
Citation 
 Permanent Link:
 http://digital.auraria.edu/AA00002226/00001
Material Information
 Title:
 Sound propagation in moving fluid
 Creator:
 Jones, Carol Sue
 Place of Publication:
 Denver, CO
 Publisher:
 University of Colorado Denver
 Publication Date:
 1996
 Language:
 English
 Physical Description:
 viii, [82] leaves : illustrations ; 29 cm
Subjects
 Subjects / Keywords:
 Fluid dynamics ( lcsh )
Sound  Transmission ( lcsh ) Soundwaves ( lcsh ) Fluid dynamics ( fast ) Sound  Transmission ( fast ) Soundwaves ( fast )
 Genre:
 bibliography ( marcgt )
theses ( marcgt ) nonfiction ( marcgt )
Notes
 Bibliography:
 Includes bibliographical references (leaf 82).
 Thesis:
 Submitted in partial fulfillment of the requirements for the degree, Master of Basic Science
 Statement of Responsibility:
 by Carol Sue Jones.
Record Information
 Source Institution:
 University of Colorado Denver
 Holding Location:
 Auraria Library
 Rights Management:
 All applicable rights reserved by the source institution and holding location.
 Resource Identifier:
 36949224 ( OCLC )
ocm36949224
 Classification:
 LD1190.L44 1996m .J66 ( lcc )

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Full Text 
SOUND PROPAGATION IN MOVING FLUID
by
Carol Sue Jones
B.A., Adams State College, 1976
B.S., University of Colorado at Colorado Springs, 1987
A. thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Master of Basic Science
1996
This thesis for the Master of Basic Science
degree by
Carol Sue Jones
has been approved
by
Martin M. Maltempo
John W. Wyckoff
Jones, Carol Sue (Master of Basic Science)
Sound Propagation in Moving Fluid
Thesis directed by Professor Randall P. Tagg
ABSTRACT
Sound propagates as a wave. Therefore, the derivation of the three
dimensional wave equation, with and without background motion, is
developed to provide a complete basis for further investigation. In
the limit that the wavelength is small compared to the distance over
which changes occur in the medium of propagation, ray acoustic theory
can also be utilized in sound propagation. An important concept in
this theory is the eikonal equation. This is a partial differential
equation that defines the surfaces of constant phase or the
wavefronts. A section is devoted to its derivation. Next, several
examples are presented to provide theoretical predictions of sound
propagation in moving fluid showing how velocity gradients deflect
sound in a manner that can be compared to refraction due to
variations in, say, the fluid density. Lastly, an experiment is
proposed to test these predictions.
This abstract accurately represents the content of the candidate's
thesis. I recommend its publication.
Signed
iii
DEDICATION
I dedicate this thesis to my husband, Barry, and daughters, Sherri
and Melissa, who provided me with support and encouragement
throughout this entire degree process.
ACKNOWLEDGMENT
My thanks to the staff of the Physics Department for their support
and patience in my efforts to pursue this degree. A special thanks
to my advisor, Dr. Tagg, for his devotion, support, understanding and
encouragement during the pursuit of this degree.
CONTENTS
Chapter
1. Introduction......................................................1
2. 3Dimensional Wave Equation.......................................2
2.1 Continuum Hypothesis...........................................2
2.2 Mass Conservation Derivation...................................3
2.3 NavierStokes Derivation.......................................5
2.4 Equation of State.............................................13
2.5 Equation for Small Disturbances in the Presence of Background
Flow.........................................................15
3. Eikonal Equation.................................................20
3.1 Eikonal Equation for Stationary Fluid but Variable Wavespeed.20
3.2 Ray Equations.................................................25
3.2.1 Arc Length Parameterization................................25
3.2.2 HamiltonJacobi Relations..................................27
3.2.3 Finding Explicit Solutions for Ray Trajectories and the
Eikonal Equation..........................................32
3.3 Eikonal Equation Derivation for a Moving Fluid...............34
3.4 Example.......................................................41
4. Applications.....................................................54
4.1 Shear Flow Problem............................................54
4.1.1 Part IA...................................................54
4.1.2 Part IB...................................................63
4.1.3 Part II....................................................70
5. Proposed Experiment..............................................72
vi
Appendix
A. Further Derivations
References
FIGURES
Figure
2.1 Schematic Variation of Average Energy of Molecules with Length
Scale (Tritton 1977)...........................................3
2.2 Definition Sketch for Derivation of Continuity Equation
(Tritton 1977).................................................3
2.3 Deformation of Rectangular Element and Relative Motion of Fluid
on Either Side of Arbitrary Line through the Element (Tritton
1977)..........................................................7
2.4 Pouring of Viscous Liquid (Tritton 1977).......................8
3.1 Ray Curves....................................................24
3.2 Tangent to Ray................................................24
3.3 Pictorial Representation of Sound Propagating through Constant
Density Fluid in Uniform Motion.............................38
3.4 Propagation of Waves in Shear Flows of Different Refractive
Indices.......................................................50
3.5 Distance Between Wavefronts in Shear Flows of Different
Refractive Indices............................................51
4.1 Graphical Representation of Shear Flow.......................55
4.2 Graphical Representation of Part I of Problem................55
4.3 Graphical Representation of Part 2 of Problem................70
5.1 Linear Shear Flow.............................................72
5.2 Graphical Representation of Ray Trajectory...................73
viii
1. Introduction
Sound propagates as a wave. Therefore, the derivation of the
threedimensional wave equation, with and without background motion,
is developed in the second chapter. The basic equations utilized in
this derivation are the mass conservation equation, NavierStokes
equation and the equation of state between pressure and density.
When a wavelength is small compared to the distance over which
changes in the medium of propagation occur, the field is expected to
look like a plane traveling wave. In this case, all acoustic field
quantities (pressure, density and velocity) vary with time and with
some cartesian coordinate, r, but are independent of position along
planes normal to the r direction (Pierce 1981). This concept of a
wavefront is important in geometrical or ray acoustics and leads to
the eikonal equation which is a partial differential equation
defining these surfaces. Chapter 3 is devoted to the derivation of
the eikonal equation.
The next chapter presents a solution to a linear shear problem
which determines the distance the sound propagates. This problem
also demonstrates how velocity gradients deflect sound in a manner
that can be compared to refraction due to variations in, say, the
fluid density. The result is then utilized in the last chapter which
proposes an experiment to test these predictions.
Sound propagation has a wide range of applications in a variety
of fields of study. Some areas where sound propagation may be of
interest are atmospheric acoustics, underwater acoustics, musical
acoustics, ultrasonics, architectural acoustics, aeroacoustics,
nonlinear acoustics, environmental acoustics and noise control
(Pierce 1981).
1
2. 3Dimensional Wave Equation
This chapter derives the 3dimensional wave equations for sound
waves with and without background motion. This is examined in order
to provide a basis of understanding for the next chapter on the
eikonal equation.
The conservation of mass and Newton's laws of motion are used
to derive the 3dimensional wave equation. These equations involve
physical and mechanical quantities such as velocity, density,
pressure and temperature which will vary continuously from point to
point throughout the fluid. To define these quantities at a point
requires the applicability of continuum mechanics or the continuum
hypothesis which is discussed in Section 1. The derivation of the
mass conservation or continuity equation is discussed in Section 2.
Section 3 derives Newton's second law of motion for a fluid, or the
NavierStokes equation. Lastly, the 3dimensional wave equation is
derived in Section 4 using the mass conservation and NavierStokes
equations.
2.1 Continuum Hypothesis
The continuum hypothesis supposes that an association can be
made with any volume of fluid and those macroscopic properties
involving a fluid in bulk. At each point there is a particle of
fluid having a certain velocity, density, pressure and temperature.
A large volume of fluid therefore consists of a continuous aggregate
of such particles.
It must be noted that this hypothesis is not correct at the
molecular scale. The various macroscopic properties are defined by
averaging over a large number of molecules (Tritton 1977) None of
these averaging processes is meaningful unless the averaging is
carried out over a large number of molecules. A fluid particle must
thus be large enough to contain many molecules. It must still be
effectively at a point with respect to the flow as a whole.
The above requires a length scale in order for the continuum
hypothesis to be valid. This length scale will be the size of the
fluid particle. The intent is for the volume to be large enough to
contain many molecules and have the fluctuations of the macroscopic
quantities negligibly small. If the volume is too small, it will
contain only a few molecules and there will be large fluctuations in
the macroscopic quantities. If the volume is too large, it will
extend into regions where the macroscopic quantity is significantly
different and there will be an increase or decrease in the average.
In other words, the applicability of the continuum hypothesis depends
on there being a significant plateau (Tritton 1977) in the
macroscopic quantity. There will thus be a length scale that can be
regarded as an infinitesimal distance as far as macroscopic effects
2
are concerned and the equations can be formulated ignoring the
behavior on still smaller length scales. Refer to Figure 2.1.
Physical
Quantity
Â£e.g.
density)
Figure 2.1. Schematic Variation of Average Energy of
Molecules with Length Scale (Tritton 1977)
For common fluids in normal laboratory conditions, the
continuum hypothesis is valid for both the mass conservation and
NavierStokes equations. For example, the fluid velocity u(r,t) at
Â£ is defined as the massweighted local average particle velocity or,
equivalently, as the average momentum per unit mass in the vicinity
of r (Pierce 1981). For both the mass conservation and NavierStokes
equations, the concept that at each point there is a particle of
fluid having a certain velocity, density, pressure and temperature is
used.
2.2 Mass Conservation Derivation
The derivation of the mass conservation or continuity equation
begins by considering an arbitrary volume V fixed relative to
Eulerian coordinates and entirely within the fluid. Refer to Figure
2.2.
Figure 2.2. Definition Sketch for Derivation of Continuity
Equation (Tritton 1977)
The equation is more readily formulated using the Eulerian
specification, coordinates fixed in space, instead of the Lagrangian
3
specification, coordinates that move with the particle. The reason
behind this is that the Lagrangian specification does not immediately
indicate the instantaneous velocity field on which depend the
stresses acting between fluid particles (Tritton 1977).
The Eulerian notation for velocity is u = u(r, t) where r is
the position in an inertial frame of reference and't is time. Values
of u at the same position f but different t do not correspond to the
same fluid particle.
Fluid moves into or out of this volume at points over its
surface. If dS is an element of the surface, whose magnitude is the
area of the element and whose direction is the outward normal, and
U is the velocity at the position of this element, then the
component of u parallel to dS is what transfers fluid out of V.
Thus, the outward mass flux (mass flow per unit time) through the
element is pfi dS, where p is the fluid density. Hence, the
rate of loss of mass from V is J pu dS (2.1)
which is negative if the mass in V is increasing. Also, the
total mass in volume V is
1
pdV.
(2.2)
Therefore, the rate of loss of mass from V becomes
_d_
dt
dS.
(2.3)
Using the continuum hypothesis, the mass balance is at a point rather
than that over an arbitrary finite volume. Hence, V is allowed to
shrink to an infinitesimal volume and, since the integration in
J (3p/ dt )dV
is redundant,
equation 2.3 becomes
 lim [ f pu dS/V .
dt v>o L"*8
That is,
dp
= divpu
dt
by definition of the div operator,
zero yields
(2.4)
(2.5)
Setting equation 2.5 equal to
4
dp
dt
+ V (pu) = 0
(2.6)
which is the general expression for the mass conservation of a fluid
in which both u and p are functions of position. This equation is
often referred to as the continuity equation.
2.3 NavierStokes Derivation
Newton's second law of motion states that the rate of change of
momentum of a fluid particle is equal to the net force acting on it.
First, an expression for the rate of change of momentum of a fluid
particle is considered. It would not be correct to equate the rate
of change of momentum at a fixed point to the force, because
different particles are there at different times (Tritton 1977).
Therefore, consider the general question of rates of change
following the fluid. In general, the small change SB produced by a
small change 8t in time and small changes 8x, 8y, 8z in Cartesian
position coordinates is
SB
dB dB dB SB
St + 8x + 8y + 8z
dt dx dy dz
(2.7)
and a rate of change can be formulated by dividing by 8t which gives
Sb 3b da 8x <9b 8y 9Â§ 8z
8t 9t dx 8t 9y 8t 9z 8t
Choosing Sx, 8y, 8z to be the components of the small distance moved
by a fluid particle in time 8t, then in the limit 8t > 0 this is
the rate of change of a vector quantity B following a fluid
particle. In the same limit, 8x/8t, 8y/8t and 8z/8t are then the
three components of the velocity of the particle u, v, and w
respectively. Then equation 2.8 becomes
DB
Dt
9b 9b 9b 9b
h U1 V1 W
9t dx dy dz
(2.9)
In general D/Dt denotes the rate of change (of whatever quantity it
operates on) following the fluid. This operator is known as the
substantive derivative (Tritton 1977) or as the convective
derivative.
Equation 2.9 can be rewritten as
5
(2.10)
DB
Dt
9b _ 
+ u VB.
at
If B = u, equation 2.10 gives the rate of change of velocity
following a fluid particle,
Du au __
= + u Vu. (2.11)
Dt at
The quantity u now enters in two ways, both as the quantity that
changes as the fluid moves and as the quantity that governs how fast
the change occurs. Mathematically, however, it is just the same
quantity in both its roles (Tritton 1977).
In the general case when both u and p are variables, the rate
of change of momentum would be D(pu)/Dt. According to Tritton
(1977), the only reason why a particular bit of fluid is changing its
momentum is that it is changing its velocity. If it is
simultaneously changing its density, this is not because it is
gaining or losing mass, but because it is changing the volume it
occupies. This change is therefore irrelevant to the momentum
change. Therefore, for this derivation, the rate of change of
momentum is pD(u)/Dt and equation 2.11 becomes
Du da _
p = p + pu Vu. (2.12)
Dt St
From equation 2.12, pD(u)/Dt is the lefthand side of the
dynamical equation which represents Newton's second law of motion.
The righthand side is the sum of the forces (per unit volume) acting
on the fluid particle. In order to complete the equation, the nature
of these forces must be considered.
Some forces are imposed on the fluid externally, and are part
of the specification of the particular problem. An example of a body
force is to specify the gravity field in which the flow is occurring.
A uniform gravitational force (per unit volume) pgz may be
eliminated from the equations replacing the pressure p with
pf pgz so that Vp' = Vp + pgz. In the absence of fluid motion,
Vp' = 0 so Vp = pgz. This means p = p0 pgz, the hydrostatic
pressure, where po is some reference pressure established at z = 0.
Other forces, pressure and viscous action, are related to the
velocity field. These surfaces forces or stresses are thus intrinsic
parts of the equations of motion and have to be considered here.
Both the pressure and viscous action generate stresses acting across
6
any arbitrary surface within the fluid; the force on a fluid particle
is the net effect of the stresses over its surface.
In the following discussion, some of the physical concepts
underlying viscous action are considered and then the mathematical
formulation is presented. According to Tritton (1977) viscous
stresses oppose relative movements between neighboring fluid
particles. Equivalently, they oppose the deformation of fluid
particles. The difference between these statements lies only in the
way of verbalizing the rigorous mathematical concepts, as illustrated
by Figure 2.3. The change in shape of the initially rectangular
region is produced by the ends of one diagonal moving apart and the
ends of the other moving together. As the whole configuration is
shrunk to an infinitesimal one, it may be said either that the
particle shown is deforming or that particles on either side of AB
are in relative motion. The rate of deformation depends on the
velocity gradients in the fluid. The consequence of this behavior is
the generation of a stress (equal and opposite forces on the two
sides) across a surface such as AB. This stress depends on the
properties of the fluid as well as on the rate of deformation.
A
Figure 2.3. Deformation of Rectangular Element and Relative Motion
of Fluid on Either Side of Arbitrary Line through the
Element (Tritton 1977)
The stress can have any orientation relative to the surface
across which it acts (Tritton 1977). Consider a simple example in
which the viscous stresses normal to the surface govern the behavior.
A pictorial representation (refer to Figure 2.4) shows a viscous
liquid poured from a container and the falling column which is
produced. In this case, since there are no sidewalls, transverse
viscous stresses cannot be generated as they are in the case of
channel flow. The reason the fluid at P does not fall with an
acceleration of g is because of the viscous interaction with the more
slowly falling fluid at Q.
7
Figure 2.4. Pouring of Viscous Liquid
In general, the stress is a quantity with a magnitude and two
directions associated with it. The two directions are the direction
in which it acts and the normal to the surface. Therefore, it is a
secondorder tensor. The stresses acting across surfaces of
different orientations through the same point are not, of course,
independent of one another (Tritton 1977). The rateofstrain tensor
is also a secondorder tensor and describes the rate of deformation.
From the considerations above velocity gradients are involved. Not
all distributions of velocity variation lead to deformation. An
example is the rotation of a body of fluid as if it were rigid. The
rateofstrain tensor selects the appropriate features of the
velocity field (Tritton 1977) .
At this point, the form of the rateofstrain tensor is
examined. Then the consequences of a linear relationship between the
stress tensor and the rateofstrain tensor will be determined.
Consider two material points separated by a distance &Â£ which
have components 6xi Then
)2 =
(2.13)
where the summation convention for repeated suffixes applies. The
rate of change of equation 2.13 following the fluid is
p(8l)a = 28x p(8xj
Dt 1 Dt
= 2Â§x18xj
28xi8u1
8xf8x,
duj. dUj
5x, dx
1 J
(2.14)
According to Tritton (1977), it is thus the symmetrical combinations
of the velocity gradients that give rise to distortions opposed by
viscous action. (A motion in which
8
(2.15)
du
dx.
dui
5xf
for all i and j does not involve any relative movement between two
neighboring points for any 8xt.) Moreover, these symmetrical
combinations represent fractional rates of change of the separation
for various orientations; for example if
8x2 = 8x3 = 0,
1 d(80 3%
8Â£ Dt 3xx
(2.16)
They are thus rates of strain of the fluid. The rateofstrain
tensor ei:j is formulated as the symmetrical part of the velocity
gradient tensor,
=
_ dut
dx,
i, fta +c,i).
(2.17)
[Parenthetically, the antisymmetric part
Hu = \ fe, 5,1 )
(2.18)
corresponds to the vorticity, (D = V x u where
V x 11I = lim (jfcurve u dÂ£ / 8A. Physically the vorticity
SA>0 Jsurronding
area element 8A
corresponds to the rotational motion of the fluid. Thus equation
2.18 is
^lij ~ 2 6iik k ^
(2.19)
The viscous stress tensor is given by defining Ti;j as the
force per unit area acting in the idirection across a surface normal
to the jdirection. For a Newtonian fluid this is linearly related
to ej :
Tij AijJcl ekl *
(2.20)
9
The physical processes must be independent of the orientation
and handedness of the axes. Ai;jkl must thus be an isotropic tensor
and the most general form it can take is
Aijki = ^5i:j 8,^ + Â£5ik 8^ + 5jk . (2.21)
This gives
*ij = ^8ij kk + U + X)eij (2.22)
(since ). There are thus two arbitrary constants involved;
these are physical properties of the particular fluid. From the
particular case e12 = ^ Su/dy and all other equal to zero, we
can identify that
4 + X = 2U (2.23)
where (x is the coefficient of viscosity. Also,
e
kk
dxk
V u
(2.24)
and substituting equations 2.17, 2.23 and 2.24 into equation 2.22
yields
*11 = ^
[dxi
dx
+ X8i;j V u.
{2.2b)
The net viscous force on a fluid particle is given by the
differences in the stresses acting across opposite faces (Tritton
1977). The ith component of the force per unit volume is given by
dtij/dxj .
The stress tensor depends on the rateofstrain tensor and on
the properties of the fluid. A Newtonian fluid can now be defined
rigorously as one in which the stress tensor and the rateofstrain
tensor are linearly related (Tritton 1977) .
According to Tritton (1977), the remaining ideas contained in
the derivation of the viscous term of the dynamical equation are
simply symmetry considerations. For example, a mirrorimage flow
pattern must generate a mirrorimage stress distribution. And the
analysis of a flow configuration using different coordinates must
give the same result.
10
The xcomponent, in Cartesian coordinates, of the viscous force
per unit volume of equation 2.25 is
d dx 2n + XV dx u a + ay f du. av") 11 1 ^ ay dx J
a + dz (aw au")
Similar expressions for the y and zcomponents are given by
appropriate permutations and are
(2.26)
dv a au dv ^
2 ii + XV u + 1 +
9y ax Kdy dx j
a i i ' dw av"
dz H s. ay dz,
(2.27)
d_
dz
aw . a 'aw dvN
2 (i ^ + XV u + 1* j
dz dy l ay dz)
a _L aw au"
dx I1 ^ dx dz
(2.28)
As stated earlier, n is the coefficient of viscosity and A, is a
second viscosity coefficient. The existence of this second
coefficient, which is independent of the first, is explained by the
fact that there are two independent elastic moduli. The constant i
is associated with friction that occurs as a fluid element undergoes
a pure strain, while X is associated with friction that occurs as a
fluid element undergoes a change in volume (compression or
expansion). The statement that a fluid is Newtonian usually means
that n is observed to be independent of the rateofstrain and that X
is assumed to be so too. According to Tritton (1977) X is difficult
to measure experimentally and is not known for the variety of fluids
for which there are values of p.. Often, it is better to identify
pressure as the mean normal stress. Let peq be the thermodynamic
pressure associated with the state of a fluid element that is
considered to be in local thermodynamic equilibrium. The mean normal
stress p is given by
P = Peq *" ~ (CT11 CT22 CT33 )
11
1 ^ 0U _ 1 f 0V ^
+ 2 u + XV u +  2 u. + XV u
3 V 0X ) 3 l dY J
1
+
3 ^
2\i y + XV u
P = Peq + H
P = Peq +
2
3
f 2
0u 0v 0w 1 .
