AN ABSTRACT OF THE THESIS OF
Franz Hugo Helfenstein
in
presented on
Mathematics
Doctor of Philosophy
for the degree of
May 1. 1986.
A Free Boundary Value Problem Modeling Streambed Erosion
Title:
Redacted for Privacy
Abstract approved:
Ronald B. Guenther
to derive
Our purpose is twofold,
equations of our model.
bounded
model
of
and to develop a solution procedure to solve the
erosion
streambed
2dimensional
a
The flow domain,
which varies in time,
is
above by a free surface and is bounded below by an erodable
streambed.
An initial flow and streambed
configuration
given
are
and we would like to determine the streambed evolution.
Our
interaction of
viscous
a
flow
with
which
constraints
boundary
an erodable streambed.
represent
over
Flow
evolution.
flow
approach to modeling this process is to combine the
basic
a
bed of sand causes the bed to undergo a distinct
At slow flows,
velocity,
the
bed
the bed remains
passes
flat.
increasing
With
through a set of stages:
smooth,
ripples, dunes, flat with chaotic flow, waves and finally antidunes.
One of our main
purposes
is
to
obtain
a
model
which
has
the
potential of addressing this evolutionary process.
We
obtain
a
model
which
exhibits
the
general
nature
of
streambed erosion.
Specifically,
we obtain classic dune migration.
One indirect but significant result is that
flow
near
vorticity
of
the
the streambed is shown to be a fundamental factor in the
erosion process.
questions
the
which
Another indirect result
have
arisen
as
is
the
wealth
a result of this work.
of
open
A Free Boundary Value Problem Modeling Streambed Erosion
by
Franz Hugo Helfenstein
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed May 1, 1986
Commencement June 1986
APPROVED:
Redacted for Privacy
Professor of Mathematics i
charge of major
Redacted for Privacy
Head of department of Mathematics
Redacted for Privacy
Dean of Gra
School
Date thesis is presented
May 1, 1986
Typed by Franz Hugo Helfenstein for
Franz Hugo Helfenstein
Al.:14017LEDGELIENTS
I would like to thank Ron Guenther for
and
help
he
thank my wife,
gave
Lea,
his
me in reaching this goal.
for her patience,
support she gave me along the way.
patience,
insight
I would also like to
understanding,
and special
TABLE OF CONTENTS
INTRODUCTION
The Problem
The General Approach
1
BACKGROUND OF FLUID DYNAMICS AND SEDIMENTATION
Fluid Dynamics
The Navier-Stokes Equation
Boundary Conditions
Initial Conditions
Dimensional Analysis
The Reynolds Number and Other Parameters
Vorticity and Potential Flow
The Law of the Wall and Boundary Layers
Energy Dissipation
Concluding Remarks- Fluid Dynamics
Sedimentation
Empirical Results- Sediment Transport
Empirical Results- Boundary Evolution
A Brief History of Sedimentation Theory
The Forces on Sediment Particles
Concluding Remarks- Sedimentation
6
1
2
6
6
9
10
11
12
13
15
18
19
21
21
23
25
27
28
THE EROSION MODEL
The Flow Field
Streambed Erosion as a Boundary Condition
Concluding Remarks- The Model
30
31
35
SPECIAL CASE FLOWS
Almost Steady, Uniform Flow
Potential Flow in 2 With Law of the Wall in r
Concluding Remarks- Special Case Flows
46
46
49
52
A GENERAL FLOW
The Flow Domain
Solution Procedure- Reducing to a System of Initial Value
Problems
Solution Procedure- Solving the Initial Value Problems
Numerically
Results- Graphical
Concluding Remarks- The General Flow
55
55
43
56
68
70
72
SUMMARY, OPEN QUESTIONS AND CONCLUDING REMARKS
Summary
Some Remarks The Model and its Solutions
Improvements The Model
Improvements The Numerical Scheme
Open Questions
Concluding Remarks
73
73
75
BIBLIOGRAPHY
80
APPENDIX
The Inner Products
83
76
77
77
78
(i)
LIST OF FIGURES
Figure_
Page
1.0
The Erosion Problem Schematic
2.0
Potential Flow
14
2.1
The Law of the Wall
17
2.2
The Streambed Close up
17
2.3
Classification of Sediment Transport Mechanisms
22
2.4
'Major' and 'Minor' Streambed Evolutionary Stages
24
2.5
Classical Force Diagram for a Sediment Particle
28
3.0
The Flow Regions 0 and
r
33
3.1
Sediment Particle Transport Mechanism
37
3.2
The Erosion Model- An Overview
44
4.0
Steady Uniform Flow
46
4.1
Almost Steady Uniform Flow- A Rippled Bed
47
4.2
Dune Migration- Almost Steady Uniform Flow
49
4.3
Potential Flow in 0 with Law of the Wall in 1
50
4.4a
Dune Migration- Potential Flow/Law of the Wall
53
4.4b
Dune Migration- Potential Flow/Law of the Wall
54
4.4c
Dune Migration- Potential Flow/Law of the Wall
54
5.0
The Grid on R
64
5.1a
Dune Migration- A General Flow
71
5.1b
Dune Migration- A General Flow
71
1
A List of the Symbols Used
In general,
while
variables subscripted with s refer to the sediment
those subscripted with f refer to the fluid.
is used
for
a
critical
empirical constants.
value.
Numbers
constant; inflow at x=a
A
matrix block in A
are
critical efficiency factor
large matrix
ei
1d basis element
constant; outflow at x=b
eij
2d basis element
matrix block in A
kinetic energy
wave speed; c=c(y*i)
energy dissipation rate of
fluid due to friction
ci
constant, empirical
cz
constant, empirical
bed packing ratio
sediment concentration
Ea
Ec
Ef
net rate energy available for
erosion work rate
critical energy rate level
rate energy convertable to
erosion work rate
matrix block in A
general function
function; d={i'+yh')/h
vector valued function
da
element of area
2
Fr
dir
local flow direction
dr
dr=[1+(e) ]
2 -1
fs
Froude number; Fr=U /Lg
force on sediment particle
magnitude of gravity
ds
sediment diameter
gravity vector
ds
for
efficiency factor
ec
cs
subscripts
No subscript indicates a generic variable.
a
A
as
The subscript c
element of arc/ength
flow depth; h=ui
flow domain; D=D(t)=B+1
rate of deformation
tensor
V. Karmen constant;
k.
constants
=.417
ks
conversion factori12
k =c c El+(e)
S
characteristic length
qf
qs
Re
arbitrary volume
Vc
critical velocity for
sediment movement
Vs
sediment velocity
neglecting resistence
PS
arbitrary function;
K=K(x,i(x))
P1
V
unit normal vector
test function
pressure, variable
set of test functions
nonfundamental parameters
volumetric fluid discharge
Ws
x
weight of particle
coordinate in
a
sediment discharge
coordinate in R
resistance term
coordinate in 0
Reynolds number; Re=UL/v
y
sediment particle
curve parameter
normal distance from 4
zo
time, variable
stress tensor
coordinate in R
stationary flow depth
z=(a41,1,,e)
a
coeficient for u; u=a..eij
ii
horizontal component of u
a
s
u*
U0
sediment velocity along
shear velocity; u*5riTirf
stream gradient, degrees
mean fluid velocity
particle rotation rate
mean flow
9c
Uc
Us
Ua
..... a
)
II
ij
coeficient for
fluid velocity, vector
U
a=(a
angle of repose
critical flow
aV
surface of V
mean sediment velocity
8
coeficient for A; A=Siei
A
del operator; A=V
fluid velocity along a
2
vertical component of u
( iv )
Ax
Ay
grid width
Xs
grid height
mainstream flow domain; 2=0(t)
coeficiera for (15;
gradient operator
direction; down
streambed flow domain;
A
r=r(t)
map; A:0>R
upper surface; ri=11(x,t)
streambed; i=i(x,t)
do
initial bed form
ir
3./4/59...
density; p=pf+ps
pf
ps
a
fluid density
sediment density
common boundary of 0 and
shear stress
TO
c
boundary shear stress
critical shear
dynamic viscosity
kinematic viscosity; V=4/Pf
vorticity, scalar
coeficient for h; h=4ie1
X
dune wave length
coefficient for w; w=X..e
A FREE BOUNDARY VALUE PROBLEM MODELING STREAMBED EROSION
INTRODUCTION
11( x, t)
flow
OW
MEM
soms,
NOM
wk.
IN=
.444
NNW
unlit
I
a(x,t)
ss.
.
,
..
.
.
.
'
The Erosion Problem Schematic
Fig 1.0
The Problem
Our purpose is twofold,
and
develop
to
model.
to
derive a model of streambed erosion
a solution procedure to solve the equations of our
The 2dimensional flow domain,
consists
region,
11=11(x.t),
of
two distinct regions,
0=0(0,
while
D = 0
is bounded above by
the
D(t),
an
+ r.
i=i(x,t).
The mainstream flow
air/fluid
eroding boundary region,
below by an erodable streambed,
which varies in time,
free
r=r(t),
surface,
is bounded
0 and r share the common
2
a, which separates the mainstream flow from the transport
boundary,
At time to,
(see Fig 1.0).
region.
for
t > to.
We
refer
region
the
given and we would like to determine the flow and
Do
in
the fluid velocity
is
D(t)
to the above problem as 'the problem' or as
'the erosion problem'.
The General Apuroach
Our
basic approach in modeling streambed erosion is to combine
the interaction of a viscous flow with
is
The boundary condition
by the NavierStokes equation.
which
The viscous flow is characterized
an erodable streambed.
represent
constraints
boundary
obtained
by balancing sediment mass, momentum, and energy.
In order to examine the erosion process mathematically, we must
first
have
discussion.
under
concise terminology
which will be common throughout our
Streambed erosion,
as a result of
the general category of fluid dynamics.
fluid
falls
we begin with
Thus,
the background and terminology of fluid dynamics as
flow,
it
relates
to
the problem.
Though
dimensions
the
for
problem
all
is
set
derivations.
in two dimensions,
All
ambiguity
is
we use three
removed
by
considering the problem to be constant in the third dimension.
The
theory
of
viscous fluids is presented in chapter II.
We
wish to describe flows in an open channel.
We explain why we choose
the NavierStokes equation to describe our
flows.
In
general,
we
require some initial conditions as well as some boundary conditions.
3
How and why we choose such conditions is included in our discussion.
theory.
The second part of chapter II highlights sedimentation
results
Empirical
and we describe various phenomena
outlined
are
The major
associated with streambed erosion.
evolutionary
and
streambed
minor
A brief history of sedimentation
changes are defined.
theory is included.
over
Flow
bed of sand causes the bed to undergo a distinct
a
increasing
With
flat.
the bed remains
At slow flows,
evolution.
flow velocity, the bed passes through a set of stages: ripples, then
dunes,
Past
antidunes.
which have not considered a general flow,
models,
One of our main purposes
cannot address these evolutionary changes.
is
to
have
a
potentially
model
finally
and
then waves
then flat again with chaotic flow,
capable
of
this
addressing
evolutionary process.
shape evolution.
streambed
we discuss in chapter II,
dispersion.
Due
the
to
The various stages of transport,
are rolling,
saltation
and
transport
We make a distinction between sediment particle
which
suspension/
and
mathematical difficulties associated with
turbulence we shall concentrate our study on erosion resulting
from
rolling and saltation.
The
formation
irregularities
or
of
ripples
fluctuating
is
usually
turbulence.
attributed
It
is
to
bed
our hope that,
bifurcation
eventually,
our model will show this evolution to be a
phenomenon.
We note model requirements which will achieve this aim.
In chapter III we derive our model.
version of the NavierStokes system
to
First, we give the precise
be
used
to
describe
Our
4
second
The
flows.
half of our modeling process involves depicting
We derive
the flow/streambed interaction.
The
condition.
result
is
a
boundary
streambed
the
boundary value
free
2dimensional,
problem with two free boundaries.
Generally,
models
past
have
been
formed from a large variety of nondimensional parameters.
relations
These models usually attempt to match a
Our
with
empirical
heavily
model
strictly
derived
is
Our empirical
from balance laws.
We obtain a
relations are left undetermined and constants are used.
streambed
model which exhibits the general nature of
emphasis
towards
is
streambed.
the
interaction
Our goal is to have a model
between
Our
erosion.
and the
flow
the
can
which
process.
erosion
specific
studied
be
to
better understand this interaction.
that
the
streambed
is
homogeneous
and
We allow the transport
to vary but the concentration of sediment is assumed to remain
constant.
broad
small,
of
consists
noncohesive, relatively spherical particles.
rate
we assume
the process of obtaining the boundary condition,
In
We show that under these assumptions,
enough
that
we obtain some nice results.
remains
the model
Specifically,
we
obtain classic dune movement.
