time = x – c2t

t = =

y – c1t = y – c1 = y

27

1c1 2c2

1 2

F1 F2

x

G2G1

When impedances are equal then there will no reflection.

Specific acoustic impedance (Very Important):

Specific acoustic impedance = ratio of pressure & velocity amplitudes. Assuming a

Linear Relationship acoustic pressure (P) and the normal component [into the surface,

out of the fluid]. is the fluid velocity along a non-moving surface so, different

frequency components of P & are uncoupled.

Acoustic impedance =

Hence since we are dealing with pressure, specific acoustic impedance.

, are complex amplitudes. Z (w) = R + iX, X = Specific acoustic reactance

R = Specific acoustic resistance. Acoustic admittance = ,

Non-dimensional admittance.

Flames sometimes drive waves called Acoustic Driving

Impedance Tube, 1-D Case.

Physical meaning: rate of change of acoustic energy= divergence of acoustic intensity

for a constant volume. Also for a finite control volume

Integral (p2u2 – p1u1) A

28

Microphone

Tube of rigid walls

Acoustic termination

Solid tube

P = Aeikx + Be-ikx. The solution to the Helmoltz equation A, B are complex coefficients,

To extract the real part, split A & B as aei1, bei2 since the outcomes depend on the

phase between the two, it is sufficient to consider the real parts only.

P = Aei(kx+1) + Be-i(kx-

2)

= a [cos (kx + 1) + i sin (kx + 1)] + b[cos (kx - 2) + i sin(kx - 2)

= [a cos (kx - 11) + b cos (kx - 2) + I [a sin (kx + 1) – b sin (kx - 2)]

|p|2 = a2cos2 (kx + 1) + b2 cos2 (kx - 1) + 2ab cos (kx + 1) – cos (kx-2)

+ a2sin2 (kx + 1) + b2 sin2 (kx - 1) + 2ab sin (kx + 1) – b sin (kx-2)

= a2 + b2 + 2ab cos (2kn + 1 - 2)

Q: How to find out (1 - 2)?

Soluton:Find the places where there are maxima & minima.

|p| max2 = (a+b)2 , (a+b) = |pmax|, |p| min

2 = (a – b)2 , (a – b) = |pmin|

a = , b =

cos (2kxmin + 2 - 1) = as , 2kxmin + (2 - 1) = .

Q: Why did we choose minima?

Sol: (a) At minima the amplitude is sharp & not distributed as in the case of

maxima.

(b) At minima the particles are oscillating in different directions on either side

at minima, there is 180o shift.

Reflection at a surface of finite impedance

,

@ y = 0, the combined pressure is

p =

vin = - , z = = constant.

For similar waves, the algebra becomes simple. Hence we say fr = Rfi &

t - i = r = , Also

29

Applying the simplifications,

Now let us consider solutions in the harmonic domain.

,

=

= kx x – ky y.

kx = sin = k sin , ky = cos = k cos .

Define amplitude reflection co-efficient.

Normal velocity for the incident wave.

=

=

Normal velocity for the reflected wave, Vin,r =

= , Negative sign shows that normal velocity is away from

the surface.

30

= ,

zs (w) =

@ y = 0,

z =

At normal incidence, = 0

Continued---------------.

Impedance tube:

P = Aeikx + Be-ikx

= aeikx + be-I[(kx-)]

= a cos kx + i a sin kx + b cos (kx - ) – i b sin(kx - )

= [a cos kx + b cos (kx - )] + I [a sin kx – b sin (kx - )]

or , acoustic powered flow

, ,

= ,

.The velocity & pressure need not save the same phase.

They are in phase in case of traveling waves.

I =

31

= ,

,

I = yr intensity + ve energy is consuming into the control

volume.

Intensity –ve energy goes out of the control volume

The sign of intensity vector gives us the direction of the energy flow.

Time averaged intensity . - Average over a cycle.

= ,

=

R =

@ is a complex quantity say it is as: R= .

Hence

@ R = 1, or or .

This is the same as the one derived from closed end.

@ R = -1 this happen when the waves are 180o out of phase canceling each other at x

= 0.

Then, .

When p = 0 no impedance.

32

Tip:

But this rule holds good for order of magnitude.

Determination of Acoustic Velocity

To get the velocity amplitude, we use the momentum

equation.

here

,

=

y = = admittance, Non – dimensionalise admittance using (c) factor.

Define Y =

Y = - =

Y = Yreal + iYimaginary =

Y (x = 0) = -

33

Z = Y = 0

R = -1 rigid termination Reflection in phase

R = -1, open endR = (1)eikx, reflection180o out of phase.

Z = 0Y =

=

= = Yreal + Y imaginary

Yreal = Yin =

Write Yreal and Yimaginary interms of |R| & ,

Say Yr = 0.1, Yimag =0.

Yr = - 0.1, Yimag = 0

For any material that absorbs some Yreal = negative.

= =

34

P / pmax

P / pmax

Phase

Yr = 0.1, Yimag =0.

Phase

Yr = -0.1, Yimag = 0.

