1. Sediment Transport
Ben Watson
April 25, 2014
A project of 30 credit points at level M
Supervised by Andrew J. Hogg
Abstract
A differential equation is constructed to model the distribution of
suspended sediment in a turbulent flow. The turbulent fluctuations
are expressed as a diffusive flux, where the sediment diffusivity can be
determined empirically. Steady solutions to a fully-developed flow are
found, where the settling velocity is considered as a function of the
concentration. We gain an understanding of how the concentration of
sediment changes over time by considering the stability of each steady
state. The accuracy and the time scale of the steady solutions are
found to be strongly dependent on the Rouse number, a dimensionless
parameter which provides a measure of the flow’s ability to suspend
particles.
2. Acknowledgement of Sources
For all ideas taken from other sources (books, articles, internet), the source
of the ideas is mentioned in the main text and fully referenced at the end of
the report.
All material which is quoted essentially word-for-word from other sources
is given in quotation marks and referenced.
Pictures and diagrams copied from the internet or other sources are la-
belled with a reference to the web page or book, article etc.
Signed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Date . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
3. List of Figures
1 A diagram showing three different modes of sediment trans-
port: Bed load, suspended load and wash load. . . . . . . . . . 8
2 A diagram showing the basic setup of suspended sediment in
a uniform flow. . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 The variation of the drag coefficient with the Reynolds number
for natural sand. (Taken from Fredsoe and Deigaard (1992)) . 13
4 A constant, linear and parabolic representation of the eddy
diffusivity. The constant fc has been chosen as 1 for simplicity,
and fl, fp are set to be very small ( 1) for reasons stated
earlier. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5 The vertical distribution of concentration in a river for con-
stant diffusivity, where different magnitudes of the Rouse num-
ber have been considered. . . . . . . . . . . . . . . . . . . . . 23
6 The vertical distribution of concentration in a river for linear
diffusivity, where different magnitudes of the Rouse number
have been considered. . . . . . . . . . . . . . . . . . . . . . . . 25
7 The vertical distribution of concentration in a river for parabolic
diffusivity, where different magnitudes of the Rouse number
have been considered. . . . . . . . . . . . . . . . . . . . . . . . 26
8 Two comparisons of the vertical distribution of sediment in
a river for constant, linear and parabolic diffusivity. In (1),
the Rouse Number has been fixed, and in (2), the mass of
sediment distributed is fixed. . . . . . . . . . . . . . . . . . . . 27
9 A plot of the two functions given in (72), taking the Rouse
Number P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . 33
10 A plot of the two functions given in (75), taking the Rouse
Number P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . 34
11 Left: A graph showing the difference in the asymptotic ap-
proximation of ∆n and the actual value of ∆n, defined as
∆∗
n −∆n. Right: A graph showing the difference in the asymp-
totic approximation of λn and the actual value of λn, defined
as λ∗
n − λn. Note that the Rouse number has been taken as
P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3
4. 12 Four separate examples displaying the accuracy of the solution
for t = 0, where the number of eigenvalues included in the
summation varies. The initial concentration is set as P ˆcr,0 =
0.4 and the Rouse number P = 2. . . . . . . . . . . . . . . . . 42
13 A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 0.37/P, and Rouse number P = 2. . . 43
14 A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 0.01/P, and Rouse number P = 2. . . 44
15 A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 1.8/P, and Rouse number P = 2. . . 45
16 (1), (2), (3) and (4) show the time evolution of concentration
for P = 0.5, 1, 2, 3 respectively. The initial concentration is
fixed at P ˆcr,0 = 0.5 in all cases. . . . . . . . . . . . . . . . . . 47
17 (1) A plot showing the relationship between the Rouse number
and the first (lowest) eigenvalue ∆1. (2) A plot showing the
relationship between the Rouse number and the first (lowest)
eigenvalue λ1. . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
18 A comparison between the upward flux (φu) and concentra-
tion (c), where the parameter n varies between 1 and 6. The
blue lines correspond to the typical values for n (determined
empirically). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
19 A plot of the two steady solutions, cs,1 and cs,2, with three
different magnitudes of the Rouse number. Note that the left
solutions (low concentration) correspond to cs,1 and the right
solutions (high concentration) correspond to cs,2. . . . . . . . . 54
20 A plot of the two steady solutions, cs,1 and cs,2, where the
Rouse number is set as P = 2, and the initial concentration is
ˆch,0/cmax = 0.5. . . . . . . . . . . . . . . . . . . . . . . . . . . 56
21 A plot of the steady solution cs,1 with the Rouse number taken
as P = 2. The initial concentration is set as ˆch,0/cmax = 0.01. . 57
22 A plot of the steady solution cs,2 with the Rouse number taken
as P = 2. (The solution cs,1 is not shown). The initial con-
centration is set as ˆch,0/cmax = 0.79. . . . . . . . . . . . . . . . 58
4
5. 23 A plot showing a relationship between Clim and P, where we
have taken cmax = 6. . . . . . . . . . . . . . . . . . . . . . . . 59
24 A plot of B+ (blue), and B− (red) against cmax where the
Rouse number has been fixed at P = 2. . . . . . . . . . . . . . 64
25 A plot showing the relationship between FB− (∆) := M11M22 −
M12M21, and ∆. The Rouse number is taken as fixed at 2 and
cmax is fixed at 2.1. . . . . . . . . . . . . . . . . . . . . . . . . 66
26 A plot showing the overall behaviour of the hindered concen-
tration equation for n = 3. The arrows show the direction
in which the solution evolves over time, where solutions for t
close to 0 haven’t been included. The Rouse number is taken
as P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
27 (1) A plot showing a relationship between Clim and P, where
we have taken cmax = 6. (2) A plot showing how the solution
behaves when we start the initial concentration at ˆch,0/cmax =
0, along with the Rouse number P = 1.4, and cmax = 6. . . . . 69
5
8. 1 Introduction
The transport of sediment in turbulent flows has been extensively researched
for many years now. The best way to mathematically model the behaviour
in a uniform flow continues to be a highly disputed matter. The ability to
accurately predict sediment transportation is very important in coastal and
hydraulic engineering and, even in simple situations, finding a relationship
between the velocity of a horizontal flow and the mass of sediment in suspen-
sion is not straightforward. This is due to individual moving grains taking
energy from the flow, lowering the flow’s capability to transport sediment.
The mass of sediment being transported is defined as the ‘total load’. There
are three ways in which the total load can be transported:
1. Bed load
2. Suspended Load
3. Wash load
Bed load represents the sediment rolling, sliding or jumping along the bed.
The grains remain in contact with the bed for the majority of their transport.
Suspended load is the part of the total load which is not in continuous contact
with the bed. This is due to turbulent fluctuations in the flow keeping the
particles in suspension. Lastly, the wash load is made up of very fine particles
and is generally not represented in the bed. For this reason, we do not have
to include the wash load in the total mass of sediment transported and it
can be neglected.
Figure 1: A diagram showing three different modes of sediment transport:
Bed load, suspended load and wash load.
8
9. Figure 1 shows an example of all three modes of transport in a river, where
the sediment is being transported due to a horizontal turbulent flow. Note
that the river bed is made up of suspended load and bed load (left uncoloured
in the diagram). In this project, we look at how the distribution of suspended
particles changes as the flow develops, and whether it reaches a steady distri-
bution over time. This is achieved by mathematically predicting the rate at
which sediment settles or erodes from the bed, relative to variables such as
time or flow velocity. Conservation of mass can lead to accurate predictions
of fully developed concentration profiles which are surprisingly consistent
with experimental results.
Before going any further, we will discuss some important sediment prop-
erties and how to mathematically define suspended sediment in a turbulent
flow.
Figure 2: A diagram showing the basic setup of suspended sediment in a
uniform flow.
We consider non-cohesive sediment in suspension driven by a fully developed,
2-dimensional, uniform flow, bounded by the river bed at z = 0, and the free
surface at z = h. Note that the flow transporting the particles is made up
of an incompressible fluid, subject to small turbulent fluctuations. The tur-
bulent nature of the flow creates complicated interactions between the fluid
and the particles. A very common approach is to express the turbulent flux
as a diffusive flux, where the rate at which the sediment diffuses (known as
the sediment diffusivity or eddy diffusivity) is determined empirically (Dyer
and Soulsby 1988). A ‘no vertical flux’ boundary condition can be imposed
at the free surface z = h (since the grains cannot pass through the surface
9
10. of the river). At the base of the river, a boundary condition must be chosen
to represent the mass exchange due to erosion and deposition. (This will be
discussed further in Section 2.2).
In this project we have assumed that every grain is non-cohesive and spher-
ical, with equal dimensions and density. Typically, non cohesive sediment
(e.g. sand) in suspension ranges from 0.0625mm to 2mm in diameter. Like
the fluid, the sediment is also incompressible; this implies that mass is con-
served throughout the whole specified boundary. Since each particle has a
specific mass, we expect them to settle due to gravity. However, since the
particles are suspended in water, the force due to gravitational acceleration
will have less of an effect than that of air.
1.1 A Brief Overview
We will begin by constructing a differential equation which models the ver-
tical concentration of sediment over time. This equation is balanced by the
advection of the particles, the mean fluid flow, particle settling and diffusion.
Boundary conditions are introduced at the bed and free surface of the river.
We go on to consider steady concentration distributions which are fully-
developed and hence independent of time. Different representations of the
eddy diffusivity are chosen and comparisons are made between them.
We then try imposing an initial concentration distribution and mathemati-
cally calculate the evolution of the distribution over time. Using MATLAB,
we can represent these results graphically and hence determine the approxi-
mate time taken for the concentration distribution to settle to a steady state.
The accuracy of the solution is taken into account, as small errors can have
large effects in turbulent flows.
Finally, we reason that the settling velocity of a grain is dependent on the
concentration; this follows from a concept known as ‘hindered settling’. Hin-
dered settling gives rise to unstable solutions which can prevent the sediment
distribution from evolving into a steady state. Linearisation is used to anal-
yse the behaviour of each steady state. Once again, several graphs are plotted
using MATLAB to interpret the results.
All diagrams in this project were drawn by Ben Watson and every graph
has been plotted in either MATLAB or Maple and then annotated. The only
exception being Figure 3 which was taken from Fredsoe and Deigaard (1992).
10
11. 1.2 Important Definitions and Theorems
Before starting we will go over some important definitions and theorems
which will be used regularly throughout the project. These definitions are
based on that of Batchelor (2000)1
, Fredsoe and Deigaard (1992)2
, Dorrell
and Hogg (2012)3
, and Arfken, Weber and Spector (1999)4
.
1.2.1 Definition: Lagrangian Derivative [1]
The Lagrangian derivative can be defined as a link between the Eulerian and
Lagrangian description of a flow.
Eulerian description:
A description of the flow as seen by a stationary observer. We choose a
point x in space, and observe the velocity of the fluid as a function of time
t. Hence, the velocity filed is u(x, t), and if the flow is steady, then u(x).
Lagrangian description:
In this case, the observer moves with the fluid. We mark a fluid particle
with dye, and observe how it moves. To describe the whole field, we label
all material points. We then choose an initial time t0 and label particles by
a = x, where x denotes the position at time t0. Then, for t > t0, the particle
is at x = x(a, t), where x(a, t0) = a. The Lagrangian velocity is defined as
v(a, t) = ∂x
∂t
, with a fixed.