The complicated nature of the general model makes it prudent to
consider
first
some
special
consider almost uniform,
flow
cases
steady flow.
of
flows.
In
chapter IV we
We also consider a potential
in 0 combined with a logarithmic velocity profile in
r.
These
flows yield the classic dune migration behavior.
In chapter V we consider a general flow.
This case
is
solved
5
We transform our free domain
time dependent finite elements.
using
This fixed boundary value problem
to a fixed domain.
into
system
a
solved
numerically.
boundary
differential equations which are then
ordinary
of
evolution
solution
The
well
as
converted
is
as
transform is required to obtain
the
the
system
this
to
evolution.
flow
boundary
the
contains
No inverse
directly.
evolution
This 'solution' is then presented graphically.
We
fundamental
vorticity
factor
the
flow
near
have
the
potential
the
bottom
as
a
Another main point is
the erosion process.
in
will
that this model
of
model.
One main point is
also contrast our model with previous theory.
that we have the
the
upgrading
The last chapter presents some ideas on
to
address
some
open
questions which most other models cannot.
We
conclude
our
work with a discussion of some of these open
questions which have arisen as a result of this work.
How
can
the
numerical
scheme
be improved?
For
example:
How can the boundary
evolution be described as a bifurcation phenomenon?
6
BACKGROUND OF FLUID DYNAMICS AND SEDIMENTATION
The process of erosion has been around since before man and was
studied
by
primitive
societies.
erosion control on Mesa Verde.
of
experimented
Anasazi
is the father of modern
the
postulated
law
of
fluid
with
Fluid Dynamics, the scientific study
has been around for quite some time too.
flows,
dynamics,
was
it
Although Euler
Archemedies
who
buoyancy which could be considered a first
step in the study of sedimentation.
fluids
The
We use the
theory
to examine the forces which can induce erosion.
Stokes equation is the standard description
of
a
of
viscous
The Navier-
general
viscous
fluid.
A
fairly complete discussion of fluid dynamics may be found in
[18], [22],
[24] and [33].
We highlight those aspects pertinent to
the erosion problem.
Fluid Dynamics
The NavierStokes Equation
The
NavierStokes
equation
is coupled with a conservation of
mass equation which is often refered to as the continuity equation.
The continuity equation states that the net flux of mass into a
7
volume must be the same as the
rate
change
of
of
mass
that
in
volume.
We derive all our relations in
To
setting.
2dimensions,
which is set in
our problem,
avoid any ambiguities,
3dimensional
a
may be thought of as uniform in the third dimension.
D.
Let
denoted
av
denote
by x=(xl,xxs).
u=(u(x,t),v(x,t))
Generally,
we
and
p(x,t) be the
let
mass
Equating the rate of change of
density of the fluid.
Let
x=(x,y).
only
use
field
velocity
the
be
Let points in the domain be
surface of V.
the
in the fluid domain
volume,
Let V be an arbitrary, but fixed,
the
to
mass flux across the surface of V we obtain:
didt fvpdv = fv[apiatldv = fav[pui,u1da = fv[divpuldv
in
The first term is the total change of mass
term
is
the
The
pulling
of
differentiation
the
The
second
inside the
The third term represents the flux across the surface
integral.
V.
result
V.
(2.0)
last
term
is
obtained
by
of
applying the Gauss divergence
theorem.
Since V is arbitrary, we obtain the mass balance equation in D:
pt
flows
For the type of
relatively
constant.
Such
(2.1)
+ div pu = 0
we
will
be
considering,
density
is
incompressible or divergence free flows
are described by a velocity field which satisfies
(2.2)
div u = 0.
The NavierStokes equation,
derived
from
the
Balance
describe viscous flows.
of Momentum Law:
They
are
the rate of change of
momentum of an arbitrary volume must be balanced by the stress
over
8
the surface of the volume plus the body forces acting on the volume.
[33].
V
Let
an arbitrary,
be
but fixed,
V is also refered to as
in the fluid domain D.
fluid,
be the pressure field and let g(x,t) be the density
The
stresses
surface
B
Let
let p=p(x,t)
be
of
the
fluid.
are given by TU. (Cauchy's stress hypothesis)
where n is the outward normal to V
[33].
material
a
u=(u(x,t),v(x,t)) be the velocity field,
Let
volume.
moving with the
volume,
and T
stress
the
is
tensor.
Equating the
the body forces per unit mass on V.
time rate of change of momentum to the forces on V we obtain:
fv[DgmiDt]dv = favTnda + fvplidv = fv(div T + gDldv
(2.3)
where
DraDt =
ut
+ (u0V)n
(2.4)
is called the material derivative.
The first term is the momentum change of V.
the
right
of
terms
The two
on
the first equal sign are the surface forces and body
forces respectively.
The last term
is
obtained
by
applying
the
divergence theorem to the surface integral.
We assume that,
for an incompressible fluid, the stress tensor
the
T depends linearly on the pressure and on the rate of strain of
fluid D:
T = pI + 2gD,
D = (VU + VUr)/2
The first term is the normal stress and
stress.
We
assume
that
viscosity,
neglecting dependence on temperature,
boundary turbulence.
[33].
(2.5)
the
second
P,
is
sediment
is
shear
the
constant,
thereby
concentration,
The only body force for our problem
and
is
9
We
gravity.
let
g
denote
gravity
the
vector.
standard rectilinear coordinate system with
the
use a
will
We
upwards
direction
positive.
Since V is arbitrary, we obtain in D:
Dpu/Dt = div T
Taking
to
p
be
produces
constant
(2.6)
pg.
classical
the
NavierStokes
equation:
p(ut + (voN)u) = Vp + p.Au
pg
(2.7)
With suitable boundary conditions, the system (2.1), (2.6) will
describe moderate erosion
mixture
density
inducing
If
flows.
the
fluid/sediment
does not vary greatly from the pure fluid density,
much
then we can describe the flow by
the
NavierStokes
(2.7).
equations
(2.2),
simpler
Hereafter,
system will be refered to as the NS system
or
the
incompressible
this
complete
NavierStokes
equations with the density clarified by the context.
Boundary Conditions
We wish to describe flows in an open channel, that is, one with
an
unconstrained
upper surface.
We call this free surface n,
postulate that the stress is continuous across n,
of
D.
If
the
atmospheric
pressure
is
and
the upper surface
normalized to zero,
the
viscous effects of the air are neglected, and the surface tension of
the fluid is considered negligible, we obtain on n
(2.8)
Tn=0.
No constraint is placed directly on u on
n.
We obtain
n(x,t)
from:
10
(2.9)
an/at = v(x,n,t).
the standard no slip boundary condition is
On a fixed surface,
We call the 'solid' streambed boundary i.
imposed.
For fixed i, we
obtain on 4:
u=0
or
u=0=v
(2.10)
For a solid but moving object (such as a ship) the no slip condition
would still apply resulting in u equal to the object velocity.
Initial Conditions
for open
In general,
velocity
field
channel
on an initial domain.
an
The initial domain,
The pressure
required to satisfy (2.9).
require
we
flow,
field
can
initial
00,
is
calculated
be
from the velocity field.
A
that
flow
is
independent of a spatial direction is called
uniform flow and flow that is independent of time is
The NS equations
flow.
uniform or
required,
steady.
For
are
steady
called
steady
greatly simplified when the flow is
no
flows,
initial
condition
is
and the absence of an explicit expression for ap/at is no
longer troublesome.
The kinematics of steady flow are often similar to average flow
especially for slow open channel flows.
however,
even
though
the
flow
In
appears
the
case
of
erosion,
steady and uniform,
the
streambed is in motion.
We
can consider the special case where the bulk of the flow is
almost steady and uniform but we are forced to treat a general
flow
11
as
unsteady
unsteady,
and
There is always at least a region of
nonuniform.
This time dependence
nonuniform flow near the streambed.
requires that we have some initial conditions which we choose to be
u(x,y,t)=u0(x,y,0)
on
Do=[(x,y)I
(x,0)
< y ( 4(x,0))
(2.11)
Dimensional Analysis
analysis
Dimensional
is
useful
a
Processes which are not well understood,
too
complicated,
by
reformulated
flow,
the
fluid dynamics.
in
for which the physics
using
Buckingham
the
Pi
Theorem.
The
[17].
which are assumed to affect the
are mixed and matched to yield a characteristic invariant
correct
incorporated in
relation.
The
are
or for which a scaled model is required are often
dimensional variables,
independent
tool
dimensional
nondimensional
combination
and the function of the
The
units.
remaining
invariants
in
a
variables
new
of
are
functional
of the first invariant (with dimension)
invariants
remaining
(without
dimension)
will yield the fundamental form for the unknown if all the necessary
dependencies were initially included.
By way of example, suppose that the magnitude of the force,
F,
trying to roll a particle is a result of the fluid moving around the
particle with velocity,
u,
and viscosity, g.
about a disk shaped particle of diameter
ds
Taking u as constant
and
applying
Stokes'
Theorem we have
2
F=k wt(d)
(w=curl u)
where k is the appropriate constant.
(2.12)
12
suppose
Now,
instead
that
assume a form for F,
we
such as
2
F=F(w,g,ds,x1,x2,...).
(.1
where
[F]=ML/T ,
denotes
So,
.
M denotes dimensions of
dimension,
and T denotes dimensions
L denotes dimensions of length,
mass,
mass length/time
units
has
Force
of
time.
We choose three variables as the set of fundamental parameters.
variable
Each succeeding
fundamental
be
represented
w,
g,
and
Choosing
set.
yields: ML/T2=[w]
will
terms
in
of
the
as the fundamental set
ds
awbid ]c.
There is always a unique quantity formed if the parameters
dimensionally
independent.
In this case,
ML/T =[wligi[ds]
2
F=wg(ds)
f(I/
,...).
X
Where
dimensional
quantity
relation f,
which is nondimensional,
Reducing
F=k
the
wg(ds)2
using
w,
g,
ds,
is
and
x..
functional
The
empirically.
determined
functional relation to a simple constant,
as before.
Thus
.
or non
invariant
an
is
IIx
are
2
2
k,
yields
k would be
To get a final working relation,
determined empirically.
The Reynolds Number and Other Parameters
For NavierStokes flows involving sedimentation,
the
Reynolds
2
number,
Re=ULp/g,
and the Froude number, Fr=U /Lg, are taken to be
the nondimensional parameters which best characterize the
flow.
U
is a representative velocity, L is a representative length, and g is
the magnitude of g.
The Reynolds number
essentially measures
the
13
ratio
of the inertial to the viscous forces while the Fronde number
measures the ratio of the inertial to the gravity forces.
Although Re and
appearing
in
Fr
are
the
NavierStokes
the
only
nondimensional
system,
another
parameters
nondimensional
quantity is often encountered in sedimentation theory.
Let V be the kinematic viscosity
stress on the boundary.
Then,
g/p.
Let
to
be
shear
the
the 'local' Reynolds number is given
by Re*=u*L/v, where u* is called the shear velocity and L is a local
length.
Re* is assumed
u*=,/to/pf.
to
be
good
a
characteristic
parameter for several unknown functions in sedimentation theory.
usually
One
sets
simplified to u=u(y).
functions
match
L=ds
and
The bottom
empirical
data?
to=du/dy where the flow has been
line
In
how
is:
the
well
previous
do
unknown
section,
the
functional relation f qUite often becomes f=f(Re,Re*,Fr,...).
In
nondimensional form the incompressible NavierStokes system
becomes:
Du/Bt = Vp + 2D/Re
y/Fr
(2.13a)
div u = 0
(2.13b)
where y is the direction of the gravity vector.
Vorticitv and Potential Flow
The vorticity of a flow is given by w, the curl of the velocity
field (recall that in 2d flow w is a scalar).
measure
The vorticity
is
a
of the rotational strength or mixing strength (circulation)
of the flow.
Irrotational flows,
flows for which
w=0,
have
some
14
special
properties.
An irrotational flow is circulation free.
integral of u around any closed curve is zero
Thus,
Stokes'
Thoerem.
u must be the gradient of a function t or equivalently u is a
potential.
lines,
by
The
t=c are the
curves
w=0=divu.
which
along
(see Fig 2.0)
equipotential
curves
the fluid moves,
t may be found from At=0,
The
[22].
for
pressure
a
u.
stream
The
are orthogonal to t.
direct
consequence
of
be recovered using Bernoulli's
can
equation. [22].
i
t=c1
t=c,
\
I
,
t
f
t=c,
I\
166.
flow
-
1
1
I
I
I
..
