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.
Transcript
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 ofsuspended sediment in a turbulent flow. The turbulent fluctuationsare expressed as a diffusive flux, where the sediment diffusivity can bedetermined empirically. Steady solutions to a fully-developed flow arefound, where the settling velocity is considered as a function of theconcentration. We gain an understanding of how the concentration ofsediment changes over time by considering the stability of each steadystate. The accuracy and the time scale of the steady solutions arefound to be strongly dependent on the Rouse number, a dimensionlessparameter which provides a measure of the flow’s ability to suspendparticles.
Acknowledgement of Sources
For all ideas taken from other sources (books, articles, internet), the sourceof the ideas is mentioned in the main text and fully referenced at the end ofthe report.
All material which is quoted essentially word-for-word from other sourcesis 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.
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 ina uniform flow. . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 The variation of the drag coefficient with the Reynolds numberfor natural sand. (Taken from Fredsoe and Deigaard (1992)) . 13
4 A constant, linear and parabolic representation of the eddydiffusivity. The constant fc has been chosen as 1 for simplicity,and fl, fp are set to be very small (� 1) for reasons statedearlier. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 lineardiffusivity, where different magnitudes of the Rouse numberhave been considered. . . . . . . . . . . . . . . . . . . . . . . . 25
7 The vertical distribution of concentration in a river for parabolicdiffusivity, where different magnitudes of the Rouse numberhave been considered. . . . . . . . . . . . . . . . . . . . . . . . 26
8 Two comparisons of the vertical distribution of sediment ina river for constant, linear and parabolic diffusivity. In (1),the Rouse Number has been fixed, and in (2), the mass ofsediment distributed is fixed. . . . . . . . . . . . . . . . . . . . 27
9 A plot of the two functions given in (72), taking the RouseNumber P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . 33
10 A plot of the two functions given in (75), taking the RouseNumber 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, definedas λ∗n − λn. Note that the Rouse number has been taken asP = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3
12 Four separate examples displaying the accuracy of the solutionfor t = 0, where the number of eigenvalues included in thesummation varies. The initial concentration is set as P cr,0 =0.4 and the Rouse number P = 2. . . . . . . . . . . . . . . . . 42
13 A graph showing how the solution (98) evolves over time withinitial concentration c = 0.37/P , and Rouse number P = 2. . . 43
14 A graph showing how the solution (98) evolves over time withinitial concentration c = 0.01/P , and Rouse number P = 2. . . 44
15 A graph showing how the solution (98) evolves over time withinitial concentration c = 1.8/P , and Rouse number P = 2. . . 45
16 (1), (2), (3) and (4) show the time evolution of concentrationfor P = 0.5, 1, 2, 3 respectively. The initial concentration isfixed at P cr,0 = 0.5 in all cases. . . . . . . . . . . . . . . . . . 47
17 (1) A plot showing the relationship between the Rouse numberand the first (lowest) eigenvalue ∆1. (2) A plot showing therelationship 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. Theblue lines correspond to the typical values for n (determinedempirically). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
19 A plot of the two steady solutions, cs,1 and cs,2, with threedifferent magnitudes of the Rouse number. Note that the leftsolutions (low concentration) correspond to cs,1 and the rightsolutions (high concentration) correspond to cs,2. . . . . . . . . 54
20 A plot of the two steady solutions, cs,1 and cs,2, where theRouse number is set as P = 2, and the initial concentration isch,0/cmax = 0.5. . . . . . . . . . . . . . . . . . . . . . . . . . . 56
21 A plot of the steady solution cs,1 with the Rouse number takenas P = 2. The initial concentration is set as ch,0/cmax = 0.01. . 57
22 A plot of the steady solution cs,2 with the Rouse number takenas P = 2. (The solution cs,1 is not shown). The initial con-centration is set as ch,0/cmax = 0.79. . . . . . . . . . . . . . . . 58
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23 A plot showing a relationship between Clim and P , where wehave taken cmax = 6. . . . . . . . . . . . . . . . . . . . . . . . 59
24 A plot of B+ (blue), and B− (red) against cmax where theRouse 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 andcmax 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 directionin which the solution evolves over time, where solutions for tclose to 0 haven’t been included. The Rouse number is takenas P = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
27 (1) A plot showing a relationship between Clim and P , wherewe have taken cmax = 6. (2) A plot showing how the solutionbehaves when we start the initial concentration at ch,0/cmax =0, along with the Rouse number P = 1.4, and cmax = 6. . . . . 69
5.6 Solutions for higher values of n . . . . . . . . . . . . . . . . . 67
6 Conclusion 69
7 Bibliography 71
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1 Introduction
The transport of sediment in turbulent flows has been extensively researchedfor many years now. The best way to mathematically model the behaviourin a uniform flow continues to be a highly disputed matter. The ability toaccurately predict sediment transportation is very important in coastal andhydraulic engineering and, even in simple situations, finding a relationshipbetween the velocity of a horizontal flow and the mass of sediment in suspen-sion is not straightforward. This is due to individual moving grains takingenergy from the flow, lowering the flow’s capability to transport sediment.