Now, the Eulerian and Lagrangian velocity fields carry the same informa-
tion, this implies
v(a, t) = u(x(a, t), t). (1)
The Lagrangian derivative of a flow field u = (u, v, w) records the change of
a variable following a particle path and is denoted as
Du
Dt
≡
d
dt
(u(x(a, t), t)) =
∂u
∂t
+ (˙x · )u. (2)
But we know from (1) that ˙x = v(a, t) = u(x(a, t), t). Thus, the Lagrangian
derivative is defined as
Du
Dt
=
∂u
∂t
+ (u · )u. (3)
The term on the left represents the Lagrangian description of the flow, the
first term on the right represents the Eulerian description of the flow, and the
second term on the right represents the directional derivative in the direction
of u.
11
12. 1.2.2 Definition: Specific Gravity [2]
The specific gravity (s) is defined as the ratio of the density of a substance to
the density of a reference substance. (In this case the the reference substance
is taken to be water). For sediment suspended in a flow, we have the ratio
s =
ρs
ρ
, (4)
where ρ denotes the density of water at 4o
C, and ρs denotes the density of
the sediment. For natural sediments s is usually close to 2.65.
1.2.3 Definition: Settling Velocity [2]
We define the settling velocity (ws) as the terminal velocity of the sediment
under the action of gravity. The settling velocity depends on the grain size,
specific gravity, shape, the dynamic viscosity of the fluid, and the proximity
of other particles/boundaries. The drag force F on a submerged body is
given by:
FD =
1
2
cDρV 2
A (5)
where cD is the drag coefficient, ρ is the density of the fluid, V is the relative
velocity, and A is the area of the projection of the body upon a plane normal
to the flow direction.
We will consider the settling of a single spherical particle of diameter d,
(volume π
6
d3
) . By combining the density of water (ρ) and the density of the
grains (ρs), the total force due to buoyancy is given by
FB = (ρs − ρ)g
π
6
d3
, (6)
where g is the earths gravitational acceleration.
Under equilibrium, FD and FB must be equal. By taking the relative ve-
locity as the settling velocity of a single grain, and the area of projection
that of a sphere, we have:
(ρs − ρ)g
π
6
d3
=
1
2
cDρw2
s
π
4
d2
(7)
Using (4), we can rearrange this to find the settling velocity of a singe grain
yielding
ws =
4(s − 1)gd
3cD
. (8)
12
13. Now, the value of cD depends on the particle Reynolds number (R = wsd
ν
),
where ν is the viscosity of the fluid. The Reynolds number is a dimension-
less quantity which is defined as the ratio of inertial forces to viscous forces.
Figure 3 shows a plot which relates the drag coefficient to different values of
the Reynolds number.
Figure 3: The variation of the drag coefficient with the Reynolds number for
natural sand. (Taken from Fredsoe and Deigaard (1992))
The relationship in Figure 3 can be expressed by
cD = 1.4 +
36
R
, (9)
and by substituting this expression into (8) we are left with a quadratic in
ws given by
4.2w2
s +
108ν
d
ws − 4(s − 1)gd = 0. (10)
Hence, if we take the kinematic viscosity of water at 4o
C as ν = 1×10−6
m2
/s,
the particle diameter as d = 0.1mm, the specific gravity as s = 2.65, and the
earths gravitational acceleration as g = 9.81m/s2
, we obtain
ws ≈ 0.055 m/s. (11)
Note that we have taken the positive root in the solution since the settling
velocity must be positive.
13
14. 1.2.4 Definition: (Critical) Bed Shear Stress, Shear Velocity and
the (Critical) Shields Parameter [2]
We are considering a steady flow over a bed composed of non-cohesive grains.
For small flow velocities these grains will not move, but when the driving
forces on the sediment exceed the stabilising forces, the sediment will begin
to move, (this is called incipient motion). We define the bed shear stress as
the shear stress exerted on the river bed, it is found to be
τb = ρu2
∗, (12)
where u∗ is called the shear velocity. The shear velocity is related to the
mean flow velocity (¯u) by a friction factor Cf such that τb = ρCf ¯u2
.
The Shields parameter is a dimensionless number which is used to calcu-
late incipient motion. It is given by
θ =
τb
(ρs − ρ)gd
. (13)
The point of incipient motion can be found empirically as it varies depending
on the particle characteristics. We denote the point of incipient motion as
θ = θcr, where we call θcr the critical Shields parameter.
Using the critical Shields parameter and the relation given in (13), we can
go on to define the critical bed shear stress as the shear stress exerted on the
bed at incipient motion, which we will denote τcr. Hence, using (12) we have
• If τb < τcr, then the sediment on the river bed will not move,
• If τb ≥ τcr, then the sediment on the river bed begins to move.
1.2.5 Definition: Rouse Number [3]
The Rouse Number (P), is a dimensionless quantity defined by
P =
ws
κu∗
. (14)
It is a ratio between the settling velocity and the shear velocity of the flow.
The constant κ = 0.41 is known as the von Karman constant.
14
15. 1.2.6 Definition: Time Average
The time average X of a variable x(t) is defined by
¯X = lim
T→∞
1
T
t0+T
t0
x dt , (15)
where the limit X must be independent of the initial condition at t0.
1.2.7 Theorem: Sturm-Liouville (regular) [4]
Consider a real, second order differential equation in self-adjoint form
−
d
dx
p(x)
dy
dx
+ q(x)y = λw(x)y, (16)
where y = y(x), defined over the finite interval [a, b]. The values of λ are
found such that a solution to the Sturm-Liouville problem exists, we call each
λ an eigenvalue of the boundary value problem.
This is a regular Sturm-Liouville problem if p(x), w(x) > 0, and p(x), p (x), q(x),
and w(x) are continuous functions over the interval [a, b], and have separated
boundary conditions of the form
α1y(a) + α2y (a) = 0 (α2
1 + α2
2 > 0), (17)
β1y(b) + β2y (b) = 0 (β2
1 + β2
2 > 0). (18)
Assuming the Sturm-Liouville problem is regular, the Sturm-Liouville theory
states that:
• The eigenvalues λ1, λ2, λ3, ... are real and can be ordered such that
λ1 < λ2 < λ3 < ... < λn < ... → ∞. (19)
• Corresponding to each eigenvalue λn there exists a unique eigenfunction
yn(x) which has exactly n − 1 zeroes in (a, b).
• The normalized eigenfuntions form an orthonormal basis,
b
a
yn(x)ym(x)w(x) dx = δmn, (20)
where δmn is the Kronecker delta.
15
16. 2 Vertical Distribution of Sediment in a Uni-
form Flow
We consider a steady flow over a flat river bed. The velocity of the flow runs
parallel with the bed and the sediment is kept in suspension due to turbulent
fluctuations. Each grain has a settling velocity ws, and is assumed to settle
relative to the surrounding water. We can define the velocity of a single grain
as
up = u − wsˆz, (21)
where u = u(z)ˆx is the velocity of the river flow and ws is the settling velocity.
We require each grain held in suspension to be in equilibrium. This means
the vertical forces exerted on each grain must balance, i.e. upˆz = 0.
2.1 Construction of the Concentration Equation
The concentration of sediment in a river with a uniform flow changes de-
pending on the distance of the sediment from the river bed, and the point of
observation along the river; assuming we began with some initial concentra-
tion distribution at time 0. Hence, the concentration, denoted c, is a function
of vertical distance from the bed (z), horizontal distance (x), and time t, i.e
c = c(x, z, t). (22)
As we move with the flow, we want see how the concentration develops over
time, so must look at the Lagrangian derivative (Dc
Dt
). This must balance
with the the sediment settling (ws
∂c
∂z
) for the grains to be in equilibrium. So,
we have
∂c
∂t
+ u · c − ws
∂c
∂z
= 0. (23)
Since the flow is turbulent we impose that the concentration and velocity
field can fluctuate quickly over time. We can use a mathematical technique
known as Reynolds decomposition to separate the average and fluctuating
parts of a quantity. We have already introduced a time parameter (t), which
corresponds to our concentration changing over ‘slow time’. So for our fast
fluctuations in velocity and concentration we introduce a ’fast time’ variable
(t∗). In this case, we can rewrite the concentration (c) and the velocity (u)
in (23) as
c(x, y, z, t, t∗) = c(x, z, t) + c (x, y, z, t∗), (24)
u(x, y, z, t∗) = u(z) + u (x, y, z, t∗), (25)
16
17. where c and u = u(z)ˆx denote the ‘fast’ time average of c and u (known as
the steady components), and c and u = (u , v , w ) denote the fluctuating
part (or perturbation) of c and u. These perturbations are defined such that
their time average is equal to 0.
By substituting these perturbations into (23), we find
∂c
∂t
+ u · c + u · c + u · c + u · c − ws
∂c
∂z
− ws
∂c
∂z
= 0. (26)
Note that the non-linear term (u · c ) represents the turbulent nature of
the flow.
Time averaging the whole of (26) yields
∂c
∂t
+ u · c + u · c + u · c + u · c − ws
∂c
∂z
− ws
∂c
∂z
= 0. (27)
Now, we will evaluate each term in (27) separately:
∂c
∂t
=
∂c
∂t
,
u · c = u · c = u
∂c
∂x
,
u · c = u · c = 0, (by definition of c ),
u · c = u · c = 0, (by definition of u ),
u · c = · (u c ) − c ( · u ) = · (u c ), (by incompressibility),
ws
∂c
∂z
= ws
∂c
∂z
,
ws
∂c
∂z
= ws
∂c
∂z
= 0, (by definition of c ).
Finally, (27) is simplified to
∂c
∂t
+ u
∂c
∂x
− ws
∂c
∂z
+ · (u c ) = 0. (28)
17
18. Here, the first term represents the change in concentration over time, the
second term arises due to the sediment being advected by the flow, the third
term represents the settling of the particles, and the last term corresponds
to the flows turbulent behaviour.
The “gradient-diffusion” model
Mathematically modelling dilute turbulent suspensions of non-cohesive par-
ticles poses many challenges because of the absence of complete models that
fully capture the complicated interactions between the fluid and the particles
(Dyer and Soulsby 1988; Fredsoe and Deigaard 1992). A common approach
is to assume the turbulence-induced flux can be expressed as a diffusive
flux, where the sediment diffusivity can be determined empirically (Dyer and
Soulsby 1988). (Dorrell and Hogg 2012). Hence, we have
u c = −K c, (29)
where K is called the sediment diffusivity or more commonly known as the
eddy diffusivity.
This relation is extremely useful as we have expressed the non-linear tur-
bulent term as a linear, first order “gradient-diffusion” term which can lead
to predictions of fully developed flow profiles by solving (28). However, we
cannot completely rely on the empirically obtained result in (29) since the
eddy diffusivity can depend on the boundary condition at the bed and is
much harder to account for at higher concentrations.
Now, taking the divergence of each side of (29) yields
· (u c ) = − · (K c), (30)
= −
∂
∂z
Kz
∂c
∂z
−
∂
∂x
Kx
∂c
∂x
, (31)
where Kz and Kx represent the vertical and horizontal eddy diffusivities
respectively. And by substituting this relation into (28) we are left with
∂c
∂t
+ u
∂c
∂x
− ws
∂c
∂z
=
∂
∂z
Kz
∂c
∂z
+
∂
∂x
Kx
∂c
∂x
. (32)
We will call (32) the Continuity Equation. The first term on the left-hand
side represents the change in concentration over time, the second term advec-
tion with the flow, the third term settling of sediments, and the two terms
18
19. on the right-hand side represent vertical and horizontal sediment diffusion
respectively.