1
.7
1
I
I
I
-
\
I
I
//z,/ Viz
1
Potential Flow
Fig 2.0
Even
for
rotational flows,
the NavierStokes system in a
it may be advantageous to express
single
equation.
applying the curl operator to (2.13). [33].
This
is
done
by
We obtain the vorticity
15
transport equation
Dw/Dt = Aw/Re
(2.14a)
w=0 on n
(2.14b)
u=0=v on 4
(2.14c)
u is easily recovered from w.
[33].
The lower boundary
however, is in terms of u rather than w.
condition,
[33].
Although mainstream flow may be approximated by potential flow,
the flow in the boundary region cannot.
Near the boundary,
we must
either use the full NavierStokes equation, some simplified version,
or an empirical relation.
The Law of the Wall and Boundary Lavers
Our sediment transport process (that is nonconstant transport)
will be restricted to the region 'near' the streambed.
is
often considered the boundary layer.
although
[31].
Our
theory.
We refer
terminology,
similar to that of classical boundary layer theory,
slightly different interpretation.
classical
layer
region
There is a branch of fluid
dynamics devoted strictly to boundary layer theory.
to this as the classical boundary
This
boundary
layer
We reserve
the
has a
application
of
theory to the erosion problem for future
works.
One standard empirical approach to boundary layer flow is given
by the logarithmic velocity profile or the 'law
of
was developed by v.
The normal velocity
Karmen and Prandtl.
[32].
the
wall'.
It
gradient, du/dz, is taken as a function of p, T, and z, the density,
16
stress,
and
normal distance respectivly.
By dimensional analysis,
the only combination of these parameters which can occur is
du/dz=i(s/p)/kz
where
k
is
an
determined by v.
empirical constant.
Karmen.
[321.
(2.15)
This constant was empirically
Upon integration we obtain the Law
of the Wall
U
=
u*
where u* is the shear velocity mentioned
Karmen
constant
0.417,
z
is
thickness of stationary fluid.
For
of
(2.16)
In(z/zo)
the
earlier,
normal
fluid.
[28].
the
von
distance and zo is the
there is always a region
(see Fig 2.2)
20=g/9pu* is generally
used for smooth boundaries and z0=c1s/30 for sandy
ds
is
(see Fig 2.1)
a boundary composed of sediment,
stationary
K
boundaries
where
is the diameter for which 65% of the sand is finer. [28].
The obvious advantage of (2.16)
boundary shape.
is
its
independence
of
the
Note, however, that only the dominant term for u is
given while assuming laminar flow conditions.
17
=
u*
ln(z/zo)
.0*
Fig 2.2
The Law of the Wall
Fig 2.1
The Streambed Close up
Fig 2.2
18
Energy Dissivation
A discussion of energy dissipation is given by Serrin, [33].
more complete discussion
is
given
by
Lamb,
[21].
We
give
A
the
highlights here.
The kinetic energy, in an arbitrary volume, V,
of the fluid, is
given by
E
= fV [pu
2
/2)dv
(2.17)
2
where u =u.u.
Such energy is
energy.
The fluid volume has potential energy as well.
not
That is,
but
rather
converted
kinetic
to
all the energy dissipated directly into the fluid
and surroundings,
kinetic energy.
dissipated
due to frictional effects,
must
come
from
the
We have in V
E = dE/dt
fvuadv) =
fv8u2/dtdv = pfvusutdv.
(2.18)
Substituting for ut from NS we obtain
E = -IvEdiv[pli(u /2 + p/p)
The
transport
2pusD1
20:Vu1dv.
term
in
parentheses
and
is
the same as for an ideal fluid.
is
the
energy
flux
(2.19)
due to fluid
[33].
The term
2gusl) is the velocity times the stress tensor and is the energy flux
due to the frictional forces.
The last term represents
the
actual
dissipation which is lost to heat.
The energy is transfered from large eddies to smaller ones
it
and
is there that the energy is finally dissipated to heat energy or
available to the sediment.
Applying the divergence theorem to
(2.19)
and
infinity or uon=0 at finite surfaces we obtain in V
using
u=0
at
19
,
2
E = 2t1ivD:Vudv = 2g1vD dv
(2.20)
2
where
D =D:D.
(2.20),
for energy dissipation,
which is a standard form for the expression
may be rewritten as
the
BobyleffForsythe
formula
Du
z
E =
w dv
(2.21)
2gfav (TiTen)da.
In the case of irrotational flow, we have in V
E = 2gfav (u.Dn) da = 21ilv (Yu)
(2.22)
dv.
Some consequences of (2.20), (2.21) and (2.22):
In the regions of irrotational flow
there
is
energy
dissipation if and only if the flow is nonuniform.
The
greatest
dissipation occurs where the fluid
has the largest gradients.
If
the entire flow is irrotational,
then there is no
energy dissipation and hence no erosion could take place.
A
flow
which minimizes energy loss must tend towards
potential flow.
Frictional dissipation of the fluid does not allow for
energy to be transfered across a solid no slip boundary.
If the flow across
ay
is steady and uniform then the
dissipation depends solely on the vorticity in V.
Concluding Remarks Fluid Dynamics
Equations (2.4) and (2.7) with boundary
conditions
(2.9)
and
23
Rolling and saltation occur
heavier
this
particles
manner.
associated
(relatively
Fig
(see
with
2.3)
turbulent
all
in
eroding
speaking)
Due
difficulties associated with turbulence
the
primarily transported in
Suspension/dispersion
flows.
with
flows
we
to
the
shall
primarily
is
mathematical
concentrate
our
study on erosion resulting from rolling and saltation.
Empirical Results Boundary Evolution
The flow over a bed of sand will cause the
distinct
evolution.
We
distinguish
between
changes and 'minor' evolutionary changes.
remains flat.
ripples,
then waves and
changes.
Ripples
This
undergo
a
'major' evolutionary
At slow
flows,
the
bed
then dunes, then flat again with chaotic
finally
antidunes.
These
are
the
'major'
are distinguished from dunes in that dunes have a
distinctive form and
upstream.
to
With increasing flow velocity, the bed passes through
a set of stages:
flow,
bed
they
migrate
migration
and
downstream.
the
classic
Antidunes
migrate
forms which arise are
'minor' changes. (see Fig 2.4)
The
formation
ripples
of
is
usually
irregularities or fluctuating turbulence.
dune
stages,
rolling
transport
mechanisms.
increased
and
and
In
moderate
the
[28].
saltation
transition
suspension/dispersion
attributed
becomes
to
bed
In the ripple
and
are
stage,
the
saltation
nontrivial.
antidune stage, saltation and dispersion are extensive.
primary
In
is
the
24
.
:*--
////77 /,///
7%/7777771,/7
7 ' smooth
/.7/7-/
/7/).
(x
t)
streambed- steady, uniform flow
______ n(x,t)
random ripples
4(x,t)
'minor' evolution
dunes- migrating downstream as
'smooth' scoured streambed- turbulent flow
Fr - 1.0 - 1.5
n(x,t)
4(x,t)
evolution
antidunes- migrating upstream as 'minor'
'Major' and 'Minor' Streambed Evolution Stages
Fig 2.4
25
Few models have made
attempt
an
account
to
however,
their
changes.
Once
explain.
Antidunes eventually form when the
the
side
lee
particle lift.
the
dunes
form,
is easy to
migration
on
gradient
pressure
with
correlated
These stages appear to be strongly
Froude number.
The boundary shape evolves as the Froude number
We shall concentrate our study on
model
these
of the dune becomes strong enough to cause extensive
1.0-1.5.
increases with antidunes forming when Fr
Our
all
for
this
evolution
phenomenon.
will attempt to account for these evolutionary stages as
well as giving an explaination for the initial dune formation.
A Brief History of Sedimentation Theory
A fairly complete history
of
recent
sedimentation
models is given by Allen, [1], and Vanoni, [39].
transport
Bed load transport
models have generally been developed from one of five main themes:
average fluid velocity,
streambed shear stress,
probabilistic inferences,
bedform celerity relationships,
energetics.
due
Probably the first formula involving an erosion process is
to Brahms,
1753,
and involves the average stream velocity.
found the empirical relation,
1/6
V =kW
c
from
some
basic
, between a critcal velocity
s
and the weight of an object it can move.
derived
Brahms
assumptions.
This relation may also
[4].
It
is
used
be
today
26
primarily in engineering design.
Gilbert,
Muller,
1914;
1948;
Donat,
et. al.
1929;
Straub,
developed
MeyerPeter and
1939;
formulae
of the form qs=qs(Un).
2
2
qs=kU (U
Uc),
Other similar
etc.
models
assume
that
sediment
discharge properties may be characterized by the mainstream Reynolds
number,
local
Reynolds
q =q s(Re,Re*,Fr) ).
s
on
characteristic
number,
from
Froude
number
(i.e.
These are empirical formulae which depend
values
and
discharge rate correlates well
arising
and
these
not
local
values.
with
these
parameters
The
only
sediment
but
models
expressions do not deal with distinct boundary
movement but only the average effect of the stream on
the
sediment
It was the idea of a critical force which led DuBoys,
in 1879,
load.
to
derive
a
discharge, qs,
general
equation
qs=kTo(To Tc)
for
the
in terms of a critical tractive force,
Tc
is
the
bed
refinements to
shear
his
and
k
model
improving the expression for
is
an empirical constant.
exist,
Tc.
most
of
which
approach is attributed to H. A. Einstein,
flat
where To
A number of
concentrate
on
[38].
Most of the theory of sediment transport from
1947.
sediment
1942,
a
probabilistic
1952 and Kalinske,
Einstein wanted to account for the initiation of ripples on a
streambed without resorting to the irregularity arguement.
assumed normally distributed turbulence fluctuations to account
Be
for
the initial uneven erosion.
The bedform celerity models such as
Crickmore,
those
of
Kennedy,
1960;
1970; Korchokha, 1972; Willis and Kennedy, 1977; at. el.
27
are
strictly
(height,
empirical
length,
formulae
etc)
to
the
relating
dune
perhaps the most extensive and one of
dune
celerity.
his
characteristics
Kennedy's work is
models
is
given
here.
2
[20].
Xs = (27r/g)U
gives the dune wavelength,
in terms of a
Xs,
mean velocity, U. g is the magnitude of gravity.
Models based on the energy of
Bagnold,
1966.
considerations
These
and
models
are
flow
are
generally
Gilbert, 1914;
qs=kt°U.
the
largely
are
derived
from
due
to
theoretical
of the form q5=k(t0U0 scy or
MeyerPeter, 1948;
Straub, 1939;
et. al.
have developed models based on the energy slope (stream gradient).
In
summary,
we
note
that,
dimensional
analysis
employed to arrive at the fundamental dependence
Of
the
present
considered
as
difficulties
models,
vorticity
fundamental
a
associated
does
not
of
often
buried
unknowns.
parameter.
The
mathematical
with turbulence makes it difficult to deal
The
flow
field
in a constant such as the average stream velocity
allowing for no local variation.
only
the
often
appear to have been
with the true flow field of typical erosion flows.
is
is
The probabilistic models
are
the
ones which seem to deal with the boundary evolution stage from
theoretical considerations.
Forces on the Sediment Particle
The
global
forces
terms
or
on
the
very
sediment
specific
are usually expressed in either
terms.
In
global
frictional force on a smooth flat surface is given by
terms,
the
28
wro=gdu/dz
where
is
z
normal
to
the
(2.23)
boundary.
individual particles are also considered.
lift
the
particle
out
of
angle.
In
entire surface.
actuallity,
specific
They
try
forces
to
on
pivot
and
its position as shown in Fig 2.5.
The
moment is written in terms of limo,
pivot
The
the
particle
weight,
and
the
the fluid forces should apply to the
We will be using a slightly different force diagram
in our model derivation.
INNY/M
Classical Force Diagram for a Sediment Particle
Fig
2.5
Concluding Remarks Sedimentation
theory of sediment transport,
from
the
theory
though related,
of boundary evolution.
is distinct
Most of the work has been
29
concentrated
developed
in
sediment
initially
from
empirical in nature.
transport
physical
theory.
The
models,
considerations,
become
It appears that only the probabilistic
though
very
models
have addressed the question of initial dune formation arising from a
smooth
streambed.
