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Fluid dynamics of a crystallizing particle in a rotating liquid sphere
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Fluid dynamics of a crystallizing particle in a rotating liquid sphere
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Content
FLUID DYNAMICS OF A CRYSTALLIZING PARTICLE
IN A ROTATING LIQUID SPHERE
by
Channarong Asavatesanupap
A Dissertation Presented to the
FACULTY OF THE GRADUATE SCHOOL
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Fulfillment of the
Requirements for the Degree
DOCTOR OF PHILOSOPHY
(AEROSPACE AND MECHANICAL ENGINEERING)
May 2007
Copyright 2007 Channarong Asavatesanupap
ii
Acknowledgements
I would like to thank all the people who have made this dissertation possible.
First, I thank my advisor, Prof. Satwindar Sadhal, for his continuous support during my
Ph.D study. He taught me how to approach a research problem and always provided
valuable advice. Without his encouragement and constant guidance, I could not have
finished this dissertation. Also, I would like to thank the rest of my thesis committee,
Prof. Larry Redekopp, Prof. Fokion Egolfopoulos, and Prof. Katherine Shing for their
constructive and generous guidance. Besides, I am very grateful to the AME staff for
their kind assistnace, especially Merrietta Penoliar. Let me also say “ thank you” to
all my friends who made me feel like home during five years of my study abroad.
A special thanks goes to the Royal Thai Government for financial support and
the Aerospace and Mechanical Engineering Department for teaching assistant offer.
Finally, I thank my parents and my brothers who always give me not only
their love but also their unconditional support and encouragement. My last thanks goes
to Malee Santikunaporn who is always there for me.
Table of Contents
Acknowledgements ii
List of Figures v
Nomenclature vii
Abstract ix
Chapter 1: Introduction 1
Chapter 2: Particle motion and path calculation: Equatorial Plane Case 8
2.1 Statement of the problem 8
2.2 The method of solution 12
2.2.1 Axisymmetrical problem 14
2.2.2 Asymmetrical problem 17
2.3 Drag force, torque and velocity of a crystallizing particle 20
2.4 A note on the case of two concentric spheres 24
2.5 Particle motion inside the rotating drop 26
2.6 Results and discussion 28
2.6.1 Force and Torque on a translating particle 28
2.6.2 Flow field 31
2.6.3 Time histories 35
Chapter 3: Particle motion and path calculation: Off-Center Case 40
3.1 Superposition solutions 42
3.2 Results and discussion 46
Chapter 4: Transient dynamics of a rotating liquid sphere 52
4.1 Introduction 52
4.2 Problem Statement and Method of solution 53
4.3 Case of a decelerating liquid drop 55
4.3.1 Limit of 0 → δ 57
4.3.2 The general solution in integral representation form 58
4.3.3 Small-time approximation 60
4.4 Case of a rotation startup 65
4.4.1 Limit of 0 → δ 67
4.4.2 Small-time approximation 68
4.5 Results and discussion 68
4.5.1 Deceleration of a rotating drop 70
4.5.2 Drop rotation startup 75
Chapter 5: Conclusions 78
Chapter 6: Suggestions for future works 80
iii
Bibliography 83
Appendices
Appendix A. Derivation of the set of equations for 87
m
n
m
n
m
n
m
n
H C B A ..., , , ,
Appendix B. Coefficients and useful formulae in
transient dynamics problems 97
iv
List of Figures
Figure 1.1: Lysozyme crystals grown inside a levitated drop. 4
Figure 2.1: Schematic of the crystallizing particle of radius suspending
p
R
within the levitated drop of radius rotating about a horizontal
d
R
axis with angular velocity Ω. 9
Figure 2.2: Schematic of the system. 13
Figure 2.3: Configurations of the decomposed systems. 16
Figure 2.4: Bipolar coordinates. 18
Figure 2.5: Scaled drag force plotted against eccentricity e,
*
z
F
for given radius ratios, in axisymmetrical case. 29
Figure 2.6: Scaled drag force plotted against eccentricity e,
*
x
F
for given radius ratios, in asymmetrical case. 30
Figure 2.7: Scaled torque plotted against eccentricity e,
*
y
T
for given radius ratios, in asymmetrical problem. 31
Figure 2.8: Velocity field for the axisymmetrical problem
with 1 = for
p
V 5 / 1 = k. 33
Figure 2.9: Velocity field for the asymmetrical problem
with for 1 =
p
V 5 / 1 = k. 34
Figure 2.10: The trajectories of a particle inside the rotating drop after time
T =30s for the case =1/40 , k
d p
ρ ρ / =1.2, and µ =1.5x10
-3
Pa.s
(~protein solution) with constant clockwise rotations. 37
Figure 2.11: The trajectory of a particle inside the rotating drop for the case
=1/5, k
d p
ρ ρ / =1.2, and µ = 1x10
-1
s Pa ⋅ after time T =20s. 38
Figure 2.12: Time of the particle orbits corresponding to a freely rotating drop
with k =1/5,
d p
ρ ρ / =1.2, and µ = 1x10
-1
Pa.s , =0.8 rps
init
Ω
after time T =20s. 39
Figure 3.1: Sketch of a liquid drop rotating about y-axis with angular velocity Ω
and having a spherical particle positioning at coordinates ( )
where b is a small distance off the equatorial plane. 41
c b a , ,
v
Figure 3.2: Schematics of decomposed systems on a plane with the angle χ
to xy-plane (a) an axis of rotation through the line o o ′
(b) an axis of rotation perpendicular to o o ′ 43
Figure 3.3: The motion of a particle initially positioned at ,
for ,
) 3 . 0 , 05 . 0 , 0 (
d d
R R
40 / 1 = k
d p
ρ ρ / =1.2, and Ω =1.5 rps. 48
Figure 3.4: The motion of a particle initially positioned at ,
for ,
) 3 . 0 , 05 . 0 , 0 (
d d
R R
40 / 1 = k
d p
ρ ρ / =1.2, and Ω =3 rps 49
Figure 3.5: The motion of a particle initially positioned at ,
for ,
) 3 . 0 , 05 . 0 , 0 (
d d
R R
40 / 1 = k
d p
ρ ρ / =0.8, and Ω =1.5 rps 50
Figure 3.6: The motion of a particle initially positioned at ,
for ,
) 3 . 0 , 05 . 0 , 0 (
d d
R R
40 / 1 = k
d p
ρ ρ / =0.8, and Ω =3.0 rps 51
Figure 4.1: Velocity profile of a rotating water drop initially rotating
with in a gas. 70
o
Ω
Figure 4.2: Deviation of velocity profile from a solid body rotation. 71
Figure 4.3: Interfacial shear stress vs τ ˆ / t. 73
Figure 4.4: Deviation from solid-body rotation of various liquid drops at t =0.3 τ ˆ 73
Figure 4.5: Velocity fields in gaseous medium induced by a freely rotating solid
sphere ( 0012 . 0 = β). 74
Figure 4.6: Velocity fields of a liquid drop rotating under constant shear stress
θ σ sin
o
in an infinite medium ( 0186 . 0 = δ) 76
Figure 4.7: Tangential stresses at the surface of a water drop under a steady torque 76
Figure 4.8: Velocity field in a gaseous medium induced by a rotating solid sphere
under a steady shear stress θ σ sin
o
( 0012 . 0 = β ). 77
vi
Nomenclature
j
D Diffusion coefficient for component j
e eccentricity
F Force
g Gravitational acceleration
I Moment of inertia
k Radius ratio
L Laplace transform operator
-1
L Inverse Laplace transform operator
m
n
L Linear operator
n Unit normal vector
p Pressure
R Radius of a sphere
R Radius ratio
Re Reynolds number
s Unit tangential vector
s s-domain
T Torque
Ta Taylor number
U Velocity
v Velocity vector
x Position vector
α Viscous drag coefficient in x direction
vii
β Viscous drag coefficient in z direction
j
β Coefficient of solutal expansion for component j
δ Dynamics viscosity ratio
µ Dynamic viscosity
ψ Stream function
κ Kinematic viscosity ratio
σ Shear stress
Ω Angular velocity
ρ Density
ν Kinematic viscosity
2
∇ Laplace operator
τ Shear stress
τ Diffusion time scale
viii
ix
Abstract
In this investigation, fluid dynamics of a spherical particle in a rotating liquid
sphere is studied. Such problems arise during protein crystal growth by the
containerless approach in which a drop of protein solution is levitated against gravity,
and rotated steadily by a torque from an acoustic field. Viscosity of the solution is
assumed to be sufficiently high such that viscous force dominates and the second-order
effects in the acoustic field do not penetrate the drop surface. The analysis is carried
out at very small Reynolds and Taylor numbers for which the fluid flow is treated by
Stokes approximation. With a given geometric configuration, the velocity field is
calculated and used to determine the particle velocity at a given time. By integrating
the particle velocity, the particle path is obtained and used for optimizing the
rotation rate of the drop in an attempt to keep the particle fully contained in the
drop.
In addition, the transient dynamics of a rotating levitated liquid drop in
unbounded gaseous fluid is investigated. A liquid drop is considered to
experience two different fundamental transients, one when the torque is turned
off and another when the torque is turned on. The motion of the fluids in the
drop interior and surroundings is described by the Stokes equations with the
time-derivative term included. The solutions are obtained analytically by the
Laplace transform technique. The results show that the transient effects caused
by the applied torque are relatively small for the high-viscosity solution
considered. Hence, for most applications of containerless protein crystal
x
growth, the liquid drop may be assumed to rotate like a solid body while
allowing the solid particle relative motion within that framework.
1
CHAPTER 1
Introduction
It is now well established that the information about the structure of
proteins hold the key to various biological processes and plays a vital role in the
human body. By understanding the molecular structure of proteins, scientists
can learn how these proteins work in humans and other biological systems.
Therefore, protein crystal growth has gained significant scientific attention in
recent years. In order to learn how these proteins function, high-quality crystals
of selected proteins have to be produced so that scientists are able to study their
molecular structures in detail.
Among the large number of various proteins, obtaining information
about the precise molecular structure is not a straightforward task. In many
instances, larger and better-ordered crystals are highly favorable for acquiring
knowledge about their respective molecular structures [1]. However, there are
several limitations in the crystal growth process for obtaining the desired quality
crystals, caused by external conditions such as buoyancy-driven convection and
sedimentation. It is therefore very important to understand the fluid dynamics
involved in the crystal growth process.
2
For the purpose of growing large and uniform crystals, the main obstacle
is convection around a growing crystal in a solution. During protein
crystallization, first a supersaturated state of the solution is attained by allowing
the solute to be depleted by some means (by evaporation, for example). Then,
by homogeneous or heterogeneous nucleation, the crystallization process
begins. Therefore, density gradients are created at crystal-solution interface due
to mass-transport process and lead to buoyancy-driven convection in a
gravitational environment. Such effects are often undesirable and negatively
impact the quality of growing protein crystals. There is evidence that the
growth rate decreases under normal gravity conditions, and the crystal
somewhat resists further growth when it grows to a certain size[2-3]. Moreover
the convective flow enhances impurities in the crystallization process[4-5]. In
addition to buoyancy-driven convection, sedimentation of crystals seems to be
another negative factor in growing protein crystals in a gravity field. Hence, a
microgravity environment is believed to be a suitable condition for improving
crystal quality[6-7]. A review of the protein crystal growth under microgravity
and earth-gravity conditions has been given by Sadhal & Trinh [8].
In order to eliminate convection and sedimentation, microgravity offers
the optimal environment where scientists are able to conduct experiments for
growing high quality crystals. It has been recently demonstrated that protein
crystal growth in a low-gravity environment (such as on the space shuttle) has
resulted in larger and better-quality crystals for numerous proteins [9-10]. This
is owing to the suppression of buoyancy-driven flow that is often disruptive for
good-quality crystals. Although the experiments in protein crystal growth,
which were performed aboard the space shuttle, provide favorable conditions,
the high cost, the experimentation time and its accessibility limit the ability to
perform an optimum experiment.
With earth-based experiments[11-13], some aspects of low-gravity
conditions can be effectively achieved by neutralizing some elements of
gravitational effects. In an earlier study [14], microgravity simulation by a
containerless approach has been investigated and the results have shown a
significant degree of success. With the main purpose of minimizing the
convection, a rotating electrostatically levitated drop of crystal solution (about
17 l µ ) was used to simulate low gravity conditions in a time-average sense
(similar to a clinostat), and provide some aspects of low gravity. In addition to
reducing convection and sedimentation, the containerless approach offers the
avoidance of contaminants from wall contact. The crystallization process in the
levitated drop also provides an inherent symmetry not available in containered
systems.
While this procedure has been successful, it is necessary to carefully
study the effect of various control parameters such as the rotational velocity of
the drop so as to sustain long-term suspension for crystallizing particle. This
procedure of containerless crystal growth, while providing various advantages
over using crucibles, does present additional technical challenges from the fluid
dynamics point of view. For example, the long-term suspension without
tumbling and slipping motion is required for the growth of good-quality
3
crystals. To this end, the use of an additive, such as a very low concentration of
agarose, has been made and yielded a more viscous solution. Small crystals
were reportedly grown from 5 up to 300 m µ in size under the controlled system
(see Fig. 1).
FIGURE 1.1: Lysozyme crystals grown inside a levitated drop (Chung et al , 1998)
The purpose of this thesis is to provide fundamental understanding of
the dynamics of the solid particle in a rotating liquid drop. One of the goals is
to calculate the appropriate rotational velocity of the system at various stages of
crystal growth that would create a minimal level of convective transport (fluid
motion in the drop) so that the best possible conditions for crystal growth are
attained in a gravitational environment. Although the problem we are dealing
4
5
with is somewhat limited and considerably idealized, it is a starting point for
more complex problems. In particular, we consider a spherical liquid drop
holding a smaller solid particle (the crystal) while the drop is levitated
electrostatically and rotated about a horizontal axis by ultrasound forces. The
resulting fluid dynamics of the system is described by a complex mathematical
model in a rotating reference frame. Various additional challenges arise due to
the eccentric positioning of the solid particle (which is idealized as a sphere as
well) inside the drop. The aim is to obtain the fluid velocity distribution in the
rotating drop while the particle (which is slightly denser than the liquid)
experiences spiral motion in the presence of a gravitational field. The ideal
situation would be a controlled rotation rate so that the particle neither sinks to
the bottom due to gravity nor gets expelled by the rotational centrifugal force.
In this study, the fluid motion inside the drop is assumed to be in the
limit of low Reynolds number owing to the highly viscous solution and small
scale of the liquid drop. In the flow regime under consideration, the convective
transport as well as the unsteady accumulation term is negligibly small
compared to viscous term. Additionally, the coriolis force, which arises in the
rotating reference frame, is also assumed to be relatively weak compared to
other forces. Therefore, the flow in these circumstances is governed by Stokes
equations and referred as quasi-steady motion. Many studies of the motion in
rotating viscous fluid have been carried out and are available elsewhere [15-19].
Since the problem involves two non-intersecting eccentric spheres, the semi-
analytical solution in the limit of these models is found by using a bipolar
6
coordinate system. The bipolar coordinate system was used to deal with this
class of the problems by several authors. For example, Stimson & Jeffery [20]
used this type of coordinate system for Stokes flow past two spheres in 1926.
O’Neill [21] studied the motion of a viscous liquid caused by the moving solid
sphere bounded by an infinite rigid plane in 1964, followed by Goldman et al
[22] in 1966 for the treatment of the slow motion of two identical spheres
through a viscous fluid. Later on, in 1970, O’Neill [23] studied the
asymmetrical fluid motion caused by the translation or rotation of two spheres.
A fairly general class of solutions of Stokes’ equation for a sphere near plane
has been given by Lee & Leal [24] in 1980. The problems concerning
compound multiphase drops have been investigated by Sadhal & Oguz [25] and
Oguz & Sadhal [26] in 1985 and 1987 respectively. The thermocapillary
migration of a bubble has been treated by Shankar & Subramanian [27] and
Meyyappan & Subramanian [28] in 1982 and 1987, respectively. The velocity
field obtained in this coordinate system allows us to calculate the hydrodynamic
forces and the torque magnitudes on the particle. At any given time, the
instantaneous velocity of the particle is determined by the external forces and
torques on the particle and the geometric configuration of the system. With the
magnitude and direction of velocity known, incremental changes in particle
position can be obtained for small time steps. By repeating this procedure, the
particle path can be now predicted and used for optimizing the rotation rate of
the drop. Many studies of the particles motions in quasi-steady flow can be
found in the literature [29-33].
7
Another aspect of the model is the unsteady fluid motion concerning a
rotating drop. In experimental situations (conducted by Chung et al [14]), as
mentioned above, the rotation rate needs to be controlled so as to minimize the
fluid motion and also to prevent the crystals from sedimenting in the drop. This
requires the turning-on and -off of the acoustic rotational mechanism giving rise
to transients in the fluid motion. In a steady situation, a constant torque on a
homogeneous drop will maintain rotational motion like solid body. This solid-
body rotation in the steady state is used as a basis for allowing the zero shear
approximation on the drop surface in general. However, transient effects in the
liquid drop can cause some shear in the drop surface. Therefore, this aspect of
the problem is also investigated in more detail and represents the full depth of
the transient characteristics of a rotating drop.
8
CHAPTER 2
Particle Motion and Path Calculation: Equatorial Plane Case
In this chapter, the motion of a crystallizing particle in a rotating drop is
investigated. As discussed earlier in Chapter 1, the modeling of the particle
motion in solution drop is rather complex because the problem is three-
dimensional and involves acoustic streaming generated by ultrasound speakers
that are used for rotationally propelling the drop. The flow field inside the
solution drop under such circumstances occurs due to particle motion and the
acoustic field. However, as studied by Chung et al [14], adding agarose into the
protein solution increases the viscosity and therefore suppresses the acoustic
streaming flow to a large entent within the drop. With higher viscosity, the
viscous drag on the particle is sufficient to counter gravitational and centrifugal
forces under controlled rotation. We begin with the case when the particle
motion is restricted to the equatorial plane of the rotating drop. The case of the
particle off the plane is discussed in the next chapter.
2.1 Statement of the problem
The idealized system being considered is that of a single drop of the
crystal solution levitated in quiescent fluid and rotated about a horizontal axis.