The mass of sediment being transported is defined as the ‘total load’. Thereare 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 contactwith the bed. This is due to turbulent fluctuations in the flow keeping theparticles in suspension. Lastly, the wash load is made up of very fine particlesand is generally not represented in the bed. For this reason, we do not haveto include the wash load in the total mass of sediment transported and itcan be neglected.
Figure 1: A diagram showing three different modes of sediment transport:Bed load, suspended load and wash load.
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Figure 1 shows an example of all three modes of transport in a river, wherethe sediment is being transported due to a horizontal turbulent flow. Notethat the river bed is made up of suspended load and bed load (left uncolouredin the diagram). In this project, we look at how the distribution of suspendedparticles changes as the flow develops, and whether it reaches a steady distri-bution over time. This is achieved by mathematically predicting the rate atwhich sediment settles or erodes from the bed, relative to variables such astime or flow velocity. Conservation of mass can lead to accurate predictionsof fully developed concentration profiles which are surprisingly consistentwith experimental results.
Before going any further, we will discuss some important sediment prop-erties and how to mathematically define suspended sediment in a turbulentflow.
Figure 2: A diagram showing the basic setup of suspended sediment in auniform 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 freesurface at z = h. Note that the flow transporting the particles is made upof an incompressible fluid, subject to small turbulent fluctuations. The tur-bulent nature of the flow creates complicated interactions between the fluidand the particles. A very common approach is to express the turbulent fluxas a diffusive flux, where the rate at which the sediment diffuses (known asthe sediment diffusivity or eddy diffusivity) is determined empirically (Dyerand Soulsby 1988). A ‘no vertical flux’ boundary condition can be imposedat the free surface z = h (since the grains cannot pass through the surface
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of the river). At the base of the river, a boundary condition must be chosento represent the mass exchange due to erosion and deposition. (This will bediscussed 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. Likethe fluid, the sediment is also incompressible; this implies that mass is con-served throughout the whole specified boundary. Since each particle has aspecific mass, we expect them to settle due to gravity. However, since theparticles are suspended in water, the force due to gravitational accelerationwill 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 theadvection 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 theeddy 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 havelarge effects in turbulent flows.
Finally, we reason that the settling velocity of a grain is dependent on theconcentration; this follows from a concept known as ‘hindered settling’. Hin-dered settling gives rise to unstable solutions which can prevent the sedimentdistribution from evolving into a steady state. Linearisation is used to anal-yse the behaviour of each steady state. Once again, several graphs are plottedusing MATLAB to interpret the results.
All diagrams in this project were drawn by Ben Watson and every graphhas been plotted in either MATLAB or Maple and then annotated. The onlyexception being Figure 3 which was taken from Fredsoe and Deigaard (1992).
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1.2 Important Definitions and Theorems
Before starting we will go over some important definitions and theoremswhich will be used regularly throughout the project. These definitions arebased on that of Batchelor (2000)1, Fredsoe and Deigaard (1992)2, Dorrelland 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 andLagrangian description of a flow.
Eulerian description:
A description of the flow as seen by a stationary observer. We choose apoint x in space, and observe the velocity of the fluid as a function of timet. 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 particlewith dye, and observe how it moves. To describe the whole field, we labelall material points. We then choose an initial time t0 and label particles bya = x, where x denotes the position at time t0. Then, for t > t0, the particleis at x = x(a, t), where x(a, t0) = a. The Lagrangian velocity is defined asv(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 ofa 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 Lagrangianderivative is defined as
Du
Dt=∂u
∂t+ (u · ∇)u. (3)
The term on the left represents the Lagrangian description of the flow, thefirst term on the right represents the Eulerian description of the flow, and thesecond term on the right represents the directional derivative in the directionof u.
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1.2.2 Definition: Specific Gravity [2]
The specific gravity (s) is defined as the ratio of the density of a substance tothe density of a reference substance. (In this case the the reference substanceis taken to be water). For sediment suspended in a flow, we have the ratio
s =ρsρ, (4)
where ρ denotes the density of water at 4oC, and ρs denotes the density ofthe 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 sedimentunder the action of gravity. The settling velocity depends on the grain size,specific gravity, shape, the dynamic viscosity of the fluid, and the proximityof other particles/boundaries. The drag force F on a submerged body isgiven by:
FD =1
2cDρV
2A (5)
where cD is the drag coefficient, ρ is the density of the fluid, V is the relativevelocity, and A is the area of the projection of the body upon a plane normalto the flow direction.
We will consider the settling of a single spherical particle of diameter d,(volume π
6d3) . 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)gπ
6d3, (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 projectionthat of a sphere, we have:
(ρs − ρ)gπ
6d3 =
1
2cDρw
2s
π
4d2 (7)
Using (4), we can rearrange this to find the settling velocity of a singe grainyielding
ws =
√4(s− 1)gd
3cD. (8)
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Now, the value of cD depends on the particle Reynolds number (R = wsdν
),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 ofthe Reynolds number.
Figure 3: The variation of the drag coefficient with the Reynolds number fornatural 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 inws given by
4.2w2s +
108ν
dws − 4(s− 1)gd = 0. (10)
Hence, if we take the kinematic viscosity of water at 4oC as ν = 1×10−6m2/s,the particle diameter as d = 0.1mm, the specific gravity as s = 2.65, and theearths 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 settlingvelocity must be positive.