Now, since we are interested in the vertical distribution of sediment, we
can neglect the horizontal gradient in c, treating the horizontal distribution
as uniform. Hence, (32) is reduced to
∂c
∂t
− ws
∂c
∂z
=
∂
∂z
K
∂c
∂z
, (33)
where, for simplicity, we have replaced c with c, and Kz with K. As we
have neglected the horizontal derivatives, without loss of generality we can
write the concentration as a function of time and vertical distance only, i.e.
c = c(z, t). We will call (33) the unsteady concentration equation.
2.2 Boundary Conditions
We would like to impose boundary conditions to the flow at the base (z = 0),
and the top (z = h) of the river. We can rewrite (33) in the form
∂c
∂t
+
∂Φ
∂z
= 0, (34)
where Φ = −wsc − K ∂c
∂z
represents the net vertical flux of sediment. The
boundary conditions are:
(i) No flux through the free surface of the river
Since we are assuming the flow is uniform, we want to look at the behaviour
of the flow in the ˆz-direction. Since the fluid is bounded at the top by z = h,
we can impose a no vertical flux condition at the free surface z = h, i.e.
wsc + K
∂c
∂z
= 0 at z = h. (35)
(ii.a) Erosion flux at the base of the river
We cannot impose a no flux condition at z = 0 since there is a potential
for mass exchange due to erosion and deposition. Instead, we can make the
following considerations:
Just above the river bed, the net flux (Φ) must be equal to the amount
of sediment eroded from the bed into the fluid, φe, minus the amount of
sediment deposited from the fluid onto the bed, φd. Hence,
−wsc − K
∂c
∂z
= φe − φd at z = 0. (36)
19
20. Now, the amount of sediment deposited (φd) is equal to the settling velocity
multiplied by the concentration of sediment (wsc), therefore (36) reduces to
−K
∂c
∂z
= φe at z = 0.
We will use a popular empirical closure to represent the upward erosion flux
by letting φe = qe(t), where qe(t) expresses the erosion rate of the flow (Dyer
and Soulsby 1988). The erosion rate is given by
qe(t) =
me
θ(t)
θcr
− 1
p
for θ(t) ≥ θcr
0 for θ(t) < θcr
, (37)
where θ = u2
∗/(s − 1)gd, θcr is the critical shields parameter for incipient
motion, and the dimensional constant me specifies the efficiency of the flux
through z = 0 (van Rijn 1984). We typically take the constant p to be in the
range [1, 3.5] (Pritchard and Hogg 2002).
(ii.b) Reference concentration at the base of the river
Alternatively, an empirically chosen reference concentration (cref) can be used
to denote the concentration at the base of the river. So the boundary condi-
tion would be
c(z) = cref at z = 0, (38)
In this project, the boundary condition (ii.a) is our preferred choice since it
takes into account the shear stress at the bed and allows for mass exchange
which better represents the interactions at the base of the river.
3 Steady Distributions of Concentration
If we consider a fully-developed flow where the concentration has reached a
steady state, then the concentration will lose its time dependence. In this
case, our concentration can be written as a function of the vertical distance
from the bed (z) only. Hence, (33) can be reduced to
−ws
∂cs
∂z
=
∂
∂z
K
∂cs
∂z
. (39)
where we have used the variable cs to denote the time independent (steady)
concentration. We call this equation the steady concentration equation.
20
21. We will begin by finding different representations for distribution of con-
centration by varying the eddy diffusivity (K). It is possible to write the
eddy diffusivity (Souslby 1988) as:
K = κu∗hf(η) (40)
where κ is the von Karman constant, u∗ is the shear velocity, h is the height
of the river, and η = z/h.
We can change the rate of diffusivity by imposing a constant, linear or
parabolic representation of our function f(η):
Constant diffusivity: f(η) = fc
This corresponds to the concentration of sediment in the river diffusing at
the same rate throughout the river. Our constant fc is bounded below at
0 (since we cannot have negative diffusivity), and bounded above by fc(max)
(the maximum rate of diffusivity). Furthermore, we will impose fc = 0 as
this corresponds the eddy diffusivity being 0 throughout the flow. Hence,
0 < fc ≤ fc(max).
Linear diffusivity: f(η) = η + fl
This corresponds to the concentration of sediment diffusing linearly though
the river. So the further from the bed the grain is situated, the quicker it
diffuses. Once again, we impose 0 < fl ≤ fl(max), where for the linear case,
we typically choose fl to be very small since the diffusion rate at the bed is
close to 0.
Parabolic diffusivity: f(η) = η(1 − η) + fp
This corresponds to the concentration of sediment diffusing parabolically
though the river. In this case, the rate of diffusivity hits a maximum at the
midpoint of the river. Just like the linear case, we have 0 < fp ≤ fp(max),
with fp small, (same argument).
Figure 4 shows a visual representation of how the three eddy diffusivities
change with height.
21
22. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
diffusivity (K)
height(z/h)
constant diffusivity
linear diffusivity
parabolic diffusivity
Figure 4: A constant, linear and parabolic representation of the eddy diffu-
sivity. The constant fc has been chosen as 1 for simplicity, and fl, fp are set
to be very small ( 1) for reasons stated earlier.
3.1 Constant Diffusivity
We will begin by looking at the simplest case: f(η) = fc = constant. This
corresponds to the sediment diffusing at the same rate throughout the river,
shown in figure 4. The Steady Concentration equation for constant eddy
diffusivity (39) becomes
ws
∂cs
∂z
= −
∂
∂z
Kc
∂cs
∂z
, (41)
where Kc = κu∗hfc.
Integrating up, and applying boundary condition (i) yields
wscs = −Kc
∂cs
∂z
. (42)
Note that the integration constant vanishes from the no flux boundary con-
dition at z = h.
22
23. We are left with a first order, linear ODE which can be solved easily to
obtain the solution
cs(z) = c0e− ws
Kc
z
. (43)
Applying boundary condition (ii.a) gives c0 = qe
ws
, and hence, the concentra-
tion distribution for constant diffusivity is
cs(z) =
qe
ws
e− ws
Kc
z
. (44)
Figure 5: The vertical distribution of concentration in a river for constant
diffusivity, where different magnitudes of the Rouse number have been con-
sidered.
Figure 5 shows a plot of the concentration (c/c0) against the vertical dis-
tance of sediment from the riverbed (z/h) to see how the concentration is
distributed. The concentration and vertical distance have been rescaled such
that the concentration of sediment is fixed to 1 at the base of the river, and
the height is bounded between 0 and 1. We consider different magnitudes of
the Rouse number; a measure of the flow’s ability to suspend particles.
We notice that as the Rouse number (P) increases, the majority of the sed-
iment sits very close to the bed of the river. This is because the settling
velocity (ws) dominates the shear velocity (u∗) when P is large enough.
23
24. 3.2 Linear Diffusivity
Now we would like to consider the case where the diffusivity is a linear
function of the vertical distance (z). i.e. f(η) = η + fl. This time, we
can rewrite (39) as a differential equation in cs as a function of η. Using the
fact that η = z/h we have
ws
1
h
∂cs
∂η
= −
1
h
∂
∂η
Kl(η)
1
h
∂cs
∂η
, (45)
where Kl(η) = κu∗h(η + fl).
Integrating up and applying boundary condition (i) yields
−wscs = κu∗(η + fl)
∂cs
∂η
. (46)
Just as before, we are left with a first order, linear ODE with the solution
cs(η) = c0(η + fl)− ws
κu∗ . (47)
Applying boundary condition (ii.a), we find c0 = qe
ws
f
ws
κu∗
+1
l .
Finally, after replacing η with z/h the concentration distribution for linear
diffusivity is
cs(z) =
qe
ws
f
ws
κu∗
+1
l (
z
h
+ fl)− ws
κu∗ . (48)
Figure 6 shows a plot of the concentration (c/c0) against the vertical dis-
tance of sediment from the riverbed (z/h) to see how the concentration is
distributed. As before, the concentration and vertical distance have been
rescaled such that the concentration of sediment is fixed to 1 at the base of
the river, and the river’s height is bounded between 0 and 1. We consider
different magnitudes of the Rouse number; a measure of the flow’s ability to
suspend particles.
As we increase the Rouse number (P), we notice that the linear concen-
tration distribution differs from the constant case. The concentration only
deviates slightly towards the top of the river, whereas in the bottom quartile
the concentration of sediment increases more rapidly. For large enough P,
i.e P > 2, the majority of the sediment remains close to the riverbed.
24
25. Figure 6: The vertical distribution of concentration in a river for linear diffu-
sivity, where different magnitudes of the Rouse number have been considered.
3.3 Parabolic Diffusivity (Vanoni-Distribution)
Finally, we would like to consider the case where the diffusivity is a parabolic
function of the vertical distance of the grain from the riverbed (z). i.e.
f(η) = η(1 − η) + fp. This concentration profile is more commonly known
as the Vanoni-distribution. Once again, we can rewrite (39) as a differential
equation in cs as a function of η so that
ws
1
h
∂cs
∂η
= −
1
h
∂
∂η
K(η)
1
h
∂cs
∂η
, (49)
where K(η) = κu∗h[η(1 − η) + fp].
Integrating up and applying boundary condition (i) yields
−wscs = κu∗[η(1 − η) + fp]
∂cs
∂η
. (50)
Again, we are left with a first order, linear ODE with solution
cs(η) = c0
η − 1
2
+ fp + 1
4
η − 1
2
− fp + 1
4
ws
2κu∗
√
fp+ 1
4
. (51)
25
26. Applying boundary condition (ii.a), and using the fact that d
dx
|f(x)| =
|f(x)|
f(x)
f (x) we find
c0 =
qe
wsfp
1
2
− fp +
1
4
2 1
2
+ fp + 1
4
1
2
− fp + 1
4
1− ws
2κu∗
√
fp+ 1
4
. (52)
Note that c0 is valid for all values of fp within the range specified earlier.
Hence, replacing η with z/h, the concentration distribution for parabolic
diffusivity is:
cs(z) =
qe
ws
1
2
+ fp + 1
4
1
2
− fp + 1
4
− ws
2κu∗
√
fp+ 1
4
z
h
− 1
2
+ fp + 1
4
z
h
− 1
2
− fp + 1
4
ws
2κu∗
√
fp+ 1
4
(53)
Figure 7: The vertical distribution of concentration in a river for parabolic
diffusivity, where different magnitudes of the Rouse number have been con-
sidered.
Figure 7 shows a plot of the (rescaled) concentration (c/c0) against the
(rescaled) vertical distance of sediment from the riverbed (z/h) to see how
26
27. the concentration is distributed. Again, we have considered different magni-
tudes of the Rouse number.
The parabolic distribution regime follows a similar pattern to the linear
model, however the concentration of sediment tails off toward the surface
of the river. This is due to the eddy diffusivity having less of an effect nearer
the boundaries.
3.4 Comparisons of Concentration Distributions
We have seen three separate models representing the distribution of sedi-
ment concentration in a river. We would now like to compare each regime
by varying other parameters such as the Rouse number or the total mass of
sediment being distributed.
We can define the total mass of sediment in suspension (A) as the the area
under the curve. This implies
A =
1
0
cs(η)
c0
dη. (54)
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (c/c
0
)
height(z/h)
constant (A=0.22)
linear (A=0.21)
parabolic (A=0.17)
(1)
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (c/c
0
)
height(z/h)
constant (P=0.32)
linear (P=0.34)
parabolic (P=0.29)
(2)
Figure 8: Two comparisons of the vertical distribution of sediment in a river
for constant, linear and parabolic diffusivity. In (1), the Rouse Number has
been fixed, and in (2), the mass of sediment distributed is fixed.