+ + +XVU
dx dy dz)
\
~ I1 + k
^ 3
u.
The ith component of the net force on a fluid element is given by
0 0
p + CTii where repeated indices mean implied summation.
dxi
Then
dx,
d d
Peq + =
dx,
dx,
dx.
eq
dx,
( 0u 3uj ^
 +
v0Xj
dx,
dx,
dx
d
p + =
dx, rn
d
+
dx.
=
dx
eq
dx, ij
aXi
d
dxi
0 _
toijV u
(Peq + u) +
dx.
(
0Ui
+ J
0Xj 0Xi j
_ 2
2 0 f
li p < u + V
3 J 0Xj . V
du,
dn.
0Xj 0Xi
0Xi
P +
0Xj
0UA +
(1
t 0Xj J
2 0
3 dx
(iV u).
Hence Newton's second law may be written in two ways
(2.29a)
(2.29b)
In equation 1.29b the second viscosity coefficient X has been
2
effectively replaced with(1, with the understanding that the
12
pressure p now refers to mean normal stress as opposed to the
pressure peq obtained from the (local) equation of state for the
fluid. This distinction is important when processes of sound
attenuation are to be considered. For the remainder of this
discussion, however, viscosity and its effects are ignored and
p = p = peq is used when there is no viscosity.
Equation 2.29 is known as the NavierStokes equation. It is
the basic dynamical equation expressing Newton's second law of motion
for a fluid of constant density.
According to Tritton (1977), the term P represents the
contribution of those forces (such as gravity) that have to be
included in the specification of the problem. This is often known as
the body force term, because such forces act on the volume of a fluid
particle, not over its surface in the way the stresses between fluid
particles act. The reaction to a body force is remote from the fluid
particle concerned usually outside the fluid region, although
occasionally on distant fluid particles.
Substituting V = i/p into equation 2.29 gives the Navier
Stokes equation as
viscosity.
The NavierStokes equation is a nonlinear partial differential
equation in u. The nonlinearity arises from the dual role of the
velocity in determining the acceleration of a fluid particle. This
nonlinearity is responsible for much of the mathematical difficulty
of fluid dynamics, and is the principal reason why our knowledge of
the behavior of fluids in motion is obtained in many cases from
observation (both of laboratory experiments and of natural phenomena)
rather than from theoretical prediction. The physical counterpart of
the mathematical difficulty is the variety and complexity of fluid
dynamical phenomena(Mitton 1982). Without the nonlinearity the
range of these would be much more limited.
2.4 Equation of State
The continuity equation (2.6) and the NavierStokes equation
(2.30) constitute a pair of simultaneous partial differential
equations. Both represent physical laws conservation of mass and
conservation of momentum which will always apply to every fluid
particle. Together they provide one scalar equation and one vector
1 2 
Vp + vV2uvV u + F
P 3
(2.30)
where
fluid called the kinematic
13
equation effectively four simultaneous equations. There are,
however, two scalar variables (pressure and density) and one vector
variable (the velocity) effectively five unknown quantities.
Therefore, at least one more equation is needed. Start with an
equation of state relating pressure to density and as required for
simple fluids one other intrinsic variable, say the entropy per
unit volume s. Then, p = p(p,s). Unfortunately, this introduces
yet another scalar, s, and consequently requires a further equation.
In general, this equation arises from the conservation of energy.
However, sound waves typically disturb the fluid on time scales short
with respect to the time over which heat diffuses across a distance
of one wavelength of sound. So, the process may be regarded as
adiabatic. The conservation of energy equation is replaced with s =
constant = so. Thus p = p(p, SQ ) or, understanding that entropy is
held constant, simply p = p(p).
Expanding the relationship between pressure and density,
P = P(p)/ using Taylor's series gives
P = Po + (p Po)^(po)s
dp ,s
(2.31)
The subscript s means the thermodynamic derivative is
constant entropy. In general an increase of pressure
increase the density and so the gradient (Po) is a
dp
taken for
tends to
positive
constant (Dowling and Ffowes Williams 1983). The parameter c is
defined as
Substituting equation 2.33 into equation 2.32 gives the linearized
relationship for pressure and density
p' = C2p' (2.33)
where p' = p pG and p' = p po .
As an aside, the speed of sound for an ideal gas can be found
by first considering the relationship between pressure and volume.
For an adiabatic process they are related by
PVY = constant
14
where y is a constant equal to the ratio of the specific heat at
constant pressure to the specific heat at constant volume.
Differentiating the above equation gives
yPVY JdV + VYdP
yPdV
dP = 
V
0
The bulk modulus is defined as the ratio of the change in pressure to
the fractional change in volume. Therefore
^adiabatic
dP
dV/V
= yP.
The speed of sound can be defined as
c
2
B
p
which for the adiabatic case is
C
2
yP
P
2.5 Equation for Small Disturbances in the Presence of Background
Flow
In order for the 3dimensional equation to be developed in the
most general case, the effect of background flow needs to be added to
the continuity equation (2.6) and the NavierStokes equation (2.30).
Let U be a steady background flow with density pb Also, let
U = U + u', p = pb + p' and p = pb + p' where u', p' and p' are
small perturbations to the velocity, pressure and density
respectively. The entropy per unit volume can be written as
S = Sb + s', but since it is assumed that the disturbance occurs
adiabatically, s' = 0. Substituting these expressions into
equations 2.6 and 2.30 yields
+ P') + V (pb + p')(u + u') = 0 (2.34)
and
15
(2.35)
(pb + p') (u + u') + (Pb + P')(u + s') v(u + u') =
at
 V(pb + p') + v(pb + p')V2(u + u') 
 v(pb + p')v (u + u') + (pb + p')f.
Expanding equation 2.34 gives
+ v (pbu + pbu' + p'u + p'u') = o
at at
+ + pbV U + U Vpb + pbV u' + u' Vpb +
dt dt (2.36)
p'V U + U Vp' + p'V ii' + ii' Vp' = 0
Expanding equation 2.35 gives
du dn' 3u du' 
p Â¥ + Pb W + p si + p "eT + PD VD +
pbu' vu + p'u vu + p'u' vu + pbu Vu' +
pbu' Vu' + p'U Vu' + p'u' Vu' =
 Vpb c2Vp' 
\dP2 J
(Vpb)p' + vpbV2U +
(2.37)
vpbV u' + vp'V U + vp'V ii' vpbV U 
2 22 
vpbV u' vp'V U vp'V u' + pbF + p'F
Eliminating the squares of small quantities and the derivatives
of constant quantities from equations 2.36 and 2.37 yields
dp'
~dt
+ pbV U + U Vpb + pbV u' + ii' Vpb +
p'V U + U Vp' = 0
(2.38)
and
16
du' 
Pb T + pbu VU + pbu' VU + p'U VU + pbU VS'
dt
 Vpb c2Vp' 
\dp2j
(vpb)p' + vpbV2U +
 2 2
vpbV2u' + vp'V2U vpbV U vpbV u' 
vp'V U + pbF + p'F.
(2.39)
The background flow, U with density pb satisfies the
continuity equation and the NavierStokes equation. Substituting the
background flow parameters into equation 2.6 gives
dPb
dt
+ V (Pbu)
0.
(2.40)
Assuming a steady background flow,
dPb
dt
= 0, equation 2.40 yields
PbV U + U VPb = 0.
(2.41)
Substituting the background flow parameters and the assumptions that
the viscosity is zero, the flow is steady and the body force is
negligible into equation 2.30 yields
da  l
+ u vu =
dt pb
VPb
da
or since  = 0 (steady background flow),
dt
pb U VU = Vpb. (2.42)
Using equations 2.41 and 2.42 in equations 2.38 and 2.39 yields
+ pbV u' + u' Vpb + p'V U + U Vp' = 0 (2.43)
dt
and
17
(2.44)
Pb
du'
dt
+ pbu' VU + p'U VU + pbu Vu' =
 c2Vp' 
o2 \
O P
dp2)
(Vpjp'
To arrive at the 3dimensional wave equation, first
differentiate the mass conservation equation (2.43) with respect to t
(remembering that U and pb are steady in time) to get
5V
at2
+ Pbv
du' du'
dt
dt
VPb +
dp'
do
dp'
V + U V
dt dt dt
= 0
(2.45)
Next, take the divergence of the NavierStokes equation (2.44)
to get
Vp ^ + pbv ^ + vPb [(S v)u] +
pbv [(u' v)u] + vP' [(o v)u] +
p'v [(a v)a] + vPb [(a v)u'] +
( ^
pbv [(u v)a'] + c2vv + Tr V (vPbP') = o.
1 J lp J
(2.46)
Subtract equation 2.46 from 2.45 and use equation 1.41 to obtain
This is the 3dimensional equation for a disturbance in the presence
of background flow. If there is no background flow (U = 0) and
constant background density (Vpb = 0) the classical 3dimensional
wave equation is recovered from equation 2.47
d2p'
dt2
c2V2p' = 0
(2.48)
18
All small quantities describing the acoustic field (pressure,
density and velocity) satisfy the wave equation. The density has
been established in equation 2.48. Since the equation of state
(2.33) equates the pressure and density, the wave equation is
satisfied for pressure. The velocity can be shown to satisfy the
wave equation by taking the divergence of the mass conservation
equation
v* p'V2*' = o
at H
and differentiating the NavierStokes equation with respect to t
c2 at2 at
and adding these two equations to obtain
P'
 p'V2u' = 0,
or
0.
For an adiabatic condition, no viscosity and negligible body
forces, the 3dimensional wave equation ( 2.48) has been established.
The equations used for this development were the continuity equation,
equation 2.6, and the NavierStokes equation, equation 2.30. This
development will be utilized in the next chapter on the development
of the eikonal equation.
19
3. Eikonal Equation
This chapter develops the eikonal equation which defines the
wavefronts of a disturbance in a moving medium. The eikonal concept
is then applied to an example. The first section derives the eikonal
equation for stationary fluid with variable wavespeed. Section 2
presents the development of the ray equations. frequency
approximation of the eikonal equation. Then, Section 3 derives the
eikonal equation for a moving fluid. The last section develops an
example which uses the concepts presented in earlier sections.
3.1 Eikonal Equation for Stationary Fluid but Variable Wavespeed
According to Jones (1989) when the wavelength is so small that
a significant change in the medium occupies many wavelengths, a
reasonable hypothesis is that in local regions the field behaves as
if it were in a homogeneous medium. Thus, locally, the field may be
expected to look like a plane wave. Let 'F(x,t) be a field
variable being transmitted as a wave according to equation 2.48
V2 Y
n2 (x) d2 Y
(3.1)
where the wavespeed c(x) is assumed to vary with position. The
parameter c is related to a reference speed constant cQ by
c(x)
c0
n(x)
(3.2)
where n(x) is interpreted as the local index of refraction and is
assumed to vary continuously and slowly over distances comparable to
, 271
the wavelength A0 = . Let
= i/ (x)eia>t .
Substituting equation 3.3 into equation 3.1 yields
V2 \j/e
2
i(Dt H (x) 2 ia)t
+ = Ve
= 0.
(3.3)
(3.4)
Defining
20
(3.5)
and substituting it into equation 3.4 and removing the eia>t terms
yields
V2 v/ + k2n2 (x)V/ = 0
(3.6)
which is a modified form of the Helmholtz equation with nonconstant
coefficient k2n2(i).
Try a solution of the form
V/(x) = A(x, k0)e (3.7)
where it is anticipated that the amplitude part will depend on k0 as
well as i. L is known as the eikonal and has dimensions of length.
Taking the gradient of equation 3.7 gives
V\j/ = VAe llc L ik0A Vie ^ L (3.8)
and taking the divergence of equation 3.8 gives
V2i/ = V2Ae~ik 1 2ik0VA Vie~ik L
ik0AV2le"ik L k2A Vl2 e"ik L .
Now, substituting equation 3.9 into the modified Helmholz equation
(3.6) yields
V2Ae'k L 2ik VA Vle~ik L ik0AV2le"k ^
0 (3.10)
k2AVl2elk 1 + k2n2Aeik 4 = 0.
ik L
Removing the e term from equation 3.10 gives
V2A 2ik0VA Vl ikGAV2l k2AVl2 + k2n2A = 0. (3.11)
Equating real and imaginary parts (assume A is real) in equation 3.11
gives
V2A k2AVl2 + k2n2A = 0 (3.12)
21
2k0VA VL + k0AV2!/ = 0
2 VA VL + AV2L = 0.
Dividing equation 3.12 by Ak0 yields
(3.13a)
(3.13b)
^ VL2 + n2 = 0. (3.14)
k2 A
Suppose there is a distance l which is some characteristic
distance over which A changes appreciably such that
__2 A
V2A . (3.15)
n*
Substituting the first term of equation 3.14
i v2a _a_ = _jl_ = mvo2
k2 A k2 ^2A (k0Â£)2 V 27t J v 1 J
where the definition of the wavelength given
also employed. If Â£ X0, i.e. the amplitude changes significantly
(X Y
only over distances much larger than X0, then I j J << 1.
Consequently, equation 3.14 simplifies to (approximately)
Vi>2 + n2 =0, or
 Vl2 = n2 .
Another approach to the same result yields some insight into
the method for making higherorder corrections. Let
into equation 3.15 gives
(3.16)
i *  271
early, K = , is
A(ac, k0 ) = A0 (i) +
At(x)
K
+
ko
(3.17)
and substitute this expansion into equations 3.12 and 3.13b which
yields
22
(3.18)
V2A0 (x) + V2^ (x) + \ V2^2 (X)
kG k0
 k2A0(i)Vi2 k0Aj (x) Vl2 A2(x)VL2.
+ n2k2AG(x) + n2k0Ax(x) + n2A,(x) = 0
2 VA0 (x) VL + 2VA1 (x) VL + \ 2(x) VL
K K
+ A0 (x)V2i + Al (x) V2i + (x) V2i = 0.
K K
Grouping like terms in k0 in both equations 3.18 and 3.19 yields
k2[n2  Vi2 j A0 (x) + k0[n2 IVL^Ja^x)
+ [n2  Vi2 ]a2 (x)
+ [ao (x) V2L + 2VA0 (x) Vi] + V2A0 (x) (3.20)
1 1
HU (x) V2i + 2VAX (x) Vil HV% (x)
k0 kG
+ \ [a2 (x) V2L + 2 VA2 (x) Vi] + V2A2 (x) = 0.
k0 kQ
If L, Av,...and their derivatives are finite and do not alter
appreciably over a few wavelengths, the coefficients of the
individual powers of k0 may be equated to zero (Jones 1989). (Note,
this is actually an asymptotic expansion that becomes better at given
order N in the series as k => oo but does not necessarily converge
for fixed kG as N oo .) Then equation 3.20 becomes
(3.21)
2VL VA0 (x) = 0 (3.22)
AjV2L + 2VL VAj = V2^ (j = 1, 2,...). (3.23)
According to Jones (1989), equation 3.21 acts as a partial
differential equation for L, the function which defines the surfaces
of constant phase, i.e. the wavefronts. Often L is called the
eikonal and 3.21 is known as the eikonal equation. The differential
equations 3.22 and 3.23 which fix the amplitudes Aj are termed
transport equations.
VL = n2
A0(x)V2L +
23
For example, when k0L = 271, Re(y) reaches a local crest.
Note, however that A may vary along the crest. Refer to Figure 3.1.
Figure 3.1.
Ray Curves
Rays are curves that are everywhere tangent to the direction of
energy propagation. For sound traveling in a nonmoving medium, the
direction parallel to Vl/ corresponds to the normal to the
wavefronts. However, this is no longer valid for sound traveling
through a moving medium. The local vector velocity of the medium
must be added to a vector normal to the wavefront to get the
direction of energy propagation. Refer to Figure 3.2.
Figure 3.2. Tangent to Ray
Before continuing to the derivation of the eikonal equation for
a moving fluid, the eikonal for a uniform, stationary medium will be
considered in order to understand physically equation 3.21.
A known solution to the wave equation (3.1) for a uniform
stationary medium is
'J'fXrt)
= Ae
i[(0t k*]
(3.24)
24
Using equations 3.3 and 3.7, the general eikonal form is given as
i cot k L (S)
T(x,t) = A(x,k0)eL J.
Equations 3.24 and 3.25 are equivalent if
A =
k X = k0L(x) .
Solving equation 3.27 for L(x) gives
A
L(x) = x = k x
k
(3.25)
(3.26)
(3.27)
(3.28)
where k is a unit vector normal to the wavefront. Notice that
Vi(x)
f
\
and therefore
VL = k! + k. + k?
= 1.
This satisfies equation 3.21 for n = 1.
3.2 Ray Equations
3.2.1 Arc Length Parameterization
This section determines how to solve for points x that lie
along a ray, that is, along the trajectory everywhere parallel to the
local direction of energy flow. As noted above, this direction is
perpendicular to the wavefronts for stationary fluids. Since Vl> is
also perpendicular to the wavefronts, an infinitesimal displacement
(dx, dy, dz) along a ray must be parallel to VL. Then
The infinitesimal arc length, ds, is related to the infinitesimal
coordinates, dx, dy and dz by
(dx)2 + (dy)2 + (dz)2 = (ds)2 . (3.30)
25
Substituting equation 3.29 into 3.30 yields
"dL"2
(dx)2 +
\dy j
dL
y\dxy J
(dx)2 +
'(sr
'''2 rdL\2
dL
dx
ydy j
*(S
V
2 y
dir
dx) )
(dx)2 = (ds)2
dL
dx
(dx)2 = (ds)2 .
Using the definition of gradient,
(3.31)
(3.32)
(3.33)
Using equation 3.21, equation 3.33 becomes
+ ( dL' 2 rdL\
< dx J [ dy , + Kd7.j
Now, substituting equation 3.34 into equation 3.32 gives
which reduces to
(3.34)
(3.35)
(3.36)
Taking the square root of equation 3.36, for the xcomponent, and
applying similar arguments for the y and zcomponents yields
26
(3.37a)
dL
dx dx
ds n
dL
dy = dy_
ds n
dL
dz dz
ds n
(3.37b)
(3.37c)
These equations represent the arc parameterizations of a ray. Taking
the integral of equation 3.37a yields
dL
x(s) = xQ + f^ds'. (3.38a)
J n
Similar integrals for equations 3.37b and 3.37c give
dL
y(s) = y0 + p^ds' (3.38b)
Jo n
dL
z(s) = ZQ + p^ds'. (3.38c)
Jo n
Equations 3.38a c give the ray parameterization curves.
Unfortunately, they require an explicit solution for L before the
trajectory can be found.
3.2.2 HamiltonJacobi Relations
If possible, a direct method for computing the ray trajectories
without solving for L first is desired. First, however, it will be
shown that motion along a ray is described by HamiltonJacobi
equations which are identical in form to the equation describing
conservative motion of a particle in analytical mechanics!
The infinitesimal arc length ds traversed in time dt is given
Cc
by ds = c(x)dt where c(ac) = . Substituting this relationship
n(ac)
into equation 3.37a gives
27
dx
dL
dx
c(i) dt n(ac)
dL
c(x)
______dx
dx
dt
dx
dt
n(Â£)
c dL
n(x)2 dx
Now, define
k = k Vi,
where, for example, the xcomponent of k is given by
and rearranging
Si = K
dx kg '
Using the eikonal equation,
k2 = k21 Vl2 = k2n2
or, equation 3.43 becomes
k = kGn.
The xcomponent of equation 3.69 is
(3.39)
(3.40)
(3.41)
(3.42a)
(3.42b)
(3.43)
(3.44)
Substituting equation 3.42b into equation 3.40 yields
dx
dt
kx. 
n(at)2 k0.
n(x)J (n(x)kQ )
(3.45)
Using the definition for C(x) and equation 3.44, equation 3.45
becomes
dx c(x)kx
dt k
(3.46)
28
Again using the definition for c(x) and equation 3.44 into equation
3.5, 00 = cok0 gives
co = n(x)[c(x)]  = c(i)k. (3.47)
n(*)
Therefore, co can be thought of as a function Cd(x, y, z, kx, ky kz ) of
both i(t) and k(t). Taking  of equation 3.47 yields
5k.
5(0
6kx 5kx
d(o
 = c(x)
5kx dkx
8(0 1
= c(x) 
[c(5)k(x)]
5 Vkx + 4 + kf
2 k.
3kx
8(0
 = c(x) .