There
does not appear to be any accepted theory
of the actual process of dune formation.
of
coagulation
around
some
stand up to close scrutiny.
yield ripples and
evolutionary
dunes.
changes
irregularity
Increased
A model
must
The generally held
combine
reasonable streambed erosion constraint.
in the surface does not
sediment
which
a
belief
can
homogeniety
address
general
flow
the
with
still
major
some
30
THE EROSION MODEL
Our basic approach in modeling streambed erosion is to
the
interaction
of
a
flow
with
represent an erodable boundary.
general
viscous
flow
(i.e.
some boundary constraints which
The
the
combine
flow
characterized
is
NavierStokes
by
equation).
a
The
boundary condition is obtained by balancing sediment mass, momentum,
and energy.
its
The erosion model consists of open channel flow
together
boundary
give
condition
overview of the problem
assumptions
along
with
erosion.
some
We
terminology.
a
brief
The
basic
pertaining to the model are noted along with the degree
of their validity.
is
representing
with
derived.
The erosion constraint as a
Some
boundary
condition
special cases of flows are considered in chapter
IV and a general flow is considered in chapter V.
We use the term erosion to encompass both
the scouring,
or lifting up, of sediment.
D, is divided into two regions:
region
sediment transport in 0.
deposition
The sediment in 0
r.
and the
0,
We assume negligible
behaves
as
if
it
fully dissolved and does not itself affect the sediment load in
contains
a
mixture
of
this mixture as a fluid.
interface,
is
the streambed,
free,
both fluid and sediment.
The upper
surface
and is denoted by
is also free,
71.
and
The entire fluid domain,
the mainstream region,
which sediment is transported,
in
the
of
is
r. r
We will refer to
0,
the
air/fluid
The lower surface of
and is denoted by i.
'Surface'
r,
will
31
refer to i while the 'bed' will refer to the streambed,
interface is denoted by a.
The 0/r
i.
Refer to Fig 1.0 for a schematic.
The Flow Field
We assume that the erosion process is a local
result
of
the
interaction
the streambed.
take
between
phenomenon,
the
the fluid and sediment along
Erosion depends on the local
field
flow
which
we
to be governed by the NavierStokes Equations for open channel
flow.
Difficulties arise concerning the
sediment
conservation
of
mass.
The
particles are discrete rigid masses but they cannot easily
be modeled as
kinematics
dense
points
the
of
continuum.
Fortunately,
the
of the flow do not depend strongly upon small amounts of
sediment in
solution.
[28).
concentrated
near
in 0 by
incompressible
the
The
the boundary.
sediment
transport
is
assumed
This allows us to govern our flow
NavierStokes
equation
for
constant
Let x=(11,12,x3) with the upwards direction positive.
Through-
density.
out our discussion we assume the flow
direction.
Thus,
we
equate
x with
to
be
(x,y)
constant
where
in
any
the
x,
ambiguity
will be clarified.
We let u=(u(x,t),v(x,t)) be the velocity field and p=p(x,t)
the
pressure
field
in
D.
atmospheric pressure is zero.
We
normalize
the
pressure
so
be
that
We let p=p(x,t) denote the density of
the fluid (mixture) and the density of the bed sediment
is
denoted
32
by
We
ps.
streambed.
assume
constant
is
p
in 0 and ps is constant in the
The flow in 0 is then governed by:
pDm/Dt = div T + pg
div u = 0
p=0
(3.0a)
in 0(t)
(3.0b)
on 71(x,t)
TU=0
lt
uo
in 0(t)
, vo,
(3.0c)
on A(x,t)
=vI
(3.0e)
y=11
given at
o
(3.0d)
(3.00
t-0
where D/Dt:=8/8t + (u.N) is the material or spatial derivative,
T = pI + 2gD,
D = (Vu + VuT)/2
and
(3.0h)
defines the stressstrain relationship of the fluid.
is the NavierStokes
(a)
tensor
and
g
being
the
with
equation
gravity
vector.
T being
the
stress
is the Continuity
(b)
equation for constant density. (c) represents normalized atmospheric
pressure.
(d) represents continuous stress accross
auxiliary
condition
resulting
from
(d).
(f)
n.
(e)
gives
is
an
the initial
conditions.
The system (3.0) is not complete
boundary
boundary,
condition.
This
but
requires
an
additional
condition cannot be given on a,
since the flow moves accross that interface
the
04
arbitrarily.
It is the sediment, not the fluid, which is trapped in U.
We
this,
would
like
to
extend
(3.0) to include
we assume that the depth of
thickness of r and let h denote the
'small'
relative to h.
f
is 'small'.
flow
(see Fig 3.0)
depth.
f.
Let 6
We
To accomplish
denote
assume
the
6
is
It is important to note that
8 is not being used in the sense of classical boundary layer theory.
33
h >>
The Flow Regions 0 and I
Fig 3.0
We interpret the region r as follows:
The region r
region D.
be
is
infintesimal
with
respect
to
the
This is a mathematical simplification which may
physically
interpreted
as
sediment load moving
the
along the bed as an impulse load.
Thus, the sediment load
is no longer a function of the vertical
displacement
but
only a function of position along i.
The
mainstream
kinematics inherent
taken
to
process.
be
The
simplifications
the
flow
in
kinematics
the
primary
classical
and boundary layer
NavierStokes
influence
boundary
on
layer
equation
the
transport
theory
to arrive at the same effects.
are
uses
The trade
34
off
this
at
point
matching
between
is
two
sets
of
equations; one set for the main stream and one set for the
boundary
versus
layer
one
equations
of
set
and
an
infintesimal boundary layer.
There is a natural boundary condition on
Since S is 'small',
condition which we adopt.
considered
as
a
along
only
flow
'distinguish between D and 0.
dominated by the flow in 0.
the
Physically,
namely, the no slip
r
the flow in
streambed.
We
no longer
totally
the flow in D is
The system (3.0) becomes:
pDu/Dt = div T + pg
div u = 0
p=0
(3.1a)
in D(t)
(3.1b)
in D(t)
on u(x,t)
TU=0
(3.1c)
on n(x,t)
nt
u=0=v
=v1
(3.1d)
(3.1e)
y=11
(3.1f)
on i(x,t)
110, vo Do given at t
The
flow
domain
0
has
been
now completely determined by (3.1).
(3.1g)
extended
boundary constraint, (f) has been added.
to D and the no slip
For a given 4, the flow is
This no slip condition does not
preclude erosion but rather states that there is a
the bed is fixed and,
Realistically,
agitated
to the bottom.
at that level,
level
which
at
the fluid adheres to the bed.
the bed has fluid moving through
mixture
can be
it
and
r
is
an
of the fluid and sediment moving along very close
We have simplified the picture by assuming that
the
D > 0.
The
sediment moves along so close to
the
streambed
that
35
'boundary' is made up of a movable solid.
Streambed Erosion as a Boundary Condition
We now have the flow given by
condition,
model.
which
require
and
(3.1)
reflects the process of erosion,
a
boundary
to complete the
We obtain this condition from the conservation of
mass
law
for the sediment.
r.
Let V be an arbitrary volume in
of
sediment
(mass/volume)
in
Let cs be the concentration
The total sediment mass in V is
V.
given by
Iv
cs dv
(3.3)
and the rate of mass gain is given by
r
dt JV cs
For fine sediment,
is a continuous function.
cs
c
dtI Vs
Let
us
(3.4)
dv.
dv =
So,
IV 8c/at dv.
(3.5)
s
denote the velocity of
sediment
particles.
Then,
the
mass flux into V is given by
f
ay
csus.n
(3.6)
da
where 8V denotes the boundary of V and n is the outward unit normal.
By the Divergence Theorem we have
IaV cu
s
da =
I
I
div c u
V
s s
dv.
(3.7)
Since mass is conserved, we have in r
IV 8c/at dv =
S
V
div c u
s
dv.
(3.8)
s
We let 6 > 0, the width of r shrinks, and let av coincide with
i,
the
bed.
We assume that the sediment is concentrated in a thin
36
layer along i and is uniform in that layer. (3.8) becomes
sfai aus/at dv =ai
k =c c [1
S
p s
fatc u )/at + ki) dv
s
(P)
S
2 _1/2
(3.9a)
st
(3.9b)
1
t is the
where ad denotes the portion of av which coincides with 4.
unit tangent vector to 4 and
i.
refers
a/at
us
is the sediment velocity parallel to
to the change along 4.
(100% concentration) to
ks relates the bed density
the bed
for a given change in 4, by c,
cs
packing ratio.
Since V was arbitrary, ad is arbitrary and we have
ac/at = 8(u u )/at
SS
S
(3.10)
ksdt.
term on the left represents the total increase in sediment
The
mass due to locally unsteady flow.
The
term
first
the
on
right
represents the tangential flux of sediment due to locally nonuniform
(tangential)
flow.
The
last
represents
term
the normal flux of
sediment which is a source/sink term due to erosion.
We would like explicit
forms
for
cs,
and
us.
We
restrict
to quasiequilibrium erosion where changes in the sediment
ourselves
concentration are 'small'.
A significant sediment flux accross i is
always accounted for by a relative increase in the
transport
rate.
In this way, changes in the sediment concentration remain small.
this case,
cs
In
is constant and (3.10) becomes
cpdt=dus/dx
(3.11)
where d/dx=(8/8x + a/ay dy/dx)I y=d(x)*
We
cannot assume
to be constant for if we were to make such
us
an assumption, no erosion would occur since the right side of (3.11)
vanishes.
There are a variety of possibilities for
determining
us
37
including:
that the sediment rolls with the fluid.
Assume
depends on the energy dissipated
by
sediment
the
Assume the rate of work in transporting
the fluid.
(3)Usedimensionalanalysisonherep.are
parameters influencing sediment transport.
We
consider
Method
both (1) and (2) and make some remarks concerning (3).
(1):
Obtaining
expression
an
for
us
by treating the
fluid as a collection of rolling, spinning fluid volumes
Sediment Particle Transport Mechanism
Fig 3.1
The
pps.
a
sediment
They are spun,
similar
volume
made
is
up of small homogeneous particles with
rolled or agitated in much the same manner as
of
immersed in a flow would
fluid.
If
quickly
we
neglect
assume
the
resistance,
same
local
a ball
angular
38
We derive
velocity as the fluid.
rolls
us for the case where the sediment
conglomeration
of
assumed to roll and spin in a similar manner.
Let
Us
Us
denotes
= (l/nd )
S
be
Then,
Ias
uds
(3.12)
Applying Stokes' Theorem we
boundary of S.
the
8
Let
Let u be the fluid velocity, the
fluid being adhered to the particle.
as
(see Fig 3.1)
be the mean surface velocity of S.
the mean rotational velocity of S.
where
is
a particle S of mean diameter ds rolling and spinning
Consider
in the fluid.
sediment
The
volumes.
spinning
rolling,
we treat the fluid as a
That is,
spins with the fluid.
and
have
Us = (//nds)
But u =nd 0 and
s
s
(3.13)
An application of the Mean Value Theorem
0=U /nd .
s
IS curl uda
$
yields
Us
(3.14)
wds/4n
where w is the magnitude of the vorticity at (x,i(x)).
We obtained (3.14) by neglecting
resistance.
include
We
the
presence of resistance in as simple a way as possible by assuming Uc
to be a critical threshold which
must
Us
exceed.
We
obtain
the
expression for us
us
where dir =
=
(3.15)
U
dir[Us
dir denotes the direction of
us.
[.]+
denotes
the positive part of (.].
In
general,
U =U (d,
C
c
movement for the sediment.
s
e, y,...)
$
gives
the
resistance
to
That is, the resistance is a function of
39
parameters
such
as
particle
the
particle weight, the bed slope, etc.
Uc = k(4' +
e
where
angle
the
is
constant.
empirical
the
For simplicity, we take
(3.16)
ec)
the sediment and k is an
for
repose
of
the particle shape,
size,
form includes only bed slope effects.
This
more genera/ relation could
be
applying
by
obtained
A
dimensional
analysis.
Recall
Paradox which says that there is no force
D'Alembert's
on a particle in a perfect
constant which
The
takes
becomes
more
us
particle size and weight.
resistance.
sediment
the
must depend on the viscosity as well as
distinct from the fluid,
all
As
fluid.
Recall that Us was derived by
term
these
incorporate
should
Us
properties
empirical
an
account.
into
neglecting
first
The
generalization of (3.15) is
us = dir[cito
Method (2):
energy
(3.17)
c5(4' + Oc)]+.
Obtaining an expression for
us
by
equating
the
dissipated by the fluid with the work rate of sediment being
transported
The
energy dissipated by the fluid must be either converted to
heat or transfered to the sediment.