The drop shape under consideration is approximately spherical. Inside the drop
there is a crystallizing particle as shown in figure 2.1. As a preliminary model,
the shape of the crystallizing particle is assumed to be spherical and the size is
approximated to remain constant over the time period of interest. Considering
that the crystallization process is quite slow compared to the dynamics of fluid
motion, the constant-size assumption is quite reasonable. Under the controlled
environment the evaporation or condensation rate at the interface of the drop
and the fluid environment is negligibly small.
d
R
p
R
Ω
Drop
Particle
FIGURE 2.1: Schematic of the crystallizing particle of radius suspended within
the levitated drop of radius rotating about a horizontal axis with angular velocity
p
R
d
R Ω .
The objective of our study is to understand the fundamental fluid
dynamics of particle motion in a rotating drop. Of primary interest is the
prediction of the trajectory of the crystallizing particle driven by the
gravitational and centrifugal forces and the calculation of the flow field inside
the levitated drop. From a fluid dynamics point of view, it is reasonable to
assume that the density of a protein solution is constant during a time of
interest. Therefore the internal flow within the drop is described by the
continuity equation and the Navier-Stokes equation which are written in the
rotating frame of reference as follows:
9
, 0 = ⋅ v ∇ (2.1)
and v v Ω v v
v
2
1
2 ∇ + − = × + ⋅ +
∂
∂
ν
ρ
d
p
t
∇ ∇ (2.2)
where is the fluid velocity relative to the rotating frame, is the
prescribed angular velocity, and
) t , (r v Ω
ν is the kinematic viscosity of the crystal
solution. The dynamic pressure is related to the fluid pressure
d
p p by
2 2
2
1
r z g p p
d
Ω ρ ρ − + = , (2.3)
where (r, z) are position variables in the cylindrical coordinates.
For convenience, the equations of motion can be non-dimensionalized
using , , and
c
R
c
U
c
t ) / (
c c
U R ≡
c
p ) / (
c c
R U ν ρ ≡ as characteristic scales for
length, velocity, time and pressure, respectively. The equations, in
nondimensional form, become
, 0 = ⋅
*
v ∇ (2.4)
and
* * * *
*
v v i v v
v
2 *
Ta 2 Re ∇ + − = × +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅ +
∂
∂
Ω
p
t
∇ ∇ . (2.5)
The Reynolds number, ν / Re
c c
R U = , characterizes the relative importance of
inertial and viscous forces, while the Taylor number, ,
characterizes that of coriolis and viscous forces. For a solution with
ν Ω / Ta
2
c
R =
-6
10 1.53 × =
s
ν s m /
2
[34], we employ the particle diameter 100 ≈
c
R m µ as the
length scale, an estimated velocity 8 . 0 ≈
c
U s mm/ as the velocity scale and
, the Reynolds and Taylor numbers are estimated as 0.052 and
0.082, respectively. With the addition of agarose at concentrations smaller
rps Ω 2 ≈ ≅ Re
≅ Ta
10
than 0.12% (w/v), agarose solutions do not reach gel states and behave like non-
Newtonian fluids [35]. However, the solution behaves like a Newtonian fluid at
very small shear rate, i.e. Newtonian plateau, and the kinematic viscosity here is
estimated as
-4
10 1 × ≈
s
ν s m /
2
. Employing the particle diameter 300 ≈
c
R m µ
as the length scale, an estimated velocity 1 . 0 ≈
c
U s mm/ as the velocity scale
and , the Reynolds and Taylor numbers are estimated as 0.0003
and 0.0113, respectively. These Reynolds and Taylor numbers are
sufficiently small so that we can neglect inertial and coriolis terms in (2.5).
Thus the flow induced by the motion of the particle may be treated as the Stokes
flow, i.e. the so-called the creeping flow. A set of quasi-steady linear equations
(with the asterisks omitted) governing the fluid motion in the rotating drop take
the form
rps Ω 2 ≈ ≅ Re
≅ Ta
0 = ⋅ v ∇ , (2.6)
and (2.7) .
2
p ∇ = ∇ v
The boundary conditions are as follows.
(i) Zero shear stress at the droplet surface:
. 0 s τ = ⋅ (2.8)
(ii) Zero normal velocity at the droplet surface:
. 0 = ⋅n v (2.9)
(iii) Prescribed velocity at the inner interface: v
.
p
v v = (2.10)
11
Here the velocity is prescribed as an unknown constant which will be
determined later by the force balance on the crystallizing particl. The surface
shear stress is assumed to be zero because of the low viscosity of the gaseous
environment, even though a small torque is required to maintain the rotation.
This assumption is reasonable because the viscous forces due to the solid
particle motion are large, and shear stresses on the surface are relatively weak.
The dynamics of spherical interface and shear stress calculation are presented in
Chapter 4.
p
v
2.2 The method of solution
The equations (2.6) and (2.7) can be written in component form in
cylindrical coordinate system as follows:
0
1 ) ( 1
=
∂
∂
+
∂
∂
+
∂
∂
z
w v
r r
ru
r φ
(2.11)
r
p v
r
u
r ∂
∂
=
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
− ∇
φ
2 2
2
2 1
, (2.12)
φ φ ∂
∂
=
∂
∂
+ ⎟
⎠
⎞
⎜
⎝
⎛
− ∇
p
r
u
r
v
r
1 2 1
2 2
2
, (2.13)
z
p
w
∂
∂
= ∇
2
, (2.14)
where
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
≡ ∇
2
2
2
2
2 2
2
2
1 1
z r r r r φ
;
12
) , , ( w v u are the velocity components in cylindrical coordinate system. In
Stokes limit, the pressure field is harmonic and satisfies Laplace’s equation,
0
2
= ∇ p . (2.15)
In the present work, we consider the simplest non-trivial problem; namely the
motion of the particle in three-dimensional axisymmetric motion in a rotating
frame. It is convenient to represent the velocity components of the particle in a
cartesian coordinate system. A schematic of the system is shown in Fig. 3. The
center of the liquid drop is located on the z-axis at some distance from the
origin. The drop rotates with angular velocity in the non-rotating reference
frame fixed to its center. The particle is placed at the location in which the
center aligns on the z axis and locates at a distance d from the center of the drop.
The particle is assumed to translate with the velocity and rotate with the
angular velocity
y
Ω i
p
v
p
Ω with respect to the drop.
z
d
R
p
R
d
x
p
Ω
p
v
z
d
R
d
R
p
R
p
R
d
x
p
Ω
p
Ω
p
v
p
v
FIGURE 2.2: Schematic of the system.
13
The velocity components in the Cartesian coordinate system are related
to those in cylindrical coordinate system, which shares the same origin, as
follows:
Translation:
z r z x
t
p
W U U W U i i i i i v + + = + =
φ
φ φ sin cos . (2.16)
Rotation:
z p x p
r
p
x d z i i v Ω Ω − − = ) (
z p p r p
r d z d z i i i φ Ω φ Ω φ Ω
φ
cos sin ) ( cos ) ( − − − − = (2.17)
In the view of the linearity of the differential equations and boundary
conditions, the complete solutions of the problem may be written as a
superposition of two solutions as
1 0
v v v + = . (2.18)
where and represent axisymmetric and asymmetric velocity fields about z-
axis respectively.
0
v
1
v
2.2.1 Axisymmetrical problem:
We choose the coordinate system in which the line of centers of two
eccentric spheres lies on the z axis as shown in figure 2.3a. The crystallizing
particle translates along the z axis with the velocity W . It is obvious that the
solution is independent of the azimuthal angle φ , so that the equations (2.12)-
(2.14) are satisfied when the solutions are given in the following form:
0
p p = , (2.19)
14
0
0
2
u
p r
u + = , (2.20)
0 = v , (2.21)
0
0
2
w
p z
w + = , (2.22)
where , , and must satisfy the following equations:
0
p
0
u
0
w
0
0
2
0 0
2
0 0
2
0
= = = w L u L p L , (2.23)
where
2
2
2
2
2
0
1
z r r r
L
∂
∂
+
∂
∂
+
∂
∂
≡ .
In addition, the continuity equation must be satisfied. By substitution of the
velocity components ( ) into (10), we obtain w v u , ,
0 2
1
2 3
0
0 0
=
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
+
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
z
w
u
r r
p
z
z
r
r , (2.24)
15
W
z
0 ≅ τ
x
d
(a) Axisymmetrical problem, V
0
U
z
0 ≅ τ
d
x
p
Ω
(b) Asymmetric problem, V
1
FIGURE 2.3: Configurations of the decomposed systems
16
2.2.2 Asymmetrical problem:
We again choose the configuration of the problem similar to the
previous problem. However, the particle now translates perpendicular to the
line of centers and rotates with the angular velocity
p
Ω as shown in Figure 4b.
Since there is no symmetry in the problem, the solutions are given by
φ cos
1
p p = , (2.25)
φ χ γ
φ
cos ) (
2
1
2
cos
1 1
1
+ + =
p r
u , (2.26)
φ χ γ sin ) (
2
1
1 1
− = v , (2.27)
φ cos
2
1
1
w
p z
w + = , (2.28)
where ,
1
p
1
χ ,
1
γ , and must satisfy the following equations:
1
w
0
1
2
2 1
2
0 1
2
1 1
2
1
= = = = χ γ L L w L p L . (2.29)
Here we define
1 1 1
v u + = χ and
1 1 1
v u − = γ . (2.30)
Also, the operator is defined by
2
n
L
2
2
2
2
2
2
2
1
z r
n
r r r
L
n
∂
∂
+ −
∂
∂
+
∂
∂
≡ , ) 2 , 1 , 0 ( = n . (2.31)
The continuity equation is satisfied if
0 2
2
3
1 1
1 1
=
∂
∂
+
∂
∂
+ ⎟
⎠
⎞
⎜
⎝
⎛
+
∂
∂
+ ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
z
w
r r r
p
z
z
r
r
χ
γ . (2.32)
17
In order to obtain the solutions of the problems involving two eccentric spheres
in which the boundary conditions are satisfied. It is convenient to introduce the
bipolar coordinate system as shown in figure 2.4.
18
0 = η
ρ
z
η
i
ξ
i
ξ= π ξ=0
0 . const > = η
0 . const > = ξ
0 . const < = η
+∞ =
+ =
η
c z
−∞ =
− =
η
c z
0 = η
ρ
z
η
i
ξ
i
ξ= π ξ=0
0 . const > = η
0 . const > = ξ
0 . const < = η
+∞ =
+ =
η
c z
−∞ =
− =
η
c z
r
z
o
η η =
1
η η = o
c Radius η csch =
o o
c d η coth =
1
η η >
o
z
o
η η =
1
η η = o
c Radius η csch =
o o
c d η coth =
o o
c d η coth =
1
η η >
o
(a) (b)
FIGURE 2.4: Bipolar coordinates (a) a meridian plane (b) Coaxial spheres.
Introduced by Jeffery [36] in 1912, the bipolar and cylindrical coordinates are
related via the following transformations
ξ η
η
cos cosh
sinh
−
= c z and
ξ η
ξ
cos cosh
sinh
−
= c r , (2.33)
where is the distance between the points defined by c
o
η η = and 0 = η on the
z-axis. The surfaces of the drop and the particle are identified by
d
η η = and
p
η η = respectively. The geometry is completely defined by radii , , and
the distance between the centers of the drop and particle . These parameters
are given by
d
R
p
R
d
d d
c R η csch = ,
p p
c R η csch = ,
p p d d
R R d η η cosh cosh − = .
The transformation of vector components are given by [37]
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂ −
=
ξ ξ
ξ η
ξ
z
w
r
u
c
u
) cos (cosh
, (2.34)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂ −
=
η η
ξ η
η
z
w
r
u
c
u
) cos (cosh
. (2.35)
In bipolar coordinate system, the solutions of the equation (2.23) and (2.29)
may be written in the following form
[ ] ) ( ) 2 / 1 cosh( ) 2 / 1 sinh( ) (cosh ) , (
2 / 1
ζ η η ζ η η ξ
m
n
m n
m
n
m
n m
P n B n A p
∑
∞
=
+ + + − =
, (2.36)
[ ] ) ( ) 2 / 1 cosh( ) 2 / 1 sinh( ) (cosh ) , (
2 / 1
ζ η η ζ η η ξ
m
n
m n
m
n
m
n m
P n D n C w
∑
∞
=
+ + + − =
, (2.37)
[ ] ) ( ) 2 / 1 cosh( ) 2 / 1 sinh( ) (cosh ) , (
1
1
0 0 2 / 1
0
ζ η η ζ η η ξ
n
n
n n
P n F n E u
∑
∞
=
+ + + − =
, (2.38)
19
[ ] ) ( ) 2 / 1 cosh( ) 2 / 1 sinh( ) (cosh ) , (
2
2
1 1 2 / 1
1
ζ η η ζ η η ξ γ
n
n
n n
P n F n E
∑
∞
=
+ + + − =
, (2.39)
[ ] ) ( ) 2 / 1 cosh( ) 2 / 1 sinh( ) (cosh ) , (
0
1 1 2 / 1
1
ζ η η ζ η η ξ χ
n
n
n n
P n H n G
∑
∞
=
+ + + − =
, (2.40)
where and is an associated Legendre functions of the first kind
with argument
1 , 0 = m ) ( ζ
m
n
P
ξ ζ cos ≡ . The unknown coefficients , , , , ,
, , and must be determined by using the boundary conditions to
complete the solutions. (See details in the Appendix A).
m
n
A
m
n
B
m
n
C
m
n
D
m
n
E
m
n
F
m
n
G
m
n
H
2.3 Drag force, torque and velocity of an encapsulated particle
When the flow fields of the problems have been solved, it is possible to
calculate the hydrodynamic force and torque experienced by the crystallizing
particle. Hence we can calculate the force and torque from the stresses at the
particle surface by:
∫∫
⋅ =
S
dS F τ
D , (2.41)
∫∫
⋅ × =
S
dS r T τ
, (2.42)
where τ is the stress tensor acting across the surface of the particle and r is the
vector of the moment arm of the force corresponding to the stress vector. With
the flow field given in terms of bipolar eigenfunctions, the drag forces and
torques on the particle can be obtained by following Dean & O’Niell [38]. In
particular, the expressions of the force and torque components on the sphere
20
identified by positive value of η in the Cartesian coordinates for
axisymmetrical problem are given as
()( )
⎪
⎭
⎪
⎬
⎫
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + − =
= =
∑
∞
=0
0 0 0 0
, )
2
1
( 2 2
, 0
n
n n n n z
y x
D C B A n c F
F F
µ π
(2.43)
. (2.44) 0 = = =
z y x
T T T
It should be pointed out that the forces in (2.43), i.e. and , are zero.
Because the inertia effect is neglected in Stokes flow, the fluid motion will
cause the viscous drag only in the direction of particle motion. Additionally,
the torques in (2.44) are zero owing to axisymmetry of the problem.
x
F
y
F
For asymmetrical cases, the motion of a particle can be a combination of
translation and rotation. This is due to non-uniform distribution of shear
stresses during particle motion. Thus, the expressions of the drag force and
torque for two motions may be written as
Translation
()( [
⎪
⎪
⎭
⎪
⎪
⎬
⎫
+ + + + − =
= =
∑
∞
=1
1 1 1 1
) 1 ( 2
, 0
n
n n n n
t
x
t
z
t
y
H G B A n n c F
F F
µ π )]
(2.45)
{} [
{} ]
⎪
⎪
⎭
⎪
⎪
⎬
⎫
− + − + + − +
− + − + + + =
= =
+ −
∞
=
+ −
∑
, ) coth 1 2 ( ) coth 2 )( 1 ( ) 2 (
) coth 1 2 ( ) coth 2 )( 1 ( ) 2 (
2 3
2
, 0
1 1 1 ) 1 2 (
1
1 1 1 ) 1 2 (
2
n n n
n
n
n n n
n t
y
t
z
t
x
H n B D n n e
G n A C n n e
c
T
T T
η η
η η
µ π
η
η
(2.46)
21
Rotation
(2.47) , 0 = = =
r
z
r
y
r
x
F F F
{} [
{} ]
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
+ − + − + + − +
− + − + + + =
= =
+ −
∞
=
+ −
∑
.
3
8
) coth 1 2 ( ) coth 2 )( 1 ( ) 2 (
) coth 1 2 ( ) coth 2 )( 1 ( ) 2 (
2 3
2
, 0
3
1 1 1 ) 1 2 (
1
1 1 1 ) 1 2 (
3
c
H n B D n n e
G n A C n n e
c
T
T T
n n n
n
n
n n n
n r
y
r
z
r
x
µ π
η η
η η
µ π
η
η
(2.48)
The coefficients , , , , , , , and can be obtained by
solving a set of equations which satisfy conditions given in Appendix A. It is
noted that these coefficients are linear with the prescribed velocity. Thus the
non-dimensional viscous drag coefficient for a unit velocity is defined by
m
n
A
m
n
B
m
n
C
m
n
D
m
n
E
m
n
F
m
n
G
m
n
H
p
x
R
F
µ
α = , (2.49)
and
p
z
R
F
µ
β = , (2.50)
At this point we are able to calculate the instantaneous velocities U and
W using the force balance on the particle. In general, the drag force on the
particle surface must be balanced by other forces. Since we neglect inertial
terms, only viscous forces are considered in the drag calculations. However, for
the instantaneous force balance, centrifugal and buoyant forces are considered
as well. The total viscous forces on the particle are proportional to U and W in
the x and z-directions respectively. The centrifugal force is due to the rotation
of the drop with the angular velocity Ω. With the above considerations,
instantaneous force balance equations can be written as follows:
22
x-direction:
U R g R
p p
µ α θ ρ π = ∆ sin
3
4
3
(2.51)
z-direction:
W R g d R
p p
µ β θ ρ π = + Ω ∆ ) cos (
3
4
2 3
(2.52)
where θ is an angle between the z-direction and the gravitational acceleration,
α and β given in Appendix B, and ρ ∆ a density difference between a particle
and a drop, As will be demonstrated in Chapter 4, the results in figure 8 show
that the motion of such system is simply the solid-body rotation. The velocity
W under a constant angular velocity at a given time may be calculated by
d W Ω = (2.53)
In addition, we examine a case in which the applied torque is stopped. In such a
case, the drop will rotate almost freely in the ambient fluid. Since the viscosity
of ambient fluid is relatively much smaller than that of the drop, i.e.
drop amb
µ µ /
<< 1, the leading-order steady motion of the drop is rigid-body rotation
(Rednikov, Riley & Sadhal, 2003). In the limit of a very small Reynolds
number, the flow field of ambient fluid induced by the freely rotating drop is
approximated by the quasi-steady Stokes flow of a rotating solid sphere. The
rotation rate of the drop decreases slowly due to viscous damping (energy
dissipation). The unsteady flow field is, however, given by a series of steady-
state fields (quasi-steady approximation). By neglecting the transient effect
after the applied torque was stopped, the velocity W after an infinitesimal
increment of time can be obtained by employing the following equation:
23
v
i
i 1 i
i i p
i
i i
i i p
E
R
W I
d g R
R
W I
d g R +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ ∆ =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ ∆
+
+ +
+ +
2
1
1
1 1
3
2
3
2
cos
3
4
2
cos
3
4
θ ρ π θ ρ π , (2.54)
where is the viscous energy dissipation, while I is
a moment of inertia of the system. The transient effects for a freely rotating
drop are examined in Chapter 3).
t R U F E
d i amb D v
∆ Ω + =
3 2
8 πµ
2.4 A note on the case of two concentric spheres
At the instant a particle passes the center of a drop, i.e. two concentric
spheres, the calculation in the bipolar coordinates breaks down. However, the
analytical results for such situation may be obtained by employing spherical
coordinates. Here we choose a spherical coordinate ( φ θ, , r ) in which the
velocity of a particle, say U , is defined by the relations
θ cos U u
r
= , θ
θ
sin U u − = . (2.55)
Since the flow is axisymmetric, i.e. independent of the azimutal coordinate φ ,
it is convenient to solve the problem in term of the stream function which
satisfies the biharmonic equation (see Happel and Brenner [37]).