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 drivingforces on the sediment exceed the stabilising forces, the sediment will beginto move, (this is called incipient motion). We define the bed shear stress asthe 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 themean flow velocity (u) by a friction factor Cf such that τb = ρCf u
2.
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 dependingon the particle characteristics. We denote the point of incipient motion asθ = θcr, where we call θcr the critical Shields parameter.
Using the critical Shields parameter and the relation given in (13), we cango on to define the critical bed shear stress as the shear stress exerted on thebed 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 =wsκu∗
. (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.
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1.2.6 Definition: Time Average
The time average X of a variable x(t) is defined by
X = limT→∞
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 arefound such that a solution to the Sturm-Liouville problem exists, we call eachλ 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 separatedboundary conditions of the form
α1y(a) + α2y′(a) = 0 (α2
1 + α22 > 0), (17)
β1y(b) + β2y′(b) = 0 (β2
1 + β22 > 0). (18)
Assuming the Sturm-Liouville problem is regular, the Sturm-Liouville theorystates 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 eigenfunctionyn(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.
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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 runsparallel with the bed and the sediment is kept in suspension due to turbulentfluctuations. Each grain has a settling velocity ws, and is assumed to settlerelative to the surrounding water. We can define the velocity of a single grainas
up = u− wsz, (21)
where u = u(z)x 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 meansthe vertical forces exerted on each grain must balance, i.e. upz = 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 ofobservation along the river; assuming we began with some initial concentra-tion distribution at time 0. Hence, the concentration, denoted c, is a functionof 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 overtime, 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 velocityfield can fluctuate quickly over time. We can use a mathematical techniqueknown as Reynolds decomposition to separate the average and fluctuatingparts of a quantity. We have already introduced a time parameter (t), whichcorresponds to our concentration changing over ‘slow time’. So for our fastfluctuations 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)
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where c and u = u(z)x denote the ‘fast’ time average of c and u (known asthe steady components), and c′ and u′ = (u′, v′, w′) denote the fluctuatingpart (or perturbation) of c and u. These perturbations are defined such thattheir 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 ofthe 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:
Here, the first term represents the change in concentration over time, thesecond term arises due to the sediment being advected by the flow, the thirdterm represents the settling of the particles, and the last term correspondsto 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 thatfully capture the complicated interactions between the fluid and the particles(Dyer and Soulsby 1988; Fredsoe and Deigaard 1992). A common approachis to assume the turbulence-induced flux can be expressed as a diffusiveflux, where the sediment diffusivity can be determined empirically (Dyer andSoulsby 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 theeddy 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 leadto predictions of fully developed flow profiles by solving (28). However, wecannot completely rely on the empirically obtained result in (29) since theeddy diffusivity can depend on the boundary condition at the bed and ismuch 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 diffusivitiesrespectively. 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-handside 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
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on the right-hand side represent vertical and horizontal sediment diffusionrespectively.
Now, since we are interested in the vertical distribution of sediment, wecan neglect the horizontal gradient in c, treating the horizontal distributionas 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 wehave neglected the horizontal derivatives, without loss of generality we canwrite 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. Theboundary 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 behaviourof the flow in the z-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 potentialfor mass exchange due to erosion and deposition. Instead, we can make thefollowing considerations:
Just above the river bed, the net flux (Φ) must be equal to the amountof sediment eroded from the bed into the fluid, φe, minus the amount ofsediment deposited from the fluid onto the bed, φd. Hence,
−wsc−K∂c
∂z= φe − φd at z = 0. (36)
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Now, the amount of sediment deposited (φd) is equal to the settling velocitymultiplied 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 fluxby letting φe = qe(t), where qe(t) expresses the erosion rate of the flow (Dyerand 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 fluxthrough z = 0 (van Rijn 1984). We typically take the constant p to be in therange [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 usedto 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 ittakes into account the shear stress at the bed and allows for mass exchangewhich 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 asteady state, then the concentration will lose its time dependence. In thiscase, our concentration can be written as a function of the vertical distancefrom 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.
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We will begin by finding different representations for distribution of con-centration by varying the eddy diffusivity (K). It is possible to write theeddy diffusivity (Souslby 1988) as:
K = κu∗hf(η) (40)
where κ is the von Karman constant, u∗ is the shear velocity, h is the heightof the river, and η = z/h.
We can change the rate of diffusivity by imposing a constant, linear orparabolic representation of our function f(η):
Constant diffusivity: f(η) = fc
This corresponds to the concentration of sediment in the river diffusing atthe same rate throughout the river. Our constant fc is bounded below at0 (since we cannot have negative diffusivity), and bounded above by fc(max)
(the maximum rate of diffusivity). Furthermore, we will impose fc 6= 0 asthis 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 thoughthe river. So the further from the bed the grain is situated, the quicker itdiffuses. 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 isclose to 0.
Parabolic diffusivity: f(η) = η(1− η) + fp
This corresponds to the concentration of sediment diffusing parabolicallythough the river. In this case, the rate of diffusivity hits a maximum at themidpoint 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 diffusivitieschange with height.