Figure 8 (1) gives an example of each distribution where the Rouse number
(P) has been fixed. When we fix the concentration to be 1 at the bottom of
27
28. the river, we notice that the distribution corresponding to constant diffusiv-
ity has the largest amount of suspended particles. However, the majority of
the grains are situated in the lower half of the river.
Figure 8 (2) gives an example of each distribution where the mass of sediment
in suspension is fixed. The linear and parabolic diffusivity models follow a
very similar pattern; the only noticeable difference occurs close to the sur-
face of river. Here, the parabolic regime suggests that the rate at which the
concentration of sediment reduces is a lot faster. This is due to the eddy
diffusivity being very small near the free surface for the parabolic case, on
the other hand, the linear case suggests the eddy diffusivity is largest at the
top of the river.
4 The Unsteady Concentration Equation
Now we would like to consider how the concentration distribution changes
over time. We begin with the unsteady concentration equation (33) that was
derived earlier, where the concentration (c) is a function of vertical distance
(z), and time (t), and the eddy diffusivity K = Kc = constant. The problem
we wish to solve is
∂c
∂t
= ws
∂c
∂z
+ Kc
∂2
c
∂z2
, (55)
subject to boundary conditions
(i) wsc + Kc
∂c
∂z
= 0 at z = h,
(ii) − Kc
∂c
∂z
= qe(θ) at z = 0.
We will also introduce a reference concentration cr,0 which defines the initial
concentration at time t = 0,
(iii) c = cr,0 at t = 0.
We will only consider a uniform initial concentration where cr,0 = constant.
4.1 Rescaling
We wish to rewrite (33) in terms of dimensionless variables to simplify the
problem. We introduce τ, ˆc and η, along with the Rouse Number (P) written
28
29. in terms of the eddy diffusivity Kc = κu∗hfc,
η =
z
h
, τ =
t
h2
, P =
wsh
Kc
, ˆc =
Kc
qeh
c, (56)
where we have taken fc = 1 for simplicity.
Using these quantities, we can rewrite (55) as a second order, partial differ-
ential equation, with separated boundary conditions, and only one unknown
constant (P). We have
∂ˆc
∂τ
− P
∂ˆc
∂η
=
∂2
ˆc
∂η2
, (57)
subject to boundary/initial conditions
(i) P ˆc = −∂ˆc
∂η
at η = 1,
(ii) ∂ˆc
∂η
= −1 at η = 0,
(iii) ˆc = ˆcr,0 at τ = 0,
where ˆcr,0 = Kccr,0
qeh
.
4.2 Solving the Unsteady Concentration Equation
Note that some of the techniques used in this section to solve (57) were taken
from Dorrell and Hogg (2012), and Pritchard (2006).
We begin by writing the solution to (57) as a linear combination of the
steady, time independent solution cs = cs(η), and a perturbed, time depen-
dent solution ˜c = ˜c(η, τ), such that
ˆc = cs(η) + ˜c(η, τ). (58)
We would like to determine whether the final solution to (57) yields a stable
(ˆc → cs as τ → ∞) or an unstable (ˆc −→ cs as τ → ∞) solution. Hence,
stability requires ˜c(η, τ) → 0 as τ → ∞.
4.2.1 Steady Solution
The steady concentration problem is set up such that
−P
∂cs
∂η
=
∂2
cs
∂η2
, (59)
subject to boundary conditions
29
30. (i) Pcs = −∂cs
∂η
at η = 1,
(ii) ∂cs
∂η
= −1 at η = 0.
By integrating up and applying the free surface boundary condition at η = 1,
we are left with a first order ODE which one can easily solve to obtain
cs(η) =
1
P
e−Pη
, (60)
where the factor 1
P
arises from the boundary condition at η = 0. We notice
that this is exactly the same as our constant diffusivity solution (44), just in
terms of ˆc, the Rouse number (P) and η, (as it should be).
4.2.2 Unsteady Solution
Substituting ˆc = cs + ˜c into (57) yields
∂˜c
∂τ
− P
∂˜c
∂η
− P
∂cs
∂η
=
∂2
˜c
∂η2
+
∂2
cs
∂η2
, (61)
subject to boundary/initial conditions
(i) P ˜c + Pcs = −∂˜c
∂η
− ∂cs
∂η
at η = 1,
(ii) ∂˜c
∂η
+ ∂cs
∂η
= −1 at η = 0,
(iii) ˜c + cs = ˆcr,0 at τ = 0.
We can simplify both the PDE and the boundary conditions using what we
know from (59). Hence, our problem is reduced to
∂˜c
∂τ
− P
∂˜c
∂η
=
∂2
˜c
∂η2
, (62)
subject to boundary/initial conditions
(i) P ˜c = −∂˜c
∂η
at η = 1,
(ii) ∂˜c
∂η
= 0 at η = 0,
(iii) ˜c = ˆcr,0 − cs at τ = 0.
30
31. Separation of Variables
We can solve this PDE by using a method known as “separation of vari-
ables”. We begin by rewriting ˜c as
˜c = F(η)T(τ), (63)
where F and T are functions of η and τ respectively. Substituting this into
(62) we find
T
T
=
F
F
+ P
F
F
= −λ, (64)
where λ is an integration constant to be determined.
We are now left with two ODE’s to solve:
dT
dτ
+ λT = 0
(∗)
,
d2
F
dη2
+ P
dF
dη
+ λF = 0
(∗∗)
.
We can solve (∗) easily, where, without loss of generality, we leave the inte-
gration constant out as it can be absorbed into the final solution later on.
So, the solution to (∗) is simply
T(τ) = e−λτ
. (65)
Now, if we rewrite (∗∗) in the form
−
d
dη
ePη dF
dη
= λePη
F, (66)
subject to boundary conditions
(i) PF(1) + dF
dη
(1) = 0,
(ii) dF
dη
(0) = 0,
then we have a regular Sturm-Liouville problem. Hence, by the Sturm-
Liouville theorem, we must have the following properties:
• The eigenvalues λ1, λ2, λ3, ... are real and can be ordered such that
λ1 < λ2 < λ3 < ... < λn < ... → ∞. (67)
• Corresponding to each eigenvalue λn there exists a unique eigenfunction
Fn(η) which has exactly n − 1 zeroes in (0, 1).
31
32. • The normalised eigenfunctions form an orthonormal basis, i.e.
1
0
Fn(η)Fm(η)ePη
dη = δmn (68)
Now, to solve (∗∗) we seek a solution of the form: F(η) = F0eβη
, where
F0 and β are constants. By substituting our ansatz into (∗∗) we obtain a
quadratic equation for β, given by
β2
+ Pβ + λ = 0, (69)
and solving for β yields
β =
−P ±
√
P2 − 4λ
2
. (70)
Now, we have three possible cases to consider:
Case 1: (P2
− 4λ > 0)
In this case, β = −P
2
± Γ, where Γ = 1
2
√
P2 − 4λ. Note that Γ is real
and strictly greater than 0. This leaves us with the general solution
F(η) = e−P η
2 [A1sinh(Γη) + B1cosh(Γη)] . (71)
Applying boundary condition (ii) gives B1 = 2Γ
P
A1, and hence applying
boundary condition (i) leaves us with the condition
tanh(Γ) = −
4PΓ
4Γ2 + P2
. (72)
Since P is just a constant, we can solve this for Γ. We see in Figure 9 that
there is only one solution to (72), which is Γ = 0, but since we assumed Γ is
strictly positive, there cannot be a solution to (∗∗) with P2
− 4λ > 0.
32
33. −6 −4 −2 0 2 4 6
−1.5
−1
−0.5
0
0.5
1
1.5
F()
F( ) = tanh ( )
F( ) = −2 / ( 2
+1)
Figure 9: A plot of the two functions given in (72), taking the Rouse Number
P = 2.
Case 2: (P2
− 4λ = 0)
Now, P2
− 4λ = 0 leaves us with β = −P
2
, and we have the general so-
lution
F(η) = A2e−P η
2 + B2ηe−P η
2 . (73)
Applying boundary condition (ii) gives B2 = P
2
A2, and hence applying
boundary condition (i) leaves us with
P(P + 4) = 0 =⇒ P = 0, P = −4.
But P is the Rouse Number which (by definition) must be positive. So there
are no solutions to (∗∗) for P2
− 4λ = 0.
Case 3: (P2
− 4λ < 0)
In this case, β = −P
2
± i∆, where ∆ = 1
2
√
4λ − P2. Note that ∆ is real
and strictly greater than 0. This leaves us with the general solution
F(η) = e−P η
2 [A3sin(∆η) + B3cos(∆η)] . (74)
33
34. Applying boundary condition (ii) gives B3 = 2∆
P
A3, and hence applying
boundary condition (i) leaves us with
tan(∆) =
4P∆
4∆2 − P2
. (75)
Since P is just a constant, we can solve this for ∆. In Figure 10 we notice
that there are infinitely many solutions to (75) for ∆ > 0.
These solutions make up all the eigenvalues (λn) and corresponding eigen-
functions (Fn) in our Sturm-Lioville problem. Our eigenvalues are found by
rearranging the definition of ∆ so that
λn = ∆2
n +
P2
4
, (76)
where, by Sturm-Liouville’s theorem, each ∆n corresponds to a unique solu-
tion of (75) which are ordered such that
0 < ∆1 < ∆2 < ∆3 < ... < ∆n < ... → ∞. (77)
0 5 10 15
−1
0
1
2
3
4
5
6
F()
F( ) = tan ( )
F( ) = 2 / ( 2
−1)
Figure 10: A plot of the two functions given in (75), taking the Rouse Number
P = 2.
34
35. We will write each eigenfunction as
Fn(η) = e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) , (78)
where the constant A3 has been left out as it can be absorbed into the final
solution.
Our final unsteady solution can be represented by taking a superposition
of each eigenfunction and substituting them (63). We obtain an infinite sum
given by
˜c(η, τ) =
∞
n=0
Cn Tn(τ)Fn(η), (79)
where the constants drawn out from the solutions to (∗) and (∗∗) are now
represented as Cn, which varies depending on the eigenvalue λn.
All that’s left to do is find the Cn’s. Applying boundary condition (iii)
to (79) gives
ˆcr,0 − cs =
∞
n=0
Cn Tn(0)Fn(η), (80)
=
∞
n=0
Cn Fn(η). (81)
Now, multiplying each side by Fm(η)ePη
and integrating over the interval
(0, 1) yields
1
0
(ˆcr,0 − cs)Fm(η)ePη
dη =
∞
n=0
Cn
1
0
Fn(η)Fm(η)ePη
dη
I
. (82)
By Sturm-Liouville’s theorem, we can deduce that NI = δmn, where N is our
normalisation constant. To find our normalisation constant we must evaluate
the integral for the cases n = m and n = m.
35
36. If n = m, we have
I =
1
0
Fn(η)Fm(η)ePη
dη,
=
1
0
sin(∆nη) +
2∆n
P
cos(∆nη) sin(∆mη) +
2∆m
P
cos(∆mη) dη,
= 0.
Note that solving this integral is by no means straightforward, but one can
easily show this integral vanishes using Maple along with the relation found
in (75).