5k. k
2 ^jk2x + k2y + kj;
(3.48)
Comparing equations 3.46 and 3.48, the xcomponent becomes
dx 8(0
dt 5kx
Similarly, for the y and zcomponents
dy _ 8(0
dt 5ky
dz 8(0
dt &K '
Next, the time derivative of kx is given,
dkx d ^L^
K
dt dt l. dx j
dkx _ , d 5L(x) k
dt dt dx
(3.49a)
(3.49b)
(3.49c)
29
dkx
dt
dk,
dt
dkx
dt
dk,
dt
= k
= k
= kn
_d_fdL
dx v dx
d_fdL\
dx l dx J
" dx + dt A rdL' ' dy + JL r 5lV
_dy ydx ; dt
dz
dt
/
= k,
VL
.dx
d (
dx
dx
dt
dx
dt
v A
+ k
dL
dx l dy
^ + k
dt k
_d_(dL'
dx l dz
dz
dt
_ dx
VL
dt)
0 dx dt
(3.50)
V7T . dx
Now, VL is parallel to  since dx is along the ray direction.
dt
Thus
d_
dx
_ dx ^ d /  vl dx ^
=
dt J dx l dt J
(3.51)
Substituting equations 3.21, 3.49a and 3.48 into equation 3.51 gives
 vi. ^
dx v dt)
= (n(x)c(x))
dx
and using the definition of c(x) gives
t(cJ
dx { dt) dx V '
d ( dx ^
VL = 0.
dx l dt)
since C0 is a constant. Thus, substituting this result into equation
3.50 gives
^ = k0Vl
dt dx \ dt
(3.52)
Substituting equation 3.46 into 3.52 gives
dkv
^ = k0VL
dt dx
d f c(x)k
(3.53)
30
Also, using equation 3.41 in equation 3.53 (recalling that k is
considered here a variable independent of x) yields
dkx i . r 8c 'j k
dt k
dkx 8c k k
dt dx k
dkx 8c k2
dt dx k
dkx _ 8c k
dt dx
dkx dt = 8 dx (ck).
But from equation 3.47,
dkx 8(0
dt dx
dky 8(0
dt dy
dkz 8(0
dt dz *
Equations 3. 49ac and 3
dx 8(0
dt 8kx
dkx 8(0
dt dx
(3.54)
(3.55a)
(3.55b)
(3.55c)
where CD = (0(x,k) plays the role of the Hamiltonian! Note that
a constant of motion along rays, i.e.
dco
 = 0.
dt
To show this,
dco 8(0 dx 8(0 dkv 8(0 dy 8(0 dky
dt 3x dt dk^ dt dy dt 3ky dt
8(0 dz 8(0 dk7
++.
dz dt dk^ dt
(3.56)
31
Using equations 3.49ac and 3.55ac in the above equation gives
d
dt
+
5 d 5 ( 5
5x dkx 5kx ^ 5x
5 d . 5(0 ( 5
5z dkz 5k, ^ 5z
5 dco 5
5y dky 5ky
5 ^
dy >
Thus,
d
 = 0.
dt
This section showed that motion along a ray is described by
HamiltonJacobi equations. The next section finds explicit solutions
for ray trajectories and the eikonal equation.
3.2.3 Finding Explicit Solutions for Ray Trajectories and the
Eikonal Equation
In order to solve the eikonal equation, consider the following.
The two forms of the eikonal equation are represented by equations
3.21 and 3.34 which are reproduced here for convenience
Vif  n2 (3.21)
f&L' 2 r dL V 2
ydx; + > + ^)=n (3.34)
Notice that in the direction parallel to VL,
VL. (3.57)
substituting equation 3.57 into equation 3.21 gives
n (3.58)
L = L0 + J n(x, y, z)ds. (3.59)
ray path
starting
at x o, y z
To compute this, we must know the ray paths. But so far, equations
3.37ac for the rays assume that we know L! Equation 3.37a gives
dL _
ds
Thus,
d L _
ds
32
dx dL
n = .
ds dx
Taking  of the above equation
ds
d ( dx
n
ds V ds
d dL
ds dx
d rdL' dx _L d
dx ds dy
n 3. 37ac : gives
dL
A (A) dx + d
dx v dx. n dy
d (Vi) Vlr
dx
n
dlA dy  d f dL
3x ) ds dzy dx
dz
ds
SL dL
dx J n dz V 3x J n
d f dx
I n
ds V ds
VI
2
2n
Using equation 3.21 in the above yields
d(n2)
d 1 ^ ds) _ dx
ds < 2n
Al o 2n dx
ds ' V ds J 2n
Thus,
(3.60a)
(3.60b)
(3.60c)
33
or, since ds = cdt
n dx
cdt vc dt
f
n
1 dx
cQ dt
d2x 2 Co. 0n
dt2 n dx
d2y _ C20 dn
dt2 n dy
d2z fi dn
dt2 n dz
dn
dx
 drc
dx
(3.61a)
(3.61b)
(3.61c)
These are the equations that will be considered in order to solve an
example in the last section.
3.3 Eikonal Equation Derivation for a Moving Fluid
This section develops the eikonal equation when a sound wave is
superimposed on a flowing fluid. Let U be the steady background
flow with a density pb in an adiabatic condition such that from
equations 3.41 and 3.42
U Vpb + pbV U = 0 (3.41)
pb U VU = Vpb. (3.42)
As in Chapter 2, let sound waves make small perturbations so that the
velocity and density become U + u' and pb + p' respectively.
Neglecting the squares of small quantities, equations 3.43 and 3.44
are developed and are reproduced here for convenience.
do'
+ pbV u' + u' Vpb + p'V U + U Vp' = 0 (3.43)
dt
and
du' _
pb + pbu' VU + p'U VU + pbU Vu'
dt
 c2Vp' 
v ^p2 J
(Vpb)p'
(3.44)
34
According to Jones (1989), let c0 be a constant typical of the
speed of sound. It can be identified with c if c does not vary with
position but otherwise might be some average of c. Write
k0 = Cd/co for harmonic vibrations of time factor e1
frequencies substitute
P' =
and
C0
ik o Â£
r0 + +
ik
\
<
ik L
mo +
m.
ik
i(Dt
(3.62)
(3.63)
in equations 2.43 and 2.44 to obtain
d_
dt
ik o Â£
+
ik
i(Dt .
e +
PbV
ik o L _  ,
e Co mo +
m.
ik,
 ik o
e c
mi
+ +
ik,
ik o L
o
\
eimt +
Vpbeimt +
r +
U V
ik o L
ik
r +
S_
ik
V Ueimt +
J
ei
(3.64)
and
35
p
ik o L
e cn
ik o L
e c
+
V
mi
ik
y
ei
i, '
m0 + +
V
ik
VUelot +
(
eik L + Â£ Â£ o + >
Pbu v
ik o L
e c.
U VUe +
Y
m0 +
ik
+
J
i(Dt
( \
c2V eikcL r2
l iko J
ydp2 j
(Vpb)ei
ik o L
r0 +
i(Dt
e
ikn
i(Dt
(3.65)
Carrying out the derivatives in equations 3.64 and 3.65 to the first
approximation (i.e. neglecting terms of order l/k0 and higher) gives
ik0c0r0 ik0c0pbm0 VL + pbc0Vm0 + cQm0 Vpb + r0VU 
ik0r0U Vi + U Vr0 =0
(3.66)
and
ik0CoPbm0 + pbc0mQ Vu + rQU VU ik0c0pbm0U VL +
pbcQU VmQ ikor0c2VL + c2Vr0 +
(d2 ^
Vpbr0 = 0.
(3.67)
Here k0C0 has been substituted for and the common terms of
e^t kL^ have been canceled. Using equation 2.41 in equations 3.66
and 3.67 and equation 2.42 in equation 3.67 yields
ik0c0rQ ik0c0pbm0 VL ikoc0 r0U VL = 0
(3.68)
and
ik0c0pbm0 ik0c0 pbm0U VL ikQr0c2 VL = 0. (3.69)
c0 c0
36
Using k0 = /c0 in equations 3.68 and 3.69 gives
i(or0 icopbm0 Vi i r0U VL = 0
(3.70)
and
1 2
ipbmQ i pbmQU VL i r0VL = 0.
(3.71)
Let M = U/cQ and a = c/co and replace them in equations 3.70 and
3.71 and remove ia> from these equations to arrive at
rQ(l M VL) pbm0 VL = 0
and
pbm0(l M VL) a2r0VL = 0.
Rearranging equation 3.73 gives
a2rQVL
Pbm (l M VL)
and putting this into equation 3.72 yields
a2V2L = (l M VL)2 .
(3.72)
(3.73)
(3.74)
(3.75)
This is the eikonal equation for rays in a moving fluid.
Now, a physical example will be presented in order to
understand more fully equation 3.75. Let sound propagate through a
constant density fluid in uniform motion UQ relative to the lab frame
given that a disturbance is launched from a source stationary in the
lab frame oscillating with angular frequency . Refer to Figure
3.3.
37
Figure 3.3. Pictorial Representation of Sound Propagating through
Constant Density Fluid in Uniform Motion
In the fluid's frame of reference, where the fluid appears to be at
rest except for the sound disturbance, a solution of the wave
equation is given by
A,e
i to't k' *
']
(3.76)
where CO', k' and x' are frequency, wavenumber and position,
respectively, measured by an observer in the fluid frame. If
k' = 3c' L the dispersion relation is fl)' = CQk'.
Position x in the lab frame is related to 5' by
ac = ac' + Ut (Galilean transformation) Thus, in lab frame
coordinates,
>F(S, t) = V1"'* *' t>1
vis, t)  0)t * *:
(3.77)
Now define k0 =
A
k', so k' = k'k. Substituting these parameters into equation 3.77
yields
Â¥(x,t)
VUrt)
iffm'+k'kult k'kxl
A0e L' J
if(u'+k'ku)t k'kil
A0e LV J
(3.88)
38
f u;
GO' 1 + k Co>
imposed in the
must be identified
lab frame
with because
is the frequency
go' =
GO
 U ^
1 + k
'o 7
(3.89)
That is, the signal in the moving frame is doppler shifted to a lower
value if U has a component in the same direction as k and to higher
value, if U has a component in the opposite direction from k. Using
the relationship, k0 = GO/co along with k' = G0'/co
GO
Using equation 3.89 for GO' and equation 3.90
~ U
1 + k
l CoJ
Substituting equations 3.89 and 3.91 into equation 3.88 yields
(
\
'P(Srt)
Comparing equations 3.76 and 3.92 gives
(3.92)
L(X)
k x
1 + k
U ^
(3.93)
Now, it will be shown that equation 3.93 satisfies equation 3.75.
First, note that U is constant in this example. Then, computing
k X in equation 3.93 gives
39
i(i) = 7^r(kxX + kyy + k^). (3.94)
1 + k
l Co
Taking the gradient of equation 3.94 yields
VL(i) = 7
1 + k
u ^
"O /
yK. j
f ~ uv
1 + k
v co y
Now, Vl; is
VL2 =
1 + k
U ^
'o y
(k:
2 + i? + i?)
(3.95)
Since k is a unit vector, kx + ky + k^ =1.
VL =
 U ^
1 + k
co y
Next, consider the righthand side of equation 3.75
f V
(l M VL)2 =
ii
u A
1 + k
'o y y
(l M Vl)2 =
\2
1 
f u ^
1 + k
V C ) )
(3.96)
40
f
(l M Vi)2
V
1 + A k U U  k
c0 c0
f 1 A + k s o
V o o
(l M Vi)2
(3.97)
Comparing equations 3.96 and 3.97 shows that they are indeed equal.
3.4 Example
As an example, consider
n = nQ az. (3.98)
First look at
n = nQ az (3.99)
where z > 0 in equation 3.98. Equations 3.60ac for this example
become
f n 1 =
ds v ds )
ALdyl =
ds V ds )
=
ds V, ds )
Assume that this example is only in the xz plane which eliminates
equation 3.100b from consideration (i.e., no deflection in the y
Taking the integral of equation 3.100a gives
Cx . (3.101)
plane).
dx
n
ds
0 (3.100a)
0 (3.100b)
a. (3.100c)
41
Substituting equation 3.99 into equation 3.101 for n yields
/ \ dx
lno az) = Cx
ds
(n0 az)dx = Cxds. (
Squaring equation 3.102 gives
(nQ az)2dx2 = C2ds2 . (
Using the definition for ds (equation 3.30) in equation 3.102
(nQ az)2dx2 = C2(dx2 + dz2) (
and collecting terms, equation 3.104 becomes
[(n0 az)2 C2 ]dx2 = C2dz2 . (
Taking the square root of equation 3.105 gives
dx(J(n0 az)2 C2 j = Cxdz (
and solving equation 3.106 for dx yields
dx =
dx =
Cxdz
V(no ~ az)2 Cx
dz
n0 az
\2
V
 1
Expanding the squared term in equation 3.107 gives
dz
dx =
"a*
V CX J
2
Z
^ 2 an ^
c
\
2. +
/ 2 \
^1
vc2 j
Integral 162 (Korn and Korn 1961) is
3.102)
3.103)
yields
3.104)
3.105)
3.106)
3.107)
3.108)
42
X
162.
= 1
dx
Jl
ax' + bx + e
log(2ax + b + 2Jaylax2 + bx + e)
where a > 0. Using this integral to solve equation 3.108, evaluating
between 0 and z, gives
X = XQ
log
a
(
\
f V
a
VCx/
2an
z 
2a
raz nc
V
 1
cx , f f2an. 1 2a ro 2 1
log +  1
a 1 l c J Cx 11 l Cx J j
x = x0
Cx
log
a
2a
c
a
+ ^ log
a
v
2a
Cv
z 
n
+
c* K
az n
 1
+
L V
n0
vcy
 l
X = x0
log
a
f
\
a
c
n \( az nn 1 2 cx f 2a 1
z + .  1 log
Cx \ l cx J J a 1 cx )
+ log
a
f
\
n
+
V )
 1
+ ^ log
a
r
2a
vCx
43
X
= x0
Cx
log
a
a n 1 [}
Z + V(az nc) 
+ log
a
\
X = x
log^az n0 + az n0 f Cx j
 Cx
+
a
log(nG + y/no2 C2X )
x = xQ +
(
log
a
no + Vno2 C*
az
 nG + az nQ )2 Cx
To find Cx suppose that at z = ZQ, z = 0
dx
dz
= tan0o .
(3.109)
(3.110)
Also let x = X0 at z = 0 Using this assumption with equations
3.107 and 3.110 gives
tan0o
1
tan0o
1
tan20o
r \2
no
v
c
X
1
 1
V
V ^X J
/ V
v cx y
tan 0,
 1
1
1
44
/ V
Ik
< >
f \2
Ik
V Cx y
y '\2
Ik
/ V
Ik
v cx y
1 + tan20o
tan20o
i + sln20
cos20
sin 0O
COS20o
cos20o + sin20o
sin20
sin 0
C.
sin0o
and finally,
Cx = nosin0o .
Substituting equation 3.111 into equation 3.109 yields
X = x +
f
a
n0 + 7n02 n2sin20c
\
az nG + az n0 f n2sin20c
x = x +
^i^iog
a
nQ + nQ ^/l sin20o
az n + n
^azV
v no y
2az
+ 1
 sin 0,
.111)
45
X
= X0 +
a
COS0o 1
17 \
az . az
1 + J  1
n0 V n0 J
 sin20
(3.112)
Notice that at z = , equation 3.99 becomes n = 0 which is
a
unlikely. The assumption is, that before this z is reached, the
medium with varying index comes to an end. The value of z where the
ray turns over is calculated next. Substituting equation 3.111 into
equation 3.107 gives
dx
dz
f \2
n0 az
V nosin0o ,
 1
dx + l
dz (n0 az)2 n2sin20o
i n2sin20o
dx = + sin0o
dz ( \ 2 '
2az az
1 +
1 no 1 n J
dz
dx
V
az
no J
sin20
sin0.
(3.113)
(3.114)
dz
The ray turns over when 
dx
0 and equation 3.114 becomes
1
az
n0
sin0o
46
= 1 sin0c
n
= (l sin0o ).
a
Conversely, if the medium extends to z = h, there is a critical
angle 0crit above which the ray is trapped in the medium. That
angle is given by
ah
sin0crit = 1 
n
here
Now it is time to verify equation 3.100c which is reproduced
d ( dz>
n
ds V ds
a.
First note that x is a function of z. Thus
. dx
dx = dz
dz
and substituting this into equation 3.30 gives
fdxY
ds =
ds =
V dz /
1 +
dz2 + dz2
dx
V
dz j
dz
ds = dzf+ if
dz
ds
n + i^
dz )
(3.115)
Substituting equation 3.115 into equation 3.113 gives
47
1
Multiplying both sides by n and using equation 3.99 yields
dz
n
ds
dz If
n = nc J
ds t
dz If
n = \\no
ds
1^
n
 sin20
/
= V(no az)2 noSin20c
Let equation 3.116 be defined as f(z). Then,
f(z) =
ds ds
and using this relationship in equation 3.116 gives
(3.116)
48
_d_
ds
dz
n
ds
2a(n0 az)
az)2 n2sin20o
az)2 n2sin20o
(no ~ az)
d [ dz ]
I n I = a
ds V ds i
which is exactly equation 3.100c which is what was to be verified.
Equation 3.114 is the slope of the direction of the ray at the
point (x, z). Rearranging, it becomes
dz = + (n0 az)2 .
dx ]] n2sin20o
It can be integrated to
x = x
dz
(n0 az)2
n2sin20o
 1
where n2 = (n0 az)2 and C2 =
2 2r\
n0sin 0O
(3.117)
and XQ is an arbitrary
2
constant. Regard the Cx as fixed, and the X0 as variable; then
equation 3.117 gives an infinite set of curves which satisfy the
definition of rays. If n is a constant, that is, if the sound
velocity is independent of depth, the rays of 3.117 are clearly
straight lines (Frank, Bergmann and Yaspin).
The eikonal for a shear flow is represented by equation 3.93
which is
L(5)
(3.118)
Using two different arguments, it will be shown that this form of the
eikonal equation leads to the generalized form of Snell's law. First
consider Figure 3.4 which shows the propagation of a wavefront in
shear flows with different background speeds and different refractive
indices.
49
Figure 3.4. Propagation of Waves in Shear Flows of
Different Refractive Indices
Let
P2 = Poe
 U
^2f
(k2s)
Hi + t.i
s(M)
Pi = Poe
Referring to Figure 3.4,
U2 k2 = U2 k2x = U2sin02
Ui ki = U, klx = UiSinGi
At the boundary,
i (Ot 
e L
or
(v*)
1 (Ot
e L
(3.119)
(3.120)
50
(3.121)
k2x = klx
C2 _j_ ^2^2x C1 _j_ ^Ax
C0 CQ CQ CG
Substituting equations 3.119 and 3.120 into equation 3.121 yields
sin09 sin,
 = , (3.122)
C2 U2sin02 cx UiSin!
Co CQ CG CG
which is the generalized form of Snell's law. If U2 = Ux equation
3.122 becomes
sin02 c2
sin! cy
which is the more common form of Snell's law.
Now consider Figure 3.5 which shows the distances between the
wavefronts for the same conditions as Figure 3.4.
Figure 3.5. Distance Between Wavefronts in Shear Flows of
Different Refractive Indices
51
Let T be the period. Then, a =
d =
c2T
, b =
U,T
and
Co C0
C2k2xT U2T Q C2k2zT
V Co
C1T K
a =  b
'0
UXT
C0 J
(
and d =
Similarly for the bottom shear flow,
\
c:klxT UiT ^ CiklzT
+ , 0, lz
V C0
y
between wavefronts is
c2k2xT U,T
, 0,
c2k2xT
(^2x t ^2y r ^2z )
y
c2k2xT U2k2xT ^ c2k2zT
The distance
or
c2T + U2 kT
c0 cG
Similarly,
CiT Ui kT
+ 
Co C0
Using the sin law
sin02
C.T + U2 kT
c0 cQ
, equations 3.123 and 3.124 become
sinBt
CiT U. kT
+ 
C0 C0
(3.123)
(3.124)
and using equations 3.119 and 3.120, the above equation gives
sin02 sin0x
c2 U2sin02 cx UiSin!
c0 c0 C0 c0
which is the same generalized Snell's law found in equation 3.122.
In conclusion, this chapter derived the eikonal equation,
equation 3.21, which defines the surfaces of constant phase, i.e. the
wavefronts. It also determined how to compute the eikonal, L, for
52
some specific examples. The next chapter will utilize the theory
presented in chapters 2 and 3 in a practical application.
53
4. Applications
The purpose of this section is to provide a practical example of
the theory developed earlier. The shear flow problem is in two
parts. The first part derives the solution to a problem stated in
Jones (1989). The second part is an extension to the text problem
which involves predicting the distance sound travels for several
angles. This preparation leads into the experimental predictions of
Chapter 5.
4.1 Shear Flow Problem
4.1.1 Part IA
The first portion of this problem is restated here from Jones
(1989) (p. 381 #33):
In a simple model of a shear layer the only nonzero component
of U is U(z) along the xaxis. U(z) is zero for z < 0, increases
linearly to Uo between z = 0 and z = h (> 0) and has the constant
value Uo for z > h. The speed of sound is constant throughout. A
ray starts from (0, 0, d) at an angle ft to the horizontal. Show
that it passes through the layer if cosft < (1 + M0)_1, and that the
difference in abscissas of its points of entry to and exit from the
later is
/
k
h
2M0
sec01[tan01 (l + Mocos01)tan02] +
f
V
h
2M
\
secOL^os! (cosh ^ec! cosh 1
where sec ft = sec ft Mo.