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 setto 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. Thiscorresponds to the sediment diffusing at the same rate throughout the river,shown in figure 4. The Steady Concentration equation for constant eddydiffusivity (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
We are left with a first order, linear ODE which can be solved easily toobtain the solution
, and hence, the concentra-tion distribution for constant diffusivity is
cs(z) =qewse−
wsKc
z. (44)
Figure 5: The vertical distribution of concentration in a river for constantdiffusivity, 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 isdistributed. The concentration and vertical distance have been rescaled suchthat the concentration of sediment is fixed to 1 at the base of the river, andthe height is bounded between 0 and 1. We consider different magnitudes ofthe 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 settlingvelocity (ws) dominates the shear velocity (u∗) when P is large enough.
23
3.2 Linear Diffusivity
Now we would like to consider the case where the diffusivity is a linearfunction of the vertical distance (z). i.e. f(η) = η + fl. This time, wecan rewrite (39) as a differential equation in cs as a function of η. Using thefact that η = z/h we have
ws1
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)− wsκu∗ . (47)
Applying boundary condition (ii.a), we find c0 = qewsf
wsκu∗
+1
l .
Finally, after replacing η with z/h the concentration distribution for lineardiffusivity is
cs(z) =qewsf
wsκu∗
+1
l (z
h+ fl)
− wsκu∗ . (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 isdistributed. As before, the concentration and vertical distance have beenrescaled such that the concentration of sediment is fixed to 1 at the base ofthe river, and the river’s height is bounded between 0 and 1. We considerdifferent magnitudes of the Rouse number; a measure of the flow’s ability tosuspend particles.
As we increase the Rouse number (P ), we notice that the linear concen-tration distribution differs from the constant case. The concentration onlydeviates slightly towards the top of the river, whereas in the bottom quartilethe 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
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 parabolicfunction 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 knownas the Vanoni-distribution. Once again, we can rewrite (39) as a differentialequation in cs as a function of η so that
ws1
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 −
[12
+√fp + 1
4
]η −
[12−√fp + 1
4
]∣∣∣∣∣∣∣
ws
2κu∗√
fp+14
. (51)
25
Applying boundary condition (ii.a), and using the fact that ddx|f(x)| =
|f(x)|f(x)
f ′(x) we find
c0 =qewsfp
(1
2−√fp +
1
4
)2∣∣∣∣∣∣
12
+√fp + 1
4
12−√fp + 1
4
∣∣∣∣∣∣1− ws
2κu∗√
fp+14
. (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 parabolicdiffusivity is:
cs(z) =qews
∣∣∣∣∣∣12
+√fp + 1
4
12−√fp + 1
4
∣∣∣∣∣∣− ws
2κu∗√
fp+14
∣∣∣∣∣∣∣zh−[
12
+√fp + 1
4
]zh−[
12−√fp + 1
4
]∣∣∣∣∣∣∣
ws
2κu∗√
fp+14
(53)
Figure 7: The vertical distribution of concentration in a river for parabolicdiffusivity, 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
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 linearmodel, however the concentration of sediment tails off toward the surfaceof the river. This is due to the eddy diffusivity having less of an effect nearerthe 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 regimeby varying other parameters such as the Rouse number or the total mass ofsediment being distributed.
We can define the total mass of sediment in suspension (A) as the the areaunder the curve. This implies
Figure 8: Two comparisons of the vertical distribution of sediment in a riverfor constant, linear and parabolic diffusivity. In (1), the Rouse Number hasbeen 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
the river, we notice that the distribution corresponding to constant diffusiv-ity has the largest amount of suspended particles. However, the majority ofthe grains are situated in the lower half of the river.
Figure 8 (2) gives an example of each distribution where the mass of sedimentin suspension is fixed. The linear and parabolic diffusivity models follow avery similar pattern; the only noticeable difference occurs close to the sur-face of river. Here, the parabolic regime suggests that the rate at which theconcentration of sediment reduces is a lot faster. This is due to the eddydiffusivity being very small near the free surface for the parabolic case, onthe other hand, the linear case suggests the eddy diffusivity is largest at thetop of the river.
4 The Unsteady Concentration Equation
Now we would like to consider how the concentration distribution changesover time. We begin with the unsteady concentration equation (33) that wasderived earlier, where the concentration (c) is a function of vertical distance(z), and time (t), and the eddy diffusivity K = Kc = constant. The problemwe wish to solve is
∂c
∂t= ws
∂c
∂z+Kc
∂2c
∂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 initialconcentration 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 theproblem. We introduce τ , c and η, along with the Rouse Number (P ) written
28
in terms of the eddy diffusivity Kc = κu∗hfc,
η =z
h, τ =
t
h2, P =
wsh
Kc
, c =Kc
qehc, (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 unknownconstant (P ). We have
∂c
∂τ− P ∂c
∂η=∂2c
∂η2, (57)
subject to boundary/initial conditions
(i) P c = − ∂c∂η
at η = 1,
(ii) ∂c∂η
= −1 at η = 0,
(iii) c = cr,0 at τ = 0,
where cr,0 = Kccr,0qeh
.
4.2 Solving the Unsteady Concentration Equation
Note that some of the techniques used in this section to solve (57) were takenfrom Dorrell and Hogg (2012), and Pritchard (2006).