0 ) (
2 2 4
= = ψ ψ L L L (2.56)
where, in spherical coordinates,
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
+
∂
∂
=
θ θ θ
θ
sin
1 sin
2 2
2
2
r r
L .
24
In term of the stream function, the boundary conditions (2.8-2.10) can be
written as
θ ψ
2
2
sin
2
p
R U
= . at
p
R r = (2.57)
and 0 =
∂
∂
θ
ψ
, 0
2
2
2
=
∂
∂
−
∂
∂
r r r
ψ ψ
, at
d
R r = (2.58)
The solution of (2.48) which satisfies the boundary conditions is
θ ψ
2 4 2
sin
⎟
⎠
⎞
⎜
⎝
⎛
⋅ + ⋅ + ⋅ + = r D r C r B
r
A
. (2.59)
where
3 2 2
3 6
) )( 2 2 )( ( 2
p d p p d d p d
p d
R R R R R R R R
U R R
A
− + + +
= ,
3 2 2
5 5
) )( 2 2 )( ( 2
) 2 3 (
p d p p d d p d
p d p d
R R R R R R R R
U R R R R
B
− + + +
+
= ,
3 2 2
5 5
) )( 2 2 )( ( 2
) 2 3 (
p d p p d d p d
p p d
R R R R R R R R
U R R R
C
− + + +
+
= ,
3 2 2
3
) )( 2 2 )( ( 2
p d p p d d p d
p d
R R R R R R R R
U R R
D
− + + +
= .
The corresponding velocity components can be obtained from
θ cos 2 2
2 2
2
3
⎟
⎠
⎞
⎜
⎝
⎛
⋅ + + + = r D C
r
B
r
A
u
r
(2.60)
and θ
θ
sin 4 2
2
3
⎟
⎠
⎞
⎜
⎝
⎛
⋅ − − − = r D C
r
B
r
A
u (2.61)
The drag force exerted by the fluid on the particle is
B F
D
µ π 8 − = (2.62)
where B is the stokeslet term (linear in r) in equation (2.59).
25
2.5 Particle motion inside the rotating drop
The solution obtained above may be employed to investigate the motion
of the particle inside the drop. Since the particle density is only slightly higher
than that of the drop, the buoyant and centrifugal forces are quite weak and the
particle relative velocity is therefore small. Thus it is reasonable to assume that
particle inertia is negligibly small. Hence, the particle in Stokes flow regime is
assumed to instantaneously adapt to a change in velocity because the particle
moves a relatively small distance as compared to the particle dimension. In
addition, the time scale for evolution of a velocity field from one steady state to
another, characterized by a diffusive time scale , is relatively small
compared to the time scale for a change in the boundary movement,
according to the quasi-steady assumption. Therefore, the velocity field at given
time appears to be fully developed and steady with respect to the present
configuration. As a consequence, the quasi-steady Stokes problem can be
viewed as a sequence of steady Stokes-flow problems at each time step. The
configuration after an infinitesimal change of time may be simply obtained by
integrating the velocities with respect to time. Here the modified Euler
predictor-corrector method [39] was used for integration and produced a
second-order accurate solution. By this integration method, a new position
vector is determined at the time-level (
ν /
2
c
R
c c
U R/,
2 / 1 + n ) and used to evaluate the
corrector velocity vector at this intermediate position. The corrector is then
used to calculate a new position vector at the time-level ( 1 + n ). The procedure
of this method may be written as follows:
26
27
2 Predictor , (2.63) /
2 1
t
n n / n
∆ ⋅ + =
+
v x x
Corrector , (2.64) t
n n n
∆ ⋅ + =
+ + 2 / 1 1
v x x
where is a position vector of the particle, the corresponding velocity
vector of the particle and
x v
t ∆ is the size of a single time step. The superscript
denotes the time level of integration. By repeating this procedure, we can
obtain the time history of the particle motion. The process is stopped when the
particle gets very close to the drop surface.
2.6 Results and Discussion
2.6.1 Force and torque on a translating particle
The viscous drag and torque on the particle can be calculated by
formulae given in (2.43)-(2.44) and (2.45)-(2.48) for axisymmetrical and
asymmetrical problems, respectively. However, it is convenient to represent
forces and torque on the particle in dimensionless form. Since the fluid motion
under consideration is the Stokes flow, the force and torque can be
appropriately scaled by
R
F
F
πµ 6
*
=
, (2.65)
and
2
*
8 R
T
T
t
t
πµ
=
, or
3
*
8 R
T
T
r
r
πµ
=
. (2.67)
For a translating particle with a unit scaled velocity, the results are calculated as
a function of the eccentricity
d
R d e / = for various radius ratios .
As mentioned earlier, it must be recognized that the calculation in the bipolar
coordinates breaks down for the cases in which two spheres are concentric or in
contact. However, we are able to obtain the analytical result for two concentric
spheres by employing spherical coordinates as stated in § 2.4.
d p
R R k / =
For the axisymmetrical problem, the viscous drag is plotted against the
eccentricity for the cases with radius ratio = , , and in
figure 2.5. It is noted that because of linearity (and therefore, reversibility) of
the Stokes flow, the results here may represent either the case of the particle
k 5 / 1 9 / 1 20 / 1 40 / 1
28
moving toward to the center of the drop or that towards the surface of the drop.
For the range of eccentricity shown in the figure, it is clear that the viscous drag
experienced by the particle increases monotonically as the eccentricity increases
and appears to increase exponentially when the particle is close to the drop
surface. The viscous torque is exactly zero due to the symmetry of the motion
along the line of centers.
0.00
0.50
1.00
1.50
2.00
2.50
3.00
0 0.2 0.4 0.6 0.8 1
Eccentricity, e
|Fz*|
k=1/5
1/9
1/20
1/80
FIGURE 2.5: Scaled drag force plotted against eccentricity e, for given
radius ratios, in axisymmetrical case.
*
z
F
On the other hand, the drag for the asymmetrical problem decreases
drastically with increasing the eccentricity as shown in figure 2.6; however, the
drag tends to gradually decrease as the particle approaches close to the drop
surface. This is contrary to what would happen if outer surface had a no-slip
condition. Since the asymmetry condition causes an imbalance of forces in the
x direction, this leads to a torque exerted by of the fluid motion on the particle.
29
Figure 2.7 is the plot of the torque for the asymmetrical problem. The negative
sign of the torque indicates that the torque vector points into the positive y axis.
It is apparent that the magnitude of the torque increases rapidly as the
eccentricity increases when d is less than the radius of the particle. Then, after
reaching a maximum, the torque decreases as the particle approaches the drop
surface.
0.0
0.4
0.8
1.2
1.6
2.0
0.0 0.2 0.4 0.6 0.8 1.0
Eccentricity, e
|Fx*|
k=1/5
1/20
1/9
1/40
FIGURE 2.6: Scaled drag force plotted against eccentricity e, for given
radius ratios, in asymmetrical case.
*
x
F
30
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
0.0 0.2 0.4 0.6 0.8 1.0
Eccentricity, e
Ty*
k=1/5
1/9
1/20
1/40
FIGURE 2.7: Scaled torque plotted against eccentricity e, for given radius
ratios, in asymmetrical problem
*
y
T
2.6.2 Flow field
In this section the fluid motion inside the liquid drop is presented.
Figure 2.8 shows the velocity fields of the axisymmetrical case for with
and . The particle translates upward to the positive z axis
with a unit velocity. As expected, internal circulation is observed due to the
symmetry of the configuration. In figure 2.9, the velocity field of the
asymmetrical problem are plotted for given values of and e. As shown in
the previous section, a torque is generated as the particle translates in x-
direction for the asymmetrical case. The particle under consideration is
however able to rotate freely about its center. Therefore, we initially apply a
certain angular velocity to the particle accompanied by a unit translating
5 / 1 = k
, 1 . 0 = e , 2 . 0 35 . 0
k
31
32
velocity (we may specify any velocity) in order to obtain the condition that the
torque about the particle center be zero.
The torque on the particle should be zero because of the shear-free
condition on the outer surface representing a freely suspended drop. An
observation of the velocity field shows that the motion of the solution fluid
interestingly turns out to be a solid-body rotation as shown in Figure 2.9. The
result here shows a behavior similar to the result at leading order of the liquid
drop in acoustic fields proposed by Rednikov et al [40]. The explanation of
such behavior is that the shear-free condition allows surface mobility so that the
liquid drop, under the Stokes flow regime, will rotate as the particle begins to
move. Evidently, the velocity of the fluid solution is linearly proportional to the
prescribed velocity on the particle surface and the surface velocity on the drop
surface becomes larger as the eccentricity decreases.
(a) e=0.1
(b) e=0.2
(c) e=0.35
FIGURE 2.8: Velocity field for the axisymmetrical problem with for 1 =
p
V 5 / 1 = k
33
(a) e=0.1
(b) e=0.2
(c) e=0.35
FIGURE 2.9: Velocity field for the asymmetrical problem with for 1 =
p
V 5 / 1 = k
34
2.6.3 Time histories
The motion of a spherical particle within a liquid drop rotating about a
horizontal axis was investigated in this analysis. Figure 2.10 shows the
trajectories of a particle suspended in the rotating drop at various angular
velocities. The case of clockwise rotation about the y-axis was examined and
the particle trajectories generally exhibit spiral motions; the particle orbits
around its centre displaced slightly to the left of the drop center and migrates
outward slowly. The spiral rotation rate is the same as the rotation rate of the
drop owing to the Stokes flow assumption. For the case Ω=0.5 rps, the
trajectory of the particle is approximately circular but off-center from the drop.
This indicates the mobility of the particle when a time-averaged acceleration on
the particle is dominated by the gravitational acceleration. Consequently, the
convective motions are induced by the particle motion inside the drop. At
Ω =1.0 rps, the rotating drop seems to better simulate low gravity conditions.
The particle spirals around the center of the drop and migrates outward gently.
The induced convective motion is very weak. For the case in which Ω =1.5 rps,
the centrifugal acceleration influences the particle motion at high rotation rate,
and the particle migrates outward at higher rate. Therefore, the particle would
reach the drop boundary faster because of the stronger centrifugal effect.
To demonstrate the particle motion inside a freely rotating drop, we
examine the case in which =1/5, k
d p
ρ ρ / =1.2 and µ = . The
comparison between a constant rotation rate and a freely rotating drop is shown
in Figure 2.11. Figure 2.11a shows the trajectory of the particle rotating with a
1
10 1
−
× s Pa ⋅
35
constant rotation rate, Ω =0.8 rps. The result represents the particle motion in
the same manner as the previous case. While the trajectory of the particle after
the applied torque was stopped and the freely rotating drop is represented in
Figure 2.11b. At the beginning, the particle spirals inward while the drop
rotates without the applied torque. Such interesting behavior is due to the non-
uniformity of the rotation rate. Gravity increases the rotation rate when the
particle is positioned for downward motion, and slows it during upward motion.
Moreover, the angular velocity decreases gradually in each spiral rotation owing
to viscous dissipation of the relative translation of the particle.
We examined the time that the particle takes during each cycle of the
spiral rotation (see Figure 2.12). It shows that the particle expends more time
when traveling within the upper hemisphere of the drop than in the lower.
Subsequently, the gravitational force tries to pull the particle downward, i.e.
toward the drop center while in the upper hemisphere. Some time later, the
particle is not able to complete a cycle of the rotation because of loss of energy
(kinetic plus potential) due to dissipation. Analogous to the under-damped
pendulum, the particle swings back and forth as the magnitude of oscillation
decreases gradually. While such oscillatory motion is not desirable for the
crystal growth process, we study it here to acquire a fundamental understanding
of the dynamics.
36
(a) =0.5 rps Ω
37
(b) Ω =1.0 rps
Ω
30
210
60
240
90
270
120
300
150
330
180 0
Ω
30
210
60
240
90
270
120
300
150
330
180 0
Ω
30
210
60
240
90
270
120
300
150
330
180 0
Ω
30
210
60
240
90
270
120
300
150
330
180 0
30
210
60
240
270
120
300
150
330
180 0
(c) Ω =1.5 rps
FIGURE 2.10: The trajectories of a particle inside the rotating drop after time t
=30s for the case =1/40 , k
d p
ρ ρ / =1.2, and µ =1.5x10
-3
Pa.s (~protein solution)
with constant clockwise rotations
90
Ω
30
210
60
240
270
120
300
150
330
180 0
90
Ω
a c
d b
a c
b d
30
210
60
240
270
120
300
150
330
180 0
90
Ω
30
210
60
240
270
120
300
150
330
180 0
90
a c
b d
Ω
c
d
a
b
Ω
30
210
60
240
90
270
120
300
150
330
180 0
Ω
30
210
60
240
90
270
120
300
150
330
180 0
(a)
30
210
60
240
90
270
120
300
150
330
180 0
Ω
30
210
60
240
90
270
120
300
150
330
180 0
Ω
(b)
FIGURE 2.11: The trajectory of a particle inside the rotating drop for the case
=1/5, k
d p
ρ ρ / =1.2, and µ = 1x10
-1
s Pa ⋅ after time t =20s (a) The particle
spirals slowly outward due to a constant clockwise rotation Ω =0.8 rps. (b) The
particle migrates toward the drop center when the applied torque is stopped and
the drop is allowed to rotate freely.
38
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 2 3 456
Cycle
Time(s)
Lower Hemisphere
Upper Hemisphere
FIGURE 2.12: Particle orbit times corresponding to a freely rotating drop with
=1/5, k
d p
ρ ρ / =1.2, and µ = 1x10
-1
Pa.s , =0.8 rps after time t =20s.
init
Ω
39
40
CHAPTER 3
Particle Motion and Path Calculation: Off-Equatorial Plane
In chapter 2, we studied the dynamics of the particle in a rotating drop
when the particle orbits (or spirals) on the equatorial plane. The flow field
inside the drop is assumed to be in Stokes limit. The results illustrate some
aspects of the qualitative picture of the problem. However the stability (about
the plane of motion) of a particle orbiting on the equatorial plane may be
questionable. In general, the particle may be positioned off the equatorial plane
of the drop. This would cause an imbalance of mass distribution about the axis
of rotation and could lead to wobbly motion due to the moment caused by the
particle. To stabilize the drop motion, the strength of acoustic fields needs to be
increased such that the applied torque is strong enough to eliminate the
undesirable wobbly motion. Under such circumstance, the secondary flow
causing by acoustic fields would become significant and make a contribution to
the flow inside the drop. This is a considerably complex problem and needs to
be solved numerically.
In order to get insight into the dynamics of such a problem, a case in
which the particle is positioned at a small distance off the equatorial plane is
examined (see Figure 3.1). With a small distance off the equatorial plane, the
moment due to the mass asymmetry would be negligibly small because the
densities of the solution and the particle are slightly different. This would
require a small degree of additional acoustic field strength. Therefore, it is still
reasonable to assume that the secondary flow is very weak and the flow under
such circumstances is governed by Stokes’ approximation. In particular, we
consider the approach which consists of decomposing the asymmetric problem
into two rotational problems due to the linearity of the system. One has the axis
of rotation through the line of centers between the particle and the drop while
the other is perpendicular to this axis.
41
FIGURE 3.1: Sketch of a liquid drop rotating about y-axis with angular velocity
Ω and having a spherical particle positioning at coordinates ( ) where b
is a small distance off the equatorial plane.
c b a , ,
Ω
c
b
a
F
G
χ
ϕ
z
Drop
o ′
Particle
r
y
O
x
3.1 Superposition Solutions
Consider a spherical particle in a levitated drop rotating about y-axis as
shown in figure 3.1. With the ( z y x , , ) are coordinates fixed on the drop at its
center. The particle is initially positioned at ( ). Here is a distance off
the equatorial plane of the rotating drop and is assumed to be very small
compared to the radius of the drop (i.e.
c b a , ,
o
y
d
R b < < ). As mentioned earlier, due to
linearity of the Stokes flow approximation, we may consider the problem as a
superposition of two rotational problems. First, we consider a plane in which
the centers of the drop and the particle lie, with χ denoting a plane angle
between this plane and xy-plane:
⎟
⎠
⎞
⎜
⎝
⎛
=
−
a
c
1
tan χ . (3.1)
By decomposing the rotating axis into this plane, one may have two rotational
problems where the first problem has the axis of rotation through the line of
centers of two spheres (i.e. o o ′), and the second is a case in which the rotating
axis perpendicular to that of first problem. The angle between o o ′ and y-axis
is given by:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
−
b
c a
2 2
1
tan ϕ . (3.2)
The schematic of these two problems is shown in Figure 3.2. Here r is the
distance between the centers of the drop and the particle, and the decomposed
angular velocity and gravity components are defined as follows.