Show that if z > h
(
L = dcosec! 
(tan02
2M0 )
A graphical representation
 tan0i) + (z h)cosec02
of the shear is shown in Figure 4.1.
54
Solution:
A graphical representation of the problem is shown in Figure 4.2.
Using equation 20.4 (Jones, 1989)
dx
M,
\
 = COS01 + s
dz v a )
sec0.
(4.1)
55
where for this problem
a = 1
M
x c
sec03 =  = Jr = .
cos03 cos(9O0!J sin!
Substituting the above into equation 4.1 gives
dx ( U0 ^ 1
 = cos! + .
dz V, c ) sm0j
Now, using equation 20.8 (Jones, 1989)
(Ci + C2) /,2acosec03 + CjM* = 1
where for this problem
a = 1
m =a
lJX
C
c2 = 0
cosec03 =  = c = .
sin03 sin(90 0X) cos61
Substituting the above into equation 4.3 gives
Cl
' 1 '
+ cx
COS0! j
Initially, let
2 I = 1
V C
0! = 0!!
UG = 0.
Substituting these initial conditions into equation 4.
f 1 ^
1 COS0!
C! = COS0! .
= 1
As the ray advances into the moving fluid, equation 4,
(4.2)
(4.3)
(4.4)
4
4 becomes
' 56
COS0n r i U1
^COS! CJ
1 U 1
COS0! c cos01:]
1 1
COS0! COS0U c
sec0! = sec!! U c
COS0j 1
= 1
sec0u 
(4.5)
The ray will emerge if 0lf > 0 or COS0lf < 1 .
emerging ray, equation 4.5 yields
Then, for the
< 1
sec0u
c
sec0u > 1
c
1 u
> 1 +
COS0
li
COS0U < 1 +
1
or, in the terminology of Jones (1989),
COS01 < (l + M0 ) 1.
For the emerging ray, equation 4.2 becomes
dx
dz
U
COS0! H
c ) sin!
Substituting the trigonometric identity
sin! = yl COS0!
and equation 4.5 into equation 4.6 gives
(4.6)
57
dx
dz
sec0u U/c
U
+
c
V1 (sec0u U/c)
(4.7)
Substituting U = 0 + UQ into equation 4.7 yields
dx
dz
sec0u
" (uo/c)(z/h)
* frit)
Jl ~ [sec0n (uo/c)(z/h)] 2
dx
(uo/c)(^h)
dZ ^[secG^ (uo/c)(z/h)]2 1 ^1 [secG^ (uo/c)(z/h)]
dx
dZ ^[secGu (u0/c)(z/h)]2 1
(uo/c)(z/h)
y[secGu (uo/c)(z/h)]2 ^[secGu (uo/c)(z/h)]2 1
dx _______________1______________ (uo/c)(z/h)[secGli (uo/c)(z/h)]
dz
dx =
Ax =
^[secGu (uo/c)(z/h)]2 1 ^[secGu (uo/c)(z/h)]2 1
(uo/c)(l/h)_________ (Uo/c)2(z/h2)[sec8li (uo/c)(z/h)]
U0/c
^[secGu (uo/c)(z/h)]2 1 ^[secGu (uo/c)(z/h)] 1
dz
h 1 U fh
Uo. r
ih Jo
dz
U/C) ch ^[secGn (uo/c)(z/h)]2 1
fTsecG^ [h . ZdZ
^[secGu (uo/c)(z/h)]2 1
' h '
(4.8)
V Uo/C J
< h '
V Uo/c J
Tr
ch )
z2dz
^[secGu (uo/c)(z/h)]2 1
In order to make some simplifications, expand
58
[sec01 (uo/c)(z/h)]2 1 =
1 z2 2sec0lifIz + sec2,* 1
\ch ) \ ch )
Use the following trigonometric identity in the above equation
2 2
sec 0U 1 = tan 0U
to get
[sec0u (uo/c)(z/h)]2 1 = z2 2sec01^jz + tan20u.
Substituting this into equation 4.8
Ax =
Uo/c
Uo fh
Jc
dz
ch
^  z2 2sec0
fe)
z + tan 0U
f h '
yVo/cj
( v. A
*0 sec0u Jh_5==
ch ) f U
zdz
ch
 1 2 2sec0u z + tan2^
ch.
li
U0/c
z2dz
z2 2sec01:L[ ^ ]z + tan20u
(4.9)
Let
a =
Uo
V. ch J
b = 2sec0u[
V ch
e = tan2 0U .
Again, to further simplify, substitute the above into equation 4.9
Ax
dz
Vaz2 + bz + e
rh
asec0u J
zdz
z2dz
Vaz2 + bz + e Vaz2 + bz + e j
(4.10)
From Korn and Korn (1961) the following are the pertinent integrals
for this problem
162
I
dx
Vax2 + bx + c VÂ£
= i=log(2ax + b + 2VaVax2 + bx + c) for a >
59
164
I
I
xdx
V
ax + bx + c
ax + bx + c , 
a 2 a Vax2 + bx + c
I
dx
V
165
n , n1
x dx x /5
= = Vax + bx + c 
ax2 + bx + c sn
nl j
x dx
b(2n l) _
2 an Vax2 + bx + c
c(n l)
I
J
xn_2dx
an Vax2 + bx + c
Using integral 164 and n = 2, 165 becomes
f x2dx x /2; 3b p
, = Vax + bx + c
J V,2 j. ^ ia J
xdx
ax2 + bx + c 2a 4 a Vax2 + bx + c
f
2a J Vs
dx
/ax + bx + c
Now, substituting the appropriate equations into equation 4.10 gives
Ax
dz
bz + e
1
Vi
asecB
ii
1 /2Z b fh
Vaz + bz + e
a 2a
dz
2a JV
az2 + bz + e
i= aVa IVaz2 + bz + e + j= aVafVlVaz2 + bz
Vi UaJ Vi UaAaJ
+ e
VTwli7JJ^fbzTe +
= 4[^r, /2 V
Va ^ Vaz2 + bz + e J
i (sec0uVaz2 + bz + e) sec0n f
Va Va ^ 2 JoV
A (4*
Vaz2 + bz + e
dz
az2 + bz + e
az2 + bz + e + 7= {jVaz2 + bz + e 
Vi V. 4>/i
1 (3b2 "j fh dz  1 fcVil fh dz
Vi ^8Vi A0 Vaz2 + bz + e Vi l 2 J ^Vaz2 + bz + e
60
Ax =
Va
fr b a 3fa2 cVI^
VasecGw
V
8 VI
+
dz
/
>V
+
az + bz + e
1
VI
VI
2
3b
secG, z + ==
4 VI
M
az + bz + e
Ax =
VII
VIsecG,, 
3b2 + cjV\ *
8 VI
VI
1
VI
log(2az + b + 2i/a V az2 + bz + e) +
r vi ^
l 2
3b
4VI
V
az + bz + e
a If, b A 3b2 clx
Ax = 7= 1j= sec6h  *
VI v 2VI 8a 2
log(2az + b + 2>/IVaz2 + bz + e) +
if. VI A 3b
t=  sec6uz +
V
az + bz + e.
VI 2 WI,
Evaluating the above between 0 and h yields
a If, b n 3b c 1 *
Ax = j= 1j= secG,.+ *
Va ^ 2VI 8a 2)
log(2ah + b + 2VIVah2 + bh + e) 
3; 1 53 sec6 17 + f *log(b + +
1
VI
Q VI 3b
secGnih +
\
v
1 (
VI
2 4 VI
3b '
Vah2 + bh + e 
secG1:L + VI.
4Va J
Now, substituting back in for a, b and e
61
Ax =
(do /c)
1 + sec2 Q1sec'10^ +
2 2
log
h
2h
:2 0!, + tan2 0n
2
ft)' >*ft) ft)
( . 2 3 2
(D/c)
log 2sec0u. . ,
V v ch y vch /
h r c9u
2 0n sec2 0U + ^ tan2 0n j *
(D0/c)
h
V ch J
1 + sec 0,,
2
^0 + 2
ch J
seo9 (i) ,
3 ^
sec0u sec^ ItanO^.
 2sec0u^j + tan20:
n
^)tan9) +
'2sec9,i(
+ tan2 0U 
(0./c)(2
To simplify this expression, we note that
 sec2 0U = (tan20u + l)
2 2 '
2 2
2 0h + ^ tan2 0U = 1
1 2 ^ 1 1 2
tan 0U+ tan 0ti
2 2
1  sec2 0U _______
2 1 2 2
1 2 r. 1 , 2 n 1
1sec 0,, Htan 015 = .
2 2 2
Adding the above simplifications and expanding the log function gives
11
(d0/c)
(D0/c) U
h
log f1 sec01 + J ruoi  sec0u 1 i1 1 M
/ l c J \L v c ;
, h; v fllog 2 ] [llog(tan01 sec0u)
(uo/c) v2y yL UhJJ (uo/c) v2y
(1)[(^)+ sec9]I(^)'seo9] '1 +
h
(do/c)
^(i).ecelitneu.
From Korn and Korn (1961)
62
cosh 1 z = log(z + Vz2 l).
Letting z =    sec0li the Ax equation becomes
Ax =
h
cosh
 sec!
2(uo/c)
2(u0/c)
log(tan0u secB^) 
2(0o/c) .
, ^r secB^tanG^.
2(u0/c)
sec.
 1 +
From an earlier relationship, sec! = sec0li we substitute
a
this into the Ax equation to obtain
Ax =
/ h > cosh 1 rsec0,l, k cosh 1 (sec,,) 
2(uo/c) L 1J 2(uo/c)
U
/ . v I + sec0u
2(0o/c)LU 1
tan, + r sec0,4tan0n
1 2(uo/c) 11 11
Ax = rr sec0li^cos011 cosh 1 (secGj) cosG^ cosh 1 (sec0u)j
2\U0/C/
?rc sec0n
2(u0/c)
cos0n + 1
tan0j +
Ax = r sec,, \ tan,, 
2(uo/c) 111
1 +
2(U0/c)
jcosQu tan!
secli(tan0li)
(4.11)
, ^ > sec9li{cos911^cosh~1 (sec!) cosh1 (sec li ) ] }
Wc]
This is the solution for the difference in abscissae of the points of
entry to and exit from the layer with 0ljL = 0j and 0i = 02 .
4.1.2 Part IB
The last part of the first portion of this problem is to show that
if z > h
63
L(x) =
1 + COS0,
(xcos2 + zsin2)
(4.12)
where
Z = z h
and
COS02 +
x =
tan!
+ Ax +
sin0,
(z h).
Substituting these parameters and equation 4.11 into equation 4.12
gives
(z h)sin0,
L(x) = zr11 +
COS0,
1 + COS02 1 + COS02
f d N
tan!
c0
COS0,
1 + cos2 l2(u0/c)
C0
COS02 h
1 + COS02 [2(u0/c)
Co
cos2 h
1 + COS02 [2(u0/c)
Co
COS2 h
1 + COS02 [2(u0/c)
C0
COS0, +
cos2
1 + COS02 sin2
Co 
e that
sec!
tan0x 
sec!
sec!
/
>
\
I
\
*Â£)
cos.
tan2 > +
sec0x
[cosiCosh 1 (sec2)j 
[cosiCosh 1 (sec!)j +
Uc
Cn
(z h)
(4.13)
64
COS02
1 + COS02
Co
COS02
1 + COS02
C0
COS02
1 + COS02
C0
COS02 1
1 + COS02
C0
COS0!
1 COS01
Cp
Uo f COS0L
C0 1 COS0
o o
COS0J
1 COS01
Co
1 COS0j Co u ^ t COS! Co
1 COS0,
C0 )
A
COS0n ( u. \
1 COS0.
1 COS0. V o J
o o
COS0!
(4.14)
where the given relationship sec02 = sec! has been used.
Co
Substituting equation 4.14 into equation 4.13 yields
65
L(x)
L{x)
(z h)sin02
1 + COS02
C0
(
+ COS!
f d ^
tan0x
COS0!
COS0!
COS0!
2(u0/c)
h
2(u0/c)
h
sec0!
tan0! 
\ fr
sec0!
\
1 + COS0!
tan09
sec0!
COS0a
COS0!
2(u jc} oc^! j[cosiCOsh~1 (sec02)] 
y sec! jcos0icosh 1 (sec0j)j +
cos02 +
sin0,
(z h)
= (z h)
sin0,
COS0i
COS02 +
Co J
1 + COS0,
COS0!
( u
V
dcos0i
sin0!
N (
+
sin0,
tan0: 
2(u0/c)
h
2(uo/c)
h
2(u0/c)
U
y >L
\
tan02 f +
2(u0/c)
1 +  ^COS0!
[cos0!cosh 1 (sec02)j 
[cosO^osh 1 (sec0!)j
66
L(x) = (z h)
sin,
COS0,
cos02 +
U
+
1 + COS0.
'o
(
c 2 sin02
1 + COS0,
dcosecO^os2! +
' h '
2(uo/c)
2(u0/c)
2(uo/c)
h
tan! 
2(uo/c)
L(x) = (z h)
tan0,  cos!tan2 +
cos01log(sec02 + y/sec202 l)j
cos01log(sec01 + Jsec2! l)j
sin202 + cos202 + cos02
sin0,
1 + COS0,
V Co
+
dcosec01(l sin2!) +
]cos01tan02 +
f h ^
2(u0/c)
(tan! tan02) 
[cos0!log(sec02 + tan02)] 
[cos0ilog(sec0! + tani)j
67
L(x) = dcosec! dsin! + (z h)cosec02 +
(tan0x tan02) ^^cosltan2 +
[cos01log(sec02 + tan 02 + 1 l)] 
[cos01log(sec01 + tan 02 + 1 l)]
L(x) = dcosecGi dsinGi + (z h)cosec02 +
(tan! tan02) ^^cosltan2 +
[cos01log(l + tan 02 )log(sec02 l)] 
[cos01log(l + tan 0j )log(sec01 l)].
Using the power series expansion for the log function (Korn and Korn
1961)
ig(i + z) = z_t + t_T "'
2 3 4
in equation 4.15 in linear terms only gives
L(x) = dcosec! dsin! + (z h)cosec02 +
(tan0a tan02) ^^cos01tan02 
(4.15)
[cosGztanGzSecGz ] +
[cos! tan GjsecG!]
68
L(x) = dcosec! dsin0! + (z h)cosec02 +
(tan0x tan02) ^^cosltan2 
cos01tan02 ^sec! ^
tan 0X
L(x) = dcosec! dsin0! + (z h)cosec02 +
(tan! tan02) ^^cosO^an
[cosOjtanO^ec! ] +
tan 0l
< h ^
2(U/c)
V c
cos01tan02
+
L(x) = dcosec0x dsin0x + (z h)cosec02 +
' h '
2(uo/c)
(tan0x tan02) +
' h '
2(u0/c)
(tan 0X tan02 )
[']cos01tan02 + Icos0itan02
\2) K2) 1 2
L(x) = dcosec! r (tan02 tan0x) + (z h)cosec02
(u0/c;
dsin!.
This is the result from above except for the dsin0x.
This section solved the problem in Jones (1989). The next
section will expand on the result of the first section.
69
4.1.3
Part II
The second portion of this problem is to find the .equation
which determines the distance from the point of entry to the point of
exit of the ray trajectory. The graphical representation of this
portion is shown in Figure 4.3.
Figure 4.3. Graphical Representation of Part 2 of Problem
The solution to the problem from Jones (1989), equation 4.11,
gives the distance from a sound source that penetrates the layer
(refer to Figure 4.2). The angle passed through the layer if
COS0! < .
1 +
c0
Therefore, in order for the ray to travel in a parabolic path in the
medium,
COS0! >
1 +
u
The distance the ray travels is give by
D =
tanGL
+
t7 secG, ftanG, cosG.cosh 1(sec01 )1
(uo/c) L . J
(4.16)
70
This is the solution for the second part of the problem.
This chapter has explored some applications to theory which
will be useful in the proposed laboratory experiment in the following
chapter.
71
5. Proposed Experiment
This last chapter is devoted to proposing a laboratory
experiment to predict the distance of the sound propagation for a
simple model of a shear layer as used in Chapter 4. Figure 5.1
depicts this linear shear flow.
The propagation of the ray in the medium is shown in Figure
5.2.
72
Figure 5.2. Graphical Representation of Ray Trajectory
The equation to be used is from Chapter 4, equation 4.16, which
D = + tr secOj^ftan! cos01cosh'1(sec01)l (5.1)
tan! (U0/c) L J
The parameters for this equation are as follows
d = 0.2 m
h = 0.4 m
U0 = 5 m/s
c0 = 343.2 m/s (Speed of sound in Denver, Colorado).
0i is determined from the critical angle calculation
COS0J
(5.2)
Substituting the appropriate parameters into equation 5.2 gives
COS 01 = 
1 + :
343. 2
1
COS 01 = 
1.014568764
73
cos 01 = 0.985640436
0i = 9.721411176 degrees.
Now, substituting all parameters into equation 5.1 yields
D =
D =
0. 2m
+
0. 4m
sec(9. 721411176) *
tan(9. 721411176) (5/343.2)
{tan(9. 721411176) cos(9. 721411176)cosh_1[sec(9. 721411176)]}
 (1. 014568765) *
0.171317769 (0.014568764)
[0.171317769 (0. 985640436)cosh_1(l. 014568765)]
D = 1.167421226m + (27. 45600107m)(l. 014568765) *
[0.171317769 (0.985640436)(0.170490627)]
D = 1.167421226m + (27. 8560011m) *(0.171317769 0.168042455)
D = 1.167421226m + (27.8560011m) *(0.003275313)
D = 1.167421226m + 0. 091237124m
D = 1.25865835m
D = 1.26m
Another set of parameters for equation 5.1 are
d = 0.2 m
h = 0.4 m
U0 = 10 m/s
c0 = 343.2 m/s (Speed of sound in Denver, Colorado).
Again, 0i is determined from the critical angle calculation using
equation 5.2. Substituting the appropriate parameters gives
COS 01 =
1 +
10
343. 2
1
cos 0i = 
1.029137529
cos 01 = 0.971687429
0i = 13.66650012 degrees.
Now, substituting these parameters into equation 5.1 yields
D
______0. 2m______
tan(l3.66650012)
{tan(l3.66650012)
+ . '4in sec(l3. 66650012) *
(10/343. 2)
 cos(l3. 66650012)cosh1 [sec(13. 66650012)]}
74
0. 2m
0. 4m
D =  +  (1. 029137529) *
0.243154382 (0.029137529)
[0.243154382 (0.971687429)cosh'1(l. 029137529
D = 0. 822522704m + (13.72800006m)(l. 0129137529)
)]
[0.243154382 (0.971687429)(0. 240819928)]
D = 0. 822522704m + (13. 90528005m) *(0.243154382 0. 234001696)
D = 0.822522704m + (13. 90528005m) (0. 009152685)
D = 0.822522704m + 0.127270652m
D = 0.949793356m
D = 0.95m
These theoretical results are to be compared in a future
publication to the actual experimental results. Sound propagation
has useful applications in various fields of study. The parabolic
approximation in this form has been applied to study many aspects of
underwater acoustic propagation (McDaniel, 1979.) Other areas of
interest are vorticity problems and chaotic problems to name a few.
75
Appendix
A. Further Derivations
Conservation of Mass Equation
For a fixed volume V inside a fluid, the rest mass in V at any time t
can be taken as the volume integral of density p(ac, t) representing
a local average (or expected value) of mass per unit volume in the
vicinity of a spatial point ac Conservation of mass requires the
time rate of change of the mass to equal the net mass per unit time
entering (minus that leaving) the volume V through the confining
surface S. The net mass per unit time leaving through a small area
element AS with outward unit normal vector n (acs ) and centered at
point on S is identified acs as
p(acs t)v(acs, t) iiAS . (A.11)
v(x, t) is the fluid velocity at ac, defined as the massweighted
local average particle velocity or as the average momentum per unit
mass in the vicinity of ac. (acs refers to a point on the surface.)
The net mass leaving V per unit time is the surface integral over S
of pv n, so the conservation of mass requires
^Ilfv PdV = If. P*'*dS
(A.12)
The surface integral can be written as a volume integral by using
Gauss' theorem
J A ndS = JJJv V AdV
E
dt
 V (pv)
dV = 0.