We begin by writing the solution to (57) as a linear combination of thesteady, time independent solution cs = cs(η), and a perturbed, time depen-dent solution c = c(η, τ), such that
c = cs(η) + c(η, τ). (58)
We would like to determine whether the final solution to (57) yields a stable(c → cs as τ → ∞) or an unstable (c 6−→ 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
=∂2cs∂η2
, (59)
subject to boundary conditions
29
(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
Pe−Pη, (60)
where the factor 1P
arises from the boundary condition at η = 0. We noticethat this is exactly the same as our constant diffusivity solution (44), just interms of c, the Rouse number (P ) and η, (as it should be).
4.2.2 Unsteady Solution
Substituting c = cs + c into (57) yields
∂c
∂τ− P ∂c
∂η− P ∂cs
∂η=∂2c
∂η2+∂2cs∂η2
, (61)
subject to boundary/initial conditions
(i) P c+ Pcs = − ∂c∂η− ∂cs
∂ηat η = 1,
(ii) ∂c∂η
+ ∂cs∂η
= −1 at η = 0,
(iii) c+ cs = cr,0 at τ = 0.
We can simplify both the PDE and the boundary conditions using what weknow from (59). Hence, our problem is reduced to
∂c
∂τ− P ∂c
∂η=∂2c
∂η2, (62)
subject to boundary/initial conditions
(i) P c = − ∂c∂η
at η = 1,
(ii) ∂c∂η
= 0 at η = 0,
(iii) c = cr,0 − cs at τ = 0.
30
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︸ ︷︷ ︸
(∗)
,d2F
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
)= λePηF, (66)
subject to boundary conditions
(i) PF (1) + dFdη
(1) = 0,
(ii) dFdη
(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 eigenfunctionFn(η) which has exactly n− 1 zeroes in (0, 1).
31
• 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, whereF0 and β are constants. By substituting our ansatz into (∗∗) we obtain aquadratic equation for β, given by
β2 + Pβ + λ = 0, (69)
and solving for β yields
β =−P ±
√P 2 − 4λ
2. (70)
Now, we have three possible cases to consider:
Case 1: (P 2 − 4λ > 0)
In this case, β = −P2± Γ, where Γ = 1
2
√P 2 − 4λ. Note that Γ is real
and strictly greater than 0. This leaves us with the general solution
F (η) = e−Pη2 [A1sinh(Γη) +B1cosh(Γη)] . (71)
Applying boundary condition (ii) gives B1 = 2ΓPA1, and hence applying
boundary condition (i) leaves us with the condition
tanh(Γ) = − 4PΓ
4Γ2 + P 2. (72)
Since P is just a constant, we can solve this for Γ. We see in Figure 9 thatthere is only one solution to (72), which is Γ = 0, but since we assumed Γ isstrictly positive, there cannot be a solution to (∗∗) with P 2 − 4λ > 0.
32
−6 −4 −2 0 2 4 6−1.5
−1
−0.5
0
0.5
1
1.5
K
F(K
)
F(K) = tanh (K)F(K) = −2K / (K2 +1)
Figure 9: A plot of the two functions given in (72), taking the Rouse NumberP = 2.
Case 2: (P 2 − 4λ = 0)
Now, P 2 − 4λ = 0 leaves us with β = −P2
, and we have the general so-lution
F (η) = A2e−Pη2 +B2ηe−
Pη2 . (73)
Applying boundary condition (ii) gives B2 = P2A2, 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 thereare no solutions to (∗∗) for P 2 − 4λ = 0.
Case 3: (P 2 − 4λ < 0)
In this case, β = −P2± i∆, where ∆ = 1
2
√4λ− P 2. Note that ∆ is real
and strictly greater than 0. This leaves us with the general solution
F (η) = e−Pη2 [A3sin(∆η) +B3cos(∆η)] . (74)
33
Applying boundary condition (ii) gives B3 = 2∆PA3, and hence applying
boundary condition (i) leaves us with
tan(∆) =4P∆
4∆2 − P 2. (75)
Since P is just a constant, we can solve this for ∆. In Figure 10 we noticethat 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 byrearranging the definition of ∆ so that
λn = ∆2n +
P 2
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
6
F(6
)
F(6) = tan (6)F(6) = 26 / (62 −1)
Figure 10: A plot of the two functions given in (75), taking the Rouse NumberP = 2.
34
We will write each eigenfunction as
Fn(η) = e−Pη2
[sin(∆nη) +
2∆n
Pcos(∆nη)
], (78)
where the constant A3 has been left out as it can be absorbed into the finalsolution.
Our final unsteady solution can be represented by taking a superpositionof each eigenfunction and substituting them (63). We obtain an infinite sumgiven by
c(η, τ) =∞∑n=0
Cn Tn(τ)Fn(η), (79)
where the constants drawn out from the solutions to (∗) and (∗∗) are nowrepresented 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
cr,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
(cr,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 ournormalisation constant. To find our normalisation constant we must evaluatethe integral for the cases n 6= m and n = m.
35
If n 6= m, we have
I =
1∫0
Fn(η)Fm(η)ePη dη,
=
1∫0
(sin(∆nη) +
2∆n
Pcos(∆nη)
)(sin(∆mη) +
2∆m
Pcos(∆mη)
)dη,
= 0.
Note that solving this integral is by no means straightforward, but one caneasily show this integral vanishes using Maple along with the relation foundin (75).