ϕ sin Ω Ω =
⊥
, (3.3)
42
43
(a) (b)
FIGURE 3.2: Schematics of decomposed systems on a plane with the angle χ to
xy-plane (a) an axis of rotation through the line o o ′ (b) an axis of rotation
perpendicular to o o ′
ϕ cos
//
Ω Ω = , (3.4)
and
) cos(
1
χ
G G
F F = , (3.5)
) sin( ) sin(
21
ϕ χ
G G
F F = , (3.6)
) cos( ) sin(
22
ϕ χ
G G
F F = , (3.7)
where
g R
p G
ρ π ∆ =
3
3
4
F . (3.8)
o
F
G1
F
G22
o ′
//
Ω
F
C
F
G21
r
o o ′
⊥
Ω
Y
Y
Z
Z
The instantaneous velocity of the particle can be considered as a
superposition of two problems and given by
⊥
+ = u u u
//
(3.9)
Here, is the velocity vector calculated from the first problem (the axis of
rotation through the line
//
u
o o ′) and is that from the second problem (the axis
of rotation perpendicular to
⊥
u
o o ′ )
As discussed in previous chapter, the instantaneous velocities of the
system under Stokes assumptions can be obtained via the viscous drag which
is calculated by balancing the net forces on the particle. From the results of the
particle motion inside the rotating drop, the force perpendicular to the line of
two centers leads to the motion of the drop like solid-body rotation. However,
this motion would be suppressed by the applied acoustic torque. The motion of
the particle under such circumstance is approximated by the Stokes drag which
is valid for the radius ratio, k<<1. Hence, the instantaneous velocity of the
particle for the first problem in the coordinates ( z y x ′ ′ ′ , , ) affixed to the drop on
the plane with angle χ is given by
//
u = , (3.10)
y x
V U
′ ′
+ e e ˆ ˆ
// //
where
p
G
R
U
πµ 6
1
//
F
= , (3.11)
p
G
R
V
πµ 6
22
//
F
= . (3.12)
44
In the second problem, the particle motion is considered as the asymmetric case
of the rotating drop, and the instantaneous velocity is given by
⊥
u = () (3.13)
z x
W U
′ ⊥ ′ ⊥
+ e e ˆ ˆ
where
r Ω U
⊥ ⊥
= , (3.14)
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
+ ∆
=
⊥
⊥
p
p
R
g r Ω R
W
βµ
ϕ χ ρ π sin sin (
3
4
2 3
. (3.15)
By superimposing the solutions of these two problems, we are able to
obtain the instantaneous velocities of the particle which may be expressed in the
coordinates ( z y x , , ) as follows:
u = (3.16)
z y x
w v u e e e ˆ ˆ ˆ + +
where
θ φ χ ϕ χ cos sin cos cos sin W V U u + − = , (3.17)
θ φ ϕ sin sin sin W V v + = , (3.18)
φ χ ϕ χ cos sin cos cos W V U w + − = . (3.19)
Here ⎟
⎠
⎞
⎜
⎝
⎛
=
−
a
b
1
tan θ and ⎟
⎠
⎞
⎜
⎝
⎛
=
−
r
c
1
cos φ . Using the modified Euler predictor-
corrector method given by (2.63) and (2.64), one can integrate the velocities
with respect to time to obtain the motion of the particle. By repeating this
procedure, the trajectories of the particle inside the drop for various conditions
can be obtained.
45
3.2 Results and Discussion
We consider a particle in a rotating drop with radius ratio 1/40. The
properties of the liquid drop are taken to be those of a typical protein solution.
Initially, the particle is positioned at (0, 0.05 , 0.3 ). The drop is rotated
about y-axis with angular velocity
d
R
d
R
Ω in the clockwise direction, i.e.
(see Fig 3.1). The trajectories of the particle are calculated for 30 sec.
of rotation. The numerical calculations of this system illustrate some interesting
results. In figure 3.3, we present a result for the motion of the particle with
y
Ωe Ω ˆ − =
Ω=1.5 rps. The particle, as can be anticipated, spirals outward due to
centrifugal forces and simultaneously migrates away from the equatorial plane.
Figure 3.3b shows that the particle is driven toward the equatorial plane when
migrating to the upper hemisphere of the drop and expelled when moving to the
lower. The reason is that the net hydrodynamic force on the particle points
toward the center of the drop in the former, and outward in the latter. In
addition, the magnitude of the net force on the particle in the latter case is
greater than that of the former because the gravitational and centrifugal forces
are in the same direction and the opposite for the former case. As a result, the
particle migrates gradually farther away from the equatorial plane.
With increasing the angular velocity, the motion of the particle is
influenced significantly by centrifugal forces. The drop rotating with Ω =3.0
rps. are illustrated in Figure 3.4. It can be seen that the amplitude of the motion
in y-direction becomes smaller and the particle migrates away farther from the
equatorial plane. This is because the particle spends less time in each cycle, and
46
therefore, it is driven by the centrifugal forces while the motion due to the
gravity becomes less significant.
Figures 3.5 and 3.6 show the motions of the lighter particle, i.e.
d p
ρ ρ < . Albeit this is not what would happen in the case of crystal growth in
the rotating drop, it is worthwhile to investigate such circumstances from a
fundamental point of view. As expected, the particle spirals inward and tends to
move toward the equator. This is because the gravitational and centrifugal
forces would act on the particle in the opposite direction when the particle has
less density than that of the liquid drop. Similarly, the particle migrates faster in
direction of rotating axis as the rotational rate increases; therefore the particle
moves closer to the equator as shown in figure 3.6b.
47
48
(a) Trajectory
0 5 10 15 20 25 30
-1
0
1
t
x
0 5 10 15 20 25 30
0
0.05
0.1
t
y
0 5 10 15 20 25 30
-1
0
1
t [sec.]
z
(b) Postion
FIGURE 3.3: The motion of a particle initially positioned at (0, 0.05 , 0.3 ),
for ,
d
R
d
R
40 / 1 = k
d p
ρ ρ / =1.2, and Ω =1.5 rps
-1
0
1
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
a c
c
a
b
d
b
.
d
49
(a) Trajectory
0 5 10 15 20 25 30
-1
0
1
t
x
0 5 10 15 20 25 30
0
0.05
0.1
t
y
0 5 10 15 20 25 30
-1
0
1
t [sec.]
z
(b) Position
FIGURE 3.4: The motion of a particle initially positioned at (0, 0.05 , 0.3 ),
for ,
d
R
d
R
40 / 1 = k
d p
ρ ρ / =1.2, and Ω =3 rps
-1
0
1
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
a
c
.
a
d b
d
b
c
-1
0
1
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
a c
50
(a) Trajectory
0 5 10 15 20 25 30
-1
0
1
t
x
0 5 10 15 20 25 30
0
0.05
0.1
t
y
0 5 10 15 20 25 30
-1
0
1
t [sec.]
z
(b) Position
FIGURE 3.5: The motion of a particle initially positioned at (0, 0.05 , 0.3 ),
for ,
d
R
d
R
40 / 1 = k
d p
ρ ρ / =0.8, and Ω =1.5 rps
a
b
c
d
b
d
.
51
(a) Trajectory
0 5 10 15 20 25 30
-1
0
1
t
x
0 5 10 15 20 25 30
0
0.05
0.1
t
y
0 5 10 15 20 25 30
-1
0
1
t [sec.]
z
(b) Position
FIGURE 3.6: The motion of a particle initially positioned at (0, 0.05 , 0.3 ),
for ,
d
R
d
R
40 / 1 = k
d p
ρ ρ / =0.8, and Ω =3.0 rps
-1
0
1
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
a c
a c
b
d
b
.
d
52
CHAPTER 4
Transient Dynamics of a Rotating Liquid Sphere
4.1 Introduction
We have been motivated to study the transient dynamics of a rotating
levitated liquid drop as a result of our preliminary study of the fluid dynamics of
containerless protein crystal growth. Our analytical investigation for the
dynamics of a rotating drop containing a solid particle has led us to a more
fundamental problem of the transient dynamics of a rotating drop which may be
manipulated by altering the intensity of the rotational torque. With the idea of
investigating this in more detail, we consider a liquid drop that is experiencing
two different transients, one when the torque is turned off, and another when the
torque is turned on. In the first case, a drop is initially rotating steadily under
the force of a constant but spatially nonuniform torque. Under such
circumstances, the drop rotates like a solid body. However, when the torque is
turned off, motion within the drop ensues due to the redistribution of shear
stress at drop surface. In the second case, as an initially stationary drop, a
torque is applied and the drop begins rotating and eventually achieves the steady
state. In such cases, we are interested in determining the flow characteristics
within the drop and in understanding the limits of the parameters under which
the basic rotation may be regarded as solid-body type for the modeling of the
more complex problem of a particle within a rotating drop.
53
In order to understand the dynamics, we consider a drop rotating steadily
with angular speed,
o
Ω , and at time 0 = t , the torque is turned off, causing the
drop to begin decelerating. The other end of the problem is the case the drop
rotation is started with a constant but nonuniform torque, θ τ sin
o
, and ends up in
steady state with a constant rotational speed. Even though, in actual
experiments the startup of an acoustic torque causes a more complex flow field
[31], our study is just an attempt to understand transients in a more fundamental
sense.
4.2 Problem Statement and Method of Solution
We consider a liquid drop of radius R levitated in an unbounded
gaseous medium and rotated about a horizontal axis with angular velocity Ω .
In our analysis, the rotation rate is assumed to be sufficiently low such that the
velocity field in the gaseous medium as well as within the rotating drop is
approximated by the Stokes flow. The fluid flow is only in the direction of
rotation, i.e. the φ direction. In a spherical coordinate system, the unsteady
Stokes equations governing the system are:
Liquid drop:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− ∇ =
∂
∂
θ
ν
φ
φ
φ
2 2
2
sin
ˆ
ˆ ˆ
ˆ
r
u
u
t
u
, (4.1a)
Gaseous environment:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− ∇ =
∂
∂
θ
ν
φ
φ
φ
2 2
2
sin r
u
u
t
u
. (4.1b)
From here on, the hat (^) is used to denote parameters in the liquid phase. The
initial and boundary conditions depend on whether the problem is for a
54
decelerating drop or a startup case. Therefore, these are discussed in section 4.3
and 4.4, respectively. We now assume that the solutions of equations (4.1) may
be written in terms of the sine function as follows
θ
φ
sin ) , ( ˆ ˆ t r u u = , (4.2a)
θ
φ
sin ) , ( t r u u = . (4.2b)
Substituting (4.2) into (4.1) and using the method of Laplace transform, one can
write the transformed equations as
ν ˆ
) 0 , ( ˆ
ˆ
2
ˆ
ˆ
2
ˆ
2 2
2
r u r
U
ν
r s
U r U r − =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ − − ′ + ′ ′ , (4.3a)
ν
r u r
U
ν
r s
U r U r
) 0 , (
2 2
2 2
2
− = ′
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ − ′ + ′ ′ , (4.3b)
where {} ) , ( ˆ ) , (
ˆ
t r u s r U L ≡ ,
{} ) , ( ) , ( t r u s r U L ≡ .
The general solution of equations (4.3) that remains bounded for both media,
takes the form
,
) 0 , ( ˆ ) 1
ˆ
( ) 1
ˆ
(
ˆ
) , (
ˆ
2
ˆ ˆ
1
s
r u
r
X e X e
C s r U
X X
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + −
=
−
R R
R R
(4.4a)
s
r u
r
X e
C s r U
X
) 0 , ( ) 1 (
) , (
2
1
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
−
R
R
. (4.4b)
where
R
r
= R , τ ˆ
ˆ
s X = and τ s X = .
Here τ ˆ and τ are the diffusion time scales with respect to the kinetic viscosity
of liquid drop and gas media, respectively, and can be defined as
55
ν
τ
ˆ
ˆ
2
R
= and
ν
τ
2
R
= .
It should be noted that the equation 4.4 consists of the homogeneous and
particular solutions which correspond to time-dependent and time-independent
solutions, respectively. The unknown coefficients
1
ˆ
C and
1
C must be
determined by boundary conditions which will be specified later.
4.3 Case of a Decelerating Liquid Drop
We begin our analysis with the case in which a liquid drop initially
rotating steadily in gaseous medium at angular velocity
o
Ω . With a constant
torque applying on a homogeneous drop, the rotating drop will maintain
rotational motion like a solid body in the steady state situation. The flow field
in gaseous medium is simply given by the steady Stokes flow solution of a
rotating sphere in infinite medium. The rotating drop is allowed to decelerate
by turning off the torque. Thus, initial and boundary conditions of such
problem are given by
θ
φ
sin ˆ
o
r u Ω =
,
θ
φ
sin
2
3
o
r
R
u
Ω
=
at 0 = t , (4.5a)
and
φ φ
u u = ˆ ,
φ φ
σ σ
r r
= ˆ at R r = . (4.5b)
The solutions of equations (4.4) satisfying the conditions (4.5) are
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∆
+ + − + Ω
−
Ω
=
−
s
X e X e X R
s
r
s r U
X X
)] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , (
ˆ
ˆ ˆ
2
o o
R R
R
R R
δ
(4.6a)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∆
+ + − + Ω
−
Ω
=
− − − + −
s
X e X e X R
s
R
s r U
X X X X
)] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , (
ˆ
) 1 (
ˆ
) 1 (
2
o
2
o
R - R
R
R R
δ
(4.6b)
56
Here )
ˆ ˆ ˆ
( )
ˆ ˆ ˆ
(
24 23
2
22
3
21
ˆ
14 13
2
12
3
11
ˆ
c X c X c X c e c X c X c X c e
X X
+ + + − + + + = ∆
−
where
2
11
δκ κ + = c ,
2
21
δκ κ − = c ,
2
12
) 1 ( 3 1 κ δ κ δ − − − = c ,
2
22
) 1 ( 3 1 κ δ κ δ − − + = c ,
) 1 )( 1 ( 3
13
− − = κ δ c , ) 1 )( 1 ( 3
23
+ − = κ δ c ,
) 1 ( 3
24 14
δ − = = c c , and µ µ δ ˆ / = , ν ν κ / ˆ = .
Thus the shear stress in the rotating drop is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∆
+ + − + − +
Ω − =
−
s
X X e X X e X
X X
r
)] 3
ˆ
3
ˆ
( ) 3
ˆ
3
ˆ
( )[ 1 (
3 ˆ
2
ˆ
2
ˆ
o
R R R R
2 R 2 R
µ σ
φ
(4.7)
At large s, the transformed equations (4.6) and (4.7) can be expanded as an
asymptotic series in negative exponentials. The series expansion equations take
the form
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + +
+ + − + Ω
−
Ω
=
∑
∞
=
+ − − −
0
14 13
2
12
3
11
ˆ
) 1 (
ˆ
) 1 (
2
o o
)] , ( [
)
ˆ ˆ ˆ
(
)] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , (
ˆ
n
n
X X
s r G
c X c X c X c s
X e X e X R
s
r
s r U
R R
R
R R
δ
(4.8a)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + +
+ + − + Ω
−
Ω
=
∑
∞
=
− − − − −
0
14 13
2
12
3
11
ˆ
2 ) 1 ( ) 1 (
2
o
2
o
)] , ( [
)
ˆ ˆ ˆ
(
)] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , (
n
n
X X X
s r G
c X c X c X c s
X e X e X R
s
R
s r U
R R
R
R R
δ
(4.8b)
and
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + +
+ + − + − +
Ω − =
∑
∞
=
−
0
14 13
2
12
3
11
2 2
ˆ
2 2
ˆ
o
)] , ( [
)
ˆ ˆ ˆ
(
)] 3
ˆ
3
ˆ
( ) 3
ˆ
3
ˆ
( )[ 1 (
3 ˆ
n
n
X X
r
s r G
c X c X c X c s
X X e X X e X R R R R
R R
µ σ
φ
(4.9)
where
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + +
+ + +
=
−
14 13
2
12
3
11
24 23
2
22
3
21
ˆ
2
ˆ ˆ ˆ
ˆ ˆ ˆ
) , (
c X c X c X c
c X c X c X c
e s r G
X
57
4.3.1 Limit of 0 → δ
One can see that when the ratio of viscosities is zero, the problem is
simply a solid sphere rotating in an infinite fluid. In the limit of 0 → δ , we can
write the velocity field induced by the rotating sphere can be obtained from the
equation (4.6b) as follows;
⎟
⎠
⎞
⎜
⎝
⎛
+
+
− Ω
Ω
+
Ω
=
− − X
e
X
X
s s
R
s
R
s r U
) 1 ( 1
2
o
2
o
1
1
] ) ( [ ) , (
- R
R
R R
(4.10)
where
) 1 ( 15 ) 5 1 (
5 1
) (
2 2 2 3
2
+ + + +
+ +
= Ω
X k X k X
k X
s
δ δ
δ
c X
p
bi a X
i n m
bi a X
i n m
−
+
+ −
−
+
− −
+
=
Here, a complex conjugate pair bi a ± and a real constant c are the roots of the
denominator of the rational function ) (s Ω . A complex conjugate pair ni m ± and
a real constant p are obtained by a partial fraction expansion. With the use of
standard inversion tables, the time-dependent solution of equation (4.10) can be
obtained for all ) , ( t r (see Appendix B). The resulting velocity field outside the
sphere is given by
()
2 1 2
o
1 ) , ( I I
R
t r u + +
Ω
=
R
(4.11)
where
14 13 12 11
) 1 (
1
1
1
) ( I I I I e
X
X
s I
X
+ + + = ⎟
⎠
⎞
⎜
⎝
⎛
+
+
Ω =
− − R 1 -
R
L ,
()
⎟
⎠
⎞
⎜
⎝
⎛
− − − − −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
− +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
τ
τ
τ τ
t
X
e t
t t
s
e
X
X
I
R R
1 -
R
R
R R
L
1 ) 1 (
2
/
/ 2
1
erfc 1
/ 2
1
erfc
1
1
Here,
{}
τ π
β
α
τ
/
Z Re
4
) 1 (
11
2
t
e
k I
t
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
R
R ,
58
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− −
−
=
⎟
⎠
⎞
⎜
⎝
⎛
− − −
τ
τ
β
α
τ
/
/ 2
) 1 (
erfc 1
1
1
12
t
t
e
c
k
I
t
R
R
R
,
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
⎟
⎠
⎞
⎜
⎝
⎛
−
−
− =
+ −
τ
τ
τ
/
/ 2
) 1 (
erfc
1
1
2
) 1 (
13
t c
t
e
c
c
k c I
t
c c
R R
R
,
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+ +
−
+ − =
+ + − +
τ
τ β
τ
/ ) (
/ 2
) 1 (
erfc ) ( Re
2
) ( ) 1 )( (
14
2 2
t bi a
t
e bi a I
t
b a bi a
R
R
where ] 1 ) ( ][ 1 ) ( )[ ( 2 Z − − − + + = bi a bi a ni m R ,
] ) 1 ( [ ) 1 ( 2 b n a m − − − = R α , ] ) 1 [(
2 2
b a + − = β .