(x) = tan-1 = tan-1

SLIDE SHOW

Instability is indefinite growth of pressure amplitudes.

WKB approximation:

High frequency approximation

- Gas properties or duct area changes over scales that are long relative to

that of the disturbance

- Amplitudes rescale to conserve flux energy flux eg. For a right running

wave.

Similarly

Energy equation.

( )

( ) +

Net convection out =

=

= dx .

Continuity.

We have to consider work done by pressure

(pAu)

35

(pAu)

=

,

From equation of state.

Wave equation derived considering pressure or velocity ended up in the same. But

here there could be a difference.

. .

Similarly desire a wave equation

for u

36

Harmonic solutions.

Momentum:

Energy:

Wave:

Trick:

Assignment:Problem: Use 4th order Range – kutta methods

P(o) ku

Duct with a non-uniform temperature.DEADLINE MAY 1, 2002 4:30PMReport is mandatory.

- Abstract- Introduction and problem statement- Background- Methodology- Results and discussion- Conclusions & Future Plan- Appendix A computer program listing - Appendix B anything else- Acknowledgement- References.

General:

Continuity: ,

Momentum: ,

Energy:

37

Jo

x

Yr,Yin

State: p = RT,

u(x,t) = u(x)+u (x,t),

p(x,t) = p(x)+p (x,t),

(x,t) = (x)+ (x,t)

Steady:

Continuity: = constant.

Momentum: P (1 + M2) = constant.

Energy:

State: Unsteady:

Unsteady:

Continuity: ,

Momentum:

Energy: ,

State: p = R (T +T)

Continuity: , if = 0.

Momentum:

Energy: ,

).

In matrix form: {z}x = [A]x {z}x (1)

[A]x =

Transmission matrix

{z}x = [T]x {z}x=0 (2)

38

And similarly for .

{z}x=0 = [T]x{z}x=0 (3)

Differentiate (2)

{z}x = [T]x [T]x-1 {z}x (4)

{z}x = [A]x {z}x (1)

Comparing (1) and (4)

[T]x [T]x-1 = [A]x

or [T]x = [A]x [T]x (5)

[T]x=0 = I(identity matrix)

What is complex frequency?

= real + iin

p = eit =

= , Where - Non oscillating part

The imaginary part indicates how the amplitude grows.

Interaction between acoustics & combinationAcoustic effects of combustion are in general unfavorable for operation of

“apparatus”. Example - Pulse combustors (later)

Combustion noise (roar)

Sound from combustion process

Combustion driven oscillation

Combustion noise broad spectrum

39

in = 0

in > 0

in < 0

Turbulent fluctuations, in combination with resonance. Sound pollution of

environment. Sound is of great interest to turbulence engineers as they believe that

sound is “fingerprint of the turbulent is the flame”.

Combustion driven oscillations:

Feedback cycle that converts chemical energy to sound.

In this case broadband spectra is not seen. CRISP SPECTRA is observed.

Pressure amplitude – very high – up to 50% of mean pressure.

1777, Higgins, “singing Flame”.

Rate of combustion in acoustics:

Combustion is a volume source of sound. This gives rise to fluctuating

densities.

Heat release fluctuations depends on the acoustic processes in the combustion

chamber, due to a feed back loop. We can get the singing flame if the heat release is

in phase with the combustion. Flame located in an acoustic resonator. Flame adds

more energy to the resonator than is dissipated, acoustic energy in the resonator grows

in time instability.

LIMIT CYCLE OSCILATIONS - Acoustic driving is balanced by energy dissipation. Then amplitude stabilities.Coupling between acoustic oscillations & flames.

- Fluctuating mass flow through the burner (modulated).

Acoustic velocity fluctuations induce density & pressure fluctuations .

Fluctuates with p heat release fluctuations

Reactions kinetic get changed. If flow instability occurs, flame area changes.

Vortex shedding/flow instability

If dominant flow frequency coincides with the natural frequency of the combustor

strong compiling.

FLUCTUATING EQUIVALENCE RATIO:

Fluctuating flame front area

40

NoiseFLAME COMBUSTION

* Droplet formation/Breakupdroplet response,

oscillatory evaporation oscillating heat release.

* Depending of chemical reaction rate on pressure & velocity.

Combustion Instability control

Combustion Dynamics Managed.

1. Component melting due to excessive heat transfer.

2. Unusually high burn rate of solid propellant due to modification of

combustion process

3. Excessive vibrations mechanical failure.

4. Interference with control system operations, electronic components – sensitive

to noise. Excessive noise, flame excitation costly system/operation failure.

Combustion – flow – Acoustic – interaction:

Radial & Tangential modes occur at high frequencies.

Pulsating operation is maintained by a feed back mechanism between the

combustion process & the systems oscillations. In the initial phases, acoustic

driving >> damping.

Exponential growth in amplitude.

Losses dependent upon amplitude.

Acoustic driving from combustion is balanced by damping over a period of

time then the amplitude stabilizes.

Linearly unstable - Oscillations start spontaneously, system is unstable w.r.t

any small amplitude oscillations.