Energy dissipation is carried
out
in
the
kinetic energy,
manner:
as they decompose,
Energy is transfered from the larger eddies,
smaller ones with little loss.
following
to
The small eddies then transfer their
usually to heat but
also
to
sediment
transport.
40
Within
large
the
vorticity
eddies,
flow
the
is almost irrotational while
dominates in the smaller eddies.
is consistent with our interpretation of
This explanation
[21].
(2.21).
From
(2.21),
we
deduced that energy dissipation was minimized by potential flows.
Let
Ef
by
be the rate energy is dissipated
the
Then,
fluid.
the rate energy becomes available to the sediment, for transport, is
given by
Ea
= [efficiency} x
Clearly,
some
properties or to
Energy
is
unavailable and lost energy].
fEf
energy
sediment/boundary
'unavailable'
due
independent of the erosion process.
friction on a cohesive boundary.
characteristics.
the
to
heating
Energy is
Obviously,
remaining energy is available to contribute
by
performed
the
sediment
due to fluid
lost
or
unavailable
is
of
(3.18)
For
example:
fluid which is
due
'lost'
to
the
only a portion of this
the
to
rate
work
of
to the 'efficiency' of the energy
due
transfer process.
Let
lost
Ec
Ec
minimum amount
be that critical,
of
energy
initially
or unavailable and let e be the efficiency of energy transfer.
accounts for the
which
energy
must
whether or not erosion takes place.
Ea
=
elEf
be
dissipated
regardless
(3.18) becomes:
Ecl
(3.19)
The total energy dissipated by the fluid is given by (2.21)
or
(2.22) and may be alternatively written as
E =
giv
2
(to
4detlifu] dv
where g is the viscosity of the fluid.
(3.20)
41
The term detVu measures the dissipation due to deforming V
and
we assume this energy to be 'unavailable', leaving:
2
Ea = e[gw
With
(3.21)
Ec]
we assume that potential energy
regards to the sediment,
changes.
changes are negligible relative to the kinetic energy
sediment
lifted
is
only
Losses due to
an infintesimal distance.
heating are absorbed in Ec and e.
all the work is
Thus,
The
contained
in transporting the sediment and is given by
(3.22)
u f
s s
where fs is the force on the sediment particles which causes them to
roll and spin.
There are a variety of ways to arrive at an expression for
We give two.
obvious
are
analysis,
fs.
First, we assume fs=fs(w,ds,1i,p1,P2....).
wids, and g
By
dimensional
choices
as
fundamental
parameters.
and g is given by
the only quantity of force using w,ds,
Thus,
gw.
(3.23)
fs=gwf
wherefrepresentsf(111,...)H.being
quantities formed from p,
w,
ds,
and
pi.
the
nondimensional
The simplest form for fs
is then
(3.24)
fs=kgw
We obtain the same expression for fs by assuming the
force
to
result of rolling a particle where the surface force is given
be
a
by
gf(u).
Applying
Stokes'
Theorem to f(u)
kill gives the force
per particle as
kigw.
(3.25)
42
The number of particles is constant so we again obtain (3.24).
we relate the work rate
With fs now given,
the
to
available
energy to obtain
(3.26)
E ] +/f
= ergo)
us
s
C
which becomes
us = dir[c w
where
we
interpret
those
(3.27)
r]+
grouped
terms
into
as a result of
r
resistance factors.
We have obtained a similar expression for
Method
Obtaining
(3):
an
expression
us
by method (1).
for
from strictly
us
empirical considerations
standard
Another
dimensional analysis.
method
for
determining
would
The only question
us
would be to use
be
the
choice
of
Consider that the vorticity plays a role in
fundamental parameters.
is a measure of the local turbulence,
the fluid energy dissipation,
is a measure of the local agitation at the bottom, and is related to
Clearly,
the rolling and spinning of the fluid and the sediment.
would be a good choice for
erosion.
a
parameter
of
streambed
However, w does not appear to have been previously used as
fundamental
implied uniform
expressions for
parameter.
fundamental
a
w
One
parameter.
flow
us
field
could
Additionally,
be
obvious explanation for this is the
most
inherent
in
obtained
using
improving
model would be best achieved by
the
replacing
w
models.
as
physical
empirical
a
Countless
fundamental
reality
of the
constants
by
43
We leave such additional
empirical relations using such parameters.
versions for later studies.
We choose
our
as
(3.17)
working
version
for
us
for
the
Our final version of our model becomes
remainder of this work.
pft/Dt = div T + pg
div u = 0
p=0
in D(t)
(3.28b)
in D(t)
(3.28c)
on n(x,t)
TU=0
(3.28a)
(3.28d)
on 11(x,t)
(3.28e)
t=vIy=11
u=0=v
(3.280
on i(x,t)
(3.28h)
cpt = dus/dx
us
where
c2(05' + 9c
(3.28g)
u0, vo, D, given at t0
(3.28h)
= dir[clw
d/dx=(a/8x + a/ay dy/dx)ly=i00-
Concluding Remarks The Model
The complete model,
boundary
condition,
consisting of a flow field and a streambed
is given by (3.28).
Given below are the high-
lights of the model and its derivation.
1.
The
erosion
process
consists
with an erodable boundary.
of a flow interacting
p=0
Du
Dt
= div T + B
div u = 0
flow
0.
11017
fi".L
001'
4 lift mi.
aft...111F
..
The Erosion Model An Overview
Fig 3.2
45
sediment
The erosion process is obtained by balancing
mass, energy and momentum.
The
erosion producing flow is governed by the Navier-
Stokes equations for open channel flow.
The sediment transport occurs in a 'thin' layer
along
the stream bed.
Our
model
is for a relatively fine,
Large discreet particles
The
the
invalidate
mass
derivation as well as the assumptions used to
balance
obtain
would
round sediment.
us.
sediment
quasiequilibrium.
boundary,
must
transport
results
be
in
a
state
of
Increased sediment flux, across the
in
an
increased
transport
rate
rather than an increased sediment concentration.
The
vorticity
of
the
fundamental factor for
flow
the
near the streambed is a
erosion
process
in
our
model.
The model will accept a
variable flow field.
Such a
flow field is necessary for variable erosion.
An overview of the complete model is shown in Fig 3.2.
46
SPECIAL FLOWS
We consider three flow cases:
almost uniform, steady flow,
potential flow in 0 with 'law of the wall' in
r,
a general flow.
Due to the complexity of a
general
flow,
(3)
is
dealt
with
in
chapter V.
Almost Steady, Uniform Flow
4=0
//1
Steady, Uniform Flow
Fig 4.0
If
we
replace the erosion condition,
(3.27),
with a fixed streambed condition,
channel
flow.
If
we
assume a steady,
(3.27g),
we
have
in our model,
standard
open
uniform flow over a smooth
bottom, 1 would be flat also. (see Fig 4.0) In this case, u=u(y) and
p=p(y) and our model, (3.27), becomes:
47
(4.0a)
pgsine
0 = pu
YY
0 = p
(4.0b)
pgcos0
u=0 on y=0
(4.0c)
Tn=0 on y=h
(4.0d)
We solve this exactly and give the solution below.
(4.1a)
u = 21(sinO)y(2hy)
2g
(4.1b)
p = pg(cose)(hy)
n(x,t)
y*
d(x,t
'small'
///
4.
Almost Steady, Uniform Flow A Rippled Bed
Fig 4.1
In
actuality,
such
flows
experience boundary turbulence.
are
unstable
and
Suppose, however,
will
at least
that we use this
flow as an approximation to the flow in the case where the streambed
is
relatively
smooth and flat;
that is,
where i'
is 'small'.
We
are, in essence, assuming this flow to be approximated by the stable
flow given by (4.1).
48
boundary.
small.
will
almost
consider
We
(see
Fig
uniform
steady,
If the flow is too slow, then
place
take
all.
at
us
rippled
a
will be zero and no erosion
flow
the
If
over
the bed gradients must be
Of necessity,
4.1)
flow
too
is
strong,
our
In between these extreems, us is
approximation will be unrealistic.
nonzero and approximated by
k, + kidu/dy.
us
Assume
now
that
(4.2)
the flow is deep enough that there exists y*
constant.
such that for y < y* the fluid discharge is approximately
Then,
i) du/dy
(112)(y*
qf
where qf is the volumetric fluid discharge
(4.3)
in
the
region
[i,y*].
This gives
i)
constantgy*
du/dy
2
.
(4.4)
Substituting (4.4) into (4.2) we obtain
)a/ax
3
k/(y*
dus/dx
(4.5)
where k is the appropriate constant.
Our erosion constraint (3.17) becomes
t
(4.6a)
+ cix = 0
3
C
= ki(y*
(4.6b)
i)
which is a quasilinear hyperbolic wave equation.
This
[41].
equation
may be solved by the method of characteristics.
We obtain a solution of the form
i=i0(x
where
40(x)
is
the
initial
downstream with speed c.
(4.7)
et)
boundary.
The
initial
dune
moves
(4.6a) indicates that the top of the dune
49
moves faster than the bottom.
at the angle
of
In actuality, the lee side sluffs off
repose of the sediment.
i=460(x - ct)
(see Fig 4.2).
actual result
Dune Migration- Almost Steady, Uniform Flow
Fig 4.2
Thus,
downstream
with
almost
uniform
steady,
flow,
assume the classical dune shape.
and
ripples
In this scenario,
if the bottom is initally flat, then no ripples would
Potential Flow in 0 with Law
We
potential
flow
the Wall in
the case where the
consider
velocity profile
of
in
a
outside
region
near
that region.
for the logarithmic velocity profile.
is denoted by
f
form.
f
flow is
the
migrate
given by a logarithmic
boundary
and
given
of
the Wall'
The region near the
boundary
We use the 'Law
and the potential flow region is denoted by 0.
a constant width S.
(see Fig 4.3).
by
f is
1
--J
t)
flow
d(x,t)
Potential Flow in 2 with Law of the Vali in r
Fig 4.3
< y < a)
r = ((x,y)1
0
(4.8)
= f(x,y)1 a < y <
a = [(x',y')i x' =
x
be/dr,
(4.9)
Y' = i(x) + 8/dr}
(4.10)
2
dr = [1 + (e) ]
flow
The
in
0
is
determined
(4.11)
in
one
of
three ways:
Method (1):
Approximate uniform flow u=u(x),
Method (2):
Conformal mapping,
Method (3):
Numerically solving At=0, where u=grad t.
If
-11
is assumed apriori and i'
is 'small' (hence &is 'small')
then we can approximate the flow on a by
51
Ua = qf/h,
a(x),
= n(x)
h
(4.12)
where qf is the volumetric fluid discharge and h is the depth of
In
general
the
However,
numerically.
for
known,
map,
case,
bed,
the
or
case,
which
one
maps
must be obtained
one
for which the conformal map is
is well approximated by a standard conformal
then we can 'conviently' use conformal mapping.
approximate
bed
the
by
In the
first
In the second case, we
the map changes.
as the bed evolves,
continuously
form
if we assume some particular geometric
particular,
in
conformal
the
0.
same
the
standard
simple
geometric form.
With
direct
the
numerical
the numerical solution
approach,
depends on whether n is assumed apriori or taken as the continuation
In the first case, we solve directly
of a streamline.
(4.13a)
in 0
At=0
subject to the boundary'conditions:
3t/8n=unormal=0
at/an=unormal=0
at A
(4.13b)
at a
(4.13c)
In the second case, we solve the same problem but iterate on n until
Bernoulli's condition,
2
u /2 + p/p + gy = constant
on
n,
(4.14)
is satisfied.
Method (3)
simplest.
For
is the
deep
most
flows
versatile
while
method
relative to ripple height,
flat and both method (1) and (3)
are
easy
to
(1)
the
n is almost
implement.
cases, some inflow and outflow constraints must be assumed.
use uniform flow, method (1).
is
In
all
Here we
52
The flow in r and has a local logarithmic form.
velocity,
The
parallel to i, is given by
U = K log[(z
where
U
zo
bed,
is the flow magnitude,
is that distance
K = IC(x,i(z))
is
where
necessary
to
+ z0)/z0]
(4.15)
z is the normal distance above the
fluid
the
match
the
velocity in
where 65% of the sediment is finer than ds.
r
However,
and
with the
d/30
zo is usually assumed to be of the order
velocity in D.
sand
movement,
begins
for
we take
z=z0 to corresponds to i.
from
Let U = U(z;x,i(x)) denote the velocity a distance z
point
Let
(x,i(x)).
the
Ua denote the velocity on a determined by the
Continuity of the flow requires
potential flow in 0.