4.3.2. The general solution in integral representation form
An examination of the set of equations (4.6) shows that there are several
fairly complex terms whose inverse Laplace transforms are not available in the
standard tables. We obtain the inverse of such terms by employing a contour
integral in the complex plane. The inverse of the first term in each of (4.6a) and
(4.6b) is simply a time-dependent solution while the remaining terms constitute
the time-dependent parts. We have
∫
∞ +
∞ −
−
∆
+ + − + Ω
− Ω =
i
i
X X
ts
ds
s
X e X e X
e
R
r t r u
γ
γ
δ )] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , ( ˆ
ˆ ˆ
2
o
o
R R
R
R R
(4.12a)
∫
∞ +
∞ −
− − − + −
∆
+ + − + Ω
−
Ω
=
i
i
X X X X
ts
ds
s
X e X e X
e
R R
t r u
γ
γ
δ )] 1
ˆ
( ) 1
ˆ
( )[ 1 ( 3
) , (
ˆ
) 1 (
ˆ
) 1 (
2
o
2
o
R - R
R
R R
(4.12b)
It can be seen that the integrands of equations (4.12) have poles and are double-
valued at 0 = s , therefore, possessing a branch point at 0 = s . Using the standard
contour integral, one can show that the integration along a small circle around
59
the branch point produces a term such that cancels out the time-independent
solution. Thus, we finally arrive at the following set of results
Flow fields:
ρ ρ τ ρ τ ρ τ
ρ
τρ
π
δ
ρ
d
B A
B A e R
t r u
t
∫
∞ −
−
+
− Ω
=
0
2 / 1 2 / 1 2 / 1
2 2 2
o
] ) ˆ ( cos ) ˆ ( ) ˆ ( [sin
) (
) ( 3
) , ( ˆ R R R
R
(4.13a)
ρ
ρ π
δ
ρ
d
B A
BD AC e R
t r u
t
∫
∞ −
+
− Ω
=
0
2 2 2
o
) (
) ( 3
) , (
R
(4.13b)
Shear stress:
ρ ρ τ ρ τ ρ τ ρ τ
ρ
τρ
π
µ
σ
ρ
φ
d
B A
B A e
t
r
∫
∞ −
+ −
+
− Ω
=
0
2 / 1 2 / 1 2
2 2 3
o
] ) ˆ ( cos ˆ 3 ) ˆ ( sin ) 3 ˆ [(
) (
) ( 3
ˆ R R R R
R
(4.14)
where
2 / 1 2 / 1
11 21 13 23
2 / 1
12 22 14 24
) ˆ cos( ) ˆ ( )] ( ˆ ) [( ) ˆ sin( )] ( ˆ ) [( ρ τ ρ τ ρ τ ρ τ ρ τ ⋅ ⋅ − ⋅ − − − ⋅ + ⋅ − + = c c c c c c c c A ,
2 / 1 2 / 1
11 21 13 23
2 / 1
12 22 14 24
) ˆ sin( ) ˆ ( )] ( ˆ ) ( ) ˆ cos( )] ( ˆ ) [( ρ τ ρ τ ρ τ ρ τ ρ τ ⋅ ⋅ + ⋅ − + + ⋅ − ⋅ − − = c c c c c c c c B ,
] ˆ ) cos (cos ) sin [(sin ˆ ) sin (sin ) cos (cos
1 2 1 2 1 2 1 2
ρ τ θ θ θ θ τρ ρ τ θ θ θ θ + − − + + + − = R C ,
] ˆ ) sin (sin ) cos [(cos ˆ ) cos (cos ) sin (sin
1 2 1 2 1 2 1 2
ρ τ θ θ θ θ τρ ρ τ θ θ θ θ + + − − + − − = R D ,
in addition to , ρ τ τρ θ ˆ ) 1 (
1
− − = R and ρ τ τρ θ ˆ ) 1 (
2
+ − = R .
60
4.3.3. Small-time approximation
The transformed equations (4.8) and (4.9) can be inverted term by term
to obtain the solutions applicable for small times, i.e. τ ˆ < < t . The inverse
Laplace transform procedure adopted here is straightforward, but rather tedious
and similar to that of the solid sphere case. First, each term which contains a
ratio of cubic function in series expansion is split into simpler ratios by using
partial fraction expansion. In carrying out this approximation, we keep only
two terms in the large-s expansions given by the summations in (4.8). Then,
employing equations (B.2.1), (B.2.2) and (B.2.3) in Appendix B, one obtains
inverses of ) , (
ˆ
s r U , ) , ( s r U and ) , ( ˆ s r
r φ
σ using the first two terms of the
summations in the set. These are given below:
Flow fields:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
+
− Ω =
∑ ∑
= =
8
1
1
2 2
4
1
0
2 2 o
ˆ
) 1 (
) 1 ( 3
ˆ
) (
3
) , ( ˆ
n
n
n
n
I I R t r u
R R
R
δκ κ
δκ δ
κ δκ
δ
(4.15a)
Here
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎦
⎤
⎢
⎣
⎡
+
−
−
+
+
− ⎟
⎠
⎞
⎜
⎝
⎛ −
+
+
− =
− + −
τ
θ θ
τ
ˆ
) ( ) 1 (
1 1 2
1
2
1
3 1 3 1
1 2
1
2
1
3 1 3 1
2
1
2
1
3 1 3 1 0
1
2
1
2
1 1
sin cos
ˆ / 4
1
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
− =
+ −
τ
τ τ
τ
ˆ /
ˆ / 4
1
erfc
ˆ / 4
1
erfc
ˆ
1
ˆ
) 1 (
1
3 0
2
2
1 1
t c
t
e
t
c
c
I
t
c c
R R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+
=
− + +
τ
θ θ
τ
ˆ
) ( ) 1 (
2 2
2
1
2
1
4 1 4 1
2
2
1
2
1
4 1 4 1
2
1
2
1
4 1 4 1 0
3
2
1
2
1 1
sin cos
ˆ / 4
1
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
+
− ⎟
⎠
⎞
⎜
⎝
⎛ +
=
+ +
τ
τ τ
τ
ˆ /
ˆ / 4
1
erfc
ˆ / 4
1
erfc
ˆ
1
ˆ
) 1 (
1
4 0
4
2
1 1
t c
t
e
t c
c
I
t
c c
R R
R
61
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+
− =
− + −
τ
θ θ
τ
ˆ
)
2
1
2
1
(
1
) 3 (
3 3
2
1
2
1
5 1 5 1
3
2
1
2
1
5 1 5 1
2
1
2
1
5 1 5 1 1
1
sin cos
ˆ / 4
3
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
− =
+ −
τ
τ τ
ˆ /
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ
1
ˆ
2
1
) 3 (
1
1
5 1
2
t c
t
e
t
c
c
I
T
t
c c
R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+ −
− =
−
−
t
e
t
b a
b b a a
t b a
b b a a b a
I
4
ˆ
2
) 3 (
2
1
2
1
6 1 6 1
2 2
1
2
1
6 1 1 6
2
1
2
1 1
3
ˆ
4
ˆ / 4
3
erfc
) (
2 ) (
2
ˆ
τ
τ π
τ
R
R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
−
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
−
−
+
−
+
− + −
3
ˆ
)
2
1
2
1
(
1
) 3 (
3
2 2
1
2
1
1 1
2
1
2
1
1
6
2
1
2
1
1
2 2
1
2
1
2
1
2
1
6
sin
) (
2 ) 3 (
ˆ
2
) 3 (
) (
θ
τ
τ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
− − − − −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
− =
+ −
−
−
τ
τ
τ τ π
τ
τ
τ
ˆ /
ˆ / 4
3
erfc ]
ˆ
2
) 3 ( 1 [
ˆ
4
ˆ / 4
3
erfc
ˆ
1
ˆ
2
1
) 3 (
1
2
1
1
4
ˆ ) 3 (
1
2
1
6 1
4
2
t c
t
e
t c
c e
t
c
t c
c
I
t c
c
t
R
R
R
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+
=
− + +
τ
θ θ
τ
ˆ
) ( ) 3 (
4 4
2
1
2
1
7 1 7 1
4
2
1
2
1
7 1 7 1
2
1
2
1
7 1 7 1 1
5
2
1
2
1 1
sin cos
ˆ / 4
3
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
+ +
τ
τ τ
τ
ˆ /
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
1
7 1
6
2
1 1
t c
t
e
t
c
c
I
t
c c
R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+ −
=
+
−
t
e
t
b a
b b a a
t b a
b b a a b a
I
4
ˆ ) 3 (
2
1
2
1
8 1 8 1
2 2
1
2
1
8 1 1 8
2
1
2
1 1
7
2
ˆ
4
ˆ / 4
3
erfc
) (
2 ) (
2
ˆ
τ
τ π
τ
R
R
+ Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
+
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
+
−
+
−
−
− + +
4
ˆ
)
2
1
2
1
(
1
) 3 (
4
2 2
1
2
1
1 1
2
1
2
1
1
8
2
1
2
1
1
2 2
1
2
1
2
1
2
1
8
cos
) (
2 ) 3 (
ˆ
2
) 3 (
) (
θ
τ
τ
t
b a a
e
b a
b a
b a
a
b
t
b a
a
b a
b a
a
R
R R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
+
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
+
−
+
−
+
− + +
4
ˆ
) ( ) 3 (
4
2 2
1
2
1
1 1
2
1
2
1
1
8
2
1
2
1
1
2 2
1
2
1
2
1
2
1
8
sin
) (
2 ) 3 (
ˆ
2
) 3 (
) (
2
1
2
1 1
θ
τ
τ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
− + − − −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
+ +
+
−
τ
τ
τ τ π
τ
τ
τ
ˆ /
ˆ / 4
3
erfc ]
ˆ
2
) 3 ( 1 [
ˆ
4
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
2
1
1
4
ˆ ) 3 (
1
2
1
8 1
8
2
1
1
2
t c
t
e
t c
c e
t
c
t c
c
I
t c
c
t
R
R
R
R
R
62
where
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
−
= Σ τ
τ
ˆ / ) (
ˆ / 4
1
erfc abs
1 1 1
t i b a
t
R
,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
−
+ + − = τ
τ
τ
θ ˆ / ) (
ˆ / 4
1
erfc arg
ˆ
2 ) 1 (
1 1 1 1 1 1
t i b a
t
t
b a b
R
R ,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
+
= Σ τ
τ
ˆ / ) (
ˆ / 4
1
erfc abs
1 1 2
t i b a
t
R
,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
+
+ + + = τ
τ
τ
θ ˆ / ) (
ˆ / 4
1
erfc arg
ˆ
2 ) 1 (
1 1 1 1 1 2
t i b a
t
t
b a b
R
R ,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
−
= Σ τ
τ
ˆ / ) (
ˆ / 4
3
erfc abs
1 1 3
t i b a
t
R
,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
−
+ + − = τ
τ
τ
θ ˆ / ) (
ˆ / 4
3
erfc arg
ˆ
2 ) 3 (
1 1 1 1 1 3
t i b a
t
t
b a b
R
R ,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
+
= Σ τ
τ
ˆ / ) (
ˆ / 4
3
erfc abs
1 1 4
t i b a
t
R
,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ +
+
+ + + = τ
τ
τ
θ ˆ / ) (
ˆ / 4
3
erfc arg
ˆ
2 ) 3 (
1 1 1 1 1 4
t i b a
t
t
b a b
R
R ,
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
+
− Ω =
∑ ∑
= =
8
1
1
2 2
4
1
0
2 2 2 o
) 1 (
) 1 ( 3
) (
3 1
) , (
n
n
n
n
I I R t r u
R R R δκ κ
δκ δ
κ δκ
δ
(4.15b)
Here
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎦
⎤
⎢
⎣
⎡
+
−
−
+
+
− ⎟
⎠
⎞
⎜
⎝
⎛ − +
+
+
− =
− + − +
τ
κ
θ θ
τ
κ
ˆ
) ( )] 1 ( 2 [
1 1 2
1
2
1
3 1 3 1
1 2
1
2
1
3 1 3 1
2
1
2
1
3 1 3 1 0
1
2
1
2
1 1
sin cos
ˆ / 4
) 1 ( 2
erfc 2
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
− +
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
− =
+ − +
τ
τ
κ
τ
κ
τ
κ
ˆ /
ˆ / 4
) 1 ( 2
erfc
ˆ / 4
) 1 ( 2
erfc
1
ˆ
)] 1 ( 2 [
1
3 0
2
2
1 1
t c
t
e
t
c
c
I
t
c c
R R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+
=
− + −
τ
κ
θ θ
τ
κ
ˆ
) ( ) 1 (
2 2
2
1
2
1
4 1 4 1
2
2
1
2
1
4 1 4 1
2
1
2
1
4 1 4 1 0
3
2
1
2
1 1
sin cos
ˆ / 4
) 1 (
erfc 2
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
63
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
−
− ⎟
⎠
⎞
⎜
⎝
⎛ −
=
+ −
τ
τ
κ
τ
κ
τ
κ
ˆ /
ˆ / 4
) 1 (
erfc
ˆ / 4
) 1 (
erfc
1
ˆ
) 1 (
1
4 0
4
2
1 1
t c
t
e
t c
c
I
t
c c
R R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
+
+
− =
− + − +
τ
κ
θ θ
τ
κ
ˆ
)
2
1
2
1
(
1
)] 1 ( 4 [
3 3
2
1
2
1
5 1 5 1
3
2
1
2
1
5 1 5 1
2
1
2
1
5 1 5 1 1
1
sin cos
ˆ / 4
) 1 ( 4
erfc 2
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
− +
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
− =
+ − +
τ
τ
κ
τ
κ
τ
κ
ˆ /
ˆ / 4
) 1 ( 4
erfc
ˆ / 4
) 1 ( 4
erfc
1
ˆ
2
1 1
)] 1 ( 4 [
1
5 1
2
t c
t
e
t
c
c
I
t
c c
R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
+
+ −
− =
− +
−
t
e
t
b a
b b a a
t b a
b b a a b a
I
4
ˆ
2
)] 1 ( 4 [
2
1
2
1
6 1 6 1
2 2
1
2
1
6 1 1 6
2
1
2
1 1
3
ˆ
4
ˆ / 4
) 1 ( 4
erfc
) (
2 ) (
2
τ κ
τ π
τ
κ
R
R
+ Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
− +
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
− +
−
+
−
−
− + − +
3
ˆ
)
2
1
2
1
(
1
)] 1 ( 4 [
3
2 2
1
2
1
1 1
2
1
2
1
1
6
2
1
2
1
1
2 2
1
2
1
2
1
2
1
6
cos
) (
2 )] 1 ( 4 [
ˆ
2
)] 1 ( 4 [
) (
θ
κ
τ
κ
τ
κ
t
b a a
e
b a
b a
b a
a
b
t
b a
a
b a
b a
a
R
R R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
− +
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
− +
−
+
−
+
− + − +
3
ˆ
)
2
1
2
1
(
1
)] 1 ( 4 [
3
2 2
1
2
1
1 1
2
1
2
1
1
6
2
1
2
1
1
2 2
1
2
1
2
1
2
1
6
sin
) (
2 )] 1 ( 4 [
ˆ
2
)] 1 ( 4 [
) (
θ
κ
τ
κ
τ
κ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
− +
− − + − − −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
− =
+ − +
− +
−
τ
τ
κ
τ
κ
τ π
τ
κ
τ
κ
τ κ
ˆ /
ˆ / 4
) 1 ( 4
erfc ]
ˆ
2
)] 1 ( 4 [ 1 [
ˆ
4
ˆ / 4
) 1 ( 4
erfc
1
ˆ
2
1
1
)] 1 ( 4 [
2
1
1
4
ˆ )] 1 ( 4 [
1
2
1
6 1
4
2
t c
t
e
t c
c e
t
c
t c
c
I
t c
c
t
R
R
R
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
+
+
=
− + − +
τ
κ
θ θ
τ
κ
ˆ
) ( )] 1 ( 2 [
4 4
2
1
2
1
7 1 7 1
4
2
1
2
1
7 1 7 1
2
1
2
1
7 1 7 1 1
5
2
1
2
1 1
sin cos
ˆ / 4
) 1 ( 2
erfc 2
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
I
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
− +
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
=
+ − +
τ
τ
κ
τ
κ
τ
κ
ˆ /
ˆ / 4
) 1 ( 2
erfc
ˆ / 4
) 1 ( 2
erfc
1
ˆ
)] 1 ( 2 [
1
7 1
6
2
1 1
t c
t
e
t
c
c
I
t
c c
R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ − +
+
+ −
=
− +
−
t
e
t
b a
b b a a
t b a
b b a a b a
I
4
ˆ )] 1 ( 2 [
2
1
2
1
8 1 8 1
2 2
1
2
1
8 1 1 8
2
1
2
1 1
7
2
ˆ
4
ˆ / 4
) 1 ( 2
erfc
) (
2 ) (
2
τ κ
τ π
τ
κ
R
R
+ Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
− +
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
− +
−
+
−
−
− + − +
4
ˆ
)
2
1
2
1
(
1
)] 1 ( 2 [
4
2 2
1
2
1
1 1
2
1
2
1
1
8
2
1
2
1
1
2 2
1
2
1
2
1
2
1
8
cos
) (
2 )] 1 ( 2 [
ˆ
2
)] 1 ( 2 [
) (
θ
κ
τ
κ
τ
κ
t
b a a
e
b a
b a
b a
a
b
t
b a
a
b a
b a
a
R
R R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
− +
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
− +
−
+
−
+
− + − +
4
ˆ
) ( )] 1 ( 2 [
4
2 2
1
2
1
1 1
2
1
2
1
1
8
2
1
2
1
1
2 2
1
2
1
2
1
2
1
8
sin
) (
2 )] 1 ( 2 [
ˆ
2
)] 1 ( 2 [
) (
2
1
2
1 1
θ
κ
τ
κ
τ
κ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
⎪
⎩
⎪
⎨
⎧
− + − − − ⎟
⎠
⎞
⎜
⎝
⎛ − +
=
− +
−
1
4
ˆ )] 1 ( 2 [
1 2
1
8 1
8
)] 1 ( 2 [ 1 [
ˆ
4
ˆ / 4
) 1 ( 2
erfc
2
c e
T
t
c
t c
c
I
t
R
R
R
κ
π τ
κ
τ κ
64
⎪
⎭
⎪
⎬
⎫
⎟
⎠
⎞
⎜
⎝
⎛
+
− +
−
+ − +
τ
τ
κ
τ
τ
κ
ˆ /
ˆ / 4
) 1 ( 2
erfc ]
ˆ
2
1
ˆ
)] 1 ( 2 [
2
1
2
1
1
t c
t
e
t c
t c
c
R
R
Shear stress:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
+
Ω =
∑ ∑
= =
8
1
1
2
4
1
0
o
ˆ
) 1 (
) 1 ( 3
ˆ
) 1 (
3
) , ( ˆ
n
n
n
n
J J t r
R R δκ
δκ
δκ
µ σ (4.16)
where
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎦
⎤
⎢
⎣
⎡
+
−
−
+
+
− ⎟
⎠
⎞
⎜
⎝
⎛ −
+
+
=
− + −
τ
θ θ
τ
ˆ
) ( ) 1 (
1 1 2
1
2
1
9 1 9 1
1 2
1
2
1
9 1 9 1
2
1
2
1
9 1 9 1 0
1
2
1
2
1 1
sin cos
ˆ / 4
1
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
J
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
− =
+ −
τ
τ τ τ
τ
ˆ /
ˆ / 4
1
erfc
ˆ / 4
1
erfc
ˆ / 4
1
erfc
ˆ
1
ˆ
) 1 (
1
9 0
2
2
1 1
t c
t
e
t
c
c
t
J
t
c c
R R R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+
− =
− + +
τ
θ θ
τ
ˆ
) ( ) 1 (
2 2
2
1
2
1
10 1 10 1
2
2
1
2
1
10 1 10 1
2
1
2
1
10 1 10 1 0
3
2
1
2
1 1
sin cos
ˆ / 4
1
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
J
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
+ +
τ
τ τ τ
τ
ˆ /
ˆ / 2
1
erfc
ˆ / 4
1
erfc
ˆ / 4
1
erfc
ˆ
1
ˆ
) 1 (
1
10 0
4
2
1 1
t c
t
e
t
c
c
t
J
t
c c
R R R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+
=
− + −
τ
θ θ
τ
ˆ
) ( ) 3 (
3 3
2
1
2
1
11 1 11 1
3
2
1
2
1
11 1 11 1
2
1
2
1
11 1 11 1 1
1
2
1
2
1 1
sin cos
ˆ / 4
3
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
J
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
− =
+ −
τ
τ τ τ
τ
ˆ /
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
1
11 1
2
2
1 1
t c
t
e
t
c
c
t
J
t
c c
R R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+ −
=
−
−
t
e
t
b a
b b a a
t b a
b b a a b a
J
4
ˆ ) 3 (
2
1
2
1
12 1 12 1
2 2
1
2
1
12 1 1 12
2
1
2
1 1
3
2
ˆ
4
ˆ / 4
3
erfc
) (
2 ) (
2
ˆ
τ
τ π
τ
R
R
+ Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
−
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
−
−
+
−
−
− + −
3
ˆ
) ( ) 3 (
3
2 2
1
2
1
1 1
2
1
2
1
1
12
2
1
2
1
1
2 2
1
2
1
2
1
2
1
12
cos
) (
2 ) 3 (
ˆ
2
) 3 (
) (
2
1
2
1 1
θ
τ
τ
t
b a a
e
b a
b a
b a
a
b
t
b a
a
b a
b a
a
R