(A.13)
(A.14)
This equation implies that the average value of the integrand is zero
for an arbitrary volume V, so the integrand itself must be zero
 V (pv) = 0.
dt
(A.15)
76
Conservation of Momentum Equation:
A general law of classical continuum mechanics is that the mass times
acceleration of center of mass of a fluid particle equals the net
apparent force exerted on it by its environment and by external
bodies. A fluid particle consists of all fluid within some moving
volume V (t) each point on the surface of which is moving with the
local fluid velocity v(ig,t). Since the mass in such a fluid
particle is constant, mass times center of mass acceleration is the
time rate of change of momentum (volume integral of pv) within the
particle, so
Â£ JIfv P*dV = IL dS + JJJv dv W2'11
fs is apparent surface force per unit area exerted by the particle's
immediate environment. fB is the body force, e.g., that due to
gravity per unit volume. Although gravity is always present, it has
negligible influence on acoustic disturbances at all but extremely
low frequencies (i.e. those of order or less than g/c.) Therefore,
the body force term is neglected. The classical assumption regarding
fs is that it is directed normal into the surface S, that is
fs = np (A. 22)
with the magnitude p of this force per unit area identified as the
pressure. The adoption of this relation, holding ideally for static
equilibrium (hydrostatics), implies a neglect of viscosity. The lack
of dependence of the pressure p(x,t) on the orientation of AS, that
is, the direction of n, may be regarded as a hypothesis but also
follows from a fundamental requirement that the net surface force
divided by the mass of a fluid particle on which it acts should
remain finite in the limit as the particle volume goes to zero. That
fs reverses direction when ii reverses direction is consistent with
Newton's third law.
JJs fgdA = JJs npdS = JJJv VpdV. (A. 23)
Vp is the equivalent force per unit volume due to pressure.
The time rate of change of momentum term in the first equation
(A.21) can be expressed as a volume integral
dv dv dv dx dv dy dv dz
dt dt dx dt dy dt dz dt
77
(A.24)
dv dv .
= + (v V)v
dt at
dv Dv
dt Dt
where
D 9  V7
= + v V.
Dt at
(A.25)
Thus
Jif JJJV P^dv
dt v JJJV Dt
which represents an instance of Reynolds transport theorem.
Inserting equations A.23 and A.26 into equation A.21 with a
neglect of the body force term gives
(A.26)
E (p^ + Vp)dv = o.
Since the integrand is zero
(A.27)
DV V7
ph Vp = 0
Dt
~ + (v V)
+ Vp = 0.
(A.28)
PressureDensity Relationship
The classical model of a compressible fluid presumes the existence of
some definite relation
P = P(p)
between density and pressure.
Employing a Taylorseries expansion in p'
p(p) = p(Po) + (p Po ) ~ (Po ) + ^ P^
dp 2! dp
(A.31)
(A.32)
78
(A.33)
P =
Po
+ (p Po)^(po)
dp
a V Po /
dp
c2p'.
(A.34)
(A.35)
The following are selected problems from Jones (1989)
28. Prove for a ray that
a) dL/ds = N
b) d(gradL)/ds = gradN
a) dL/ds = (gradL) s
dL/ds = (gradL) (l/N)(gradL) (Jones 1989, Eqn. 16.1)
dL/ds = (l/N)(gradL) (gradL)
dL/ds = (l/N)( gradL)2
dL/ds = (l/N)N2 (Jones 1989, Eqn. 15.4)
dL/ds = N
b) d(8L/8x)/ds = grad(5L/8x) s
d(8L/8x)/ds = grad(8L/8x) (l/N)(gradL)
(Jones 1989 Eqn. 16.1)
d(8L/8x)/ds = (l/N)grad(SL/8x) (gradL)
d(8L/8x)/ds = (i/n) 8/8x grad(L) (gradL)
d(8L/8x)/ds = (l/N)grad(L) 8/8x (gradL)
8/8x (gradL)2 = 2(gradL) 8/8x (gradL)
d(8L/8x)/ds = (l/N)(l/2) 8/8x (gradL)2
d(8L/8x)/ds = (l/N)(l/2) 8/8x N2 (Jones 1989 Eqn. 15.4)
8/8x N2 = 2N 8/8x N
d(8L/8x)/ds = (l/N)(l/2)(2N 8/Sx n)
79
d(5L/Sx)/ds = 8/8x N
Using similar arguments for 8/8y and 8/8z we find
d(8L/8x)/ds = 8/8y N
d(8L/8x)/ds = 8/8z N
Putting them together we have
d(gradL)/ds = gradN
29. If x is a point of a ray and K = kogradL prove that
a) dx/ds = K/k0N
b) dx/ds = k0gradN
a) dx/ds = gradx s
dx/ds = (l/N)(gradL) (Jones 1989 Eqn. 15.4)
dx/ds = x/k0N
b) dx/ds = d (k0gradL)/ds
dx/ds = kQd (gradL)/ds
dx/ds = kogradN (Problem 28b see above)
32. If X = kogradL for a ray in a moving fluid and
dx = (l M gradL)da show that
dx/dx = M + aic/x
What is the corresponding formula for dx/dx ?
a) dx/da = a2gradL + (l M gradL)m
(Jones 1989 Eqn. 19.13)
dx/dx = (dx/da )(da/dx)
dx/dx = [a2gradL + (l M gradL)M][l/(l M gradL)]
dx/dx = a2gradL/(l M gradL) + M
dx/dx = a2gradL/agradL + M (Jones 1989 Eqn. 19.11)
80
dx/dx = cxk/k + M
b)
dic/da = d(k0gradL)/da
dK/do = kQ d( gradL )/da
dK/dx = (dK/dcy)(da/dx)
dK/dx = k0 [d(gradL)/da][l/(l M gradL)]
From Jones (1989), equation 19.14 is
dx/dx = k0[agrad2Lgrada (l id gradL){grad(M gradL) (id grad)gradL}] *
[]/(1 M gradL)]
dK/dx = k0[(agrad2Lgrada)/(l M gradL) {grad(M gradL) (m grad)gradL}]
From Jones (1989), equation 19.11 is
dK/dx = kQ[(agrad2Lgrada)/(a]gradL) {grad(i4 gradL) (m grad)gradL}]
dK/dx = k0[(gradLgrada)(k/K) {grad(M gradL) (m grad)gradL}]
From part (a)
dK/dX = kQ[(V a)(gradLgrada)(dx/dx id) grad(M gradL) (id grad)gradL}]
81
References
Dowling AP and Ffowcs Williams JE. 1983. Sound and Sources of
Sound. New York: John Wiley & Sons.
Frank PG, Bergmann PG, and Yaspan A. n.d. Ray Acoustics, Summary
Tech Rept. Division 6, NDRC, 8, 41 49, 51 54, 59 68.
Jones DS. 1989. Acoustic and Electromagnetic Waves. New York:
Oxford University Press.
Korn GA and Korn TM. 1961. Mathematical Handbook for Scientists and
Engineers. New York: McGrawHill Book Company, Inc.
McDaniel, ST. 1979 Jan. Application of the parabolic approximation
to predict acoustical propagation in the ocean. Am J Phys; 47(1).
Mitton Von Dyke 1982. An Album of Fluid Motion. Stanford: Parabolic
Press.
Pierce AD. 1981. Acoustics An Introduction to Its Physical
Principles and Applications. New York: McGrawHill Book Company.
Tritton DJ. 1977. Physical Fluid Dynamics. New York: Van Nostrand
Reinhold Company Ltd.

Full Text 
PAGE 1
SOUND PROPAGATION IN MOVING FLUID by Carol Sue Jones B.A., Adams State College, 1976 University of Colorado at Colorado Springs, 1987 A.thesis submitted to the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Basic Science 1996
PAGE 2
This thesis for the Master of Basic Science degree by Carol Sue Jones has been approved by
PAGE 3
Jones, Carol Sue (Master of Basic Science) Sound Propagation in Moving Fluid Thesis directed by Professor Randall P. Tagg ABSTRACT Sound propagates as a wave. Therefore, the derivation of the threedimensional wave equation, with and without background motion, is developed to provide a complete basis for further investigation. In the limit that the wavelength is small compared to the distance over which changes occur in the medium of propagation, ray acoustic theory can also be utilized in sound propagation. An important concept in this theory is the eikonal equation. This is a partial differential equation that defines the surfaces of constant phase or the wavefronts. A section is devoted to its derivation. Next, several examples are presented to provide theoretical predictions of sound propagation in moving fluid showing how velocity gradients deflect sound in a manner that can be compared to refraction due to variations in, say, the fluid density. Lastly, an experiment is proposed to test these predictions. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed iii
PAGE 4
DEDICATION I dedicate this thesis to my husband, Barry, and daughters, Sherri and Melissa, who provided me with support and encouragement throughout this entire degree process.
PAGE 5
ACKNOWLEDGMENT My thanks to the staff of the Physics Department for their support and patience in my efforts to pursue this degree. A special thanks to my advisor, Dr. Tagg, for his devotion, support, understanding and encouragement during the pursuit of this degree.
PAGE 6
CONTENTS Chapter 1. Introduction ..................................................... 1 2. 3Dimensional Wave Equation ................................... 2 2 .1 Continuum Hypothesis .......................................... 2 2.2 Mass Conservation Derivation ................................. 3 2.3 NavierStokes Derivation ...................................... 5 2. 4 Equation of State ............................................ 13 2.5 Equation for Small Disturbances in the Presence of Background Flow ......................................................... 15 3. Eikonal Equation ......................................... 20 3.1 Eikonal Equation for Stationary Fluid but Variable Wavespeed.20 3. 2 Ray Equations ................................................ 25 3.2.1 Arc Length Parameterization ........................... 25 3.2.2 Relations ................................. 27 3.2.3 Finding Explicit Solutions for Ray Trajectories and the Eikonal Equation .......................................... 32 3.3 Eikonal Equation Derivation for a Moving Fluid ............... 34 3.4 Example ........................................ 41 4. Applications .................................................... 54 4.1 4 .1.1 4.1.2 4 .1. 3 Shear Flow Problem ....................................... 54 Part IA .................................................. 54 Part IB .................................................. 63 Part II .................................................. 70 5. Proposed Experiment ............................................. 72 vi
PAGE 7
Appendix A. Further Derivations ............................................. 76 References vii
PAGE 8
FIGURES Figure 2.1 Schematic Variation of Average Energy of Molecules with Length Scale (Tritton 1977) .......................................... 3 2.2 Definition Sketch for Derivation of Continuity Equation (Tritton 1977) ................................................ 3 2.3 Deformation of Rectangular Element and Relative Motion of Fluid on Either Side of Arbitrary Line through the Element (Tritton 1977) ......................................................... 7 2.4 Pouring of Viscous Liquid (Tritton 1977) ...................... 8 3.1 Ray Curves ................................................... 24 3.2 Tangent to Ray ....................................... 24 3.3 Pictorial Representation of Sound Propagating through Constant Density Fluid in Uniform Motion ............................. 38 3.4 Propagation of Waves in Shear Flows of Different Refractive Indices ...................................................... 50 3.5 Distance Between Wavefronts in Shear Flows of Different Refractive Indices ........................................... 51 4.1 Graphical Representation of Shear Flow ..................... 55 4.2 Graphical Representation of Part I of Problem ................ 55 4.3 Graphical Representation of Part 2 of Problem ................ 70 5 .1 Linear Shear Flow ............................... 7 2 5.2 Graphical Representation of Ray Trajectory ................... 73 viii
PAGE 9
1. Introduction Sound propagates as a wave. Therefore, the derivation of the threedimensional wave equation, with and without background motion, is developed in the second chapter. The basic equations utilized in this derivation are the mass conservation equation, NavierStokes equation and the equation of state between pressure and density. When a wavelength is small compared to the distance over which changes in the medium of propagation occur, the field is expected to look like a plane traveling wave. In this case, all acoustic field quantities (pressure, density and velocity) vary with time and with some cartesian coordinate, r, but are independent of position along planes normal to the r direction (Pierce 1981). This concept of a wavefront is important in geometrical or ray acoustics and leads to the eikonal equation which is a partial differential equation defining these surfaces. Chapter 3 is devoted to the derivation of the eikonal equation. The next chapter presents a solution to a linear shear problem which determines the distance the sound propagates. This problem also demonstrates how velocity gradients deflect sound in a manner that can be compared to refraction due to variations in, say, the fluid density. The result is then utilized in the last chapter which proposes an experiment to test these predictions. Sound propagation has a wide range of applications in a variety of fields of study. Some areas where sound propagation may be of interest are atmospheric acoustics, underwater acoustics, musical acoustics, ultrasonics, architectural acoustics, aeroacoustics, nonlinear acoustics, environmental acoustics and noise control (Pierce 1981). 1
PAGE 10
2. 3Dimensional Wave Equation This chapter derives the 3dimensional wave equations for sound waves with and without background motion. This is examined in order to provide a basis of understanding for the next chapter on the eikonal equation. The conservation of mass and Newton's laws of motion are used to derive the 3dimensional wave equation. These equations involve physical and mechanical quantities such as velocity, density, pressure'and temperature which will vary continuously from point to point throughout the fluid. To define these quantities at a point requires the applicability of continuum mechanics or the continuum hypothesis which is discussed in Section 1. The derivation of the mass conservation or continuity equation is discussed in Section 2. Section 3 derives Newton's second law of motion for a fluid, or the NavierStokes equation. Lastly, the 3dimensional wave equation is derived in Section 4 using the mass conservation and NavierStokes equations. 2.1 Continuum Hypothesis The continuum hypothesis supposes that an association can be made with any volume of fluid and those macroscopic properties involving a fluid in bulk. At each point there is a particle of fluid having a certain velocity, density, pressure and temperature. A large volume of fluid therefore consists of a continuous aggregate of such particles. It must be noted that this hypothesis is not correct at the molecular scale. The various macroscopic properties are defined by averaging over a large number of molecules (Tritton 1977). None of these averaging processes is meaningful unless the averaging is carried out over a large number of molecules. A fluid particle must thus be large enough to contain many molecules. It must still be effectively at a point with respect to the flow as a whole. The above requires a length scale in order for the continuum hypothesis to be valid. This length scale will be the size of the fluid particle. The intent is for the volume to be large enough to contain many molecules and have the fluctuations of the macroscopic quantities negligibly small. If the volume is too small, it will contain only a few molecules and there will be large fluctuations in the macroscopic quantities. If the volume is too large, it will extend into regions where the macroscopic quantity is significantly different and there will be an increase or decrease in the average. In other words, the applicability of the continuum hypothesis depends on there being a significant plateau (Tritton 1977) in the macroscopic quantity. There will thus be a length scale that can be regarded as an infinitesimal distance as far as macroscopic effects 2
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are concerned and the equations can be formulated ignoring the behavior on still smaller length scales. Refer to Figure 2.1. Physical Quantity (e.g. density) Length Figure 2.1. Schematic Variation of Average Energy of Molecules with Length Scale (Tritton 1977) For common fluids in normal laboratory conditions, the continuum hypothesis is valid for both the mass conservation and NavierStokes equations. For example, the fluid velocity ii(r, t) at r is defined as the massweighted local average particle velocity or, equivalently, as the average momentum per unit mass in the vicinity of r (Pierce 1981) For both the mass conservation and NavierStokes equations, the concept that at each point there is a particle of fluid having a certain velocity, density, pressure and temperature is used. 2.2 Mass Conservation Derivation The derivation of the mass conservation or continuity equation begins by considering an arbitrary volume V fixed relative to Eulerian coordinates and entirely within the fluid. Refer to Figure 2.2. Figure 2.2. Definition Sketch for Derivation of Continuity Equation (Tritton 1977) The equation is more readily formulated using the Eulerian specification, coordinates fixed in space, instead of the Lagrangian 3
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specification, coordinates that move with the particle. The reason behind this is that the Lagrangian specification does not immediately indicate the instantaneous velocity field on which depend the stresses acting between fluid particles (Tritton 1977). The Eulerian notation for velocity is ii = u(r, t) where r is the position in an inertial frame of reference andt is time. Values of ii at the same position r but different t do not correspond to the same fluid particle. Fluid moves into or out of this volume at points over its surface. If dS is an element of the surface, whose magnitude is the area of the element and whose direction is the outward normal, and u is the velocity at the position of this element, then the component of u parallel to dS is what transfers fluid out of V. Thus, the outward mass flux (mass flow per unit time) through the element is pu dS, where p is the fluid density. Hence, the rate of loss of mass from V is Is pu dS (2.1) which is negative if the mass in V is increasing. Also, the total mass in volume V is L pdV. (2.2) Therefore, the rate of loss of mass from V becomes d J, J, 8p pdV= dt v v 8t Is pu cts. (2. 3) Using the continuum hypothesis, the mass balance is at a point rather than that over an arbitrary finite volume. Hence, V is allowed to shrink to an infinitesimal volume and, since the integration in J (8pf8t}lv is redundant, equation 2.3 becomes op = 1 im [ r pu ctsjv J. 8t Js (2. 4) That is, 8p d' = 1vpu (2.5) 8t by definition of the div operator. Setting equation 2.5 equal to zero yields 4
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+ v (p;;.) = 0 I (2. 6) which is the general expression for the mass conservation of a fluid in which both u and p are functions of position. This equation is often referred to as the continuity equation. 2.3 NavierStokes Derivation Newton's second law of motion states that the rate of change of momentum of a fluid particle is equal to the net force acting on it. First, an expression for the rate of change of momentum of a fluid particle is considered. It would not be correct to equate the rate of change of momentum at a fixed point to the force, because different particles are there at different times (Tritton 1977). Therefore, consider the general question of rates of change following the fluid. In general, the small change 88 produced by a small change 8t in time and small changes 8x, 8y, 8z in Cartesian position c'oordinates is aB 08 08 aB 8B = ot + 8x + 8y + oz 8t 8x 8y 8z (2. 7) and a rate of change can be formulated by dividing by 8t which gives 88 8t aB 8t aB ox +ax 8t aB 8y aB 8z ++. By 8t 8z 8t (2. 8) Choosing 8x, 8y, 8z to be the components of the small distance moved by a fluid particle in time 8t, then in the limit 8t + 0 this is the rate of change of a vector quantity B following a fluid particle. In the same limit, 8x/8t, 8y/8t and 8z/ot are then the three components of the velocity of the particle u, v, and w respectively. Then equation 2.8 becomes DB aB OB aB OB =+u+v+w. Dt 8t Bx By 8z (2. 9) In general D/Dt denotes the rate of change (of whatever quantity it operates on) following the fluid. This operator is known as the substantive derivative (Tritton 1977) or as the convective derivative. Equation 2.9 can be rewritten as 5
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ns Dt = aB at +u VB. If B = u, equation 2.10 gives the rate of change of velocity following a fluid particle, Du Dt = au + u. Vu. at (2 .10) (2 .11) The quantity u now enters in two ways, both as the quantity that changes as the fluid moves and as the quantity that governs how fast the change occurs. Mathematically, however, it is just the same quantity in both its roles (Tritton 1977). In the general case when both u and p are variables, the rate of change of momentum would be D(pliVDt. According to Tritton (1977), the only reason why a particular bit of fluid is changing its momentum is that it is changing its velocity. If it is simultaneously changing its density, this is not because it is gaining or losing mass, but because it is changing the volume it occupies. This change is therefore irrelevant to the momentum change. Therefore, for this derivation, the rate of change of momentum is pD(u)/Dt and equatiop 2.11 becomes Du pDt au v=p+pu u. at (2 .12) From equation 2.12, pD(u)/Dt is the lefthand side of the dynamical equation which represents Newton's second law of motion. The righthand side is the sum of the forces (per unit volume) acting on the fluid particle. In order to complete the equation, the nature of these forces must be considered. Some forces are imposed on the fluid externally, and are part of the specification of the particular problem. An example of a body force is to specify the gravity field in which the flow is occurring. A uniform gravitational force (per unit volume) pgz may be eliminated from the equations replacing the pressure p with p' pgz so that Vp' = Vp + pgz. In the absence of fluid motion, Vp' = 0 so Vp = pgz. This means p = Po pgz, the hydrostatic pressure, where Po is some reference pressure established at z = 0. Other forces, pressure and viscous action, are related to the velocity field. These surfaces forces or stresses are thus intrinsic parts of the equations of motion and have to be considered here. Both the pressure and viscous action generate stresses acting across 6