If n = m, we have
I =
1
0
F2
m(η)ePη
dη,
=
1
0
sin(∆mη) +
2∆m
P
cos(∆mη)
2
dη,
=
1
0
sin2
(∆mη) +
4∆2
m
P2
cos2
(∆mη) +
4∆m
P
cos(∆mη)sin(∆mη) dη,
=
1
0
2∆2
m
P2
−
1
2
cos(2∆mη) +
2∆m
P
sin(2∆mη) +
1
2
+
2∆2
m
P2
dη,
=
∆m
P2
−
1
4∆m
sin(2∆mη) −
1
P
cos(2∆mη) + η
1
2
+
2∆2
m
P2
1
0
,
=
1
2
+
2∆2
m
P2
+
1
P
+
∆m
P2
−
1
4∆m
sin(2∆m) −
1
P
cos(2∆m). (83)
Now, using (75) we can find similar relations for sin(∆m) and cos(∆m). We
find that
sin(∆m) =
tan2
(∆m)
1 + tan2
(∆m)
=
4P∆m
(4∆2
m + P2)
, (84)
36
37. cos(∆m) =
1
1 + tan2
(∆m)
=
(4∆2
m − P2
)
(4∆2
m + P2)
. (85)
Substituting these relations into (83) and simplifying, our normalisation con-
stant N is found to be
N = I−1
=
1
2
+
2∆2
m
P2
+
8∆2
m
P(16∆2
m + P2)
−1
. (86)
Plugging this back into (82), we deduce that
Cn = N
1
0
(ˆcr,0 − cs)Fn(η)ePη
dη. (87)
To solve this integral we will split it up into two parts (denoted I1 and I2)
and evaluate each part separately. We have
Cn = N
1
0
ˆcr,0Fn(η)ePη
dη
I1
− N
1
0
csFn(η)ePη
dη
I2
(88)
So, using the fact that
1
0
eAx
sin(Bx) dx =
AeA
sin(B) − BeA
cos(B) + B
A2 + B2
, (89)
1
0
eAx
cos(Bx) dx =
BeA
sin(B) + AeA
cos(B) − A
A2 + B2
, (90)
we find
I1 = ˆcr,0
1
0
e
P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) dη, (91)
=
8ˆcr,0∆ne
P
2
4∆2
n + P2
. (92)
Note that we replaced sin(∆n) and cos(∆n) with the relations found in (84)
and (85).
37
38. Since we found cs in (60), we can evaluate I2 in a similar manner yielding
I2 =
1
P
1
0
e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) dη, (93)
=
8∆n
P(4∆2
n + P2)
. (94)
Hence substituting the solutions to I1 and I2 into (88) we have
Cn = N(I1 − I2), (95)
=
8ˆcr,0∆ne
P
2
4∆2
n+P2 − 8∆n
P(4∆2
n+P2)
1
2
+ 2∆2
n
P2 + 8∆2
n
P(16∆2
n+P2)
. (96)
Finally, since we have found the Tn, Fn and Cn’s, we can substitute them
into (79) and we are left with a solution to the perturbed, time dependent
part of our solution given by
˜c(η, τ) =
∞
n=0
Cn e−λnτ
e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) . (97)
An important observation in this expression is the fact that each contribution
from the sum decays as n increase. This is due to every λn being positive
and tending to infinity as n tends to infinity (67). Using this observation, we
deduce that as time increases ˜c(η, τ) tends to 0.
4.2.3 Final Solution
The final solution to (57) is expressed as the sum of the steady and unsteady
solutions, this turns out to be
ˆc(η, τ) =
1
P
e−Pη
+
∞
n=0
Cn e−λnτ
e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) , (98)
where the Cn’s are defined in (96) and depend on the initial concentration.
Since the perturbed part of our solution tends to 0 over time, the concen-
tration (ˆc) must approach the steady state (cs) for any initial concentration
(ˆcr,0). This tells us that our steady solution (60) must be stable.
38
39. 4.3 Asymptotic solution for P nπ
We would like to construct a method to find all positive eigenvalues from our
equation in (75). Firstly, noting that the eigenvalues are ordered as defined
in (77), we can see from Figure 10 that as n increases, the ∆n’s approach
nπ. Now, for small n, we cannot find the eigenvalues to a satisfactory degree
of accuracy. Therefore, a MATLAB solver has been used to find the first 40
eigenvalues. This can be achieved by introducing the function
f(x) := tan(x) −
4Px
4x2 − P2
, (99)
and looking for the points at which f(x) = 0, i.e. where the curve crosses
through the x-axis.
To find the ∆n’s for larger n, we use the fact that the eigenvalue approach
nπ from below as n increases. So, we seek a solution to (75) of the form
∆n = nπ + n , (100)
where n → 0+ as n → ∞.
Since we are seeking solutions for large n, we can make the assumption
nπ n. Hence, since n takes integer values only, we can use the asymp-
totic equivalence
tan(nπ + n) ∼ n, (101)
Furthermore, if we choose our Rouse Number (P) such that nπ P, we
have that 4∆2
n − P2
∼ 4∆2
n and hence
4P∆n
4∆2
n − P2
∼
P
∆n
. (102)
So, replacing (75) with the asymptotic expressions found in (101) and (102)
we obtain
n ∼
P
∆n
. (103)
But we want to take the limit as n → ∞, and n → 0+, which implies
∆n ∼ nπ. This means (103) becomes
n ∼
P
nπ
. (104)
Hence, providing P nπ, we have an asymptotic expression for our eigen-
values ∆n, given by
∆∗
n = nπ +
P
nπ
+ ... (105)
39
40. Furthermore, using the relation in (76), and the fact that ∆2
n ∼ n2
π2
for
P nπ we have
λ∗
n =
P2
4
+ n2
π2
+ ... (106)
4.3.1 Error
We can find the error in the asymptotic approximation of each eigenvalue by
simply calculating the difference between the asymptotic solution (∆∗
n, λ∗
n)
and the actual values found in MATLAB (∆n, λn). Hence, the two error
functions are defined by
E∆n = ∆∗
n − ∆n, (107)
Eλn = λ∗
n − λn. (108)
Figure 12 shows a plot of these two functions for n ranging between 1 to 40.
5 10 15 20 25 30 35 40
0
0.5
1
1.5
2
2.5
3
x 10
−3
n
∆∗
n−∆n
5 10 15 20 25 30 35 40
0
0.005
0.01
0.015
0.02
0.025
0.03
n
λ∗
n−λn
Figure 11: Left: A graph showing the difference in the asymptotic approx-
imation of ∆n and the actual value of ∆n, defined as ∆∗
n − ∆n. Right: A
graph showing the difference in the asymptotic approximation of λn and the
actual value of λn, defined as λ∗
n −λn. Note that the Rouse number has been
taken as P = 2.
We would like to determine the value of n to which the asymptotic solution
provides a result with a sufficiently small error. As seen in Figure 12 (left
graph), the decay rate of E∆n is very high as the magnitude of the error
approaches 0 very quickly. For example, the approximation to the 6th
eigen-
value ∆6 has an error of only E∆n = 0.001. However, the decay rate of Eλn
is not as high (right graph). We do not reach the same error (Eλn = 0.001)
until we hit the 40th
eigenvalue (λ40).
40
41. We deduce that for the solution to have a sufficient degree of accuracy, we
can only use the asymptotic approximation for n > 40, providing the Rouse
number does not exceed 2.
4.4 Accuracy
In this section, we aim to show how the accuracy of our solution changes
depending on the number of eigenvalues considered by plotting the concen-
tration (c) against the vertical distance (z). Now, since we rescaled both of
these parameters to simplify our initial problem, we must be careful when la-
belling the axes. The vertical distance will be labelled as η = z/h so it ranges
between 0 and 1. Similarly, the concentration will be labelled as P ˆc = c/c0
(once again this has been normalised so the concentration ranges between 0
and 1.)
To display the accuracy of our solution, we will plot the initial concentra-
tion (τ = 0) and vary the number of eigenvalues included in the summation.
From how we set up our solution, we have
ˆc(η, 0) = cs(η) + ˜c(η, 0) = ˆcr,0, (109)
where cs was found in (60), ˜c was found in (97), and ˆcr,0 was defined as our
initial reference concentration.
Since we defined the initial concentration to be constant, we expect to see
a vertical line when we plot (109) against η. Figure 12 shows four separate
cases:
Case 1: (First 5 eigenvalues)
ˆc(η, 0) = cs(η) +
5
n=0
Cn e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) (110)
In this case, we can clearly see that the first 5 eigenvalues do not provide
us with an accurate representation of the specified initial concentration. A
curved line which slowly oscillates around P ˆc = 0.4 is produced. This oscil-
latory nature of the solution comes from the trigonometric functions in the
solution to the unsteady concentration equation.
41
42. Case 2: (First 10 eigenvalues)
ˆc(η, 0) = cs(η) +
10
n=0
Cn e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) (111)
If we take the first 10 eigenvalues, the line representing the initial concentra-
tion straightens out a lot more. The vertical line oscillates more rapidly with
a smaller amplitude towards the centre. However, this is still not a desirable
amount of accuracy.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (Pˆc)
height(z/h)
(1)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (Pˆc)
height(z/h)
(2)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (Pˆc)
height(z/h)
(3)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (Pˆc)
height(z/h)
(4)
Figure 12: Four separate examples displaying the accuracy of the solution
for t = 0, where the number of eigenvalues included in the summation varies.
The initial concentration is set as P ˆcr,0 = 0.4 and the Rouse number P = 2.
Case 3: (First 40 eigenvalues)
ˆc(η, 0) = cs(η) +
40
n=0
Cn e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) (112)
After taking the first 40 eigenvalues, our line is essentially straight, however
we lose accuracy as we approach the top of the river and the riverbed. The
solution essentially represents a Fourier series, so as we increase the number
of terms used in our summation, the oscillations shorten in wavelength and
stay closer to the actual solution.
Case 4: (First 1000 eigenvalues)
ˆc(η, 0) = cs(η) +
1000
n=0
Cn e−P η
2 sin(∆nη) +
2∆n
P
cos(∆nη) (113)
42
43. If we include 960 more eigenvalues, using the asymptotic expressions found
in (105) and (106), we achieve a perfect vertical line. Note that the asymp-
totic expression can only obtain an accurate representation providing P is
sufficiently less than nπ, as this was the condition in which it was derived.
4.5 Comparison of Results
In this section, we compare the results by varying different parameters. By
changing the Rouse number or the initial concentration the solution can
change by a surprising amount. we have plotted several graphs in MATLAB
to compare the concentration distribution over time.
4.5.1 Physical Interpretation
The solution found in (98) determines how the concentration distribution
evolves over time. We set it up to begin with a specific mass of sediment
distributed evenly throughout the river.
Figure 13: A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 0.37/P, and Rouse number P = 2.
As we move with the flow, a combination of the turbulent fluctuations and
the settling of sediment under gravity allows the sediment distribution to
change over time. We discovered that irrespective of the initial mass of the
43
44. sediment in suspension, providing the initial concentration distribution is
uniform, the distribution will evolve towards the steady state (cs).
Figure 13 represents a situation where the initial (normalised) concentra-
tion is P ˆcr,0 = 0.37. This initial concentration has been chosen such that the
concentration of sediment at the midpoint of the river remains fixed (provid-
ing we take the Rouse number to be 2). After small time, the distribution is
mainly affected at the surface and the base of the river. As time increases,
the concentration in both halves of the river tends to the steady state as we
would expect. The mass of sediment lost in the in the upper half of the river
is due to sediment settling, which also adds to the mass gained in the lower
half of the river.
Figure 14: A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 0.01/P, and Rouse number P = 2.
Another situation is shown in Figure 14 where the initial (normalised) con-
centration is P ˆcr,0 = 0.01. In this case there is a small amount of sediment
in suspension at time 0. To allow the initial concentration to approach the
steady state, we can only assume that grains are eroded from the river bed.