R R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
−
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
−
−
+
−
+
− + −
3
ˆ
) ( ) 3 (
3
2 2
1
2
1
1 1
2
1
2
1
1
12
2
1
2
1
1
2 2
1
2
1
2
1
2
1
12
sin
) (
2 ) 3 (
ˆ
2
) 3 (
) (
2
1
2
1 1
θ
τ
τ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
65
−
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
−
− − − − − ⎟
⎠
⎞
⎜
⎝
⎛ −
=
+ −
−
−
τ
τ τ τ π τ
τ
τ
ˆ /
ˆ / 4
3
erfc ]
ˆ
2
) 3 ( 1 [
ˆ
4
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
2
1
1
4
ˆ ) 3 (
1 2
1
12 1
4
2
1
1
2
t c
t
e
t c
c e
t
c
t c
c
J
t c
c
t
R
R
R
R
R
⎟
⎠
⎞
⎜
⎝
⎛ −
−
τ ˆ / 4
3
erfc
t
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
Σ
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+
− =
− + +
τ
θ θ
τ
ˆ
) ( ) 3 (
4 4
2
1
2
1
13 1 13 1
4
2
1
2
1
13 1 13 1
2
1
2
1
13 1 13 1 1
5
2
1
2
1 1
sin cos
ˆ / 4
3
erfc 2
ˆ
t
b a a
e
b a
a b b a
b a
b b a a
t b a
b b a a
J
R
R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
+ +
τ
τ τ τ
τ
ˆ /
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
1
13 1
6
2
1 1
t c
t
e
t
c
c
t
J
t
c c
R R R
R
⎪
⎩
⎪
⎨
⎧
+
+
+
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
+
+ −
− =
+
−
t
e
t
b a
b b a a
t b a
b b a a b a
J
4
ˆ ) 3 (
2
1
2
1
14 1 14 1
2 2
1
2
1
14 1 1 14
2
1
2
1 1
7
2
ˆ
4
ˆ / 4
3
erfc
) (
2 ) (
2
ˆ
τ
τ π
τ
R
R
+ Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
+
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
+
−
+
−
−
− + +
4
ˆ
) ( ) 3 (
4
2 2
1
2
1
1 1
2
1
2
1
1
14
2
1
2
1
1
2 2
1
2
1
2
1
2
1
14
cos
) (
2 ) 3 (
ˆ
2
) 3 (
) (
2
1
2
1 1
θ
τ
τ
t
b a a
e
b a
b a
b a
a
b
t
b a
a
b a
b a
a
R
R R
⎪
⎭
⎪
⎬
⎫
Σ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
+
+
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
+
+
−
+
−
+
− + +
4
ˆ
) ( ) 3 (
4
2 2
1
2
1
1 1
2
1
2
1
1
14
2
1
2
1
1
2 2
1
2
1
2
1
2
1
14
sin
) (
2 ) 3 (
ˆ
2
) 3 (
) (
2
1
2
1 1
θ
τ
τ
t
b a a
e
b a
b a
b a
a
a
t
b a
a
b a
b a
b
R
R R
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
+
− + − − − ⎟
⎠
⎞
⎜
⎝
⎛ +
− =
+ +
+
−
τ
τ τ τ π τ
τ
τ
ˆ /
ˆ / 4
3
erfc ]
ˆ
2
) 3 ( 1 [
ˆ
4
ˆ / 4
3
erfc
ˆ
1
ˆ
) 3 (
2
1
1
4
ˆ ) 3 (
1 2
1
14 1
8
2
1
1
2
t c
t
e
t c
c e
t
c
t c
c
J
t c
c
t
R
R
R
R
R
+ ⎟
⎠
⎞
⎜
⎝
⎛ +
τ ˆ / 4
3
erfc
t
R
Note that the set of coefficients
m
a ,
m
b and
m
c in (4.15) and (4.16) are
introduced during partial-fraction-expansion process and given in Appendix B.
4.4 Case of a Rotation Startup
In this section, we consider the case of an initially stationary drop placed
under a steady torque with a θ sin -type distribution. The drop would eventually
achieve the steady state and rotate steadily with a constant angular velocity,
o
Ω .
66
It should be recognized that acoustic streaming using in drop levitation induces
not only the torque in time-averaged sense but also the internal circulation
within a liquid drop with velocity O( δ ). However, the viscosity of a liquid
drop is relatively higher than that of gas medium, i.e. 1 << δ . Hence the
acoustically driven internal circulation is negligibly small in this investigation.
Therefore, the appropriate initial and boundary conditions are given by
0 ˆ =
φ
u , 0 =
φ
u at 0 = t , (4.17a)
and
φ φ
u u = ˆ , θ σ σ σ
φ φ
sin ˆ
o r r
= − at R r = (4.17b)
The solutions of equations (4.4) satisfying the conditions (4.17) are
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
∆
+ + − +
=
−
s
X e X e X R
s r U
X X
)] 1
ˆ
( ) 1
ˆ
( )[ 1 (
) , (
ˆ
ˆ ˆ
2
o
R R
R
R R
µ
δ σ
(4.18a)
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
∆
+ + − +
=
− − − + −
s
X e X e X R
s r U
X X X X
)] 1
ˆ
( ) 1
ˆ
( )[ 1 (
) , (
ˆ
) 1 (
ˆ
) 1 (
2
o
R - R
R
R µ
δ σ
(4.18b)
The shear stress is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∆
+ + − + − +
=
−
s
X X e X X e X
X X
r
)] 3
ˆ
3
ˆ
( ) 3
ˆ
3
ˆ
( )[ 1 (
ˆ
2
ˆ
2
ˆ
o
R R R R
2 R 2 R
σ σ
φ
(4.19)
In similar manner, the inverse Laplace transforms of (4.18) and (4.19) using
inversion theorem which represent in terms of integral solution are given as
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
+ − =
∫
∞ −
0
2 / 1 2 / 1 2 / 1
2 2 2
o
] ) ˆ ( cos ) ˆ ( ) ˆ ( [sin
) (
) (
3
) , ( ˆ ρ ρ τ ρ τ ρ τ
ρ
τρ
π
δ
µ
σ
ρ
d
B A
A B e R
t r u
t
R R R
R
R
(4.20a)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+ − =
∫
∞ −
0
2 2 2 2
o
) (
) (
3
1
) , ( ρ
ρ π
δ
µ
σ
ρ
d
B A
AC BD e R
t r u
t
R R
(4.20b)
67
However, it is noted that, contrary to the previous case, the first term of the
equations which arises from the integration along a small circle around the
branch point survives in this case and represents a steady state solution of
problem. The shear stress of the liquid drop is obtained as
ρ ρ τ ρ τ ρ τ ρ τ
ρ
τρ
π
σ
σ
ρ
φ
d
B A
B A e
t
r
∫
∞ −
+ −
+
−
=
0
2 / 1 2 / 1 2
2 2 3
o
] ) ˆ ( cos ˆ 3 ) ˆ ( sin ) 3 ˆ [(
) (
) (
ˆ R R R R
R
(4.21)
In (4.20) and (4.21), the coefficients A, B, C, and D are those being defined in
(4.14)
4.4.1. Limit of 0 → δ
In the limit of 0 → δ , the problem can be viewed as a solid sphere rotated
in an infinite fluid by the applied torque θ σ sin
o
. The velocity field of the
ambient fluid is reduced to
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ − − +
+
=
−
) 1 ( 15 ) 5 1 (
1
) , (
2 2 2 3
) 1 (
2
o
X k X k X
X
s
e R
s r U
X
δ δ µ
σ R
R
- R
, (4.22)
Similarly, using the partial fraction expansion procedure, the equation (4.22)
may be written in the form:
⎟
⎠
⎞
⎜
⎝
⎛
−
+
+ −
−
+
− −
+
=
−
c X
p
bi a X
i n m
bi a X
i n m
s
e R
s r U
X ) 1 (
2
o
) , (
- R
R µ
σ
. (4.23)
Here, a complex conjugate pair bi a ± and a real constant c are the roots of the
denominator of the rational function ) (s Ω . A complex conjugate pair ni m ± and
a real constant p are obtained by a partial fraction expansion. The inverse-
Laplace transform of the equation (4.23) can be obtained by the use of standard
inversion tables and is given by
68
()
2 1 2
o
) , ( I I
R
t r u + =
R µ
σ
(4.24)
where
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+ −
−
+ ⎟
⎠
⎞
⎜
⎝
⎛ −
−
+
+
=
+ + − + −
τ
τ τ
τ
/ ) (
/ 2
1
erfc
/ 2
1
erfc Re 2
2
) ( ) 1 )( (
1
t bi a
t
e
t bi a
ni m
I
t
bi a R bi a
R R
,
⎟
⎠
⎞
⎜
⎝
⎛
−
−
+ ⎟
⎠
⎞
⎜
⎝
⎛ −
− =
+ − −
τ
τ τ
τ
/
/ 2
1
erfc
/ 2
1
erfc
2
) 1 (
2
t c
t
e
c
p
t c
p
I
t
c R c
R R
.
4.4.2. Small-time approximation
The inversions of (4.18) and (4.19) for small time are
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
+
=
∑ ∑
= =
8
1
1
2 2
4
1
0
2 2
o
ˆ
) 1 (
) 1 (
ˆ
) (
) , ( ˆ
n
n
n
n
I I
R
t r u
R R δκ κ
δκ δ
κ δκ
δ
µ
σ
(4.25a)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
+
=
∑ ∑
= =
8
1
1
2 2
4
1
0
2 2
o
) 1 (
) 1 (
) (
) , (
n
n
n
n
I I
R
t r u
R R δκ κ
δκ δ
κ δκ
δ
µ
σ
(4.25b)
The shear stress is
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
+
− =
∑ ∑
= =
8
1
1
2
4
1
0
o
ˆ
) 1 (
) 1 (
ˆ
) 1 (
1
) , ( ˆ
n
n
n
n
J J t r
R R δκ
δκ
δκ
σ σ (4.26)
4.5 Results and Discussion
We have carried out the velocity and shear stress calculations for various
sets of parameters. It should be noted that the expressions for velocity fields of
a rotating drop in infinite fluid is valid only in the low-Reynolds number
approximation. The numerical calculations are presented in this section. It is
worthwhile to discuss about the evaluation of the integral-representation
69
solution of such problems. The integral evaluation of the problem is not as
straightforward as one might think because of the complexity of the integrand.
Since, an analytical expression for the integral could not be found, numerical
integration is necessary. The integrand of the integral solution of such problems
contains a nearby pole, hence, giving rise to some difficulty in computing the
integration. In particular, the location of a nearby pole, say
o
ρ , must be
determined and the integration is divided into 2 parts: the first integral is from 0
to
o
ρ and the second from
o
ρ to infinity. The second integral needs to be
evaluated by making a change of variable that transforms the infinite range to
one that is finite when ρ approaches infinity. Then, the integration is performed
using six-point Newton-Cotes formula [32]. For a small time solution, the
calculation is straightforward because the solutions are expressed explicitly.
The results are compared to those of the integral-representation solutions and
the full range of solutions therefore can be established.
It is note that the results presented in this section are in dimensionless
form. Here, we choose R as a length scale, τ ˆ a time, R
o
Ω or
µ
σ R
o
a velocity,
and
o
Ω µ 3 or
o
σ a tangential stress.
70
4.5.1 Deceleration of a rotating drop
Figure 4.1 shows the transient flow field of a rotating water drop in a
gaseous medium. In steady state situation, the tangential stress due to the
applied torque is entirely taken up by the outer medium and the drop will rotate
like a solid body. When the applied torque is turned off, the shear stresses at the
drop interface are redistributed. The skin friction retards the drop rotation and
the drop is not able to maintain the constant rotational speed. As a result, the
rotation rate decreases as time progresses. As expected, the transient effect
initially takes place on the surface from inside and spreads into the drop inside
as illustrated in Figure 4.2. It shows that the liquid in region close to the drop
surface is slowed down by surface friction and the region away from the surface
still maintains solid-body-like rotation at τ ˆ 01 . 0 = t . However, the shear stress
0 0.5 1 1.5 2 2.5 3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
r/R
u/ Ω
o
R
t=0.01 τ
0.3 τ
1.01 τ
10 τ
FIGURE 4.1: Velocity fields of a water drop initially rotating with
o
Ω in a gas.
71
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3
-2.5
-2
-1.5
-1
-0.5
0
r/R
Deviation from solid body rotation(%)
1.01 τ
0.3 τ
0.05 τ
t=0.01 τ
FIGURE 4.2: Deviation of velocity profile from a solid body rotation.
gradually penetrates into the center of drop as illustrated. Consequently, the
flow inside the drop starts deviating from the solid body-like rotation. The
largest deviation apparently occurs at the drop surface and reaches the
maximum as τ ˆ 3 . 0 ≅ t , approximately. Thereafter, the shear stress inside the
drop is slowly depleted as the transient effect begins to die down. The velocity
Fluid
Silicone Oil, I 0.81352 0.00192
Water, II 0.25459 0.01863
Methanol, III 0.22274 0.03070
Acetone, IV 0.16616 0.05517
δ
κ
TABLE 4.1: Parameters of various liquid drops in air
72
profile deviates from solid-body rotation less than 0.5% after τ ˆ 01 . 1 < t . The
interfacial shear stress is shown in figure 4.4. The shear stress decreases
monotonically with time and become insignificant at τ ˆ 20 > t
In additional, various liquids in wide range of viscosities (which are
shown in Table 4.1) were examined and the results were shown in figure 4.4.
One can see that the deviation of the flow inside fluid I is much smaller than
those of other liquids. This is due to the fact that the viscosity of fluid I is
relatively high and therefore the transient effect is suppressed. In contrary, the
liquid with low viscosity such as fluid III and fluid IV is notably influenced by
the transient effect. The surface stresses penetrate into the drop and the
deviation of flow inside the drop is comparatively large. At moderate viscosity
such as fluid II, the maximum deviation is approximately 3%. Therefore the
solid-body-like rotation approximation is fairly reasonable, especially for high
viscous liquid drops, even though the flow at transient state is still affected by
the skin friction.
73
0 2 4 6 8 10 12 14 16 18 20
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
t/ τ
σ / (3 µΩ
o
)
FIGURE 4.3: Interfacial shear stress vs τ ˆ / t
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-8
-7
-6
-5
-4
-3
-2
-1
0
r/R
Deviation from solid body rotation(%)
II
I
III
IV
FIGURE 4.4: Deviation from solid-body rotation of various liquid drops at τ ˆ 3 . 0 = t
74
For a solid sphere case, when turning off the applied torque, the sphere
rotates freely and the rotational rate decreases gradually due to surface
tangential stresses. With the density ratio 0012 . 0 = β , the flow field induced by
the freely rotating sphere is shown in figure 4.5.
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
r/R
U/ Ω
o
R
t=0.01 τ
5 τ
15 τ
150 τ
FIGURE 4.5: Velocity field in gaseous medium induced by a freely rotating
solid sphere ( 0012 . 0 = β )
75
4.5.2 Drop rotation startup
The velocity development of a levitated drop under the applied steady
torque is shown in figure 4.6. It can be seen that the shear stress due to the
applied torque initially drives fluids in vicinity of the drop surface. The viscous
stress gradually spreads into the center of the drop and acts throughout. Then,
the drop begins to rotate about its center. Apparently, the rotational velocity of
the liquid drop increases under the steady torque; however, after τ 20 > t , the
rotation eventually reaches the steady state and the liquid drop steadily rotates
like a solid body with a constant rotation rate. The scaled interfacial shear
stress is plotted in figure 4.7 against scaled time. Note that the sign of the shear
stress in the plot indicates only the direction of shear stress. As expected, the
result shows that the shear stress decreases gradually after the applied torque is
exerted the drop surface. It is consistent with the velocity development shown
above; namely, the interfacial stress at liquid side vanishes as the drop rotates
like solid-body in steady state. In this state, the applied shear stress on the drop
surface is taken up entirely by the motion of the gas, and the liquid drop is able
to maintain a steady solid-body rotation.