(4.16)
Ua = U(6;x,4(x))
and K(x,i(x)) is determined from
K(x,d(x)) = Ua/log[(6
We obtain u and v in
+ z0)/z0)
(4.17)
I- by
u = U/dr,
where ds is given by (4.11).
v = Ue/dr
(4.18)
us is then easily obtained
and
hence
t.
Some solutions (graphically represented) are shown in Fig 4.4.
Concluding Remarks Special Case Flows
We
have
shown from these two cases that our model does indeed
embody some of the major facets of dune migration.
to
these
facets
as
We have
'minor' evolutionary processes.
refered
In contrast,
53
'major'
evolutionary
changes
characterization to another.
are
(e.g.
evolution
from
one
'major'
smooth bed > ripples > dunes)
(refer to Fig 2.4, page 23)
In
the
simpler flow.
above
cases
we have replaced the general flow with a
Under these simplifications,
our model is similar to
most others in that no local flow variation is allowed.
that when the local flow cannot vary,
rate.
unless
Thus,
It is clear
neither can the local erosion
although we get classical dune
shape
and
migration,
we attach a general flow to our erosion constraint our model
will not have the potential to address 'major' evolutionary
in the streambed.
Dune Migration Potential Flow with Law of the Wall
Fig 4.4a
changes
54
430
ssam.ass,...naweszsmansnmsamstm:nYM ....
.
Dune Migration Potential Flow with Law of the Wall
Fig 4.4b
°Co
Dune Migration Potential Flow with Law of the Wall
Fig 4.4c
..........
55
A GENERAL FLOW
The Flow Domain
The general model for erosion was given in chapter III, (3.28).
In considering the special case flows
assumed uniform flow at infinity.
chapter
of
IV,
tacitly
we
A similar adjustment is necessary
here also since we cannot deal numerically with a domain of infinite
For
length.
we take the domain,
the remainder of this discussion,
D=D(t), to be the finite region
D(t)=((x,y)I a < x < b,
< y < n(x,t)).
i(x,t)
(5.0)
that is,
For simplicity, we assume uniform flow outside [a,b],
outside
the irespective
inflow
and
outflow
requiring uniform flow at the entrance
respect
limiting.
to
the
inflow,
this
With respect to the
constraint is quite limiting.
not
tend
towards
and
artificial
outflow,
boundaries.
are
With
boundaries.
exit
constraint
is not very
this
however,
additional
We are excluding those flows which do
a uniform flow near the exit boundary.
flows (in particular,
We
those not tending
towards
a
For some
uniform
flow)
numerical 'reflections' off this artificial exit boundary may exist.
We choose only flows where this reflection appears to be negligible.
The
erosion
model
on a finite interval,
denoted by (*),
is
given by:
pDu/Dt = div T + pg
div u = 0
in D(t)
in D(t)
(5.1a)
(5.1b)
56
p=0
Tn=0
(
*)
(5.1c)
on m(x,t)
(5.1d)
on n(x,t)
nt
u=0=v
=v1
(5.1e)
y=11
(5.1f)
on i(x,t)
(5.1g)
opt = dus/dx
us
= dirEciw
0u/ax=0=av/ax
c2(i' +)]
(5.1h)
at x=a and x=b
(5.10
(5.1j)
no, vo, Do given at to
D /Ox + i'au
where du /dx = (On
$
s/4)1y=i(x,t).
Solution Procedure Reducing to a System of Initial Value Problems
unsteady flow.
The system (*) depicts nonuniform,
would
variables:
The
only
equation
eliminate
have
to
like
u,
V. p,
exceptions
for
the
a
time
p,
the
incompressibility.
pressure
have exclusively:
we
for each of the
and only time dependent equations.
and 1,
are
equation
dependent
Ideally
pressure variable and (b),
We
term from (a).
can
use
equation
(b)
the
to
The 'improved' system will
time dependent equations,
one for each remaining
variable.
Reducing (*) to a system of initial value problems is a
of
choosing
finite elements as our solution method.
result
There are two
main reasons for choosing finite elements over another method:
(1) We have seen from almost steady, uniform flow that our
erosion
boundary condition becomes a nonlinear hyperbolic
wave equation.
'Shocks' develop in such equations and the
57
natural approach for this phenomenon is to reformulate the
problem in its weak setting.
[23], [41]
(2) The free surface condition is easily incorporated into
our method and it simplifies the final equations.
Having
decided
finite
upon
elements,
apply
a
natural
which simplifies certain aspects
transformation, to a fixed domain,
of our problem.
we
The resulting ordinary differential equation system
requires some initial conditions.
Numerical considerations force us
to choose starting values consistent with the eventual flow.
We give a brief outline of our method for solving
Step (1):
'Remove' the
pressure
(*):
term and the continuity
the
equation, (b), by applying the divergence operator to
NavierStokes equation, (a).
Step (2)
Generate the weak form of this system.
Step (3)
Map the system from the free
domain,
to
D,
a
fixed domain R.
Step (4) Apply
time
dependent
finite
elements
problem in the R domain.
Step (5)
Determine 'suitable' initial values.
to
the
58
result
The
of
(1)(5)
steps
is
a
system of initial value
problems set in the fixed domain, R.
Step (1):
Modifying (*) to 'Remove' p and equation (5.1b)
Taking the divergence of (5.1 a), we obtain in D(t):
div[p[ut + (u.V)u]) = div(Vp + pAu
Holding
p
constant,
pg).
applying (5.1 b) and simplifying,
(5.2)
we have in
D(t):
p div[(uV)u) = Ap.
(5.3)
Thus, given u at time t, we have p implicitly given.
Equations (a) and (b)
in (*) are replaced by
+ gAu
pint + (u101)10 =
pg
in D(t).
where p is now the implicitly known function given
by
(5.4)
the
Poisson
equation:
Ap = 2p detVu
p = 0
method
(5.5b)
on u(s,t)
p=pg(ui)cose
The
(5.5a)
in D(t)
(5.5c)
on i(x,t)
ap/8x=0
at x=a
(5.5d)
8p/ax=0
at x=b
(5.5e)
of solution for (5.5) is independent of the method
for the other portions of our problem.
Problems such as
(5.5)
are
easily solved. [6], [29] Consequently, we proceed as if p is a known
function, albeit implicitly.
The system (*) has been altered to
problem which we denot by (#).
this
new
version
of
our
59
TU=0
(5.6a)
in D(t)
(5.6b)
on u(x,t)
nt
=v1
u=0=v
c pt
us
pg
+ pAu
=
PIUt +
(5.6c)
y=71
on 4(x,t)
(5.6d)
= dus/dx
(5.6e)
c2(4'
= dir[c1w
au/ax=0=8v/ax
+c
(5.6f)
at x=a and x=b
(5.6g)
uo, vo, Do given at to
where du /dx =
(au lax +eauslay)
Step (2):
(5.6h)
iy.i(x,t) and p is given by (5.5).
The Weak Form of (#)
We let <f,g>r1 be the inner product given by LTD fg da.
We also
use the equivalent notation <f,g> when no ambiguity will result.
let W be our class
of
test
on
D.
The
exact
The weak form of the problem
restrictions on W will be given later.
(*)
defined
functions
We
is given as:
For t > to, find u and v such that for all w s W
<ut,w> =
<(nrusw> + <p
<lew> = <v1
<dt,w> = <cp
div T,w> + <g,w>
,w>
(5.7a)
(5.7b)
y=11
dus/dx,w>
(5.7c)
subject to:
TU=0
on n(x,t)
(5.8a)
u=0=v
on i(x,t)
(5.8b)
us = dir[clw
8u18x=°=av/8x
c2(4' +c)]+
(5.8c)
at x=a and x=b
(5.8d)
60
(5.8e)
uo, vo, Do given at to
and where p is given by (5.5).
now
We
incorporate
Recall
the constraints (5.8) into (5.7).
Green's theorem:
(5.9)
fD(div T)w da = ID Vw T da + faD (Tn)w ds
We apply Green's theorem, (5.9), to (5.7a) and obtain
<ut,w> = <(u*V)u,w> + <jolt>
T,Vw> + <pTk,
<p
(5.10)
w>OD
hand
where the last two products are defined by the right
side
of
(5.9).
The
functions,
test
boundary.
are
w,
usually chosen to vanish on the
substitution
is
made to get zero boundary data.
does
Such a proceedure
the standard
In our case,
eliminates products of the form
procedure
a simple
For problems with nonzero boundary conditions,
not quite work.
The boundaries are free making the
awkward.
usual (constant for all time) substitution
We
achieve
a
similar end from a slightly different approach.
We consider our boundary term
<pTn,w>OD =
<pTh,w>awo +
<pTh,w)OD(4) + <pTn,w>8D(a)
(5.11)
<pTn,w>OD(b)
where
the
parentheses indicate the portion of boundary to which we
are restricting ourselves.
The constraint (5.8a) assures
Tn=0 on
but u does.
(5.5) in the same manner as (5.8),
which admits p.
Our
test
so that
(5.12a)
O.
<pTn,w>OD(m)
Th does not vanish on
71
Since we
need
not
solve
we need not choose a basis for w
functions
must
only
admit
u
and
v.
61
Therefore,
we
can
require
4.
functions in W vanish on
all
that
Thus, w=0 on 4(x,t) yields
<pTn,w>(4)
The uniformity of
determine
flow
the
and
x=a
at
the actual flow there.
(5.12b)
o.
(In fact,
x=b
allows
to
us
we will choose uniform
planar flow which we determine analytically.) We need not generate a
finite element equation for these edges since
they
only solve (*) in the horizontal interior of D.
need
to vanish at x=a
discrepancy
and
we
x=b
in
essence
By choosing w
throwing
There is no reflection when the flow becomes uniform as
the outflow boundary x=b.
We
out
the
the boundary reflection mentioned above.
by
generated
are
known.
are
it
reaches
We choose w=0 at x=a and x=b and obtain
(PTh'w>3D(a) =
(5.12c)
0 =
CpTit,w>8D(b).
The result from (5.12a), (5.12b) and (5.12c) is
(5.13)
<pTn,w>arl = O.
Our problem is now:
Let W be given by
W=fw(x,y)1 w e C1 (D),
Given uo,
vo,
Do,
w(x,4)=0, w(a,y)=0, w(b,y)=01.
find u and v such that for t > to,
(5.14)
and for all
w e W,
<ut,w> = <(usV)u,w> + <g,w>
<p
<nt,w> = <viy=71,w>
_1
<4t,w> = <c
p
du /dx,w>
1%ilw>
(5.15a)
(5.15b)
(5.150
s
where p is given implicitly by (5.5).
It is this problem, (5.15),
domain R.
which is to be mapped to the fixed
62
Mapping D(t) to R
Step (3):
The problem (5.15) has two free boundaries making it awkward to
Mapping D to a fixed domain will allow us to fix
fix a basis for W.
a basis for W.
The drawbacks are,
We can eliminate the need to invert
additional nonlinearities.
map
but
must
we
dealing with the inverse map and
suffer
consequences
the
our
additional
of
nonlinearities.
Let R be the rectangle given by
R=f(I,;)i a
<7
4< ; < n)
< b,
(5.16)
and let A :D > R be the map given by
(5.17a)
= x
The Jacobian of
Ais
4).
4)/(T1
; = (y
given by
JEAl = J =
y,
A (4(x,t)
= i(a,t).
We
need
calculation to obtain Ca,t) from R.
the
sediment
Since a=x and i is independent
that is i(x,t).
_1
of
(5.18)
= h
We are primarily interested in
which never vanishes.
boundary evolution,
(5.17b)
not
perform
a
complicated
If u is desired, however, then
such a calculation must be performed.
Our
inner products must also be mapped to R.
<f,g> > <f,g>R
where
<7';R = ffR f8 Y[A] da.
We
now
suppress
the
R
present version of our problem.
and
(5.19)
the tilda notation and give the
63
_1
<ut3.,w> = f2<v y ,w>
x
2<v ,w d>
Y Y
<vu ,w> +
+ <v d,w >)/Re + <uv ,w>
Y
Y
Y
<v ,w >
x
y
u ,w >
Y Y
<h
Y
<ph,wI>
(5.20a)
<pdh,w > + <(sinO)h,w>/Fr
Y
<vtLvi> = [<u Y,w x> + <u
+ <v dh,w >
3'
yy
<v dh,w > + <v dh,w >
y
x
x
y
,w d>
_i
2<h
<v d2h,w )
Y
Y
I
v ,w )/Re
Y Y
<vv ,w> + <p,w >
Y
Y
<uhvx,w> + <uvY.dh,w>
(5.20b)
<(cos9)h,w>/Fr
<T1t,w> = <v1
(5.20c)
,w> = <v(x01),w>
Y=11
_1
<4t,w> = cp
f<u d,w> + <i'h
(5.20d)
u '0)1y=i
where
h = h(x,t) = 1(x,t)
d = d(x,y,t) =
4(x,t)
(5.20e)
+ yh')/h
(5.20f)
and p is known implicitly.