If n = m, we have
I =
1∫0
F 2m(η)ePη dη,
=
1∫0
(sin(∆mη) +
2∆m
Pcos(∆mη)
)2
dη,
=
1∫0
(sin2(∆mη) +
4∆2m
P 2cos2(∆mη) +
4∆m
Pcos(∆mη)sin(∆mη)
)dη,
=
1∫0
((2∆2
m
P 2− 1
2
)cos(2∆mη) +
2∆m
Psin(2∆mη) +
1
2+
2∆2m
P 2
)dη,
=
[(∆m
P 2− 1
4∆m
)sin(2∆mη)− 1
Pcos(2∆mη) + η
(1
2+
2∆2m
P 2
)]1
0
,
=1
2+
2∆2m
P 2+
1
P+
(∆m
P 2− 1
4∆m
)sin(2∆m)− 1
Pcos(2∆m). (83)
Now, using (75) we can find similar relations for sin(∆m) and cos(∆m). Wefind that
sin(∆m) =
√tan2(∆m)
1 + tan2(∆m)=
4P∆m
(4∆2m + P 2)
, (84)
36
cos(∆m) =
√1
1 + tan2(∆m)=
(4∆2m − P 2)
(4∆2m + P 2)
. (85)
Substituting these relations into (83) and simplifying, our normalisation con-stant N is found to be
N = I−1 =
[1
2+
2∆2m
P 2+
8∆2m
P (16∆2m + P 2)
]−1
. (86)
Plugging this back into (82), we deduce that
Cn = N
1∫0
(cr,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
cr,0Fn(η)ePη dη
︸ ︷︷ ︸I1
−N1∫
0
csFn(η)ePη dη
︸ ︷︷ ︸I2
(88)
So, using the fact that
1∫0
eAxsin(Bx) dx =AeAsin(B)−BeAcos(B) +B
A2 +B2, (89)
1∫0
eAxcos(Bx) dx =BeAsin(B) + AeAcos(B)− A
A2 +B2, (90)
we find
I1 = cr,0
1∫0
ePη2
(sin(∆nη) +
2∆n
Pcos(∆nη)
)dη, (91)
=8cr,0∆ne
P2
4∆2n + P 2
. (92)
Note that we replaced sin(∆n) and cos(∆n) with the relations found in (84)and (85).
37
Since we found cs in (60), we can evaluate I2 in a similar manner yielding
I2 =1
P
1∫0
e−Pη2
(sin(∆nη) +
2∆n
Pcos(∆nη)
)dη, (93)
=8∆n
P (4∆2n + P 2)
. (94)
Hence substituting the solutions to I1 and I2 into (88) we have
Cn = N(I1 − I2), (95)
=
(8cr,0∆ne
P2
4∆2n+P 2 − 8∆n
P (4∆2n+P 2)
)(
12
+ 2∆2n
P 2 + 8∆2n
P (16∆2n+P 2)
) . (96)
Finally, since we have found the Tn, Fn and Cn’s, we can substitute theminto (79) and we are left with a solution to the perturbed, time dependentpart of our solution given by
c(η, τ) =∞∑n=0
Cn e−λnτe−Pη2
[sin(∆nη) +
2∆n
Pcos(∆nη)
]. (97)
An important observation in this expression is the fact that each contributionfrom the sum decays as n increase. This is due to every λn being positiveand tending to infinity as n tends to infinity (67). Using this observation, wededuce 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 unsteadysolutions, this turns out to be
c(η, τ) =1
Pe−Pη +
∞∑n=0
Cn e−λnτe−Pη2
[sin(∆nη) +
2∆n
Pcos(∆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 (c) must approach the steady state (cs) for any initial concentration(cr,0). This tells us that our steady solution (60) must be stable.
38
4.3 Asymptotic solution for P � nπ
We would like to construct a method to find all positive eigenvalues from ourequation in (75). Firstly, noting that the eigenvalues are ordered as definedin (77), we can see from Figure 10 that as n increases, the ∆n’s approachnπ. Now, for small n, we cannot find the eigenvalues to a satisfactory degreeof accuracy. Therefore, a MATLAB solver has been used to find the first 40eigenvalues. This can be achieved by introducing the function
f(x) := tan(x)− 4Px
4x2 − P 2, (99)
and looking for the points at which f(x) = 0, i.e. where the curve crossesthrough the x-axis.
To find the ∆n’s for larger n, we use the fact that the eigenvalue approachnπ from below as n increases. So, we seek a solution to (75) of the form
∆n = nπ + εn , (100)
where εn → 0+ as n→∞.
Since we are seeking solutions for large n, we can make the assumptionnπ � εn. Hence, since n takes integer values only, we can use the asymp-totic equivalence
tan(nπ + εn) ∼ εn, (101)
Furthermore, if we choose our Rouse Number (P ) such that nπ � P , wehave that 4∆2
n − P 2 ∼ 4∆2n and hence
4P∆n
4∆2n − P 2
∼ P
∆n
. (102)
So, replacing (75) with the asymptotic expressions found in (101) and (102)we obtain
εn ∼P
∆n
. (103)
But we want to take the limit as n → ∞, and εn → 0+, which implies∆n ∼ nπ. This means (103) becomes
εn ∼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
Furthermore, using the relation in (76), and the fact that ∆2n ∼ n2π2 for
P � nπ we have
λ∗n =P 2
4+ n2π2 + ... (106)
4.3.1 Error
We can find the error in the asymptotic approximation of each eigenvalue bysimply calculating the difference between the asymptotic solution (∆∗n, λ
∗n)
and the actual values found in MATLAB (∆n, λn). Hence, the two errorfunctions 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 35 400
0.5
1
1.5
2
2.5
3x 10−3
n
!! n"
!n
5 10 15 20 25 30 35 400
0.005
0.01
0.015
0.02
0.025
0.03
n
!! n"
!n
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: Agraph showing the difference in the asymptotic approximation of λn and theactual value of λn, defined as λ∗n−λn. Note that the Rouse number has beentaken as P = 2.