For a solid-sphere case, when turning on the applied torque, the sphere
starts to rotate about its center due to the applied shear stress. With the density
ratio 0012 . 0 = β , the development of the velocity induced by the rotating sphere
is shown in Figure 4.6. As one might anticipate from the results, the sphere, as
well as the induced flow field, would reach a steady state and the applied shear
stress entirely takes up by the fluid motion.
76
0 0.5 1 1.5 2 2.5 3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
r/R
u/ ( σ
o
R/ µ)
20 τ
1.01 τ
0.3 τ
t=0.01 τ
FIGURE 4.6: Velocity fields of a liquid drop rotating under constant shear
stress θ σ sin
o
in infinite medium. ( 0186 . 0 = δ )
0 5 10 15 20 25 30 35 40 45 50
-1
-0.5
0
0.5
1
1.5
2
t/T
σ
r φ
/ σ
o
Air
Water
FIGURE 4.7: Tangential stresses at the surface of a water drop under a steady torque.
77
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
r/R
u/ ( σ
o
R/ µ)
5 τ
150 τ
15 τ
t=0.01 τ
FIGURE 4.8: Velocity field in a gaseous medium induced by a rotating solid
sphere under a steady shear stress θ σ sin
o
( 0012 . 0 = β ).
78
Chapter 5
Conclusions
A simple model of the solid particle in a rotating liquid drop has been
developed. The problem is defined as a spherical liquid drop holding a smaller
solid particle while the drop is levitated electrostatically and rotated about a
horizontal axis by acoustic forces. The resulting fluid dynamics of the system is
described by a complex mathematical model in a rotating reference frame and is
valid in the limit of low Reynolds and Taylor numbers. The model is used to
predict the motion of the particle on the equatorial plane. The stability of the
particle on this plane is studied by examining the motion of the particle which is
positioned slightly off the equatorial plane.
For the particle motion on the equatorial plane, the predicted trajectories
show that the particle orbits around the center of the rotating drop and exhibits
an outwardly spiral motion. The balance between gravitational and centrifugal
forces determines the motion of the particle under such circumstances. The
particle will be expelled toward drop surface quickly due to the centrifugal
forces at high rotation rate. In contrast, the particle motion at low rotation rate
experiences relatively greater influence of gravity and shows a level of particle
mobility which creates convective motion inside the drop. Therefore, an
appropriate rotation rate needs to be obtained in order to minimize the
convective motion. Moreover, our results show that the particle will initially
spiral inward when the applied torque is turned off and the drop is allowed to
rotate freely. Therefore, one would obtain a long time suspension of the particle
by turning off the torque when the particle reaches the drop surface and
allowing the particle to migrate back to the desired position.
The stability of the particle on equatorial plane is examined in the off-
equatorial plane case. It can be seen that the denser (
d p
ρ ρ > ) particle migrates
away from the equatorial plane. The lighter (
d p
ρ ρ < ) particle migrates toward
the equatorial plane and would settle in on this plane eventually. However, it
should be recognized that the study is based on the assumption in that the
particle is positioned only slightly off the equatorial plane and the required
additional acoustic force to maintain this asymmetric (but not wobbly) motion is
therefore negligibly small.
The transient dynamics of a rotating drop is studied in the limit of the
low Reynolds number approximation. The results show that the transient effect
initially takes place at the drop when the applied torque is either turned on or
off. The transient effect will gradually spread into the drop inside and then
deplete eventually. The maximum deviation of the drop velocity takes place at
the drop surface. However, the deviation will depend on the viscosity of the
liquid drop. With highly viscous liquids, the velocity deviation is relatively
small, especially away from the drop surface. Therefore, turning on/off the
applied torque would affect the velocity field of the rotating drop very little and
it is reasonable to approximate the bulk of the drop motion to be the solid-body
rotation.
79
Chapter 6
Suggestions for Future Works
There are several prospective works can be done in the future of this
investigation. First of all, in order to improve the proposed model, some
restrictions can be relaxed. The crystallizing process should be included in the
model. When the density gradient is created in the solution drop, protein
crystallization takes place and leads to mass transfer during crystal growth
process. The unsteady transport of mass, momentum in the solution is governed
by the following non-dimensional equations.
continuity
, 0 = ⋅ ∇ v (6.1)
momentum
,
Sc
Ra
2
∑
+ ∇ + −∇ = ∇ ⋅ +
∂
∂
j
j
j
j
p
t
g v v v
v
ρ (6.2)
species
j
j
j
j
t
ρ ρ
ρ
2
Sc
1
∇ = ∇ ⋅ +
∂
∂
v (6.3)
Here
j
ρ is dimensionless species mass density defined as , where
is the initial, uniform mass density of component j in the solution. The
dimensionless Rayliegh and Schmidt numbers for component j are defined as
, and
0 0
/ ) (
j j j
ρ ρ ρ −
0
j
ρ
ν ρ β
j j j c j
g R D / Ra
0 3
=
j j
D / Sc ν = respectively.
80
By including the inertia terms in the momentum equation, the motion of a
particle is represented by the following equations:
∑
=
i
i
dt
d
m F
V
, T
I
=
dt
Ω d ) (
(6.4)
V
X
=
dt
d
, Ω
dt
d
=
θ
(6.5)
where is the particle mass, I is the inertia moment, V is the translational
velocity and
m
Ω is the angular velocity of the particle, X is the centroid position,
and θ is the orientation of the particle. F
i
and T are the forces and torque on
the particle, respectively. Obviously, the solution to these equations must be
obtained by means of numerical simulation and depends on the appropriate
intial and bounary conditions on the protein particle and the soltuion drop.
Furthermore, as dicussed in chapter 3, the full 3-D calculation needs to
be studied when the particle is positioned off the equatorial plane. The
secondary flow induced by the acoustic streaming can be included to the model.
By applying the standing waves to the outside fluid, the equations of motion
govering the flow are given by
() 0 ,
M
1
2
2
= ′ ⋅ ∇ ′ ∇ + ′ −∇ = ′ ∇ ⋅ ′ +
∂
′ ∂
v v v v
v
p
t
ε (6.6)
At large distance from the solution drop, we require
∞ → − ′ x as t t j i v ω ω sin cos ~ (6.7)
81
Here, ω is the frequency of an acoustic wave. ε is the inverse of a Strouhal
number defined as
c c
R U ω / and is the frequency parameter defined as
. In addition, we require the continuity of velocity and stresses at the
surface of a solution drop. Because of complexity of the problem, the solution
of this problem needs to be obtained numerically.
2
M
ν ω /
2
c
R
82
83
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APPENDIX A: Derivation of the set of equations for
m
n
m
n
m
n
m
n
H C B A ..., , , ,
In order to complete the solutions, the set of unknown coefficients from
equations (2.36)-(2.40) must be obtained by satisfying the continuity and
boundary conditions of the considered system. We first truncate the infinite
series solutions in the Stokes flow region and retain only the leading N terms in
each series. The coefficients that uniquely satisfy the system of linear equations
obtained from the continuity equation and boundary conditions are then found
as follows:
A.1 Continuity equation
Substituting the series solutions given by (2.36)-(340) into (2.24) and (2.32) and
using the recursion relations for the Legendre polynomials, it can be shown that
the continuity equations for the case m=0 and m=1 are satisfied if
for m=0,
0
1
0 0
1
0
1
0 0
1
) 1 ( ) 1 2 ( ) 1 (
2
1
2
5
2
1
+ − + −
+ − + + − + + + −
n n n n n n
D n D n nD A n A nA
0 ) 2 )( 1 ( ) 1 ( 2 ) 1 (
0
1
0 0
1
= + + + + − − +
+ − n n n
E n n E n n E n n , (A.1)
0
1
0 0
1
0
1
0 0
1
) 1 ( ) 1 2 ( ) 1 (
2
1
2
5
2
1
+ − + −
+ − + + − + + + −
n n n n n n
C n C n nC B n B nB
0 ) 2 )( 1 ( ) 1 ( 2 ) 1 (
0
1
0 0
1
= + + + + − − +
+ − n n n
F n n F n n F n n , (A.2)
87
for m=1,
1
1
1 1
1
1
1
1 1
1
) 2 ( ) 1 2 ( ) 1 ( ) 2 (
2
1
2
5
) 1 (
2
1
+ − + −
+ − + + − − + + + − −
n n n n n n
D n D n D n A n A A n
1
1
1 1
1
) 3 )( 2 (
2
1
) 2 )( 1 ( ) 2 )( 1 (
2
1
+ −
+ + + + − − − − +
n n n
E n n E n n E n n
0
2
1
2
1
1
1
1 1
1
= − + −
+ − n n n
G G G , (A.3)
1
1
1 1
1
1
1
1 1
1
) 2 ( ) 1 2 ( ) 1 ( ) 2 (
2
1
2
5
) 1 (
2
1
+ − + −
+ − + + − − + + + − −
n n n n n n
C n C n C n B n B B n
1
1
1 1
1
) 3 )( 2 (
2
1
) 2 )( 1 ( ) 2 )( 1 (
2
1
+ −
+ + + + − − − − +
n n n
F n n F n n F n n
0
2
1
2
1
1
1
1 1
1
= − + −
+ − n n n
H H H . (A.4)
A.2 Boundary conditions at the interfaces
1. No slip boundary at particle and liquid interface.
The prescribed velocity may be written in terms of the bipolar
eigenfunctions. We therefore expand the velocity components as
p
v
∑
=
=
1
0
) cos( ) , (
m
p
m
p p
m u u φ η ξ (A.5a)
∑
=
=
1
0
) sin( ) , (
m
p
m
p p
m v v φ η ξ (A.5b)
∑
=
=
1
0
) cos( ) , (
m
p
m
p p
m w w φ η ξ (A.5c)
88
With the velocity components given in (2.16) and (2.17), we may write the
expansion as
for m=0,
, 2
, 0
, ) (cosh ) , (
, 0 ) (cosh ) , (
, 0 ) (cosh ) , (
)
2
1
(
0
0 0
0
0
0 2 / 1 0
1
0
0 2 / 1 0
1
0
0 2 / 1 0
p
n
n
n n
n
n
n p p p
n
n
n p p p
n
n
n p p p
e W Z
Y X
W P Z w
P Y v
P X u
η
ς η η ξ
ς η η ξ
ς η η ξ
+ −
∞
=
∞
=
∞
=
=
= =
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
= − =
= − =
= − =
∑
∑
∑
(A.6)
for m=1,
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎬
⎫
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
Ω − = − =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
Ω + = − = −
= − = +
∑
∑
∑
∞
=
∞
=
∞
=
,
cosh
sin
) (cosh ) , (
,
cosh
sinh
2 2 ) (cosh ) , ( ) , (
, 0 ) (cosh ) , ( ) , (
1
0
1 2 / 1 1
0
0
1 2 / 1 1 1
2
0
1 2 / 1 1 1
ζ η
ξ
ς η η ξ
ζ η
η
ς η η ξ η ξ
ς η η ξ η ξ
p
p n
n
n p p p
p
p
p n
n
n p p p p
n
n
n p p p p
c
P Z w
d
c
U P Y v u
P X v u
[]
. 2 2
, ) 1 2 ( 2 2 2 2
, 0
)
2
1
(
1
)
2
1
( )
2
1
(
1
1
p
p p
n
p n
n
p
n
n
n
e c Z
e d n c e U Y
X
η
η η
+ −
+ − + −
Ω − =
− + Ω + =
=
(A.7)
The expressions given by (A.5)-(A.7) are satisfied if
for all m,
m
n n p
m
n n
m
n p
m
n p
C s n C s m n B n A n η η η cosh ) 2 4 ( ) ( 2 )
2
1
cosh( )
2
1
sinh(
1 1
+ + − − + + +
− −
m
n n
m
n n o
m
n n
m
n n
D c m n D c n D c m n C s m n
1 1 1 1 1 1
) 1 ( 2 cosh ) 2 4 ( ) ( 2 ) 1 ( 2
+ + − − + +
+ + − + + − − + + − η
) (
sinh ) 3 2 (
) 1 ( 2
) ( coth 2 ) (
sinh ) 1 2 (
) ( 2
1 0 1 p
m
n
o
p
m
n p
m
n
o
Z
n
m n
Z Z
n
m n
η
η
η η η
η
+ −
+
+ +
− +
−
−
− = , (A.8)
89
m
n p
m
n p
m
n n
m
n n
m
n n
m
n n
F n E n D c D c C s C s η η )
2
1
cosh( )
2
1
sinh(
1 1 1 1 1 1 1 1
+ + + + + − + −
+ + − − + + − −
) (
sinh ) 3 2 (
1
) (
sinh ) 1 2 (
1
) (
1 1 p
m
n
p
p
m
n
p
p
m
n
Z
n
Z
n
X η
η
η
η
η
+ −
+
+
−
− = , (A.9)
for m=0,
) ( )
2
1
cosh( )
2
1
sinh(
0 0 0
p n n p n p
Y H n G n η η η = + + +
, (A.10)
for m=1,
1
1 1
1
1 1
1
1 1
1
1 1
) 2 )( 1 ( ) 1 ( ) 2 )( 1 ( ) 1 (
+ + − − + + − −
+ + + − + + + + −
n n n n n n n n
D c n n D nc n C s n n C ns n
) (
sinh ) 1 2 (
) 1 (
) ( )
2
1
cosh( )
2
1
sinh(
1
1
1 1 1
p n
p
o n n p n p
Z
n
n n
Y H n G n η
η
η η η
−
−
−
+ = + + + +
) (
sinh ) 3 2 (
) 2 )( 1 (
1
1 p n
o
Z
n
n n
η
η
+
+
+ +
− , (A.11)
where
p
p
n
n
n
s
η
η
sinh ) 1 2 (
)
2
1
sinh(
+
+
= ,
p
p
n
n
n
c
η
η
sinh ) 1 2 (
)
2
1
cosh(
+
+
= .