This implies that in the time frame plotted, the boundary condition at the
river bed must adjust to allow for incipient motion. As shown in the graph,
the mass of sediment in suspension continues to increase until it reaches the
steady solution, at which time the sediment settling and shear velocity are
in equilibrium.
44
45. If we consider the contrary, where we begin with a very large amount of
grains in suspension at time 0, the settling of sediment dominates the erosive
upward flux. Figure 15 shows an example of the initial concentration having
a much larger mass than the steady state. After short time, the sediment set-
tles quickly allowing the mass of sediment close to the river bed to increase,
and the mass of sediment at the surface to decrease quickly. But as time
progresses, the distribution of concentration flattens out and steadily evolves
to the shape of our steady solution and the mass of sediment in suspension
decreases. This occurs because the grains settle onto the riverbed and remain
as bed load.
Figure 15: A graph showing how the solution (98) evolves over time with
initial concentration ˆc = 1.8/P, and Rouse number P = 2.
We have looked at what effects occur when we change the initial reference
concentration. However, we have not considered what significance the Rouse
number has on our solution. We will now go on to look at how the distribution
of concentration changes when we vary the Rouse number.
4.5.2 Time Dependence of the Rouse number
The Rouse number (P) is a ratio between the settling velocity and the shear
velocity of the individual grains. For this reason, the magnitude of the Rouse
45
46. number could have a large effect on how the concentration changes over time,
and more importantly, the rate at which the steady solution is approached.
We will now consider plotting the (normalised) concentration against the
(normalised) height, and try to determine the significance of the Rouse num-
ber with respect to time.
The rate at which the solution (ˆc) tends to the steady state can be de-
termined by defining an error function. The error function finds the error of
the final solution from the steady state after a specific time, and is defined
as
E(τ) =
1
0
|c(η, τ) − cs(η)|
|cs(η)|
dη. (114)
If we set the maximum error as Emax = 0.01, we would like to find the value
of τ = τmin at which E(τ) = Emax. We will then look at how τmin changes
for different magnitudes of the Rouse number P.
Four different magnitudes of P have been considered in Figure 16 showing
the change in concentration distribution over time with an initial normalised
concentration of P ˆcr,0 = 0.5. Note that we have only included the first 15
eigenvalues for these plots, hence the slight oscillations in the initial concen-
tration. We have the following cases:
(1) P = 0.5:
In this case, the Rouse number is quite small, hence the mass of sediment in
suspension at the steady state is larger due to the ratio of settling velocity
over shear velocity being less than 1. Using MATLAB, we can compute the
integral in (114) and hence, by setting E(τ) = Emax = 0.01 we can find the
value of τmin. We find that
τmin ≈ 10.36. (115)
(2) P = 1:
If we increase the Rouse number, the mass of sediment in suspension at the
steady state reduces. Again, we can compute the integral in (114) using
MATLAB to find the error at time τ. By setting E(τ) = Emax = 0.01 we
have
τmin ≈ 3.12. (116)
46
47. (3) P = 2:
If we take the Rouse number as 2, and after computing the integral in (114)
set E(τ) = Emax = 0.01 we have
τmin ≈ 1.61. (117)
(4) P = 3:
Finally, we consider the case where P = 3. As expected the value of τmin
decreases again. Solving (114) in MATLAB and setting E(τ) = Emax = 0.01
yields
τmin ≈ 1.61. (118)
(1) (2)
(3) (4)
Figure 16: (1), (2), (3) and (4) show the time evolution of concentration for
P = 0.5, 1, 2, 3 respectively. The initial concentration is fixed at P ˆcr,0 = 0.5
in all cases.
47
48. Hence, since τ is just a non-dimensional representation of time t, we can
deduce that for a higher the Rouse number, the rate at which we approach
the steady solution decreases.
4.5.3 Accuracy Dependence on the Rouse Number
Another interesting observation from the previous five figures is how the ac-
curacy depends on the Rouse number. We kept the number of eigenvalues in
each solution fixed to 15, however as the Rouse number increases we notice
the size of the oscillations in the initial concentration also increase. The rea-
son for this can be seen in our asymptotic solution to the eigenvalues (105)
and (106). As we increase P the size of n increases. Hence, the higher the
Rouse number, the larger the error in our solution.
A better way of understanding the relationship between the Rouse num-
ber and the decay rate of the unsteady solution is to compare the value of
the lowest possible eigenvalue to different magnitudes of the Rouse number.
Now, from our solution we know the unsteady perturbation decays at the
rate e−λnτ
. This implies the size of the lowest eigenvalue will give us an
approximate guess as to how fast the unsteady concentration tends to the
steady solution.
(1) (2)
Figure 17: (1) A plot showing the relationship between the Rouse number
and the first (lowest) eigenvalue ∆1. (2) A plot showing the relationship
between the Rouse number and the first (lowest) eigenvalue λ1.
From Figure 17 we can see that the decay rate of the unsteady solution
increases with the magnitude of the Rouse number. The relation between
∆n and λn can be found in (76).
48
49. 5 Hindered Settling
We began by taking the settling velocity (ws) of each particle to be constant,
as calculated in its definition. However, in reality the settling velocity will
vary depending on the concentration of the flow. For this reason, we intro-
duce a concept known as hindered settling which allows the settling velocity
to change depending on the concentration of the river.
One of the main reasons hindered settling arises is due the “return flow”
effect. Falling particles create an upward directed return flow which affects
the fall velocity of other particles in the near vicinity, decreasing the overall
effective settling velocity (Dankers 2006). Another highly contributing fac-
tor is that at higher concentrations, the effective viscosity increases. Each
individual particle falls in the remainder of the suspension with increased vis-
cosity, thus decreasing the effective settling velocity of all particles (Dankers
2006). Other processes, e.g. particle collisions, are known to affect hindered
settling, however the contribution is minimal so we will not consider them.
Hindered settling tends to occur for higher levels of concentration, since
there is a higher risk of affecting a grains settling velocity when the sediment
in suspension is more condensed. We will identify a function to represent the
settling velocity which changes with the concentration.
5.1 Settling Velocity of Particles at High Concentra-
tions
Experiments have suggested that the settling velocities of particles is lower
at high concentrations, a factor commonly used to represent this (Richardson
and Zaki 1954) is:
ws(c) = ws,0(1 − c)n
, (119)
where ws,0 represents the terminal settling velocity of each particle, (the
quantity used in previous sections for the settling velocity), and the parame-
ter n is a constant which depends on the particle Reynolds number. The con-
stant n was experimentally determined to lie in the range (2.4, 4.65), where
the Reynolds number decreases as n increases (Baldock, Tomkins, Nielsen
and Hughes 2003).
We can now rewrite our unsteady concentration equation (33) in terms of
49
50. the new hindered settling velocity, given by
∂c
∂t
− ws,0
∂
∂z
(c(1 − c)n
) =
∂
∂z
K
∂c
∂z
, (120)
subject to boundary conditions
(i) ws,0c(1 − c)n
+ K ∂c
∂z
= 0 at z = h,
(ii) − K ∂c
∂z
= qe(θ) at z = 0.
As before, we introduce a reference concentration cr,0 which defines the initial
concentration distribution. So the initial condition is
(iii) c = ch,0 at t = 0,
where we only consider a uniform initial concentration where cr,0 = constant.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.05
0.1
0.15
0.2
0.25
c
φu=c(1−c)n
Figure 18: A comparison between the upward flux (φu) and concentration
(c), where the parameter n varies between 1 and 6. The blue lines correspond
to the typical values for n (determined empirically).
50
51. Since the settling velocity changes with concentration, we can plot the up-
ward flux (φu = c(1 − c)n
) against the concentration (c). Figure 18 shows
this plot for different values of n, we notice that φu vanishes at c = 0 and
c = 1. We would clearly have no upward flux if the concentration was 0 since
the particles would have instantly diffused into the flow, hence there would
be no grains in suspension at any time. Similarly, the upward flux is 0 when
the concentration is 1 as this is the maximum concentration for non-cohesive
sediment; any higher and the particles would begin to overlap.
As our concentration cannot exceed a specified maximum, a further con-
straint has been created such that
0 < c < cmax, (121)
where cmax denotes the maximum concentration for non-cohesive sediment.
Another important observation is that the upward flux hits a maximum
as the concentration changes between 0 and 1. Figure 18 shows how the
maximum reduces as we increase the parameter n. We can easily find the
concentration at which the upward flux is maximum by simply differentiating
the upward flux with respect to c and setting it to 0. We have that
∂φu
∂c
= (1 − c)n−1
(1 − c(n + 1)). (122)
Setting this to 0 tells us the upward flux is maximum at
c =
1
n + 1
. (123)
Hence, we have created another constraint on our differential equation if we
wish to pick up a solution. We must have
0 < φu < φu,max, (124)
where the upward flux φu = c(1−c)n
, and the maximum upward flux φu,max =
1
n+1
1 − 1
n+1
n
5.2 Rescaling
Once again, we would like to rewrite (120) in terms of dimensionless variables
to simplify the problem. We will use the same variables used before in (56)
51
52. with an extra parameter cmax which arises due to our new settling velocity.
We make the transformations
η =
z
h
, τ =
t
h2
, P =
wsh
Kc
, ˆc =
Kc
qeh
c, cmax =
Kc
qeh
. (125)
Using these quantities, we can rewrite (120) as a non-linear, second order par-
tial differential equation, with separated boundary conditions, and only two
unknown constants (P and cmax). Our rescaled hindered equation becomes
∂ˆc
∂τ
− P
∂
∂η
ˆc 1 −
ˆc
cmax
n
=
∂2
ˆc
∂η2
, (126)
subject to boundary/initial conditions
(i) P ˆc 1 − ˆc
cmax
n
= −∂ˆc
∂η
at η = 1,
(ii) ∂ˆc
∂η
= −1 at η = 0,
(iii) ˆc = ˆch,0 at τ = 0,
where ˆch,0 = cmaxch,0.
5.3 Solving the Hindered Concentration Equation
Since our differential equation is non-linear, we expect to pick up two solu-
tions in the interval 0 < c < cmax. This can be seen in Figure 18, providing
the upward flux remains in the interval 0 ≤ φu ≤ φu,max. We can write the
solutions to (126) as a linear combination of the steady, time independent
solution cs = cs(η), and a perturbed, time dependent solution ˜c = ˜c(η, τ).
(i.e ˆc = cs + ˜c). We would like to determine whether the final solutions to
(126) yields a stable (ˆc → cs as τ → ∞) or an unstable solution (ˆc −→ cs as
τ → ∞). Hence, stability requires ˜c(η, τ) → 0 as τ → ∞.
We will consider different values of the parameter n, and compare the solu-
tions obtained. Let us begin with the most simplest case (n = 1).
5.3.1 Steady Solution (n = 1)
The steady hindered concentration problem for n = 1 is set up such that
−P
∂
∂η
cs 1 −
cs
cmax
=
∂2
cs
∂η2
, (127)
Subject to boundary conditions
52
53. (i) Pcs 1 − cs
cmax
= −∂cs
∂η
at η = 1,
(ii) ∂cs
∂η
= −1 at η = 0.
By integrating up and applying the boundary condition at η = 1, we are left
with a non-linear, first order ODE given by
−Pcs 1 −
cs
cmax
=
∂cs
∂η
. (128)
Separation of variables yields
−P dη =
dcs
cs
+
dcs
cmax − cs
, (129)
where partial fractions have been used to find the two terms on the right-
hand side. Now, solving these integrals gives rise to an integration constant,
and after some careful rearrangement we find the solution to be
cs =
cmax
1 + BePη
, (130)
where B is a constant to be determined.