We would like to determine the value of n to which the asymptotic solutionprovides a result with a sufficiently small error. As seen in Figure 12 (leftgraph), the decay rate of E∆n is very high as the magnitude of the errorapproaches 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 Eλnis not as high (right graph). We do not reach the same error (Eλn = 0.001)until we hit the 40th eigenvalue (λ40).
40
We deduce that for the solution to have a sufficient degree of accuracy, wecan only use the asymptotic approximation for n > 40, providing the Rousenumber does not exceed 2.
4.4 Accuracy
In this section, we aim to show how the accuracy of our solution changesdepending on the number of eigenvalues considered by plotting the concen-tration (c) against the vertical distance (z). Now, since we rescaled both ofthese 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 rangesbetween 0 and 1. Similarly, the concentration will be labelled as P c = c/c0
(once again this has been normalised so the concentration ranges between 0and 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
c(η, 0) = cs(η) + c(η, 0) = cr,0, (109)
where cs was found in (60), c was found in (97), and cr,0 was defined as ourinitial reference concentration.
Since we defined the initial concentration to be constant, we expect to seea vertical line when we plot (109) against η. Figure 12 shows four separatecases:
Case 1: (First 5 eigenvalues)
c(η, 0) = cs(η) +5∑
n=0
Cn e−Pη2
[sin(∆nη) +
2∆n
Pcos(∆nη)
](110)
In this case, we can clearly see that the first 5 eigenvalues do not provideus with an accurate representation of the specified initial concentration. Acurved line which slowly oscillates around P c = 0.4 is produced. This oscil-latory nature of the solution comes from the trigonometric functions in thesolution to the unsteady concentration equation.
41
Case 2: (First 10 eigenvalues)
c(η, 0) = cs(η) +10∑n=0
Cn e−Pη2
[sin(∆nη) +
2∆n
Pcos(∆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 witha smaller amplitude towards the centre. However, this is still not a desirableamount of accuracy.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentrat ion (P c)
height(z
/h)
(1)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentrat ion (P c)
height(z
/h)
(2)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentrat ion (P c)
height(z
/h)
(3)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentrat ion (P c)
height(z
/h)
(4)
Figure 12: Four separate examples displaying the accuracy of the solutionfor t = 0, where the number of eigenvalues included in the summation varies.The initial concentration is set as P cr,0 = 0.4 and the Rouse number P = 2.
Case 3: (First 40 eigenvalues)
c(η, 0) = cs(η) +40∑n=0
Cn e−Pη2
[sin(∆nη) +
2∆n
Pcos(∆nη)
](112)
After taking the first 40 eigenvalues, our line is essentially straight, howeverwe lose accuracy as we approach the top of the river and the riverbed. Thesolution essentially represents a Fourier series, so as we increase the numberof terms used in our summation, the oscillations shorten in wavelength andstay closer to the actual solution.
Case 4: (First 1000 eigenvalues)
c(η, 0) = cs(η) +1000∑n=0
Cn e−Pη2
[sin(∆nη) +
2∆n
Pcos(∆nη)
](113)
42
If we include 960 more eigenvalues, using the asymptotic expressions foundin (105) and (106), we achieve a perfect vertical line. Note that the asymp-totic expression can only obtain an accurate representation providing P issufficiently 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. Bychanging the Rouse number or the initial concentration the solution canchange by a surprising amount. we have plotted several graphs in MATLABto compare the concentration distribution over time.
4.5.1 Physical Interpretation
The solution found in (98) determines how the concentration distributionevolves over time. We set it up to begin with a specific mass of sedimentdistributed evenly throughout the river.
Figure 13: A graph showing how the solution (98) evolves over time withinitial concentration c = 0.37/P , and Rouse number P = 2.
As we move with the flow, a combination of the turbulent fluctuations andthe settling of sediment under gravity allows the sediment distribution tochange over time. We discovered that irrespective of the initial mass of the
43
sediment in suspension, providing the initial concentration distribution isuniform, the distribution will evolve towards the steady state (cs).
Figure 13 represents a situation where the initial (normalised) concentra-tion is P cr,0 = 0.37. This initial concentration has been chosen such that theconcentration 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 ismainly 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 wewould expect. The mass of sediment lost in the in the upper half of the riveris due to sediment settling, which also adds to the mass gained in the lowerhalf of the river.
Figure 14: A graph showing how the solution (98) evolves over time withinitial concentration c = 0.01/P , and Rouse number P = 2.