Step (4):
Applying Time Dependent Finite Elements
The general idea behind finite elements
is
the
RitzGalerkin
method. [7], [29] Our variables should be C (D).
The finite version
'arbitrarily
close' to the true
of
our
variables
should
become
variables as the number of basis
is
of
elements
increase.
course desirable as is compact support.
Finally,
chosen must suffice for both the test functions and
We
choose
a
special
finite
basis
for
W.
Orthogonality
the
the basis
variables.
This is the basis of
64
piecewise linear functions.
The 'problem' is written in terms of the finite basis elements.
The test functions now contain only a finite number of
do
the variables.
that,
as
the coefficients will also be time
In our case,
The hope is
dependent.
unknowns
as
number
the
of
basis
elements
the finite version of the variables converges to the true
increase,
Under certain conditions,
weak solution.
such convergence
can
be
proven. [7]
We proceed as follows:
choose a 'suitable' finite basis for W
write the entire problem in terms of the finite basis
and
evaluate the inner products
form
a
system
of
ordinary differentia/ equations
Step (4a):
The Grid on R
y=1
--0
y=0
flow >
1-7171A,,r;
x=a
i=1
i=2
-171Ai=I
i=3
The Grid on R
Fig 5.0
j=1so.
x=b
65
We let ei denote the following basis for i=1,...I
1i + x/Ax,
=1
ei
x/Ax,
1+i
x e [(i-1)Ax,iAxl
x e EiAx,(i+1)Axl
(5.21)
otherwise
0
where Ax is (ba)/(I+1).
We let eij denote the following basis for i=1,...I; j=1,...J
1+j-i
1+j 1+i 1+i-j
1-j +
1-i +
eij(x,y)
+ x/Ax - y/Ay,
y/Ay,
x/Ax,
- x/Ax + y/Ay,
y/Ay,
x/Ax,
x
x
x
x
x
x
e region
e region
e region
e region
e region
e region
otherwise
1
2
3
4
5
6
(5.22)
where Ay is 1/J.
There are numerous properties of this
basis.
A
typical
such
example being
mu
ij
e
e
Such
step.
properties
The
= 0 if m > 1+1 or respectively n > j+1
will
following
be
used
without explicitly explaining each
representation
for
our
variables
is
used
throughout where we employ the standard summation notation.
u(x,y,0=a..(t) eij
(5.23a)
v(x,y,t)=P..(t) e
11(x,t)=Si(t) ei
(5.23c)
i(x,t)=6.(t) ei
(5.23d)
(71-4)=h=ti(t) ei
(5.23e)
w(x,y)=Xij e3
(5.23f)
i=1,...I; j=1,...J
All our functions are piecewise linear.
tion may
be
easily
carried
out.
Piecewise differentia-
The following notation will be
common during the remaining discussion.
66
au/ax = a.. aeij/ax = a,. e
13
13
(5.24)
ij
X
(5.25)
au/at = aa../at eij = aij eij
13
Greek letters as coefficients may be thought of
We shall use the notation a =
po, Po,
vector.
ali,...a/s).
so, find a, p, 6, a such that for all X
X<IP1>i
where
a
The Discrete Version of Our Problem
Step (4b):
Given ao,
(a11,a12
as
z=(a,6,6,e).
The
full
IP2>
=
(5.26)
form of the inner products <IP1> and
<IP2>, in their discreet version, are given in appendix 1.
Step (4c):
Evaluation of the inner products to form the system
of ordinary differential equations
Our evaluation is analogous to a much
simpler
case
which
we
explain first.
Given the following sort of 1dimensional inner product,
must hold for all a1
We could simplify in this way:
bi<ei,ei>
<aiei,biej> =
=
which
bi+1<ei.ei+1>1
aij.)
The terms in the brackets,
(5.27)
may all be evaluated in terms of b
and this example reduces to the form
aEb = at.)
no summation for a.
where E represents the matrix formed from evaluating
terms in (5.26).
(5.28)
1
the
bracketed
67
Since the a. (test functions) appear on both sides of the final
equation, we end up with a linear system to solve.
(5.29)
El = [.)
b
If
was
dependent,
time
coefficients in (5.23) with time
result
would
the
be
replace
would
we
dependent
the
coefficients.
constant
The
end
system of ordinary differential equations in
place of (5.29).
El;
(5.30)
= C.)
Our problem is analogous to this example except
four
that
we
have
If we
instead of one and we are set in 2d domain.
variables
momentarily denote z = (a,A,b,e) then we will ultimately obtain
the
form:
(5.31)
Al = C.)
where A is the counterpart to E above.
We skip all the intermediate steps and give only the final form
of the problem:
0\
I-`
a
A, 6
sA
A
0
0
A
0
0
A
f2(a,0,6,e)
0
0
B
0
8
(
(0)
f3
0
0
C,/
0
if
(a,
1
,
(5.32)
0
\el
\f4(a'ft'b,e)I
or
Al = f
where A, B, C, and U. are given in appendix 1.
x
Our
problem
has
now
been
reduced
to
solving the ordinary
68
differential equation system
(5.32).
need
We
only
suitable
add
initial conditions to complete the problem.
The Initial Condition for the Initial Value Problems
Step (5):
Presumably,
system
could
we
Although
(5.32).
attach
most
any
fluid
(numerically) eventually smooth out.
to
initial
flow
is
conditions to our
unstable
it
may
In practice, it is impractical
use initial conditions which are too incompatible with the flow.
we take for our initial conditions the
With this in mind,
solution
the steady flow generated when we momentarily
to a related problem:
fix the sediment boundary.
We fix the
resulting
bottom
flow
steady
with
some
initial
shape
and
solve
the
This solution becomes our initial
problem.
condition at the time we turn the sediment loose.
We refer to these
initial conditions simply as no, po and Do.
Solution Procedure Solving the Initial Value Problems Numerically
Initial
vlaue
various methods.
as
(5.32) may be attacked with
[7], [29], [35].
The method of choice should
problems
[5],
such
deal well with nonlinear equations.
it
is
an
We choose AdamsBashforth since
implicit/explicit multistep method.
more stable than the convenient,
These methods are
explicit, single step methods such
as RungaKutta.
We give the outline for solving (5.32) numerically.
69
Generate the initial conditions,
Set up initial steps of multistep start,
Use AdamsBashforth predictor/corrector,
Find the new pressure distribution,
Go to (3).
We give the highlights of the procedure.
Step (1): Generating the Initial Conditions
Generating
the
initial
conditions is a two step process.
This allows
first read in all the initial data which predetermined.
problem which
us to set the parameters for solving the steady flow
is the second initialization step.
We use
and solve the fixed boundary problem by finite
we
reduce
to
domain
predetermined
a
We
That
elements.
is,
Our output is the initial
a linear equation system.
flow, the initial domain being already known.
Step (2): Set Up Initial Steps of Multistep Start
The
multistep
method
requires
periods.
These are generally
explicit
method.
In
our
the
obtained
case,
by
two time
solution
for
using
single
a
step,
we want to avoid the single step,
The
steady
explicit methods.
Our method benefits us
flow
is supposed to gradually go into an eroding flow.
(fixed
bed)
here
twice.
Thus, by using the steady flow at both t=0 and t=ti,
we satisfy the
multistep requirements as well as making a smooth transition.
70
Step (3): Using AdamsBashforth
Applying AdamBashforth is easy at this stage since we have the
initial
values
at
function evaluation.
The
tn
and
tn-1.
The
slopes are given by simple
We simply run these values through
our
loop.
only nonstandard calculation is the implicit calculation of the
pressure after each pass through the loop.
Step (4): Find the New Pressure Distribution
The pressure distribution is necessary in evaluating the
slopes.
This is obtained by solving the Poisson equation for the
pressure in terms of un,n, and in.
We used finite elements here.
Thus, we must solve the linear equation system.
Results Graphical
The results are presented graphically in Fig 5.1a and Fig 5.1b.
71
'30
Dune Migration A General Flow
Fig 5.1a
usarnesmumsuatalaLsw"...."1"...."661
i3o
ewe"
ISHMONI11111.11.1MINNIMMOSMINSW"*"'
Dune Migration A General Flow
Fig 5.1b
72
Concluding Remarks The General Solution
We have developed a
model
for
general flow.
a
procedure
'reasonable'
solving
for
Given an initial dune,
our
similar to the
special case flows' initial dune, we get a similar migration.
An equilibrium
However,
flow
also
was
used
for
the
The difficulty there lies with the
large variation in time scales between the evolution
to
flow.
in that case, the flow either becomes unstable or does not
undergo any appreciable erosion.
flow
initial
rate.
promote
If
rate
and
the
we increase the flow and the erodability of the bed
evolution'
and
even
instability at an increased rate.
up'
the
Certainly,
rates,
then
we
get
the first improvement
is a finer grid for faster flows.
Our solution method for the general flow has
some
limitations
which we note here.
The flow rate must not be too excessive.
The initial configuration must be fairly smooth.
The outflow must tend towards uniform flow.
73
SUMMARY, OPEN QUESTIONS AND CONCLUDING REMARKS
Summary
Our
erosion.
streambed
In chapter III we derived a model for
basic approach was to combine the interaction of a viscous flow with
boundary
The
which represented an erodable streambed.
constraints
viscous flow was characterized by the Navier-Stokes equation and the
boundary
condition
was
momentum, and energy.
The
erosion
by
obtained
balancing
sediment
mass,
The highlights of the model are:
process
consists of a flow subject to a
boundary condition which represent an erodable streambed.
The general erosion producing flow is governed by
the
Navier-Stokes equations for open channel flow.
The
boundary
condition
is
obtained
by
balancing
sediment mass, energy and momentum.
The sediment transport occurs in a 'thin' layer
the streambed.
This is
not
the
classical
along
boundary
layer.
Our model is for relatively fine,
noncohesive,
geneous, spherical sediment particles.
homo-
Large discrete
particles would invalidate the mass balance derivation
as well as the assumptions
sediment transport velocity.
used
to
obtain
us,
the
74
transport
sediment
The
quasiequilibrium.
must
Increased
be
in
sediment
of
state
a
across
flux,
results in an increased transport
the boundary,
rate
rather than an increased sediment concentration.
The vorticity
of
near
flow
the
bottom
the
is
a
fundamental factor for the erosion process.
The erosion model we obtained was a 2dimensional free boundary
value problem with two free boundaries.
flows
well as a general flow.
as
The
second
case
was:
A
examined
special
some
The special flows were discussed
The first case was:
in chapter IV.
We
Almost
flow.
flow
profile
velocity
logarithmic
uniform
steady,
in
conjunction with potential flow.
Results
from the first special
shock nature of the model equations.
case
flow
show
Also, from this case we, found
the flow to be too restrictive to allow for initial dune
However,
inherent
the
formation.
classical dune migration pattern for an existing dune
the
is obtained.
Results from the second special case flow are
first
case.
The
'shock'
account
for
to
the
forms as the angle of repose is attained
and the solution begins to breakdown.
cannot
similar
initial
dune
As in the previous
formation
case,
we
but the classic dune
migration patterns are obtained.
The
main
difference
between case one and case two is that we
are using an empirical flow in case
two.
The
importance
of
this
case is that it would allow us to upgrade our empirical constants to
empirical relations.
75
In chapter V we considered a general
migration
our
but
unstable flow
changes.
method
while
some limitations which produces an
has
attempting
attributed
We
this
between flow rate and bed
dune
achieved
We
flow.
achieve
to
to:
evolution
time scale variation
the
one,
and
rate
evolutionary
'major'
two,
relative
the
of the grid for high flows or flows with an artificially
coarseness
high erodability.
Some Remarks The Model and its Solutions
One
of this model aside from the
distinctions
major
the
of
general flow field is the appearance
near
vorticity,
the
of
the
streambed, as a fundamental parameter affecting erosion.
The largest particle a stream can move is called the
midstream
1/6
Recall Brahms' formula, V =kW
of the stream.
c
velocity
to
competence
, which relates the
s
the weight of the particle.