To find the particular solution, we can apply the boundary condition (ii)
at η = 0, this leaves us with two possible values for B written as
B± =
Pcmax
2
− 1 ± Pcmax
Pcmax
4
− 1 . (131)
Hence, we find that the two steady solutions to (127) are
cs,1 =
cmax
1 + B+ePη
, (132)
cs,2 =
cmax
1 + B−ePη
. (133)
For these steady solutions to exist, they must be real. This means we must
choose the Rouse number (P) such that the constant B given in (131) is real.
So we have the condition
P ≥
4
cmax
. (134)
Figure 19 shows a plot of the two steady solutions to our hindered concen-
tration equation for three different values of the Rouse number (P = 1, 2, 3).
53
54. We notice that one of our steady solutions (cs,1) arises for low concentrations,
and happens to match the solution we picked up when solving the steady con-
centration problem (59). However, due to hindered settling, another steady
solution has been introduced. This occurs at much higher concentrations and
corresponds to the solution cs,2 found earlier.
Figure 19: A plot of the two steady solutions, cs,1 and cs,2, with three dif-
ferent magnitudes of the Rouse number. Note that the left solutions (low
concentration) correspond to cs,1 and the right solutions (high concentration)
correspond to cs,2.
As P decreases, the two steady solutions move closer together. However, we
set a lower limit for P in (134), and if P drops below this limit, we do not
obtain any real solutions. If P = 4/cmax, the two steady solutions collide
and create one solution (repeated root).
We wish to determine the stability of the two steady concentration solu-
tions, (132) and (133). To do this, we must seek a solution to the perturbed
concentration (˜c) and check whether it tends to 0 as time tends to infinity.
54
55. 5.3.2 Unsteady Solution (n = 1)
By substituting ˆc = cs + ˜c into (126), we have
∂˜c
∂τ
− P
∂˜c
∂η
− P
∂cs
∂η
+
2P
cmax
˜c
∂˜c
∂η
+ cs
∂cs
∂η
+ cs
∂˜c
∂η
+ ˜c
∂cs
∂η
=
∂2
˜c
∂η2
+
∂2
cs
∂η2
,
(135)
subject to boundary/initial conditions
(i) −P ˜c − Pcs + P
cmax
c2
s + 2P
cmax
cs˜c + P
cmax
˜c2
= ∂˜c
∂η
+ ∂cs
∂η
at η = 1,
(ii) ∂˜c
∂η
+ ∂cs
∂η
= −1 at η = 0,
(iii) ˜c + cs = ˆch,0 at τ = 0.
We can simplify both the PDE and the boundary conditions using what we
know from (127). Hence, the problem is now reduced to
∂˜c
∂τ
− P
∂˜c
∂η
+
2P
cmax
˜c
∂˜c
∂η
+ cs
∂˜c
∂η
+ ˜c
∂cs
∂η
=
∂2
˜c
∂η2
, (136)
subject to boundary/initial conditions
(i) −P ˜c + P
cmax
(2cs˜c + ˜c2
) = ∂˜c
∂η
at η = 1,
(ii) ∂˜c
∂η
= 0 at η = 0,
(iii) ˜c = ˆch,0 − cs at τ = 0.
The first problem which arises is the fact that our partial differential equa-
tion in ˜c is non-linear. This is due to the ˜c∂˜c
∂η
term in the PDE, and the ˜c2
term in boundary condition (i). These non-linear terms make it impossible
to solve the differential equation analytically, therefore a MATLAB solver
called pdepe has been used to produce graphical results.
The solver pdepe solves initial-boundary value problems for systems of parabolic
and elliptic PDEs in the one space variable x and time t. The ordinary dif-
ferential equations resulting from discretisation in space are integrated to
obtain approximate solutions at times specified (MATLAB Version 8.1).
5.4 Comparison of Results
In a similar fashion to before, we would like to compare the results of our so-
lutions by varying specific parameters. By changing the initial concentration
55
56. distribution or the Rouse number, we will see how the solutions behave over
time. To simplify the solution, we will normalise the vertical distance (z)
and rescaled concentration (ˆc) by dividing them by η and cmax respectively.
The results will provide us with a better understanding of the stability of the
two solutions.
5.4.1 Physical Interpretation
We will begin by briefly describing what the hindered settling equation phys-
ically represents. We set up a differential equation which describes how the
distribution of sediment in suspension varies as we move with the flow. We
found that two steady states exist: one for low concentrations (cs,1), and one
for high concentrations (cs,2). At time 0, we set up the suspended sediment
such that it is distributed uniformly throughout the river. However, we take
into account that the initial mass of sediment in suspension is free to change.
We would like to discover whether a specific initial concentration (ch,0) will
evolve into a steady state over time, and if not how does it evolve?
Figure 20: A plot of the two steady solutions, cs,1 and cs,2, where the Rouse
number is set as P = 2, and the initial concentration is ˆch,0/cmax = 0.5.
We begin with fixing the Rouse number (P) to be 2, and beginning the initial
concentration distribution as ˆch,0/cmax = 0.5. From (123), this happens to
56
57. be the concentration profile at which the upward flux (φu) is a maximum.
We see the described set up in Figure 20. As time increases, more and more
suspended sediment settles onto the riverbed and we move closer to the steady
state for small concentrations. This suggests cs,1 is a stable solution.
Figure 21: A plot of the steady solution cs,1 with the Rouse number taken
as P = 2. The initial concentration is set as ˆch,0/cmax = 0.01.
Figure 21 shows how the solution behaves when we set the initial concentra-
tion as ˆch,0/cmax = 0.01. Once again, over time the concentration distribution
tends to the steady concentration state cs,1. This confirms that we pick up
a stable solution for low concentrations. This solution yields the same result
that was seen when we took the settling velocity to be constant.
Now, to understand more about the behaviour of the solution at higher
concentrations, we start the initial concentration close to the steady solu-
tion cs,2. Figure 22 shows an example for ˆch,0/cmax starting at 0.79. As time
increases, the solution begins to tend towards the steady state cs,2, however
if we follow the behaviour for long enough, we begin to move away from the
steady state. This is an interesting observation as it suggests that the steady
solution for higher concentrations could be unstable.
57
58. Figure 22: A plot of the steady solution cs,2 with the Rouse number taken
as P = 2. (The solution cs,1 is not shown). The initial concentration is set
as ˆch,0/cmax = 0.79.
The steady solution cs,2 seems to have similar behaviour to an unstable saddle,
this is because we initially tend towards the state but get repelled away
before hitting the solution. If the initial concentration is low enough, the
stable behaviour of cs,1 allows the solution to evolve into a steady state.
An example of this is shown in Figure 20 where the evolution consists of
sediment in suspension being deposited onto the bed. However, if we choose
a sufficiently high initial concentration, we have a situation in which the
solution will not approach the stable state cs,1. Instead, the erosion flux
seems to dominate the downward flux, allowing more sediment to be picked
up off the bed into suspension. We will define the point at which the initial
concentration no longer tends to a steady state as
ˆch,0
cmax
= Clim(P). (137)
The stable concentration limit Clim will vary depending on the Rouse number,
since we know our steady solution for high concentrations changes depend-
ing on how we choose P. Using an iterative method in MATLAB, we can
determine the value of Clim for a given Rouse number. Figure 23 shows this
relationship.
58
59. Figure 23: A plot showing a relationship between Clim and P, where we have
taken cmax = 6.
For smaller magnitudes of the Rouse number, the size of Clim decreases
rapidly. Furthermore, if we choose P = 4/cmax, which is defined in (134)
as the lower limit for P for which we obtain real solutions, then Clim = 0.
This is exactly as we expected as if P < 4/cmax we cannot find an initial
concentration such that we reach a stable solution. On the other hand, as
we increase the Rouse number, Clim seems to plateau and approach an upper
limit given by
C∗
lim = lim
P→∞
Clim(P) (138)
Given the maximum concentration cmax, and choosing P to be sufficiently
large, one can determine an accurate approximation of this limit.
We would now like to investigate what happens if we do not approach a
steady solution over time. From earlier calculations, if we choose the initial
concentration such that
ˆch,0
cmax
> Clim(P), (139)
then we do not approach the steady state cs,1. This situation is shown in
Figure 22.
59
60. As we move away from the steady solutions, the concentration of sediment
in suspension increases. However, the concentration can never exceed cmax
since this is the maximum concentration which is physically possible. For
this reason, we must enforce that the (rescaled) up flux (ˆφu) vanishes for
ˆc ≥ cmax. So the up flux becomes
ˆφu = ˆc 1 −
ˆc
cmax
H 1 −
ˆc
cmax
, (140)
where H(x) =
0 if x < 0
1 if x > 1.
is the Heaviside step function.
Hence, we can rewrite the hindered settling problem (126) for c ≥ cmax
as
∂ˆc
∂τ
=
∂2
ˆc
∂η2
, (141)
subject to boundary/initial conditions
(i) ∂ˆc
∂η
= 0 at η = 1,
(ii) ∂ˆc
∂η
= 0 at η = 0,
(iii) ˆc = ˆch,0 at τ = 0,
where the boundary condition at the free surface has altered due to φu = 0
for c ≥ cmax, and furthermore, the boundary condition at the bed has been
changed so the erosion flux is also 0. This is because no more sediment can
be eroded into suspension if the concentration at the bed is already at a
maximum.
The result is simply the well known heat equation. At the point when
ˆc = cmax, no more mass of sediment can be added to the concentration
distribution. So as a result of the heat equation, we would expect the line
representing the distribution of sediment at this point to flatten out. How-
ever, this is beyond the scope of this project so we will not pursue this case
any further.
5.5 Linearisation (n = 1)
One possible method of analytically finding the stability of our two solu-
tions (cs,1 and cs,2) would be to linearise around each solution and deter-
mine whether the eigenvalue problem yields positive or negative eigenvalues.
60
61. These eigenvalues should give us a good understanding of the stability of
each steady state.
Going back the rescaled hindered problem in (126), and taking n = 1, we
have
∂ˆc
∂τ
− P
∂
∂η
ˆc 1 −
ˆc
cmax
=
∂2
ˆc
∂η2
, (142)
Subject to boundary/initial conditions
(i) P ˆc 1 − ˆc
cmax
= −∂ˆc
∂η
at η = 1,
(ii) ∂ˆc
∂η
= −1 at η = 0,
(iii) ˆc = ˆch,0 at τ = 0.
This differential equation leaves us with two steady (time independent) states,
denoted cs. To determine the stability of these states, we will use a technique
known as linearisation. Linearisation is used to find a linear approximation
of a function at a given point, hence, we can linearise around our steady
states to yield a linear approximation of (126). We begin by taking a small
perturbation around cs, given by
ˆc = cs(η) + ˜c(η, τ), (143)
where epsilon is defined such that 0.
Now, substituting (143) into our differential equation, and only considering
the terms of O( ), we have
∂˜c
∂τ
− P
∂
∂η
˜c 1 −
2cs
cmax
=
∂2
˜c
∂η2
, (144)
subject to boundary/initial conditions:
(i) P ˜c 1 − 2cs
cmax
= −∂ˆc
∂η
at η = 1,
(ii) ∂˜c
∂η
= 0 at η = 0,
(iii) ˜c = ˜ch,0 at τ = 0.
We are now left with a linear PDE which can be solved using the method
called separation of variables (as was used to solve the unsteady concentration
equation). We begin by rewriting ˜c as
˜c = F(η)T(τ), (145)
61
62. where F and T are functions of η and τ respectively.
Substituting this into (144) yields
T
T
=
F
F
+ P
F 1 − 2cs
cmax
F
= −λ, (146)
where λ is an integration constant to be determined.