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any arbitrary surface within the fluid; the force on a fluid particle is the net effect of the stresses over its surface. In the following discussion, some of the physical concepts underlying viscous action are considered and then the mathematical formulation is presented. According to Tritton {1977) viscous stresses oppose relative movements between neighboring fluid particles. Equivalently, they oppose the deformation of fluid particles. The difference between these statements lies only in the way of verbalizing the rigorous mathematical concepts, as illustrated by Figure 2.3. The change in shape of the initially rectangular region is produced by the ends of one diagonal moving apart and the ends of the other moving together. As the whole configuration is shrunk to an infinitesimal one, it may be said either that the particle shown is deforming or that particles on either side of AB are in relative motion. The rate of deformation depends on the velocity gradients in the fluid. The consequence of this behavior is the generation of a stress {equal and opposite forces on the two sides) across a surface such as AB. This stress depends on the properties of the fluid as well as on the rate of deformation. A Figure 2.3. Deformation of Rectangular Element and Relative Motion of Fluid on Either Side of Arbitrary Line through the Element {Tritton 1977) The stress can have any orientation relative to the surface across which it acts {Tritton 1977). Consider a simple example in which the viscous stresses normal to the surface govern the behavior. A pictorial representation {refer to Figure 2.4) shows a viscous liquid poured from a container and the falling column which is produced. In this case, since there are no sidewalls, transverse viscous stresses cannot be generated as they are in the case of channel flow. The reason the fluid at P does not fall with an acceleration of g is because of the viscous interaction with the more slowly falling fluid at Q. 7
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Q p ___________ ::: __ "'"'=1Figure 2.4. Pouring of Viscous Liquid In general, the stress is a quantity with a magnitude and two directions associated with it. The two directions are the direction in which it acts and the normal to the surface. Therefore, it is a secondorder tensor. The stresses acting across surfaces of different orientations through the same point are not, of course, independent of one (Tritton 1977). The rateofstrain tensor is also a secondorder tensor and describes the rate of deformation. From the considerations above velocity gradients are involved. Not all distributions of velocity variation lead to deformation. An example is the rotation of a body of fluid as if it were rigid. The rateofstrain tensor selects the appropriate features of the velocity field (Tritton 1977) At this point, the form of the rateofstrain tensor is examined. Then the consequences of a linear relationship between the stress tensor and the rateofstrain tensor will be determined. Consider two material points separated by a distance of which have components Oxi. Then where the summation convention for repeated suffixes applies. The rate of change of equation 2.13 following the fluid is (2 .14) According to Tritton (1977), it is thus the symmetrical combinations of the velocity gradients that give rise to distortions opposed by viscous action. (A motion in which 8
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aui = auj axj axi (2.15) for all i and j does not involve any relative movement between two neighboring points for any Bxi.) Moreover, these symmetrical combinations represent fractional rates of change of the separation for various orientations; for example if 1 o( B.e) 8u1 = They are thus rates of strain of the fluid. The rateofstrain tensor eij is formulated as the symmetrical part of the velocity gradient tensor, [Parenthetically, the antisymmetric part corresponds to the vorticity, 00 = '\J X U where (2 .16) (2.17) (2 .18) J:.urve U d / BA hurronding Physically the vorticity IV X uj = lim corresponds to 2.18 is 1 area element ISA the rotational motion of = E. 'k O)k ] 2 l.J the fluid. Thus equation (2 .19) The viscous stress tensor is given by defining as the force per unit area acting in the idirection across a surface normal to the jdirection. For a Newtonian fluid this is linearly related to eij : (2.20) 9
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The physical processes must be independent of the orientation and handedness of the axes. Aijkl must thus be an isotropic tensor and the most general form it can take is (2.21) This gives (2 .22) (since eij = eji ) . There are thus two arbitrary constants involved; these are physical properties of the particular fluid. From the particular case e12 = oufoy and all other eij equal to zero, we can identify that + 'X = 2J.L (2.23) where is the coefficient of viscosity. Also, Vu (2.24) and substituting equations 2.17, 2.23 and 2.24 into equation 2.22 yields u. (2. 25) The net viscous force on a fluid particle is given by the differences in the stresses acting across opposite faces (Tritton 1977). The ith component of the force per unit volume is given by mij;axj The stress tensor depends on the rateofstrain tensor and on the properties of the fluid. A Newtonian fluid can now be defined rigorously as one in which the stress tensor and the rateofstrain tensor are linearly related (Tritton 1977). According to Tritton (1977), the remaining ideas contained in the derivation of the viscous term of the dynamical equation are simply symmetry considerations. For example, a mirrorimage flow pattern must generate a mirrorimage stress distribution. And the analysis of a flow configuration using different coordinates must give the same result. 10
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The xcomponent, in Cartesian coordinates, of the viscous force per unit volume of equation 2.25 is [ 2J.l au + t..v u] + [J.l( au + av )] ax ax By ay ax + [J.l( fJw + au)] oz ox oz Similar expressions for the yand zcomponents are given by appropriate permutations and are a [ av ] a [ (au 2J.1+'A.Vu +J.1ay ay ax ay + [J.l( Ow oz fJy + :)] + :)] ew, + t..v. u] + + av)J az az ay ay az + [J.l( Ow + ou )] ox ox oz (2.26) (2. 27) (2.28) As stated earlier, is the coefficient of viscosity and A is a second viscosity coefficient. The existence of this second coefficient, which is independent of the first, is explained by the fact that there are two independent elastic moduli. The constant is associated with friction that occurs as a fluid element undergoes a pure strain, while A is associated with friction that occurs as a fluid element undergoes a change in volume (compression or expansion) The statement that a fluid is Newtonian usually means that is observed to be independent of the rateofstrain and that A is assumed to be so too. According to Tritton (1977) A is difficult to measure experimentally and is not known for the variety of fluids for which there are values of Often, it is better to identify pressure as the mean normal stress. Let Peq be the thermodynamic pressure associated with the state of a fluid element that is considered to be in local thermodynamic equilibrium. The mean normal stress p is given by 1 ( ) P = Peq + crn + O'zz + 0'33 3 11
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p = p +"AVu) eq 3 Bx 3 By aw + "AV. u) 3 Bz p = Peq + ( au + Bv + Bw ) + "A V u 3 Bx By Bz P = Peq + ( + A) V U. The ith component of the net force on a fluid element is given by a a Peq + cr1j where repeated indices mean implied summation. Bx1 Bxj Then a +"A8 .. V u a 1) xj a ( ) a [ ( &u. &uj J J = Peq + "AV u + Jl __ l. + Bx1 Bxj Bxj Bx1 a ( 2 ) a [ ( au. &uj J ] = P . u + 1 + Bx1 3 8xj Bxj Bx1 a a [ ( au. &uj J ] 2 a ( ) P + 1 + . u axi axj axj axi 3 axi .a a a Peq + .. crij Xi UXJ Hence Newton's second law may be written in two ways IP "ii = Vp,. + J.1V'u + ..v u + pFI (2.29a) I Du nn22 n p = v p + v u 3 v u + pF Dt (2 .29b) In equation 1.29b the second viscosity coefficient A has been 2 effectively replaced with with the understanding that the 3 12
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pressure p now refers to mean normal stress as opposed to the pressure Peq obtained from the (local) equation of state for the fluid. This distinction is important when processes of sound attenuation are to be considered. For the remainder of this discussion, however, viscosity and its effects are ignored and p = p = Peq is used when there is no viscosity. Equation 2.29 is known as the NavierStokes equation. It is the basic dynamical equation expressing Newton's second law of motion for a fluid of constant density. According to Tritton (1977), the term F represents the contribution of those forces (such as gravity) that have to be included in the specification of the problem. This is often known as the body force term, because such forces act on the volume of a fluid particle, not over its surface in the way the stresses between fluid particles act. The reaction to a body force is remote from the fluid particle concerned usually outside the fluid region, although occasionally on distant fluid particles. Substituting V = Jl/P into equation 2. 29 gives the Navier Stokes equation as 1 n n22 n = vp + Vv U Vv p 3 _uL u Vu where V = Jl/P is a property of the fluid called the kinematic viscosity. (2. 30) The NavierStokes equation is a nonlinear partial differential equation in u. The nonlinearity arises from the dual role of the velocity in determining the acceleration of a fluid particle. This nonlinearity is responsible for much of the mathematical difficulty of fluid dynamics, and is the principal reason why our knowledge of the behavior of fluids in motion is obtained in many cases from observation (both of laboratory experiments and of natural phenomena) rather than from theoretical prediction. The physical counterpart of the mathematical difficulty is the variety and complexity of fluid dynamical phenomena(Mitton 1982). Without the nonlinearity the range of these would be much more limited. 2.4 Equation of State The continuity equation (2.6) and the NavierStokes equation (2.30) constitute a pair of simultaneous partial differential equations. Both represent physical laws conservation of mass and conservation of momentum which will always apply to every fluid particle. Together they provide one scalar equation and one vector 13
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equation effectively four simultaneous equations. There are, however, two scalar variables (pressure and density) and one vector variable (the velocity) effectively five unknown quantities. Therefore, at least one more equation is needed. Start with an equation of state relating pressure to density and as required for simple fluids one other intrinsic variable, say the entropy per unit volume s. Then, p = p ( p, s). Unfortunately, this introduces yet another scalar, s, and consequently requires a further equation. In general, this equation arises from the conservation of energy. However, sound waves typically disturb the fluid on time scales short with respect to the time over which heat diffuses across a distance of one wavelength of sound. So, the process may be regarded as adiabatic. The conservation of energy equation is replaced with s = constant = S0 Thus p = p ( p, S0 ) or, understanding that entropy is held constant, simply p = p(p). Expanding the relationship between pressure and density, p p(p), using Taylor's series gives P = Po + { P pJ :: { pJ Is + The subscript s means the thermodynamic derivative is taken for constant entropy. In general an increase of pressure tends to increase the density and so the gradient dp (p0 ) is a positive dp (2.31) constant (Dowling and Ffowes Williams 1983) The parameter c is defined as (2. 32) Substituting equation 2.33 into equation 2.32 gives the linearized relationship for pressure and density (2. 33) where p' p Po and p' = P Po As an aside, the speed of sound for an ideal gas can be found by first considering the relationship between pressure and volume. For an adiabatic process they are related by PV1 = constant 14
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where y is a constant equal to the ratio of the specific heat at constant pressure to the specific heat at constant volume. Differentiating the above equation gives yPVy1dV + yY dP = 0 dP = yPdV v The bulk modulus is defined as the ratio of the change in pressure to the fractional change in volume. Therefore Badiabatic dP = = yP. dV/V The speed of sound can be defined as B p which for the adiabatic case is 2.5 Equation for Small Disturbances in the Presence of Background Flow In order for the 3dimensional equation to be developed in the most general case, the effect of background flow needs to be added to the continuity equation (2.6) and the NavierStokes equation (2.30). Let U be a steady background flow with density pb Also, let u = U + u', p = Pb + p' and p = pb + p' where u', p' and p' are small perturbations to the velocity, pressure and density respectively. The entropy per unit volume can be written as s = sb + s', but since it is assumed that the disturbance occurs adiabatically, s' = 0. Substituting these expressions into equations 2.6 and 2.30 yields ( Pb + p') + V ( Pb + p' )( U + u') Ot. and 15 0 (2.34)
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( Pb + P1 ) ( U + iii) + ( Pb + P1 )( U + ii1 ) V ( U + ii1 ) = Ot v ( Pb + PI ) + v ( Pb + pI ) Y'2 ( u + u I ) v(pb + (u + li1 ) + (pb + P1)F. 3 Expanding equation 2.34 gives a a I + +V(pbo+pbu1+P1U+p1u1 ) = o a a I + ______ + p V ij + ij Vpb + pbV ii1 + ii1 Vpb + at at b p1V ij + ij Y'p1 + p1V u1 + u1 Y'p1 = 0 Expanding equation 2.35 gives vu + pbul vu + plu vu + P1U1 vu + pbu + pbu1 Vu1 + p1u Vu1 + p1u1 Vu1 = Vp. c'Vp'( :;, }vp. )p' + vp.V'U + 2 vpb V2u1 + vp1Y'2u + vp1V2u1 3 vpb V u 2 t7 I 2 l' vp v u vp v u 3 b 3 2 vp1Y' u1 + pbF + p1F 3 (2.35) (2 .36) (2.37) Eliminating the squares of small quantities and the derivatives of constant quantities from equations 2.36 and 2.37 yields a I + pb V ij + ij Vpb + Pb V u1 + U1 Vpb + (2.38) p1V ij + u V p1 = 0 and 16
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ou' Pb at + pbu vu + pbu' vu + p'u vu + pbu Vu' = Vpb c'Vp' ( }vpb )p + vpb V'U + 2 z2 2 vpb v u' + vp'V' u 3 vpb v u 3 vpb v u' 2 vp'V' u + pbF + p'F. 3 (2. 39) The background flow, U with density pb satisfies the continuity equation and the NavierStokes equation. Substituting the background flow parameters into equation 2.6 gives (2. 40) Assuming a steady background flow, = 0, equation 2.40 yields (2.41) Substituting the background flow parameters and the assumptions that the viscosity is zero, the flow is steady and the body force is negligible into equation 2.30 yields au +UV'U= at or since = 0 (steady background flow), at Using equations 2.41 and 2.42 in equations 2.38 and 2.39 yields 8p' + pb V' u' + u' Vpb + p'V U + u V'p' = 0 at and 17 (2.42) (2. 43)
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Pb at + pbul vu + plu vu + pbu Vu1 c'Vp' ( }vrb )p' (2.44) To arrive at the 3dimensional wave equation, first differentiate the mass conservation equation (2.43) with respect to t (remembering that U and pb are steady in time) to get ap1 au ap1 + Vpb + V + U V = 0 at UL. at at at Next, take the divergence of the NavierStokes equation (2.44) to get + p v + Vpb [(u1 V)u] + at b at Pb v [ ( u I v ) ij] + v P1 [ ( ij v ) ij ] + p1V [ ( U V) U] + Vpb ( ( U V) U1 ] + pbV ((ii V)uj + c'V'p' + ( )v (Vpbp') = 0. (2.45) (2 .46) Subtract equation 2.46 from 2.45 and use equation 1.41 to obtain a2 I a; vpb [Cui v)u]pbv. [(u1 v)u] V p1 [ ( ij V) ij] p1V [ ( ij V) ij] v Pb [ ( ij v ) u I ] pb v [ ( ij v) u I] (2.47) c2V2p' ( ) V'pbp' ( ) Vpb Vp' = 0. This is the 3dimensional equation for a disturbance in the presence of background flow. If there is no background flow (U = 0) and constant background density (Vpb = 0) the classical 3dimensional wave equation is recovered from equation 2.47 (2. 48) 18
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All small quantities describing the acoustic field (pressure, density and velocity) satisfy the wave equation. The density has been established in equation 2.48. Since the equation of state (2.33) equates the pressure and density, the wave equation is satisfied for pressure. The velocity can be shown to satisfy the wave equation by taking the divergence of the mass conservation equation n op' tn2, 0 vpvu = at and differentiating the NavierStokes equation with respect to t p' azu., c2 Otz op' V= 0 Ot and adding these two equations to obtain For an adiabatic condition, no viscosity and negligible body forces, the 3dimensional wave equation ( 2.48) has been established. The equations used for this development were the continuity equation, equation 2.6, and the NavierStokes equation, equation 2.30. This development will be utilized in the next chapter on the development of the eikonal equation. 19
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3. Eikonal Equation This chapter develops the eikonal equation which defines the wavefronts of a disturbance in a moving medium. The eikonal concept is then applied to an example. The first section derives the eikonal equation for stationary fluid with variable wavespeed. Section 2 presents the development of the ray equations. frequency approximation of the eikonal equation. Then, Section 3 derives the eikonal equation for a moving fluid. The last section develops an example which uses the concepts presented in earlier sections. 3.1 Eikonal Equation for Stationary Fluid but Variable Wavespeed According to Jones (1989) when the wavelength is so small that a significant change in the medium occupies many wavelengths, a reasonable hypothesis is that in local regions the field behaves as if it were in a homogeneous medium. Thus, locally, the field may be expected to look like a plane wave. Let 'f' ( x, t) be a field variable being transmitted as a wave according to equation 2.48 where the wavespeed c(x) is assumed to vary with position. parameter c is related to a reference speed constant Co by c(x) = _s__ n(x) The (3 .1) (3.2) where n(x) is interpreted as the local index of refraction and is assumed to vary continuously and slowly over distances comparable to 21t the wavelength A0 = Let ko 'f' (x, t) = \11 (x)eiCilt (3.3) Substituting equation 3.3 into equation 3.1 yields (3.4) Defining 20
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(3.5) and substituting it into equation 3.4 and removing the eiCilt terms yields ol (3. 6) which is a modified form of the Helmholtz equation with nonconstant coefficient ( x). Try a solution of the form (3.7) where it is anticipated that the amplitude part will depend on k0 as well as x. L is known as the eikonal and has dimensions of length. Taking the gradient of equation 3.7 gives ik L ik L V\jl = VAe o ik0A VLe o (3. B) and taking the divergence of equation 3.8 gives (3. 9) 2 ik L 2 I 12 ik L l.k0AV Le 0 k0A VL e o Now, substituting equation 3.9 into the modified Helmholz equation (3.6) yields (3.10) 2 I 12 ik L 2 2 ik L 0. k0A V L e o + k0n Ae o ik L Removing the e o term from equation 3.10 gives (3.11) Equating real and imaginary parts (assume A is real) in equation 3.11 gives (3.12) 21