We show that a
fundamental
similar expression is obtained using to as a
parameter.
gw is
Let fs be the force trying to roll a particle of diameter ds.
proportional
of the same order as To and To is known to be
square of the midstream velocity, V.
(V)o
C
Flow
over
evolution.
a
gw
s
the
So,
1/3
W
d
f
to
s
(4.17)
s
bed of sand causes the bed to undergo a distinct
At slow flows,
the bed remains
flat.
With
increasing
flow velocity, the bed passes through a set of stages: ripples, then
dunes,
then
antidunes.
flat
again with chaotic flow,
then waves and finally
One of our main purposes was to have a model capable
of
76
addressing these 'major' evolutionary changes.
Erosion producing flows are generally turbulent, or at the very
unsteady
least
and
existence
system,
nonuniform.
For
uniqueness
and
turbulent
existence
flows,
upon
depend
configuration and the Reynolds number.
is,
unsteady NavierStokes
the
the
boundary
Large Reynolds number,
that
and complex boundaries restrict the period of
and uniqueness.
It is just such
flows
which occur during
the 'major' evolutionary changes.
We
have
imposed
restrictions
on the class
of flows
with the
goal of attaining a better mathematical understanding of the erosion
process
over
precisely.
the goal
of
attaining a model which matches data more
This restricted class
of
flows
still
yields
classic
erosion patterns.
The results of our model and its solutions are summarized here.
Our special case flows have a
flow
field similar
to
many models.
Our model can accept a general flow.
We
obtain
flows
dune
migration
and the general
in
both the special case
flow.
A fundamental parameter in our model is the
vorticity
near the streambed.
Imurovements The Model
The main improvements which we could make to the model are:
1.
Replace
empirical
constants with empirical relations
7'7
in the derivation of the boundary condition
Consider the case
cs
nonconstant (this could
be
done
by assuming a simple exchange law)
Consider
more involved expressions for the resistence
terms
Take the pressure lift force into account
Allow for nonconstant density in
Improvements The Solution
The
main
improvements
to
the
solution
procedure as it now
stands are:
Use a finer, variable mesh
Use a more sophisticated method in the elimination
of
the pressure variable and the continuity equation
Open Questions
The open questions which arise from this problem are of two
types:
Those which deal with the model and its implications and
those which deal with the solution procedures.
Some of the more
interesting open questions are given below.
Can
the
model display the other 'minor' evolutionary
processes, namely antidune migration?
Can the model display 'major' evolutionary stages?
Can the model be used to show
that
the
evolutionary
78
stages are, in fact, bifurcation phenomenon?
Can
apply the classical boundary layer theory and
we
retain enough flow realism and still get
effects?
bifurcation
the
Note that in ELT one often assumes a boundary of
constant curvature.
What
Is there another
is the best solution approach?
reasonable solution method?
This
problem
unique
is
so
that there is no set solution method.
How
does one deal with the outflow boundary condition
when the flow does not become uniform?
How can the difference in natural time scales for
streambed
versus
the
region
flow
be
dealt
the
with more
profitably?
Concluding Remarks
We
have
derived a new mode/ for streambed erosion.
behavior.
exhibits classical dune migration
general
flow field.
accepts
a
The pressence of the general flow is necessary
for certain aspects of
nonuniform
potentially
of
capable
model
The
The model
erosion.
addressing
Thus,
erosion
as
this
a
model
is
bifurcation
phenonenon.
The fundamental parameter in this model, which distinguishes it
from most others,
is
basic
to
is the vorticity near the streambed.
vortex
sheeting
initiation of the ripples.
which
is
The model can be
possibly
easily
Such a term
related to the
'improved'
by
79
replacing empirical constants with relations.
80
BIBLIOGRAPHY
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McGraw Hill (1978).
Daugherty, R. L., Basic Equations of Fluid Flow in Handbook of
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81
Holt, M, Dimentional Analysis in Handbook of Fluid Mechanics,
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Mechanics, Publish or Perish, Inc. (1976).
loose, G. and Joseph, D. D., Elementary Stability and
Bifurcation Theory, SpringerVerlag (1980).
Kennedy, J. F., Mechanics of Dunes and Antidunes in Erodible
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Wesley (1959).
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I251
MeyerPeter, E. and Muller, R., Formulas for Bedload
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Pal, S. I., Laminar Flow in Handbook of Fluid Mechanics,
1 St Ed., V. L. Streeter Editor, McGraw Hill (1961).
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APPENDIX
33
APPENDIX
The Inner Products
A6 = q11
412
q13
q14
q21
q22
q23
q24
q211
"4-
q15
416
q17
qpl
q25
q26
q27
q28
429
qp2
C=441
B8=q31
q42
h./2 h./12
airAxAy
0/12
0/2
h./12
1
1
1
h./12
h./2 h./12
1
i= ,...I, j=1,...3
0/12
0/2
1
1
a11
A=
a12
aij 1
a..41
aIJ 1 aIJ
q210
84
2/3 1/6
1/6 2/3 1/6
1/6 2/3 1/6
B.
ii
116 2/3 1/6
= Ax
2/3 1/6
1/6 2/3 1/6
1/6 2/3
1/6
2/3
1/6
2/3
1/6
1/6
C..
= Ax
1j
2/3
1/6
1/6
2/3
1/6
Axa.. [-p.1-1j-1 + 5 i-13 - 25
.
.
+ 25..
- 5.141.3 + 0.1+1.j+1 ibs
.
q
=
11
AxaiJ (-0i-1J-1 + 0i-1J
i=1,...I; j=1,...J
1/6
2/3
25iJ-1 + 20iJ)/6
.
85
Ax3
+ a
fai-lj-1
'12
i-lj
- 2a
ij-1
i=1,...I;
[-ai_11 +
ai-17 - 2aiJ-1
+ a.
1+1j+11/6
+ 2a. }/6
j=1,...J
Pii-i
n13
aij+1 - ai+lj
=
Ax.
'
+ 2
5i+ii+1)/2
°i.j+1
=
PiJ-1
1-13i-1J-1
i=1,...I;
Ax
AY
°ill/2
j=1,...J
+ 2p
g.. t-p
13
1j-1
- p
ij
1
q14 =
1
Ax
Ay giJ
i=1,...I;
1
1
)
ij+1
j=1,...J
gij = [
5iJi
[sin. -
2/ix
ei-1 + jAy(ti.o.
ti_i)]
86
-64 1-13ij-1
°ij+1)/
213ij
=
'15
Ax (-3*7-1
Ay
i=1,.../:
°iJilti
j=1,...J
f-Pi-lj-1
°i-lj
Pi+141.112
pij-1
(116
i1-1
i=1,...I;
i.+11/12
j=1,...J
isin(0) ti AxAy)/Fr
[sin(8) ti AxAy)/2Fr
i=1,...I;
i=1,...I;
j=1,...3
87
fli = {e2
ci + (j-1)Ay(t2
ei_i + (j-1)Ay{ti+1
fij = lei+1
fIj = CEI
ti))/Ax
(j-1)Ay{t1
EI-1
tI-1))/Ax
7ijlti + Yij+1ti
aij = 1.YtT1-13-1ti-1 + 271-1j41-1
bij = dx{y.1-1j-1 f.1-1j-1
aiI =
.f.
/-13 1-13
4171-13-11-1 + Ti-13.1-1
biJ = Ax{y i-1J-1f. 1-1J
c.
ij
y.
= {y.
1-1j-1
ciJ = {y1-1J 1
y.
1-1j
+ 2y
17-1
+ 2y.
+ 2y.
f.
ij-1 ij-1
71J-1ti
f
tS-1 iI-1
+ 2y
+ 2y..1 + y
3.
+ 2y.-11.7f iI +
1+13
2y.. .f..
13+1 13+1
Ti+1Ii+1)/6
+ y.
2y..
1+1j
13+1
ij-1
41+1341+1
1+1I 1+1J
1+13+1
)/6
a.. + b.. + c..
13
q
13
13
=
pl
a
+b.iJ
+ c .
1J
i=1,...I; j=1,...7
Ayaiii
+ 2pi+li + Oi+ii+1)/6
+ 1314-1
q21
Axaisti fi-1J-1
i=1,...I;
j=1,...1
1-17
Pi1-1
131+111/6
38
g a.ij [-p.
1-1j-1
-
+ p.
25
ij-1
+ 2p.
ij+1
i+lj
+ p
i+lj+1
}/6
c122
g a.1J f-P.1-1J-1
g"11 = 8'1+1 E.1
1-
+i-17
- 20
i=1,...I; j=1,...J
(4i+11-1)(j-1)Ay/2
exp.ij f-p.1-1j-1 + p.1-ij
-
+ 201.1.1/6
- p
+ 20.
+ p
I/6
i+lj+1
i+lj
ij+1
2p
q23 =
Axpij [-pi_1j-1 + pi_lj -
i=1,...I;
j=1,...J
{-a 1-1j-1
n
=
+ 213u1/6
+a+a-a-"1-4.(2--a.
1-1j
ij-1
1j
ij+1
1+1j
[-a1-1J-1 + a.11+ ai1-1 - aiJ1/2
i=1,...I;
j=1,...J
.
1+13+1
1/2
89
Ax
Ay
'25
g..
)
i j [-a..
ij-1 + 2a.. - a..
1.1+1
1.1
=
Ax
Ay
g.. = [
+a
.
g.13 (-a 0-1
)
"i+1 - 8i-1
1
2Ax
jAY(i+1
1=1,...I; j=1,...J
VI,
p
2ij
[-A
i-1j + 20..
ij
[
ReAx
q26
2ReAx
-Pi-1j
5i+1j)
i+1 j
j=1,...J
gij
-a27
Pij-1+2*5ij+Pijil.
Pi+lj+Pi+lj+1)
=
g.j rP. 1j-1 41
g.. =
[
ei+1
°J.j-1
'ij
+
A_o
i+1.1
ei-1 + (j-1)AyQi+1 -i-1) ]/(4AxRe)
i=1,...I; j=1,...J
i-1)]
90
°i-li
gij 1-5i-lj-1
Pi+li
giJ 1ilj-1
gij =
e.1+1
°i+ii+11
°ij-1
4
ei-1 + (j-1)dy(i+1
1-1 )
)/4AxRe
j=1,...J
q29=
h.lj
Ib.ij-1
h
iJ
tb ij-1ii
+ b..)
ci_i
gij =
b.1j+1 I
+ 2*b.1j
h..=(g..) /4AxAyRe4.1
(j-1)AYQi4.1
i=1,...I; j=1,...J
Ay
(-0..
13
ij-1 + 20..
p..+1
1J
4210 =
Ax
Ay
[-P iY-1
i=1,...I; j=1,...3
+ p. )i.
i/.
91
[cos(0) 4i AxAy)/Fr
-'211
i=1,...I; j=1,...J 1
=
[cos(9)
all
alj
alJ
q2 =
all
aij
aiJ
aIl
aIj
aIJ
i=1,...I; j=1,...J
AxAy)/2Fr
1=1,.. .1;
j=1,...J
92
all =
y11f11/2 + yijkl + y13f13/2 - Y21k2f'21
y31f31/2
als = -y1s..1f11/2 + y1J-1kl + y1Jf13/2 - y _12f'
2.)
2J
-
y3Jf3J/2
=£=
-71j-1f1j-1/2 + yljkl + ylj+1f lj+1
=-Yi-11fi-11/2
Yilk2f'il
/2
Yi+11f1+11/2
Y1+12k1 - yi+12fi+12/2
aij = 71+1j-1fi+lj-1/2
aiJ
+y.
'kl
= yi-1Jfi-1J/2 --1k2f*
-y i+lj+
1+
+ yi+13-1f1+17-1/2 +
i+1J-1kl - yi+1Jfi+1J/2
/2
all = -y
aij
f
j-1 Ij-1
YI112fiI1
/2 +Yl -.k1
i
aIJ = YI-1JfI-1J/2
j
YIj-lfIj-1/2
Y,
ij+1fIj+1/2
YIJfIJ/2
93
f'lj =
1.0 + fe2
(j-1)dy(0+1
= [1.0 + tei+1
(ei+1
el + (j-1)dy(0+1
0))(0+1
0) (i+1
2
0))/4(Ax) )
0-1))(0+1
0-1))/4(Ax) 11/(4i)
f'1j = 1.0 + [ei + ei]. + (j-1)dy(-0+0-1)
(j-1)dy(0
20 + 0-1)
2ei + ei-1 + (j-1)dy(0+1
ei-1 + (j-1)dy(0+1
0-1))(0
ei +
0-1))/4(Ax)
(ei
I
ei-1 +
2