2. Zero normal velocity and zero shear stress at the drop surface.
Under a controlled environment, we assume that there is no evaporation or
condensation on the liquid drop surface. This is a reasonable approximation
because any evaporation to create favorable crystallization conditions would be
quite weak and have little impact on the fluid dynamics. For the relatively high
90
viscosity of the liquid drop, we assume that shear stresses acting on the drop
surface are negligibly small. Therefore we require
0 ) 1 cos (cosh ) sinh (sin = − + = w u u
d d
ξ η η ξ
η
, (A.12)
[ ] 0 ) sinh (sin ) 1 cosh (cos = − −
∂
∂
=
=
d
w u
d d
η η
ηξ
η ξ η ξ
η
τ , (A.13)
0 sinh ) cos (cosh = +
∂
∂
− = v
v
d d
η
η
ξ η τ
ηφ
(A.14)
The substitution of ( ) into (A.12)-(A.14) and use of the recursion
relations for the Legendre polynomials lead to the following equations:
w v u , ,
for m=0,
0
) 3 2 (
) 2 )( 1 (
) 1 2 (
) 1 (
) 6 4 (
) 1 (
2 4
0
1
0
1
0 0
1
0
1
=
+
+ +
+
+
−
− +
+
+
+
−
+ − + − n n n n n
n
n n
n
n n
n
n
n
n
C C B A A , (A.15)
where
⎥
⎦
⎤
⎢
⎣
⎡
+ + + =
d n d n d n
n B n A η η η )
2
1
cosh( )
2
1
sinh( sinh
0 0 0
A
⎥
⎦
⎤
⎢
⎣
⎡
+ + + +
d n d n d
n D n C η η η )
2
1
cosh( )
2
1
sinh( cosh 2
0 0
,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + − =
d n d n n
n D n C η η )
2
1
cosh( )
2
1
sinh(
0 0 0
B ,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + =
d n d n d n
n F n E η η η )
2
1
cosh( )
2
1
sinh( sinh
0 0 0
C ,
0
1
0
1
0
2
0 0
2
) 3 2 (
1
-
) 1 2 (
1
) 3 2 (
) 2 (
) 3 2 )( 1 2 (
) 1 2 (
) 1 2 (
) 1 (
+ + −
+ −
+
+
+
−
+ −
+
+
−
−
n n- n n n
n n n
n
n n
n
n
n
E E D D D
0
1
0
2
0
1
) 4 2 )( 3 2 (
) 3 )( 2 (
) 3 2 )( 1 2 (
) 3 2 2 (
) 3 2 )( 1 2 (
) 1 )( 2 (
+
+ +
+ +
+
+ −
− +
+
− −
− −
+
n n n-
n n
n n
n n
n n
n n
n n
F F F
0
) 3 2 (
) 2 (
) 1 2 (
) 1 (
0 0
1
0
1
= +
+
+
+
−
−
+
+ n n n-
n
n
n
n
H G G , (A.16)
91
with
0 0
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh) 2
n d d d d n
A n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + = η η η η D
0
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 2
n d d d d
B n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
0
)
2
1
cosh( sinh ) 2 4 ( )
2
1
sinh( cosh 4
n d d d d
C n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
0
)
2
1
sinh( sinh ) 2 4 ( )
2
1
cosh( cosh 4
n d d d d
D n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η ,
0 2 0
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( cosh sinh 3
n d d d d d n
A n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η η E
0 2
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( cosh sinh 3
n d d d d d
B n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η
0 2
)
2
1
cosh( cosh sinh ) 2 4 ( )
2
1
sinh( ) cosh 6 2 (
n d d d d d
C n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η
0 2
)
2
1
sinh( cosh sinh ) 2 4 ( )
2
1
cosh( ) cosh 6 2 (
n d d d d d
D n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η ,
0 0
)
2
1
cosh( cosh ) 2 4 ( )
2
1
sinh( sinh 4
n d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η F
0
)
2
1
sinh( cosh ) 2 4 ( )
2
1
cosh( sinh 4
n d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η ,
0 2 0
)
2
1
cosh( ) cosh 1 )( 2 4 ( )
2
1
sinh( cosh sinh 6
n d d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + = η η η η η G
0 2
)
2
1
sinh( ) cosh 1 )( 2 4 ( )
2
1
cosh( cosh sinh 6
n d d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + + η η η η η ,
0 0
)
2
1
cosh( cosh ) 2 4 ( )
2
1
sinh( sinh 2
n d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η H
0
)
2
1
sinh( cosh ) 2 4 ( )
2
1
cosh( sinh 2
n d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η ,
92
for m=1
1
1
1
1
1 1
1
1
1
) 3 2 (
) 3 )( 2 (
) 1 2 (
) 1 )( 2 (
) 3 2 (
) 2 (
) 1 2 (
) 1 (
+ − + −
+
+ +
+
−
− −
− +
+
+
+
−
−
n n n n n
n
n n
n
n n
n
n
n
n
C C B A A
0
) 3 2 (
1
) 1 2 (
1
1
1
1
1
=
+
−
−
+ n n-
n n
D D , (A.17)
where
⎥
⎦
⎤
⎢
⎣
⎡
+ + + =
d n d n d n
n B n A η η η )
2
1
cosh( )
2
1
sinh( sinh
1 1 1
A
⎥
⎦
⎤
⎢
⎣
⎡
+ + + +
d n d n d
n D n C η η η )
2
1
cosh( )
2
1
sinh( cosh 2
1 1
,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + − =
d n d n n
n D n C η η )
2
1
cosh( )
2
1
sinh( 2
1 1 1
B ,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + =
d n d n d n
n F n E η η η )
2
1
cosh( )
2
1
sinh( sinh
1 1 1
C ,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + =
d n d n d n
n H n G η η η )
2
1
cosh( )
2
1
sinh( sinh
1 1 1
D ,
and
1
2
1 1
2
) 5 2 )( 3 2 (
) 3 )( 2 (
) 3 2 )( 1 2 (
) 1 ( 2
) 1 2 )( 3 2 (
) 1 )( 2 (
+
+ +
+ +
−
+ −
+
+
− −
− −
−
n n n-
n n
n n
n n
n n
n n
n n
E E E
1
1
2
1
3
) 3 2 )( 1 2 )( 3 2 (
) 2 (
) 1 2 )( 3 2 )( 5 2 (
) 1 )( 2 )( 3 (
−
+ − −
+
+
− − −
− − −
−
n n-
n n n
n n
n n n
n n n
F F
1
3
1
1
2
) 7 2 )( 5 2 )( 3 2 (
) 4 )( 3 )( 2 (
) 5 2 )( 3 2 )( 1 2 (
) 2 ( ) 1 (
+ +
+ + +
+ + +
−
+ + −
+ +
+
n n
n n n
n n n
n n n
n n
F F
1
1
1
1
) 3 2 (
) 3 )( 2 (
) 1 2 (
) 1 )( 2 (
+
+
+ +
+
−
− −
−
n n-
n
n n
n
n n
G G
1
2
1 1
2
) 5 2 )( 3 2 (
) 4 )( 3 )( 2 (
) 3 2 )( 1 2 (
) 2 )( 1 ( 3
) 1 2 )( 3 2 (
) 1 )( 2 )( 3 (
+
+ +
+ + +
+
+ −
+ −
+
− −
− − −
−
n n n-
n n
n n n
n n
n n
n n
n n n
H H H
93
1
1
2
1
3
) 3 2 )( 1 2 )( 3 2 (
) 9 3 )( 1 )( 2 (
) 1 2 )( 3 2 )( 5 2 (
) 1 )( 2 )( 3 )( 4 (
−
+ − −
− − − −
−
− − −
− − − −
−
n n-
n n n
n n n n
n n n
n n n n
I I
1
3
1
1
2
) 7 2 )( 5 2 )( 3 2 (
) 5 )( 4 )( 3 )( 2 (
) 5 2 )( 3 2 )( 1 2 (
) 5 5 )( 3 )( 2 (
+ +
+ + +
+ + + +
+
+ + −
− + + +
+
n n
n n n
n n n n
n n n
n n n n
I I
1
1
1
1
) 3 2 (
1
) 1 2 (
1
+
+
+
−
−
n n-
n n
J J
1
2
1 1
2
) 5 2 )( 3 2 (
) 2 (
) 3 2 )( 1 2 (
1
) 1 2 )( 3 2 (
) 1 (
+
+ +
+
−
+ −
+
− −
−
+
n n n-
n n
n
n n n n
n
K K K
1
1
2
1
3
) 3 2 )( 1 2 )( 3 2 (
) 3 (
) 1 2 )( 3 2 )( 5 2 (
) 1 )( 2 (
−
+ − −
− +
+
− − −
− −
+
n n-
n n n
n n
n n n
n n
L L
0
) 7 2 )( 5 2 )( 3 2 (
) 3 )( 2 (
) 5 2 )( 3 2 )( 1 2 (
) 3 (
1
3
1
1
2
=
+ + +
+ +
−
+ + −
− +
−
+ + n n
n n n
n n
n n n
n n
L L (A.18)
with
1 2 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( cosh sinh 3
n d d d d d n
A n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η η E
1 2
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( cosh sinh 3
n d d d d d
B n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η
1 2
)
2
1
cosh( cosh sinh ) 2 4 ( )
2
1
sinh( ) cosh 6 2 (
n d d d d d
C n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η
1 2
)
2
1
sinh( cosh sinh ) 2 4 ( )
2
1
cosh( ) cosh 6 2 (
n d d d d d
D n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η η ,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh 2
n d d d d n
A n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + = η η η η F
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 2
n d d d d
B n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
1
)
2
1
cosh( sinh ) 2 4 ( )
2
1
sinh( cosh 4
n d d d d
C n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
94
1
)
2
1
sinh( sinh ) 2 4 ( )
2
1
cosh( cosh 4
n d d d d
D n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η ,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh
n d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η G
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh
n d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η ,
1 2 1
)
2
1
cosh( ) cosh 1 )( 1 2 ( )
2
1
sinh( cosh sinh 3
n d d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + = η η η η η H
1 2
)
2
1
sinh( ) cosh 1 )( 1 2 ( )
2
1
cosh( cosh sinh 3
n d d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + + η η η η η ,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh 2
n d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η I
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 2
n d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η ,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh
n d d d d n
G n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η J
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh
n d d d d
H n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η ,
1 2 1
)
2
1
cosh( ) cosh 1 )( 1 2 ( )
2
1
sinh( cosh sinh 3
n d d d d d n
G n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + = η η η η η K
1 2
)
2
1
sinh( ) cosh 1 )( 1 2 ( )
2
1
cosh( cosh sinh 3
n d d d d d
H n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + + η η η η η
,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh 2
n d d d d n
G n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − = η η η η L
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 2
n d d d d
H n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + − + − + η η η η
.
95
1
1
1
1
1
1
1
1
) 3 2 (
1
) 1 2 (
1
) 3 2 (
) 3 )( 2 (
) 1 2 (
) 1 )( 2 (
+ +
+
+
−
−
+
+ +
+
−
− −
−
n n- n n-
n n n
n n
n
n n
N N M M
1
2
1 1
2
) 5 2 )( 3 2 (
) 4 )( 3 )( 2 (
) 3 2 )( 1 2 (
) 2 )( 1 ( 3
) 1 2 )( 3 2 (
) 1 )( 2 )( 3 (
+
+ +
+ + +
−
+ −
+ −
−
− −
− − −
+
n n n-
n n
n n n
n n
n n
n n
n n n
O O O
0
) 5 2 )( 3 2 (
) 2 (
) 3 2 )( 1 2 (
1
) 1 2 )( 3 2 (
) 1 (
1
2
1 1
2
=
+ +
+
−
+ −
+
− −
−
+
+ n n n-
n n
n
n n n n
n
P P P
. (A.19)
Here,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh 3
n d d d d n
E n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + = η η η η M
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 3
n d d d d
F n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
,
1 1
)
2
1
cosh( cosh ) 1 2 ( )
2
1
sinh( sinh 3
n d d d d n
G n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + = η η η η N
1
)
2
1
sinh( cosh ) 1 2 ( )
2
1
cosh( sinh 3
n d d d d
H n n n
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + + η η η η
,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + =
1 1 1
)
2
1
sinh( )
2
1
cosh( ) 1 2 (
n d n d n
F n E n n η η O
,
⎥
⎦
⎤
⎢
⎣
⎡
+ + + + =
1 1 1
)
2
1
sinh( )
2
1
cosh( ) 1 2 (
n d n d n
H n G n n η η P
.
96
APPENDIX B: Coefficients and useful formulae in transient
dynamics problems
B.1 Coefficients
m m m
c b a , ,
The set of the coefficients , and in (4.15) and (4.16) can be
obtained by the following procedure. First, finding the roots of the cubic
functions in gives
m
a
m
b
m
c
n
G
i b a Ζ i b a Z c X Z X Z X
c
c
X
c
c
X
c
c
X
1 1 1 1 1 1 1 1 1
11
14
11
13 2
11
12 3
and where ),
ˆ
)(
ˆ
)(
ˆ
(
ˆ ˆ ˆ
− = + = − − − = + + +
,
. and where
2 2 2 2 2 2 2 2 2
21
24
21
23 2
21
22 3
),
ˆ
)(
ˆ
)(
ˆ
(
ˆ ˆ ˆ
i b a Ζ i b a Z c X Z X Z X
c
c
X
c
c
X
c
c
X − = + = − − − = + + +
Then, the fractions of cubic function contained in series expansion are
simplified by partial fraction expansion. Performing the expansions yields the
following results:
for n =0
liquid drop:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
+ +
+
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− +
+ − − −
)
ˆ
)(
ˆ
)(
ˆ
(
) 1
ˆ
)( 1
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
(
) 1
ˆ
)( 1
ˆ
(
1 1 1
ˆ
) 1 (
1 1 1
ˆ
) 1 (
c X Z X Z X
X X
s
e
c X Z X Z X
X X
s
e
X X
R R
R R
κ κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
−
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
=
+ − − −
1
4
1
4
1
4
ˆ
) 1 (
1
3
1
3
1
3
ˆ
) 1 (
ˆ ˆ ˆ ˆ ˆ ˆ
c X
c
Z X
Z
Z X
Z
s
e
c X
c
Z X
Z
Z X
Z
s
e
X X R R
97
gas:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
+ +
+
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− +
− + − − −
)
ˆ
)(
ˆ
)(
ˆ
(
) 1
ˆ
)( 1
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
(
) 1
ˆ
)( 1
ˆ
(
1 1 1
ˆ
2 ) 1 (
1 1 1
ˆ
) 1 (
c X Z X Z X
X X
s
e
c X Z X Z X
X X
s
e
X X X
R R
R R
κ κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
=
− + − − −
1
4
1
4
1
4
ˆ
2 ) 1 (
1
3
1
3
1
3
ˆ
) 1 (
ˆ ˆ ˆ ˆ ˆ ˆ
c X
c
Z X
Z
Z X
Z
s
e
c X
c
Z X
Z
Z X
Z
s
e
X X X R R
Shear stress:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
+ + +
−
⎥
⎦
⎤
⎢
⎣
⎡
− − −
+ − +
− −
)
ˆ
)(
ˆ
)(
ˆ
(
) 3 3 )( 1
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
(
) 3 3 )( 1
ˆ
(
1 1 1
2 2
ˆ
1 1 1
2 2
ˆ
c X Z X Z X
X X X
s
e
c X Z X Z X
X X X
s
e
X X
R R R R R R
R R
κ κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+ −
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+ =
+ − − −
1
10
1
10
1
10
ˆ
) 1 (
1
9
1
9
1
9
ˆ
) 1 (
ˆ ˆ ˆ
1
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
s
e
c X
c
Z X
Z
Z X
Z
s
e
X X R R
for n =1,
liquid drop:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − − +
− −
2
1
2
1
2
1
2 2 2
ˆ
) 3 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 1
ˆ
)( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X
s
e
X
R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − + +
+
+ −
2
1
2
1
2
1
2 2 2
ˆ
) 3 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 1
ˆ
)( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X
s
e
X
R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ =
− −
2
1
6
2
1
6
2
1
6
1
5
1
5
1
5
ˆ
) 3 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X R
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ +
+ −
2
1
8
2
1
8
2
1
8
1
7
1
7
1
7
ˆ
) 3 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X R
98
gas:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − − +
+ −
2
1
2
1
2
1
2 2 2
ˆ
1 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 1
ˆ
)( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X
s
e
X )
R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − + +
+
− + −
2
1
2
1
2
1
2 2 2
ˆ
4
ˆ
) 1 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 1
ˆ
)( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X
s
e
X X
R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ =
+ −
2
1
6
2
1
6
2
1
6
1
5
1
5
1
5
ˆ
) 1 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X R
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ +
− + −
2
1
8
2
1
8
2
1
8
1
7
1
7
1
7
ˆ
4 ) 1 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X X R
Shear stress:
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − + − +
− −
2
1
2
1
2
1
2 2 2
2 2
ˆ
) 2 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 3 3 )( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X X
s
e
X
R R R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
− − −
− − − + + +
−
+ −
2
1
2
1
2
1
2 2 2
2 2
ˆ
) 2 (
)
ˆ
( )
ˆ
( )
ˆ
(
)
ˆ
)(
ˆ
)(
ˆ
)( 3 3 )( 1
ˆ
(
c X Z X Z X
c X Z X Z X X X X
s
e
X
R R R
R
κ
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ =
− −
2
1
12
2
1
12
2
1
12
1
11
1
11
1
11
ˆ
) 2 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X R
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
+
+
+
+
+
+
+
+
+ −
+ −
2
1
14
2
1
14
2
1
14
1
13
1
13
1
13
ˆ
) 2 (
)
ˆ
( )
ˆ
( )
ˆ
(
ˆ ˆ ˆ
1
c X
c
Z X
Z
Z X
Z
c X
c
Z X
Z
Z X
Z
s
e
X R
Here , i b a Z
m m m
+ = i b a Z
m m m
+ = and is a real constant.
m
c
It should be noted that the values of the coefficients to in liquid
drop and gas expressions are different even though the same notations are used
in the expressions.
3
a
3
b
3
c
8
a
8
b
8
c
99
100
B.2 Useful formulae
Inversion of Laplace transforms
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−
−
2 / 1
1
) ( 2
erfc
at
x
s
e
x z
L B.2.1
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
+
−
− 2 / 1
2 / 1 2 / 1
1
) (
) ( 2
erfc
1
) ( 2
erfc
1
) (
2
at k
at
x
e
k at
x
k k z s
e
t k a x k
x z
L B.2.2
+ ⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
−
− at x
x z
e
at
k at
x
k k z s
e
4 /
2 / 1
2 / 1 2 2
1
2
2
) ( 2
erfc
1
) ( π
L
t k a x k
e at k
at
x
t k a x k
k
2
2 / 1
2 / 1
2
2
) (
) ( 2
erfc ) 2 1 (
1
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ − − − B.2.3
where, ; a and x are positive real
2 / 1
) / ( a s z =
Complimentary error function, erfc
Define as:
∫
−
− =
x
t
dt e x
0
2
2
1 ) ( erfc
π
B.2.4
For complex argument, the complex characteristics are given as
Real part:
{}
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− − +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− + = +
2
2
2
2
erfc erfc
2
1
) erfc( Re
x
y
x x
x
y
x x yi x
∑
∞
=
+
+
−
+
−
− =
0
1 2
2 2
) (
)! 2 2 (
) 1 ( 2
) erfc(
2
k
k
k k
x
x H
k
y
e x
π
. B.2.5
Imaginary part:
{}
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− + −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− − − = +
2
2
2
2
2
2
erfc erfc
2
) erfc( Im
x
y
x x
x
y
x x
x
y
y
x
yi x
∑
∞
=
+
−
+
−
− =
0
2
1 2
) (
)! 1 2 (
) 1 ( 2
2
k
k
k k
x
x H
k
y
e
π
. B.2.6
Absolute value:
{}
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− + ⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
− − = +
2
2
2
2
erfc erfc ) erfc( Abs
x
y
x x
x
y
x x yi x . B2.7
Argument:
{} { } { } () ) erfc( Im , ) erfc( Re tan ) erfc( Arg
1
yi x yi x yi x + + = +
−
B2.8
where
) (x H
n
is the Hermite polynomial function.
Two-argument inverse tangent, ) , ( tan
1
y x
−
Define as:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
− =
−
2 2
1
log ) , ( tan
y x
iy x
i y x B2.9
101
Abstract (if available)
Abstract
In this investigation, fluid dynamics of a spherical particle in a rotating liquid sphere is studied. Such problems arise during protein crystal growth by the containerless approach in which a drop of protein solution is levitated against gravity, and rotated steadily by a torque from an acoustic field. Viscosity of the solution is assumed to be sufficiently high such that viscous force dominates and the second-order effects in the acoustic field do not penetrate the drop surface. The analysis is carried out at very small Reynolds and Taylor numbers for which the fluid flow is treated by Stokes approximation. With a given geometric configuration, the velocity field is calculated and used to determine the particle velocity at a given time. By integrating the particle velocity, the particle path is obtained and used for optimizing the rotation rate of the drop in an attempt to keep the particle fully contained in the drop.
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Asset Metadata
Creator
Asavatesanupap, Channarong
(author)
Core Title
Fluid dynamics of a crystallizing particle in a rotating liquid sphere
School
Viterbi School of Engineering
Degree
Doctor of Philosophy
Degree Program
Mechanical Engineering
Publication Date
02/01/2007
Defense Date
12/11/2006
Publisher
University of Southern California
(original),
University of Southern California. Libraries
(digital)
Tag
fluid dynamics,levitated liquid drop,liquid sphere,OAI-PMH Harvest,rotating sphere,Stokes flow
Language
English
Advisor
Sadhal, Satwindar S. (
committee chair
), Egolfopoulos, Fokion N. (
committee member
), Redekopp, Larry G. (
committee member
), Shing, Katherine S. (
committee member
)
Creator Email
asavates@usc.edu
Permanent Link (DOI)
https://doi.org/10.25549/usctheses-m238
Unique identifier
UC192959
Identifier
etd-Asavatesanupap-20070201 (filename),usctheses-m40 (legacy collection record id),usctheses-c127-159667 (legacy record id),usctheses-m238 (legacy record id)
Legacy Identifier
etd-Asavatesanupap-20070201.pdf
Dmrecord
159667
Document Type
Dissertation
Rights
Asavatesanupap, Channarong
Type
texts
Source
University of Southern California
(contributing entity),
University of Southern California Dissertations and Theses
(collection)
Repository Name
Libraries, University of Southern California
Repository Location
Los Angeles, California
Repository Email
cisadmin@lib.usc.edu
Tags
fluid dynamics
levitated liquid drop
liquid sphere
rotating sphere
Stokes flow