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2k0VA VL + k0AV2L = 0 2VAVL+AV2L=O. 2 Dividing equation 3.12 by Ak0 yields 1 V2A k2 A 0 (3.13a) (3.13b) (3.14) Suppose there is a distance f which is some characteristic distance over which A changes appreciably such that 2 A VA2 f (3.15) Substituting the first term of equation 3.14 into equation 3.15 gives 1 V2A 1 A k2 A k 2 f2A 0 0 1 = )' (3 .16) 27t where the definition of the wavelength given early, A = is 0 k 0 also employed. If f >> A0 i.e. the amplitude changes significantly only over distances much larger than Ao, then ('; J << 1. Consequently, equation 3.14 simplifies to (approximately) Another approach to the same result yields some insight into the method for making higherorder corrections. Let (3.17) and substitute this expansion into equations 3.12 and 3.13b which yields 22
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V2A0 (x) + 2_ (x) + V2Az (x) ko ko (x)IVLI2 k0Adx)IVLI2 Az (x)IVLI2 + n2k!Ao (x) + n2k0A1 (x) + n2Az (x) = 0 2VA0 (x) VL + 2_ (x) VL + 2VAz (x) VL ko ko + A0(x)V2L + + = 0. ko ko (3.18) (3.19) Grouping like terms in k0 in both equations 3.18 and 3.19 yields [ n2 IVLI2 ]Ao (x) + k0 [ n2 IVLI2 ]A1 (x) + [n2 IVLI2]A2(x) + [A0(x)V2L + 2VA0(X) VL] + V2A0(X) If L, Ao, ... and their derivatives are finite and do not alter appreciably over a .few wavelengths, the coefficients of the (3.20) individual powers of k0 may be equated to zero (Jones 1989). (Note, this is actually an asymptotic expansion that becomes better at given order N in the series as k0 oo but does not necessarily converge for fixed k0 as N oo ) Then equation 3. 20 becomes I1VLI2 = nzj A0 (x)V2L + 2VL VAo (x) = 0 V2L + 2 VL V Aj = V2 (j = 1, 2, ... ). (3. 21) (3.22) (3.23) According to Jones (1989), equation 3.21 acts as a partial differential equation for L, the function which defines the surfaces of constant phase, i.e. the wavefronts. Often L is called the eikonal and 3.21 is known as the eikonal equation. The differential equations 3.22 and 3.23 which fix the amplitudes Aj are termed transport equations. 23
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For example, when k0L = 21t, Re (y) reaches a local crest. Note, however that A may vary along the crest. Refer to Figure 3.1. L(x) 0 L(x) Figure 3.1. Ray Curves Rays are curves that are everywhere tangent to the direction of energy propagation. For sound traveling in a nonmoving medium, the direction parallel to vo/ corresponds to the normal to the wavefronts. However, this is no longer valid for sound traveling through a moving medium. The local vector velocity of the medium must be added to a vector normal to the wavefront to get the direction of energy propagation. Refer to Figure 3.2. Figure 3.2. Tangent to Ray Before continuing to the derivation of the eikonal equation for a moving fluid, the eikonal for a uniform, stationary medium will be considered in order to understand physically equation 3.21. A known solution to the wave equation (3.1) for a uniform stationary medium is nJ() =A i[llltkx] T X, t 0e 24 (3. 24)
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Using equations 3.3 and 3.7, the general eikonal form is given as Equations 3.24 and 3.25 are equivalent if A=Ao k x = k0L(x) Solving equation 3.27 for L(x) gives L(x) k x = k x where k is a unit vector normal to the wavefront. Notice that VL(;<) and therefore In 12 2 2 2 V L = kx + ky + kz = 1 This satisfies equation 3.21 for n 1. 3.2 Ray Equations 3.2.1 Arc Length Parameterization (3.25) (3.26) (3.27) (3.28) This section determines how to solve for points x that lie along a ray, that is, along the trajectory everywhere parallel to the local direction of energy flow. As noted above, this direction is perpendicular to the wavefronts for stationary fluids. Since VLis also perpendicular to the wavefronts, an infinitesimal displacement (dx, dy, dz) along a ray must be parallel to VL. Then dx dy dz = (:) (:J (3.29) The infinitesimal arc length, ds, is related to the infinitesimal coordinates, dx, dy and dz by (3.30) 25
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Substituting equation 3.29 into 3.30 yields (:) 2 2 (dx)2 + (dx)2 + (dx)2 = (ctsl (3.31) (3.32) Using the definition of gradient, (3.33) Using equation 3.21, equation 3.33 becomes (3.34) Now, substituting equation 3.34 into equation 3.32 gives (3.35) which reduces to (BL)2 (dx)2 = ds n (3. 36) Taking the square root of equation 3.36, for the xcomponent, and applying similar arguments for the yand zcomponents yields 26
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8L dx ax = (3.37a) ds n 8L dy By = (3.37b) ds n 8L dz az = ds n (3.37c) These equations represent the arc parameterizations of a ray. Taking the integral of equation 3.37a yields BL x(s) = xo + rs axds'. Jo n (3.38a) Similar integrals for equations 3.37b and 3.37c give y(s) = Yo + BL rs By ds' Jo n, 8L z(s) = zo + Jo n (3.38b) (3.38c) Equations 3.38a c give the ray parameterization curves. Unfortunately, they require an explicit solution for L before the trajectory can be found. 3.2.2 Hami1tonJacobi Re1ations If possible, a direct method for computing the ray trajectories without solving for L first is desired. First, however, it will be shown that motion along a ray is described by HamiltonJacobi equations which are identical in form to the equation describing conservative motion of a particle in analytical mechanics! The infinitesimal arc length ds traversed in time dt is given by ds = c(x)dt where c(x) into equation 3.37a gives n(x) 27 Substituting this relationship
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oL 1 dx c(x) dt n(x) oL dx c(x)ox = dt n(x) dx co oL dt n(x)2 ox Now, define where, for example, the xcomponent of k is given by fJL kx = ko OX and rearranging fJL ax Using the eikonal equation, or, equation 3.43 becomes The xcomponent of equation 3.69 is Substituting equation 3.42b into equation 3.40 yields dx dt Using the definition for c(x) and equation 3.44, equation 3.45 becomes dx dt 28 (3.39) (3.40) (3. 41) (3.42a) (3.42b) (3. 43) (3.44) (3.45) (3. 46)
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Again using the definition for c(x) and equation 3.44 into equation 3.5, (J) = coko, gives ro = n ( x) [ c ( x ) ] n (x ) = c ( x) k (3.47) Therefore, co can be thought of as a function ro {X, y, z, kx, ky kz ) of a both x(t) and k(t). Taking of equation 3.47 yields akx aro a [ J = c(x)k(x) akx akx aro a 2 2 +K = akx c(x) kx + ky akx aro 2kx = akx c(X) !:_ 2 k 2 X + k2 y +K aro () kx = C X. akx k Comparing equations 3.46 and 3.48, the xcomponent becomes dx aro = Similarly, for the yand zcomponents dy dt dz aro = ctt (3.48) (3. 49a) (3.49b) (3.49c) Next, the time derivative of kx is given, using equation 3.42a, by dkx = d ( k aL) dt dt 0 ax dkx aL(x) = dt 0 dt ax 29
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dkx = k { [ a ( aL ) ] dx [ a ( aL ) ] dy [ a ( aL)] dz } dt o ox ox dt + By ox dt + oz ox dt dkx = k [ a ( aL)] dx k [ a ( aL)] dy k [ a ( aL)] dz dt o ox ox dt + 0 ox By dt + 0 ox oz dt dkx (a ) ctX = k VL dt 0 ax dt dkx a ( ctX) a ctx = kVLkVL. (3.50) dt 0 ax dt 0 ax dt dX Now, V Lis parallel to since dX is along the ray direction. dt Thus a ( ctX) VLox dt (3.51) Substituting equations 3.21, 3.49a and 3.48 into equation 3.51 gives a(o ctx) a( __ ) vL = n(x)c(x) ox dt ox and using the definition of c(x) gives ox dt ctX) = o. ox dt since C0 is a constant. 3.50 gives Thus, substituting this result into equation (3.52) Substituting equation 3.46 into 3.52 gives dkx = k VL. (c(x)k) dt 0 ox k (3.53) 30
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Also, using equation 3.41 in equation 3.53 (recalling that k is considered here a variable independent of X) yields dkx = k. ( oc) k dt ox k dkx oc k. k = dt OX k dkx oc k2 = dt ox k dkx oc = k dt OX dkx 0 = (ck). dt OX (3. 54) But from equation 3.47, ro ck. Thus, equation 3.54 becomes dkx om = dt ox (3.55a) dky om = dt oy (3.55b) dkz om = oz dt (3.55c) Equations 3.49ac and 3.55ac are just like Hamilton's equations dx am = dt okx dkx = om dt ox where m = m (x, k) plays the role of the Hamiltonian! Note that ro is a constant of motion along rays, i.e. dm = 0. (3.56) dt To show this, dm dt om dx om dkx am dy om dky = + + + ox ctt ctt 8y ctt aky ctt am dz om dkz ++. az dt dt 31
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Using equations 3.49ac and 3.55ac in the above equation gives dro dt Thus, dro = 0 dt This section showed that motion along a ray is described by HamiltonJacobi equations. The next section finds explicit solutions for ray trajectories and the eikonal equation. 3.2.3 Finding Explicit Solutions for Ray Trajectories and the Eikonal Equation In order to solve the eikonal equation, consider the following. The two forms of the eikonal equation are represented by equations 3.21 and 3.34 which are reproduced here for convenience Notice that in the direction parallel to VL, dL ds = VL. (3.21) (3.34) (3.57) Thus, substituting equation 3.57 into equation 3.21 gives dL ds = n J n(x, y, z)ds. ray path starting atX0,y0,Z0 To compute this, we must know the ray paths. 3.37ac for the rays assume that we know L! 32 (3.58) (3.59) But so far, equations Equation 3.37a gives
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dx fJL n= ds fJx d Taking of the above equation ds d fJL ds ox a ( fJL) dx a ( fJL) dy a ( fJL) dz ox ox ds + fJy ox ds + oz ox ds Using equation 3.37ac gives fJL fJL fJL = a ( oL) ax a ( oL) ay a ( oL) az ax ax ;+ ay ax 7 + az ax ;(VL) VL ox = n dx) = ds ds 2n Using equation 3.21 in the above yields a(n2 ) d ( dx) ox ds n ds = 2n an d ( dx) 2nox ds n ds = 2n Thus, (3.60a) (3.60b) (3. 60c) 33
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or, since ds = edt d (n dx) 8n edt c dt ax n d ( 1 dxJ 8n co dt c0 dt ax d2x dt2 d2y de d2z dt2 = c! 8n n ax c2 8n = _on8y c2 an = _Q__ n az. (3.61a) (3.61b) (3.61c) These are the equations that will be considered in order to solve an example in the last section. 3.3 Eikonal Equation Derivation for a Moving Fluid This section develops the eikonal equation when a sound wave is superimposed on a flowing fluid. Let U be the steady background flow with a density ph in an adiabatic condition such that from equations 3.41 and 3.42 (3.41) (3.42) As in Chapter 2, let sound waves make small perturbations so that the velocity and density become U + u' and pb + p' respectively. Neglecting the squares of small quantities, equations 3.43 and 3.44 are developed and are reproduced here for convenience. ap' + pb V u' + u' Vpb + p'V U + u Vp' = 0 at and BU' Pb at + pbu' Vu + p'U Vu + pbu Vii' = c'Vp'( }vpb)p' 34 (3. 43) (3.44)
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According to Jones (1989), let c0 be a constant typical of the speed of sound. It can be identified with c if c does not vary with position but otherwise might be some average of c. Write k0 = rojc0 for harmonic vibrations of time factor eiwt At high frequencies substitute p' = eik o L ( rl J iwt r + + e 0 'k 1 0 and ik o L ( + :iiil + J iwt = e mo :k e co 1 0 u' in equations 2.43 and 2.44 to obtain a [ ik 0 L ( + rl + J J iwt + e r .. e ot 0 ike 'rJ [ ikoL (:iiil JJ iwt + Pb v e co mo + ike + ... e and 35 (3. 62) (3. 63) (3.64)
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a [ ikoL (iii1 )] imt p e c m + + e + bat 0 0 .k l 0 [ ikoL (iii.l )] nimt Pb e co m0 + iko + vUe + [ e"' ( r. + +)]u VUe'"' + n [ ik o L ( iiil ) ] imt p U v e c m + + e b 0 0 k l 0 (3.65) 2n [ ik o L ( rl ) ] imt c v e ro + iko + e ( :; }vp, )e"' ( r. + Carrying out the derivatives in equations 3.64. and 3.65 to the first approximation (i.e. neglecting terms of order 1/ko and higher) gives ik0c0r0 ik0copbiii.o VL + pbco Viii.0 + C0iii0 Vpb + r0 Vu ik0r0U VL + U Vro = 0 and + pbcoiii.o VU + roo VU ikocopbiii.oU VL + p,c.U VIii. ik.r.c'VL + c'Vr. + ( :; )vp,r. 0. Here k0C0 has been substituted for ro and the common terms of (3. 66) (3.67) i(mtkoL) h b 1 d , 2 41 , 3 66 e ave een cance e . and 3.67 and equation 2.42 in equation 3.67 yields (3.68) and (3. 69) 36
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Using k0 = rofco in equations 3.68 and 3.69 gives and 2 "VLiro c2 r0VL = 0. co (3.70) (3. 71) Let M = fl/co and a. = cfc0 and replace them in equations 3. 70 and 3.71 and remove iro from these equations to arrive at (3.72) and (3.73) Rearranging equation 3.73 gives a.2r0VL = (1 M VL) (3.74) and putting this into equation 3.72 yields 2 2 ( )2 a. V L = 1 M "VL (3.75) This is the eikonal equation for rays in a moving fluid. Now, a physical example will be presented in order to understand more fully equation 3.75. Let sound propagate through a constant density fluid in uniform motion U0 relative to the lab frame given that a disturbance is launched from a source stationary in the lab frame oscillating with angular frequency ro. Refer to Figure 3.3. 37
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Figure 3.3. Pictorial Representation of Sound Propagating through Constant Density Fluid in Uniform Motion In the fluid's frame of reference, where the fluid appears to be at rest except for the sound disturbance, a solution of the wave equation is given by __ A ei[ ro't it' x' ] 'P(x',t) o (3.76) where ro', k' and x' are frequency, wavenumber and position, respectively, measured by an observer in the fluid frame. If k' = lit'l, the dispersion relation is ro' = cok'. Position X in the lab frame is related to x' by X = x' + Ut (Galilean transformation). Thus, in lab frame coordinates, 'P (x, t) 'P(x,t) i[ ro't k' (ll fit)] = A0e i[(ro'+k' ii")tit' ll] = A0e A (3.77) Now define k0 = rofco Also let k be a unit vector in the direction k', so k' = k'k. Substituting these parameters into equation 3.77 yields 'P(x,t) = A0e i[(ro'+k'kii" )t k'kx] 'P (x, t) = A0e i[(ro'+k'kii" )t k'kx] 'P(x, t) = A i[( ro'+ kii" )t k'kx] oe ( u) iro' l+kt k'kx 'P(x,t) = A co (3.88) oe 38
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ro'( 1 + k be identified with m because m is the frequency imposed in the lab frame ro' = ro (3. 89) + k. That is, the signal in the moving frame is doppler shifted to a lower value if U has a component in the same direction as k and to higher A value, if U has a component in the opposite direction from k. Using the relationship, ko = rofco along with k' = ro'fco k' ro' = ko ro k' co' = ko. (3. 90) ro Using equation 3.89 for ro' and equation 3.90 (3.91) Substituting equations 3.89 and 3.91 into equation 3.88 yields ( 3. 92) Comparing equations 3.76 and 3.92 gives A L(x) = kx (3.93) Now, it will be shown that equation 3.93 satisfies equation 3.75. First, note that U is constant in this example. Then, computing A k x in equation 3. 93 gives 39
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L(:i) Taking the gradient of equation 3.94 yields VL(X) = ( l + :. ( J = A k Now, I VLI2 is 2 = ( 1 ) (k! Since k is a unit vector, k! + + k; = 1. 2 Next, consider the righthand side of equation 3.75 2 A (1M VLY u k = 1. (1+ k. J co 2 0 A k (1M VL)2 1co = (l+k 40 (3.94) (3. 95) (3' 96)
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2 .... 1+k u .... k 2 (1M VL)2 = 1 (3.97) Comparing equations 3.96 and 3.97 shows that they are indeed equal. 3.4 Example As an example, consider n = no aizl. (3.98) First look at n = no az (3.99) where z > 0 in equation 3.98. Equations 3.60ac for this example become d ( dx) ds n ds = 0 (3.100a) d ( dy) ds n ds = 0 (3.100b) d ( dz) ds n ds = a. (3 .lODe) Assume that this example is only in the xz plane which eliminates equation 3.100b from consideration (i.e., no deflection in theyplane). Taking the integral of equation 3.100a gives dx nds (3.101) 41
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Substituting equation 3.99 into equation 3.101 for n yields ) dx az = e ds x az)dx = exds. (3.102) Squaring equation 3.102 gives (3.103) Using the definition for ds2 (equation 3.30) in equation 3.102 yields (3.104) and collecting terms, equation 3.104 becomes (3.105) Taking the square root of equation 3.105 gives (3.106) and solving equation 3.106 for dx yields dx exdz = azY e2 X dx dz = n, az J (3.107) 1 Expanding the squared term in equation 3.107 gives dx dz (3.108) ( a )2 2 ( 2 an0 ) z2z ex ex Integral 162 (Korn and Korn 1961) is 42
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162. J dx x = .J ax2 + bx + e = 1 * ..Ja log{2ax + b + 2Ja .Jax2 + bx +e) where a > 0. Using this integral to solve equation 3.108, evaluating between 0 and z, gives ex ( ( a. )2 ( 2 a.no ) 2 a a.z no )2 ) log 2 z2 +1 a. ex ex ex ex ex 2a.no 2a. no (() + log + ex 1 ) ex (2a.) 1 ex ( () ex no no ex 2 a. + log + 1 + log a. ex ex a. ex 43
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X = : log ( az n0 + az nJ2 e; ) +ex log(no + Jno2 e;) a xo + Slog( no + Jno2 ez CJ X a az no + J( az no Y To find ex suppose that at Z = Z0 dxl = tan90 dz o (3.109) z = 0 (3.110) Also let X = X0 at z = 0 Using this assumption with equations 3.107 and 3.110 gives tan90 1 =lfG tan90 1 = 1 tan290 ( J 1 = + 1 tan290 44
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1 + tan2 8 0 = tan2 8 0 28 1 s1n o J + 2 = cos eo 28 Sln 0 cos2 8 0 ( J cos2 8 0 + 28 = s1n o 28 s1n o J 1 = 28 s1n o no 1 ex sin80 and finally, (3.111) Substituting equation 3.111 into equation 3.109 yields X = xo + n 0Sin80 no + 2 28 J log( nosln 0 a. a.z no + a.z no y n!sin280 n 0sin80 no + no 28 log s1n o a. [ ( J 2a.z + 1] 28 a.z no + no Sln 0 no 45
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cos80 1 (3.112) a a.z 1 + ( 1 r sin20, Notice that at equation 3.99 becomes n = 0 which is a. unlikely. The assumption is, that medium with varying index comes to ray turns over is calculated next. equation 3.107 gives before this z is reached, the dx = dz dx = dz dx = dz dz = dx 1 ( r n a.z n:sin80 1 1 (no a.z)2 2 28 nosln 0 2 28 nosln 0 sin80 2a.z an end. The value of z where the Substituting equation 3.111 into (3.113) [ 1+ J] 28 s1n o no (1 az )' 28 Sln 0 no sin80 (3.114) dz The ray turns over when dx = 0 and equation 3.114 becomes 1 a.z = sin80 no 46
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a.z = 1 sin8o Conversely, if the medium extends to z = h, there is a critical angle 8 above which the ray is trapped in the medium. That cn.t angle is given by sin8crit a.h = 1Now it is time to verify equation 3.100c which is reproduced here First note that x is a function of z. Thus dx = dx dz dz and substituting this into equation 3.30 gives ds2 = ( :: J ctz2 + ctz2 ds2 = [1 + (::)']ctz' ds = dz 1 + (::J dz 1 = ds (::J 1 + Substituting equation 3.115 into equation 3.113 gives 47 (3.115)
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dz ds dz ds dz ds = = = 1 2e 1 + s1n 0 2e s1n o 1 (1 az J sin280 2e + s1n o no ( 1 )' 2e s1n o 1 ( 1 J sin'e, Multiplying both sides by n and using equation 3.99 yields dz nds = n (1 a.z) ( 1 J sin'e, 0 n0 ( a.z )2 1dz n= ds no Let equation 3.116 be defined as f(z). Then, f(z) = f' dz ds ds and using this relationship in equation 3.116 gives 48 (3 .116)
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(n dz) = a ds ds which is exactly equation 3.100c which is what was to be verified. Equation 3.114 is the slope of the direction of the ray at the point (x, z). Rearranging, it becomes dz (no azr 1. = dx 2 29 nosln 0 It can be integrated to (3.117) 2 ( )2 c2 2 29 where n = n0 az and x = n0sln 0 and X0 is an arbitrary constant. Regard the c; as fixed, and the X0 as variable; then equation 3.117 gives an infinite set of curves which satisfy the definition of rays. If n is a constant, that is, if the sound velocity is independent of depth, the rays of 3.117 are clearly straight lines (Frank, Bergmann and Yaspin) The eikonal for a shear flow is represented by equation 3.93 which is L(x) kn x = (3.118) en + kn un co co Using two different arguments, it will be shown that this form of the eikonal equation leads to the generalized form of Snell's law. First consider Figure 3.4 which shows the propagation of a wavefront in shear flows with different background speeds and different refractive indices. 49
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/\ U2ex c2 y = 0 z = 0 A ul ex cl Figure 3.4. Propagation of Waves in Shear Flows of Different Refractive Indices Let Referring to Figure 3.4, A 01 = U1 k1x = U1sin91 At the boundary, or 50 (3.119) (3 .120)
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k2x = klx (3.121) Cz + U2k2x cl + Ulklx co co co co Substituting equations 3.119 and 3.120 into equation 3.121 yields sin92 = sin81 (3.122) Cz + U 2sin92 .s_ + U1sin81 co co co co which is the generalized form of Snell's law. If Uz = ul' equation 3.122 becomes which is the more common form of Snell's law. Now consider Figure 3.5 which shows the distances between the wavefronts for the same conditions as Figure 3.4. Figure 3.5. Distance Between Wavefronts in Shear Flows of Different Refractive Indices 51
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Let T be the period. c2T Then, a = b co Similarly for the bottom shear flow, The distance or (3.123) Similarly, A c1T U1 kT + (3.124) co co Using the sin law, equations 3.123 and 3.124 become :::::: and using equations 3.119 and 3.120, the above equation gives Cz which is the same generalized Snell's law found in equation 3.122. In conclusion, this chapter derived the eikonal equation, equation 3.21, which defines the surfaces of constant phase, i.e. the wavefronts. It also determined how to compute the eikonal, L, for 52