Transition to pulsating operation will occur when the amplitude of disturbance

will exceed a – certain threshold

41

Pressure -amplitude

Heat addition at amplitude maximum. It drives the amplitude, “Acoustic driving”

Conclusion: Adding heat in phase with the pressure oscillation drives the oscillation.

What happens if heat addition occurs in between?

The time period decreases 1/n or frequency increase.

In this case the time period increases.

Summary:

Phase of heat supply Effect on amplitude Effect on frequency

Relative to phase of Pressure

In phase Increase No

Out of phase Decrease No

Quarter period before No Increase

Quarter period after pack No Decrease

Driving: .The volume should include all points at which

heat addition occurs. Losses ------

42

Amplitude damping by adding heat minima

Net driving > net Damping

T

How would we adjust T so that it would decrease amplitude?

- Atomize the fuel move.

To decrease frequency increase the length of the duct.

RAYLIGH CEITERION: Analytical considerations

closed “closed”

“Closed” acoustically closed (i.e the acoustic wave gets reflected)

Assumptions:

(1) Uniform steady state gas properties along a tube as Q = 0. In a rocket

combustor,

T = const is not a bad assumption, but p = const. is a bad assumption.

(2) Gas is inviscid and non-heat conducting

(3) Oscillations are ID (good if L >> D)

(4) Cp, Cv & are constants.

(5) Fluctuations of u, p, T, & are small.

Equations:

Continuity: (1)

Momentum: (2)

Energy: (3)

where oscillatory heat addition per unit mass

Multiply (2) with u & (3) with to get

43

, Closed duct.

Term on R.H.S. will be positive which be positive when the energy will

rise.

p = , = cos (cot +)

=

if = 150o, the combustion process will absorb all of the energy. If = 0. In phase

driving

If = 90o, no driving / damping.

Effect of oscillatory heat release

Momentum: ,

,

Small combustion none: L << l,

,

Acoustic pressure does not jump across a flame it is thin

44

Velocity jumps but pressure does not jump.

Flame transfer function.

Derivation of wave equation

Continuity: (1)

Momentum: (2)

Energy: (3)

Perform: to get,

(4)

This is somewhat an INHOMEGENEOUS HEMOLTZ EQUATION.

Solution by green’s function technique: To obtain the Green’s function in terms of the

natural acoustic modes of the cobuster,

Solid wall B.C (7)

@ x = 0 & x = L

45

x = n / L, n = 0,1,2,….

px , x building blocks of solution for sound in a duct with oscillating great release.

Momentum: (1)

Energy: (2)

Assume

, (3)

=

(4)

Differentiate the momentum equation

(5)

Find a fundamental solution f.

,

For x , ln -

ln + ,Taking continuity

c sin k + D cos k = A sin k + B cos k

(C-A) Sin k + (D-B) cosk = 0 , or

, let (D-B) = V & (C-A) = u

usink + V cosk = 0 , Vsink - u cosk = -

46

, hence

Or ,

then C = A + U & D = B + V, let L =

, L(f) = {}

=

=

, = Assist x + B cosk

x l

Time scales involved s i or fractions of us

acoustic period of 100 Hz oscillations 0.01 sec

500 Hz –u- 0.002 sec

47

chemical kinetic ignition delay = 0.7 ~ 0.001 sec

- = 1.0~ 0.0003 sec

convection of a disturbance through 10cm ~ 0.01 sec. at 10 m/sec

convection of a disturbance thorugh 10cm

at 50 m/sec

convection of a disturbance through 10cm

at 50 m/sec

evaporation of a 10m dia hydrocarbon gyoplet ~ 0.0003 sec

50 droplet ~ 0.008 sec

propagation of an acoustic disturbance through 10cm at 330 m/s C.0003 sec

breakup time for a 10mm dia liquid jet ~ 0.125 sec.

Processes

1) Rate of fuel delivery

2) Atomization of fuel droplets

3) Reaction rate

4) Mixing between fuel & air

If the time scales of two processes are very close to each other then they

“adjust” to form one common time scale.

There may be lot of acoustic driving but if there is equal dissipation, all will be

futile.

0.1 0.3 0.9 1.2 1.4 1.6

for small velocities small mean flow.

Acoustic velocity is continuous

48

Chemical

if the fraction is such that XXX coincides with then we have acoustic driving, By

changing the airflow length we can change the convective time scale.

Aerodynamic sound: Lightmill analogy. Powell – vortex sound, if the vortex is

perpendicular to acoustic velocity, it will create sound

Momentum equation in the traditions flow

Using,

= acceleration corresponding to coriolis force.

Angular velocity =

Continuity:

49

u’,p’ e.t.c are small

Let

(1)

(2)

Wave equation

References for further study

1) P.A. Nelson, N.A Halliwell & P.A.Doak (1983)

Journal of sound & vibration, Vol. 91(3). 375- 402

2) “Fluid Dynamics of a fluid flow exited resonance.

Part II; Flow Acoustic interaction”.

3) “An introduction to acoustics”

www.win.tue.nl/n sjocrdr, Rienstra & Hirschberg

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