Hence, after some manipulation, we are left with two ODE’s to solve which
are given by
dT
dτ
+ λT = 0
(∗)
d2
F
dη2
+ P
d
dη
F 1 −
2cs
cmax
+ λF = 0
(∗∗)
. (147)
We can solve (∗) easily since it is simply a first order, linear differential
equation. The solution is
T(τ) = e−λτ
, (148)
where the integration constant has been left out as we can absorb it into the
F solution.
Since (∗∗) is an eigenvalue problem as before, there must exist n solutions
(Fn(η)), known as the eigenfunctions. Corresponding to each eigenfunction,
there must exist a unique eigenvalue (λn) which usually can be found by
applying the boundary conditions. The idea is to substitute our steady so-
lutions (cs,1 and cs,2) into (∗∗), and use the eigenvalues to determine the
stability of each solution. We have two cases:
Case 1: All the λn’s are positive.
In this case, the perturbed part of our solution (˜c) will vanish as we increase
the time parameter (τ). This implies the steady concentration solution must
be stable.
Case 2: At least one of the λn’s is negative.
If one or more of our eigenvalues is negative, then the perturbed solution
will not vanish as we increase the rescaled time (τ). Due to the exponential
in (148), the perturbed concentration will exponentially increase with time.
This implies the steady concentration solution must be unstable.
62
63. Since (∗∗) is a linear, second order ODE, we can assume the general solution
will be of the form
Fn(η) = αf1,n(η) + βf2,n(η), (149)
where α, β are constants, and f1, f2 are functions to be determined.
The functions f1 and f2 are difficult to find analytically, however using Maple
to solve (∗∗) we find that
f1,n(η) =
B±(2∆n + P)e−1
2
(2∆n−3P)η
+ (2∆n − P)e−1
2
(2∆n−P)η
(1 + B±ePη)2
, (150)
f2,n(η) =
−B±(2∆n − P)e
1
2
(2∆n+3P)η
− (2∆n + P)e
1
2
(2∆n+P)η
(1 + B±ePη)2
, (151)
where B± is defined in (131), and the transformation λn = P2
4
−∆2
n has been
made to simplify the expression.
Now, the boundary conditions for (∗∗) can be determined from boundary
conditions (i) and (ii) in (144), they take the form
(i) P 1 − 2cs(1)
cmax
F(1) + F (1) = 0,
(ii) F (0) = 0.
Applying these boundary conditions to the general solution (149) will enable
us to determine α and β in terms of ∆n, B± and P. Using Maple, we obtain
two simultaneous equations of the form
M11(∆n, B±, P)α + M12(∆n, B±, P)β = 0, (152)
M21(∆n, B±, P)α + M22(∆n, B±, P)β = 0. (153)
This is simply an eigenvalue problem such that
Mv = µv, (154)
where µ = 0 is the eigenvalue of M =
M11 M12
M21 M22
, and v =
α
β
the
corresponding eigenvector.
It is a fundamental result of linear algebra that an equation has a non-zero
63
64. solution if, and only if, the determinant of the matrix is zero. So, to have
non-zero solutions for α and β we must have
M11M22 − M12M21 = 0. (155)
In order to solve (155) for ∆n, we must fix the Rouse number P, and cmax (in
doing so determining B±). We can do this since the hindered concentration
equation must hold for all P and cmax, providing
P >
4
cmax
. (156)
Figure 24 shows how B± varies for different sizes of cmax when the Rouse
number is fixed to P = 2. Since (24) implies cmax > 2, we always find two
solutions as expected. The blue line corresponds to B+, and the red line
corresponds to B−. (155) for ∆n.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2
2.5
3
3.5
4
cmax
B±
B
+
B−
Figure 24: A plot of B+ (blue), and B− (red) against cmax where the Rouse
number has been fixed at P = 2.
5.5.1 Determining the Stability
We wish to determine the stability of cs,1 and cs,2. Earlier we found that the
stability of each solution depends on the eigenvalues λn. Since we made the
64
65. transformation λn = P2
4
− ∆2
n, the stability conditions on the eigenvalues ∆n
are as follows:
• The steady solution is stable if all the ∆n’s are less than P
2
or purely
imaginary.
• The steady solution is unstable if there exists a ∆n such that ∆n > P
2
.
Since P, cmax and B± are arbitrary constants, we can fix them in order to
find the ∆n’s. Let us set
P = 2, and cmax = 2.1 =⇒ B+ ≈ 1.56, and B− ≈ 0.64. (157)
This situation is shown in Figure 24.
Using Maple to solve (155) for ∆n, with P, cmax and B+ fixed as given
above, we obtain
∆n = ±0.64. (158)
The stability conditions state that the steady state is stable if ∆n < P/2.
Since we took P = 2 we deduce that the steady solution corresponding to
B+ must be stable.
Furthermore, using Maple to solve (155) for ∆n with P, cmax and B− fixed
as given above, we obtain
∆n ≈ ±1.25. (159)
The stability conditions state that the steady state is unstable if ∆n > P/2.
Since we took P = 2 we deduce that the steady solution corresponding to
B− must be unstable.
Figure 25 shows a graphical result of the solutions to (155) for B− by defining
the function
FB− (∆) := M11M22 − M12M21, (160)
and plotting it against ∆. The points at which the function crosses the ∆-
axis are the solutions to (155). Now, the solutions at ±1 corresponds to the
trivial solution λ = 0, and the solution at 0 corresponds to λ = 1. But we
see there are two more solutions at ±1.25, these correspond to λ < 0 and
hence confirm the fact that cs,2 is unstable.
65
66. Figure 25: A plot showing the relationship between FB− (∆) := M11M22 −
M12M21, and ∆. The Rouse number is taken as fixed at 2 and cmax is fixed
at 2.1.
Generalisation for P = 2
Now, using Figure 24 we can generalise our result. If we fix the Rouse num-
ber as 2, we can determine the type of solutions we obtain for ∆n as cmax
and B± vary. We have the following cases:
Case 1:
If cmax > 2, we obtain 2 solutions. The solution cs,2 corresponds to B− which
ranges between 0 and 1. In this case the ∆n’s are strictly greater than 1, this
implies cs,2 is unstable. The solution cs,1 corresponds to B+ which ranges
between 1 and ∞. For 1 < B+ < 2 the ∆n’s are strictly less than 1, and for
B+ ≥ 2 the ∆n’s are purely imaginary. This implies that the solution cs,1 is
stable.
Case 2:
If cmax < 2, we do not obtain any solutions, so there are no stabilities to
determine. This corresponds to the case where the up flux is 0 which was
redefined in (140) in terms of the Heaviside step function.
66
67. Case 3:
If cmax = 2, we obtain one (repeated) solution corresponding to B± = 1. In
this case, the ∆n’s are
∆n = 1 + ln
e
e2
= 0, (161)
∆n = 1 + ln
−e
e2
= iπ, (162)
where we have used Euler’s identity (eiπ
+ 1 = 0) with i =
√
−1 in (162).
So, since the solution must hold for any P > 0, we have that
• cs,1(η) is a STABLE steady solution.
• cs,2(η) is an UNSTABLE steady solution.
5.6 Solutions for higher values of n
When constructing the hindered settling equation (126), we stated that the
parameter n is typically in the range (2.4, 4.65). Let us consider how the
behaviour of our solution differs if we take n = 3. The problem we wish to
solve is
∂ˆc
∂τ
− P
∂
∂η
ˆc 1 −
ˆc
cmax
3
=
∂2
ˆc
∂η2
, (163)
subject to boundary/initial conditions
(i) P ˆc 1 − ˆc
cmax
3
= −∂ˆc
∂η
at η = 1,
(ii) ∂ˆc
∂η
= −1 at η = 0,
(iii) ˆc = ˆch,0 at τ = 0.
Note that from (123) the upward flux (φu) is a maximum at
c =
1
n + 1
=
1
4
(164)
We cannot explicitly find the two steady solutions in the range [0, cmax],
instead, by varying the initial concentration and using the MATLAB solver
pdepe to show the how concentration distribution changes through time, we
67
68. are left with an accurate description of where the two solutions lie. Figure 26
shows the two solutions cs,1 (low concentration) and cs,2 (high concentration),
where the arrows point in the direction of increasing time.
Figure 26: A plot showing the overall behaviour of the hindered concentration
equation for n = 3. The arrows show the direction in which the solution
evolves over time, where solutions for t close to 0 haven’t been included. The
Rouse number is taken as P = 2.
If we compare the result to the case where n = 1, we notice that the position
of the unstable solution, cs,2, has moved closer to the stable solution, cs,1.
As a matter of fact, as n increases, the two steady solutions move closer and
closer together. We see once again that if we choose the initial concentration
to be high enough, then we reach a point, denoted Clim, where we do not
approach a stable solution. Using an iterative method in MATLAB we can
once again determine the size of Clim for different magnitudes of the Rouse
number (P). This relationship is shown in Figure 27 (1).
The behaviour of Clim follows a very similar pattern to the case where we
took n = 1. However, if we hold the value of cmax constant, the size of P
at which Clim hits 0 will increase. This is due to the up flux (φu) having a
different distribution over c depending on the parameter n.
68
69. 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
Rouse Number (P )
Clim
(1)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentration (ˆc/cmax)height(z/h)
(2)
Figure 27: (1) A plot showing a relationship between Clim and P, where we
have taken cmax = 6. (2) A plot showing how the solution behaves when we
start the initial concentration at ˆch,0/cmax = 0, along with the Rouse number
P = 1.4, and cmax = 6.
In Figure 27 (2) we see an example of how the solution behaves if we take
P low enough such that Clim drops below 0. The initial concentration is set
to 0, and as time increases we notice that the steady solution is completely
bypassed.
6 Conclusion
In this project, we constructed a differential equation which describes the
vertical distribution of sediment in a uniform flow. The turbulent nature of
the flow was modelled using an empirically found relationship with the eddy
diffusivity (Dyer and Soulsby 1988). We first considered fully-developed flow
profiles with different representations of the eddy diffusivity.
We then went on to consider how the concentration develops over time. Given
an initial reference concentration, we could determine how the distribution of
sediment evolves. This was achieved by solving a Sturm-Liouville eigenvalue
problem where asymptotic representations of the eigenvalues were found for
sufficiently small Reynold’s numbers. This provided an accurate solution
which compared reasonably well with experimental results. Further investi-
gations were taken to determine what affect the Rouse number had on the
rate of evolution and accuracy.
69
70. Finally, a concept known as hindered settling was imposed which takes the
settling velocity’s dependence on the concentration into account. Hindered
settling tends to occur at higher concentrations since the interactions between
particles would be more likely. We found that there were two solutions to the
hindered settling equation and using Maple we could analytically determine
their stability. We discovered that the solution for lower concentrations was
stable and the solution at higher concentrations was unstable, as a result of
hindered settling. Once again, the Rouse number played a vital role in how
the distribution of concentration evolves and whether it approaches a steady
state.
There were many parts of this project that could have undergone further
investigations. For example, we could have modelled the evolution of the
concentration given a non-uniform initial concentration. Or linearisation
could have been trialled for larger n, since n typically takes values in the
range (2.4, 4.65) (Baldock, Tomkins, Nielsen and Hughes 2003). However,
this was beyond the scope of this project.
Many uncertainties still remain when describing the behaviour of cohesive
sediment in a uniform flow. This is because of the difficulty in mathe-
matically modelling turbulence, with additional unpredictabilities arising for
higher concentrated flows.
70
71. 7 Bibliography
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