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some specific examples. The next chapter will utilize the theory presented in chapters 2 and 3 in a practical application. 53
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4. Applications The purpose of this section is to provide a practical example of the theory developed earlier. The shear flow problem is in two parts. The first part derives the solution to a problem stated in Jones (1989). The second part is an extension to the text problem which involves predicting the distance sound travels for several angles. This preparation leads into the experimental predictions of Chapter 5. 4.1 Shear Flow Problem 4.1.1 Part IA The first portion of this problem is restated here from Jones (1989) (p. 381 #33): In a simple model of a shear layer the only nonzero component of U is U(z) along the xaxis. U(z) is zero for z < 0, increases linearly to Uo between z = 0 and z = h (> 0) and has the constant value Uo for z > h. The speed of sound is constant throughout. A ray starts from (0, 0, d) at an angle fA to the horizontal. Show that it passes through the layer if cos01 < (1 + M0 ) I, and that the difference in abscissas of its points of entry to and exit from the later is (1 + M0cos9Jtan92 ] + 2M0 ( sec91 [ cos91 (cosh 1 sec91 :cosh 1 sec92)] 2M0 where sec fh = sec fA Mo. Show that if z > h L dcosec91 tan92 tan91 ) + (z h)cosec92 2M0 A graphical representation of the shear is shown in Figure 4.1. 54
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h Figure 4.1. Graphical Representation of Shear Flow Solution: A graphical representation of the problem is shown in Figure 4.2. Figure 4.2. Graphical Representation of Part I of Problem Using equation 20.4 (Jones, 1989) (4 .1) 55
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where for this problem a= 1 c 1 1 = cos(90 el) sin91 Substituting the above into equation 4.1 gives dx ( u) 1 = cos91 + ____.. dz c sin91 (4.2) Now, using equation 20.8 (Jones, 1989) (4. 3) where for this problem a = 1 Mx uo c c2 = 0 cosec93 = 1 1 1 = = sin(90 el) Substituting the above into equation 4.3 gives 1. ( 4. 4) Initially, let Substituting these initial conditions into equation 4.4 As the ray advances into the moving fluid, equation 4.4 becomes 56
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cos9u ( 1 + 1 cosel 1 u 1 += cosel c cos eli 1 1 u = cosel coseli c secel seceli u = c 1 u c The ray will emerge if 91f > 0 or cos91f < 1 Then, for the emerging ray, equation 4.5 yields 1 < 1 seceli u c seceli u > 1 c 1 u >1+cos eli c coseli < ( 1 + Jl or, in the terminology of Jones (1989), For the emerging ray, equation 4.2 becomes dx ( U) 1 = cosel +. dz c s1n81 Substituting the trigonometric identity and equation 4.5 into equation, 4.6 gives 57 ( 4. 5) (4. 6)
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Substituting U = 0 + U0 ( into equation 4. 7 yields dx [( 1 J + ( Uo dz sec9u ( U0jc )( zjh) c h [ [sec9u (u.;c)(zfh)f] dx 1 (uofc)(zjh) dz = seceli (uofc){zjh)r1 + [ seceli (uofc){zjh)r dx 1 dz = J[ sec91i ( U0jc)(zjh) r 1 + (uofc){zjh) In order to make some simplifications, expand 58
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Use the following trigonometric identity in the above equation 2 2 sec eli 1 = tan eli to get Substituting this into equation 4.8 = ( J:;=( =cuho=)=z=zz===d=z=(=)==== + 2seceli z + tan2 eli ( h )( uo J e r zdz Uofc ch sec u o ( J z' 2sec9u (;:;, )z + tan' 9u ( )( J Let a = b = 2sece (51_) h. ch 2 e = tan eli. ( 4. 9) Again, to further simplify, substitute the above into equation 4.9 1 ( c1h dz 1h zdz 1 h z2dz ) Ax = ca "Va o 'az2 + asec9H ;:::.======= a "Va 'V + bz + e o.Jaz2 + bz + e o.Jaz2 + bz + e (4 .10) From Korn and Korn (1961) the following are the pertinent integrals for this problem 162 I dx = ;:_a log(2ax + b + 2.Ja.Jax2 + bx +c) for a > 0 .Jax2 + bx + c "Va 59
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164 165 J xdx = .Jax2 + bx + c ]:_ .J ax2 + bx + c _.E._ J dx a 2 a .J ax2 + bx + c n1 = .J ax2 + bx + c an b(2n 1) J xn1dx 2an .Jax2 + bx + c c(n 1) J xn2dx an .J ax2 + bx + c Using integral 164 and n = 2, 165 becomes Now, substituting the appropriate equations into equation 4.10 gives 1 (.ra. r dz ) = .ra a o.Jaz2 +bz + e + J:_ [asec9u .Jaz2 + bz + e _E._ r dz )] a 2a o.Jaz2 + bz + e }a 2za ).Jaz2 + bz + e + }a ).Jaz2 + bz + e 1 r:( 3b2 )lh ctz 1 c( c )1h dz ava +ava 8a2 o.Jaz2 + bz + e .ra 2a o.Jaz2 + bz + e 1 r dz ) = a 0 .Jaz2 + bz + e + ;:._ { sec911 .J az2 + bz + e) :;.._ ( b sec9u J.h .J dz ) va va 2 az2 + bz + e Ja ( z ).Jaz2 + bz + e + }a ( ).Jaz2 + bz + e 1 ( 3b2 )r ctz 1 dz .ra a.J"a 0 .Jaz2 + bz + e + .ra 20 .Jaz2 + bz + e 60
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L1x = (.Jab sed)l. 3b2 + c.Ja)r dz + .Ja 2 s.Ja 2 o.Jaz2 + bz + e ( secSli ;; z + )v'az 2 + bz + e L1x = (.Jab seeS. 3b2 + c.Ja) .,fa 2 11 s.Ja 2 log(2az + b + 2.Ja.Jaz2 + bz +e)+ ( secSu ;; z + 4 3"Ja )v'az 2 + bz + e 1 ( b 3b2 c) L1x = .Ja 1 2Ja. secSli + 2 log(2az + b + 2.Ja.Jaz2 + bz +e)+ 1 ( .Ja 3b )"" 2 Ja secSli 2 z + 4Ja. az + bz + e. Evaluating the above between 0 and h yields 1 ( b 3b2 c) L1x = .Ja 12Ja. secSli + 2 log(2ah + b + 2Ja.Jah2 + bh +e)(1___E._ secSu 3 b 2 + log(b + 2Ja..Je) + .Ja 2Ja. Sa 2 .Ja h + + bh + e .Ja h 2 4Ja. }a ( secSu + 4 3"Ja ).Je. Now, substituting back in for a, b and e 61
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Ax = ( u:jc) ( 1 + sec2 ali % sec2 ali + tan2 ali) log[ 2h( 2seceu( + 2( ( )' 2seceu( + tan' au]To simplify this expression, we note that 1 2 sec 8. 2 h 1 2 8 1sec 2 h 1 2 1sec e. 2 11 1 2 1 2 +tan eli = 1tan e. 2 2 h 1 2 1 +tan eli = 2 2 1 1 2 +tan e. 2 2 h Adding the above simplifications and expanding the log function gives Ax= (u:jc) + (u:;c) mlog[ seceu + seceu r 1] secaa)+ secali J secali r1 + From Korn and Korn (1961) 62
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1 cosh z Letting z = ( ) sec91i the Ax equation becomes = ( h ) cosh1 [( uo) sec9H J ( h ) log( tan9u sec911 ) 2 U0/C C 2 U0/C + seceli] seceli r1 + h u From an earlier relationship, sec91 = sec9li we substitute a this into the Ax equation to obtain h 1 [ ] h 1 ( ) = ( I ) cosh sec91 ( I ) cosh sec9H 2 uo c 2 uo c ( h ) [( uo) + seceli Jtan91 + ( h ) sec91itaneli 2 U01c c 2 U01C = 2 ( seceli { taneli [ 1 + ( )cos9u Jtan91 } + ( h ) seceli{coseu[cosh1 (seceJcosh1 (sec9u)]}. 2 U01c (4 .11) This is the solution for the difference in abscissae of the points of entry to and exit from the layer with 91i = 91 and 91 = 92 4 .1.2 Part IB The last part of the first portion of this problem is to show that if z > h 63
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L(x) = where z = z h and X u cos92 + ___Q_ _d_ + .L\x + co ( z h). tan91 s1.n82 (4 .12) Substituting these parameters and equation 4.11 into equation 4.12 gives ( + cos92 ( d J L(x) = U + 1 + ___Q_ cos92 1 + Uo cos92 tanel co cos92 ( h ) u secel tanel 1+___Q_cos9 2(Uo/c) 2 Notice that u cos92 + ___Q_ ____ c=o ( z h) sin92 64 (4.13)
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= = = u 1 + ____Q_ case2 co case2 1 + uo easel co 1 uo case 1 co easel 1uo case 1 co 1 uo case 1 co where the given relationship sece2 = secel uo has been used. co Substituting equation 4.14 into equation 4.13 yields 65 (4 .14)
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L( ) (z h)sin82 8 ( d ) X = +COS 1 + 1 + Uo cos8z tan81 co cos81 ( ( h ) sec81 )tan81 2 U0/C cose1 ( 2 ( u:;c) sec81 ){[ 1 + ( )cose1 ]tane,} + cos81 ( 2 ( u:;c) sece1 )l cose1 cosh 1 ( sece,) l cos81 ( 2( u:;c) sece1 )l cos81cosh1 ( sece1) l + u cos82 + _o ____ c.:=o ( z h) sin82 L(x) = (z h) 66
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cos92 (cosB2 + L(x) = (z h) sin9z + o + 1 + cos9, sin92 ( 1 + cos92 ) dcosec91cos291 + ( ( h ) )tan91 2 Uo/c ( 2( u:jc) }an9, e )cos9,tan9, + ( 2 ( u: /c)) [ cos911og( sec92 + sec'e, 1} ]( z( u:jc)) [ cos911og { sec91 + sec'e1 1}] sin292 + cos292 + Uo cos92 L (X) = ( z h) + Co + sin92 ( 1 + cos92 ) dcosec91 (1 sin291 ) + ( ( h ) )( tan91 tan92 ) 2 U0/c ( )cos91tan92 + ( ( h ))[cos91log(sec92 +tan 92)]2 U0/C ( 2 ( u:jc)) [ cos911og( sec91 + tan 91 ) ] 67
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L(x) = dcosec91 dsin91 + (z h)cosec92 + ( 2 (u:jc) }tane,tane,)+ ( 2 ( u:jc) )[ cos6,log(sec6, + tan e, + 1 1))( ( h ))[cos91log(sec91 + tan91 + 11)] 2 U0/C L(x) = dcosec91 dsin91 + (z h)cosec92 + ( 2 ( u:jc)} tane, tane, ) ( )cosO,tane, + ( ( h ))[cos91log(1 + tan92)log(sec92 1)]2 Uo/c ( ( h ) )[ cos91log(1 + tan el )log(secel 1) ]. 2 U0/C ( 4 .15) Using the power series expansion for the log function (Korn and Korn 1961) log(l + z) zz zJ z4 = z+2 3 4 in equation 4.15 in linear terms only gives L(x) = dcosec91 dsin91 + (z h)cosec92 + ( 2 (u:jc) }tane, tane,))cos6,tan6, ( ( h ) )[ cos91tan92sec92 ] + 2 U0/C ( ( h ) )[ cosel tan elsecel] 2 U0/C 68
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L(x) = dcosec81 dsin81 + (z h)cosec82 + ( 2 ( } tan9, tan92 ) (:) cos9, tan92 ( 2 ( cos9,tan92 ( sec9, + ( 2 ( tan 91 L(x) = dcosec81 dsin81 + (z h)cosec82 + ( 2( u:;c)} tan9, tan9,) (: )cos9,tan9, ( 2 ( U:/c)) cos9,tan9,sec9,] + ( 2 ( )[ ( )cos9,tan92 ] + ( 2 ( U:/c)) tan 9, L(x) = dcosec81 dsin81 + (z h)cosec82 + ( ( h ) ) ( tan81 tan82 ) + ( ( h ) ) (tan 81 tan82 ) 2 U0/C 2 U0/C ( )cos81tan82 + ( )cos81tan82 L(x) = dcosec81 ( h ) ( tan82 tan81 ) + (z """ h)cosec82 Uo/c dsin81 This is the result from above except for the dsin81 This section solved the problem in Jones (1989) The next section will expand on the result of the first section. 69
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4 .1.3 Part II The second portion of this problem is to find the.equation which determines the distance from the point of entry to the point of exit of the ray trajectory. The graphical representation of this portion is shown in Figure 4.3. Figure 4.3. Graphical Representation of Part 2 of Problem The solution to the problem from Jones (1989), equation 4.11, gives the distance from a sound source that penetrates the layer (refer to Figure 4.2). The angle passed through the layer if cos81 < 1 + uo 1 Therefore, in order for the ray to travel in a parabolic path in the medium, The distance the ray travels is give by (4 .16) 70
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This is the solution for the second part of the problem. This chapter has explored some applications to theory which will be useful in the proposed laboratory experiment in the following chapter. 71
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5. Proposed Experiment This last chapter is devoted to proposing a laboratory experiment to predict the distance of the sound propagation for a simple model of a shear layer as used in Chapter 4. Figure 5.1 depicts this linear shear flow. h ' Figure 5.1. Linear Shear Flow The propagation of the ray in the medium is shown in Figure 5.2. 72
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Figure 5.2. Graphical Representation of Ray Trajectory The equation to be used is from Chapter 4, equation 4.16, which is The parameters for this equation are as follows d=0.2m h=0.4m Uo 5 m/s c0 = 343.2 m/s (Speed of sound in Denver, Colorado). al is determined from the critical angle calculation 1 = 1 + uo co Substituting the appropriate parameters into equation 5.2 gives cos a1 cos e1 1 + 1 5 343.2 1 1. 014568764 73 (5 .1) (5.2)
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cos e1 = o.985640436 el = 9.721411176 degrees. Now, substituting all parameters into equation 5.1 yields D = 0 2 m + 0 4 m sec{9. 721411176) tan(9.721411176) { tan{9. 721411176) cos(9. 721411176)cosh1 [ sec(9. 721411176)]} D = 0 2 m + ( 0 4 m ) {1. 014568765) 0.171317769 0. 014568764 [ 0.171317769{0. 985640436)cosh1(1. 014568765)] D = 1.167421226m + (27. 45600107m)(l. 014568765) [ 0. 171317769 ( 0. 985640436)( 0. 170490627)] D = 1.167421226m + (27. 8560011rn) (0.171317769 0.168042455) D = 1.167421226m + (27. 8560011rn) (0. 003275313) D = 1.167421226m + 0. 091237124rn D = 1. 25865835m D = 1.26m Another set of parameters for equation 5.1 are d=0.2m h=0.4m Uo = 10 m/s c0 = 343.2 m/s (Speed of sound in Denver, Colorado). Again, 81 is determined from the critical angle calculation using equation 5.2. Substituting the appropriate parameters gives 1 + 1 10 343.2 1 1. 029137529 cos e1 o.971687429 el = 13.66650012 degrees. Now, substituting these parameters into equation 5.1 yields 0 = 0 2m + 0 4rn sec(13. 66650012) tan{13. 66650012) (10/343. 2) { tan{13. 66650012) cos(13. 66650012)cosh1 [ sec(13. 66650012)]} 74
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D = 0 Zm + 0 4 m (1. 029137529) 0.243154382 (0.029137529) [o. 243154382(0. 971687429)cosh1(1. 029137529)] D = 0. 822522704m + (13. 72800006m)(l. 0129137529) [0.243154382(0.971687429)(0.240819928)] D = 0. 822522704m + (13. 90528005m) (0. 243154382 0. 234001696) D = 0. 822522704m + (13. 90528005m) (0. 009152685) D = 0.822522704m + 0.127270652m D = 0.949793356m D = 0.95m These theoretical results are to be compared in a future publication to the actual experimental results. Sound propagation has useful applications in various fields of study. The parabolic approximation in this form has been applied to study many aspects of underwater acoustic propagation (McDaniel, 1979.) Other areas of interest are vorticity problems and chaotic problems to name a few. 75
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Appendix A. Further Derivations Conservation of Mass Equation For a fixed volume V inside a fluid, the rest mass in V at any time t can be taken as the volume integral of density p(x, t) representing a local average (or expected value) of mass per unit volume in the vicinity of a spatial point x Conservation of mass requires the time rate of change of the mass to equal the net mass per unit time entering (minus that leaving) the volume V through the confining surface S. The net mass per unit time leaving through a small area element As with outward unit normal vector ii (x5 ) and centered at point on S is identified x5 as (A.l1) v ( x, t) is the fluid velocity at x, defined as the massweighted local average particle velocity or as the average momentum per unit mass in the vicinity of X. (X5 refers to a point on the surface.) The net mass leaving V per unit time is the surface integral over s of pv ii, so the conservation of mass requires ddt fffv pdv = ffs pv. nds. (A.l2) The surface integral can be written as a volume integral by using Gauss' theorem Jf5A iidS = JJJv V AdV JJfv V (pv) Jdv = o. (A.l3) (A.l4) This equation implies that the average value of the integrand is zero for an arbitrary volume V, so the integrand itself must be zero ap V(pv) = o. at 76 (A.l5)
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Conservation of Momentum Equation: A general law of classical continuum mechanics is that the mass times acceleration of center of mass of a fluid particle equals the net apparent force exerted on it by its environment and by external bodies. A fluid particle consists of all fluid within some moving volume V(t), each point on the surface of which is moving with the local fluid velocity v(x8,t). Since the mass in such a fluid particle is constant, mass times center of mass acceleration is the time rate of change of momentum (volume integral of pv) within the particle, so (A.21) f8 is apparent surface force per unit area exerted by the particle's immediate environment. f8 is the body force, e.g., that due to gravity per unit volume. Although gravity is always present, it has negligible influence on acoustic disturbances at all but extremely low frequencies (i.e. those of order or less than g/c.) Therefore, the body force term is neglected. The classical assumption regarding f8 is that it is directed normal into the surface S, that is (A.22) with the magnitude p of this force per unit area identified as the pressure. The adoption of this relation, holding ideally for static equilibrium (hydrostatics), implies a neglect of viscosity. The lack of dependence of the pressure p(x, t) on the orientation of As, that is, the direction of n, may be regarded as a hypothesis but also follows from a fundamental requirement that the net surface force divided by the mass of a fluid particle on which it acts should remain finite in the limit as the particle volume goes to zero. That fs reverses direction when n reverses direction is consistent with Newton's third law. Jfs f8dA = Jfs npdS = Jff v VpdV. Vp is the equivalent force per unit volume due to pressure. The time rate of change of momentum term in the first equation (A.21) can be expressed as a volume integral ctv dt = av at av dx +ax dt av cty ++ 8y dt av dz az dt 77 (A.23)
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ctv ov V)v = + (v dt Ot ctv ov = ctt Dt where D a v v. + Dt Ot Thus ct ov dt JJJ vvdV = JJJ v p Dt dV which represents an instance of Reynolds transport theorem. Inserting equations A.23 and A.26 into equation A.2l with a neglect of the body force term gives Since the integrand is zero ov p+ Vp = 0 Dt p[: +(vV)v]+Vp = o. PressureDensity Relationship (A.24) (A.25) (A. 26) (A.27) (A.28) The classical model of a compressible fluid presumes the existence of some definite relation p = p(p) (A.3l) between density and pressure. Employing a Taylorseries expansion in p' (A.32) 78
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P = Po + ( P pJ :: ( pJ cz = : (pJ The following are selected problems from Jones {1989) 28. Prove for a ray that a) dL/ds = N b) d(gradL)/ds = gradN a) dL/ds = ( gradL) s {A.33) {A.34) {A.35) dL/ds = ( gradL) (1/N )( gradL) dL/ds = (1/N)(gradL) (gradL) {Jones 1989, Eqn. 16.1) dL/ds = (1/N)(gractLl dL/ d s = ( 1/N) N2 dL/ds = N {Jones 1989, Eqn. 15.4) b) ct(8Lj8x)/cts = grad(8L/8x) s d(8Lj8x)/cts = grad(8Lj8x) (1/N)(gradL) {Jones 1989 Eqn. 16.1) ct(8Lj8x)/cts = (1/N)grad(8L/8x) (gradL) d(8Lj8x)/cts = (1/N) 8/8x grad(L) (gradL) d( 8Lj8x)/cts = (1/N) grad(L) 8/8x ( gradL) 8j8x ( gradL)2 = 2( gradL) 8/8x ( gradL) ct(8L/8x)/cts = (ljN)(1j2) 8/8x (gradL)2 d(oLjox)/cts = (ljN)(1j2) oj8x N2 (Jones 1989 Eqn. 15.4) 8j8x N2 = 2 N 8/8x N ct( 8Lj8x)/cts = (1/N )(1j2)(2N 8j8x N) 79
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ct(8L/8x)/cts = 8/8x N Using similar arguments for 8/8y and 8/8z we find d(8Lj8x)/cts = 8/8y N d(8L/8x)/cts = 8/8zN Putting them together we have d(gradL)/ds = gradN 29. If x is a point of a ray and 1C = k0gradL prove that a) dxjds = K/k0N b) dK/ds = k0gradN a) dx/ds = gradx s dx/ds = (1/N) ( gradL) (Jones 1989 Eqn. 15.4) dxjds = Kjk0N b) dK/ds = d (k0gradL)/cts dK/ds = k0d ( gradL)/ds dK/ds = k0gradN (Problem 28b see above) 32. If 1C = k0gradL for a ray in a moving fluid and dr = (1 M gradL)dcr show that ctxjctr = :M + aiC/IKI What is the corresponding formula for dK/dr ? a) dxjdcr = a2gradL + (1 M gradL)M {Jones 1989 Eqn. 19.13) dx/d't = ( dxjdcr )( dcr/dT) dx/dr = [ a2gradL + (1M gradL)M ][ 1/(1M gradL)] dx/dr = a2gradL/(1 M gradL) + M dxjdr = a2gradL/algradLI + M {Jones 1989 Eqn. 19.11) 80
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b) dK/dcr = d(k0gradL)/ctcr dK/dcr = k0 d(gradL)/dcr dK/dr = ( dK/dcr )( dcrfdr) dK/dr = k0 [ d(gradL)/dcr ][ 1/(1M gradL)] From Jones (1989), equation 19.14 is dtc/d't = k0 [ agrad2Lgrada. (1M gractL){ grad(M gradL) (:M grad)gractL}] [J/(1M gradL)] dtc/d't = k0[(a.grad2Lgracta)/(1M gradL){grad(M gradL)(:M grad)gractL}] From Jones (1989), equation 19.11 is dtc/d't = k0 [ ( a.grad2Lgrada. )/( a.J gradLj) { grad(M gradL) (:M grad)gractL}] dtc/d't = k0[(gradLgracta.)(i/JtcJ){gract(:M gradL)(:M grad)gractL}] From part (a) dtc/d't = k0[{1fa.){gradLgrada.)(ctxjd't:M){gract(:M gradL)(:M grad)gractL}] 81
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References Dowling AP and Ffowcs Williams JE. 1983. Sound and Sources of Sound. New York: John Wiley & Sons. Frank PG, Bergmann PG, and Yaspan A. n.d. Ray Acoustics, Summary Tech Rept. Division 6, NDRC, 8, 41 49, 51 54, 59 68. Jones DS. 1989. Acoustic and Electromagnetic Waves. New York: Oxford University Press. Korn GA and Korn TM. 1961. Mathematical Handbook for Scientists and Engineers. New York: McGrawHill Book Company, Inc. McDaniel, ST. 1979 Jan. Application of the parabolic approximation to predict acoustical propagation in the ocean. Am J Phys; 47(1). Mitton Von Dyke 1982. An Album of Fluid Motion. Stanford: Parabolic Press. Pierce AD. 1981. Acoustics An Introduction to Its Physical Principles and Applications. New York: McGrawHill Book Company. Tritton DJ. 1977. Physical Fluid Dynamics. New York: Van Nostrand Reinhold Company Ltd.