Another situation is shown in Figure 14 where the initial (normalised) con-centration is P cr,0 = 0.01. In this case there is a small amount of sedimentin suspension at time 0. To allow the initial concentration to approach thesteady 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 theriver 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 thesteady solution, at which time the sediment settling and shear velocity arein equilibrium.
44
If we consider the contrary, where we begin with a very large amount ofgrains in suspension at time 0, the settling of sediment dominates the erosiveupward flux. Figure 15 shows an example of the initial concentration havinga 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 timeprogresses, the distribution of concentration flattens out and steadily evolvesto the shape of our steady solution and the mass of sediment in suspensiondecreases. This occurs because the grains settle onto the riverbed and remainas bed load.
Figure 15: A graph showing how the solution (98) evolves over time withinitial concentration c = 1.8/P , and Rouse number P = 2.
We have looked at what effects occur when we change the initial referenceconcentration. However, we have not considered what significance the Rousenumber has on our solution. We will now go on to look at how the distributionof 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 shearvelocity of the individual grains. For this reason, the magnitude of the Rouse
45
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 (c) tends to the steady state can be de-termined by defining an error function. The error function finds the error ofthe final solution from the steady state after a specific time, and is definedas
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 valueof τ = τmin at which E(τ) = Emax. We will then look at how τmin changesfor different magnitudes of the Rouse number P .
Four different magnitudes of P have been considered in Figure 16 showingthe change in concentration distribution over time with an initial normalisedconcentration of P cr,0 = 0.5. Note that we have only included the first 15eigenvalues 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 insuspension at the steady state is larger due to the ratio of settling velocityover shear velocity being less than 1. Using MATLAB, we can compute theintegral in (114) and hence, by setting E(τ) = Emax = 0.01 we can find thevalue 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 thesteady state reduces. Again, we can compute the integral in (114) usingMATLAB to find the error at time τ . By setting E(τ) = Emax = 0.01 wehave
τmin ≈ 3.12. (116)
46
(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 τmindecreases again. Solving (114) in MATLAB and setting E(τ) = Emax = 0.01yields
τmin ≈ 1.61. (118)
(1) (2)
(3) (4)
Figure 16: (1), (2), (3) and (4) show the time evolution of concentration forP = 0.5, 1, 2, 3 respectively. The initial concentration is fixed at P cr,0 = 0.5in all cases.
47
Hence, since τ is just a non-dimensional representation of time t, we candeduce that for a higher the Rouse number, the rate at which we approachthe 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 ineach solution fixed to 15, however as the Rouse number increases we noticethe 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 εn increases. Hence, the higher theRouse 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 ofthe lowest possible eigenvalue to different magnitudes of the Rouse number.Now, from our solution we know the unsteady perturbation decays at therate e−λnτ . This implies the size of the lowest eigenvalue will give us anapproximate guess as to how fast the unsteady concentration tends to thesteady solution.
(1) (2)
Figure 17: (1) A plot showing the relationship between the Rouse numberand the first (lowest) eigenvalue ∆1. (2) A plot showing the relationshipbetween the Rouse number and the first (lowest) eigenvalue λ1.
From Figure 17 we can see that the decay rate of the unsteady solutionincreases with the magnitude of the Rouse number. The relation between∆n and λn can be found in (76).
48
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 willvary depending on the concentration of the flow. For this reason, we intro-duce a concept known as hindered settling which allows the settling velocityto 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 affectsthe fall velocity of other particles in the near vicinity, decreasing the overalleffective settling velocity (Dankers 2006). Another highly contributing fac-tor is that at higher concentrations, the effective viscosity increases. Eachindividual particle falls in the remainder of the suspension with increased vis-cosity, thus decreasing the effective settling velocity of all particles (Dankers2006). Other processes, e.g. particle collisions, are known to affect hinderedsettling, however the contribution is minimal so we will not consider them.
Hindered settling tends to occur for higher levels of concentration, sincethere is a higher risk of affecting a grains settling velocity when the sedimentin suspension is more condensed. We will identify a function to represent thesettling 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 lowerat high concentrations, a factor commonly used to represent this (Richardsonand Zaki 1954) is:
ws(c) = ws,0(1− c)n, (119)
where ws,0 represents the terminal settling velocity of each particle, (thequantity 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), wherethe Reynolds number decreases as n increases (Baldock, Tomkins, Nielsenand Hughes 2003).
We can now rewrite our unsteady concentration equation (33) in terms of
49
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 initialconcentration 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 10
0.05
0.1
0.15
0.2
0.25
c
!u=
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 correspondto the typical values for n (determined empirically).
50
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 showsthis plot for different values of n, we notice that φu vanishes at c = 0 andc = 1. We would clearly have no upward flux if the concentration was 0 sincethe particles would have instantly diffused into the flow, hence there wouldbe no grains in suspension at any time. Similarly, the upward flux is 0 whenthe concentration is 1 as this is the maximum concentration for non-cohesivesediment; 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 maximumas the concentration changes between 0 and 1. Figure 18 shows how themaximum reduces as we increase the parameter n. We can easily find theconcentration at which the upward flux is maximum by simply differentiatingthe 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 wewish 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
)n5.2 Rescaling
Once again, we would like to rewrite (120) in terms of dimensionless variablesto simplify the problem. We will use the same variables used before in (56)
51
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
, c =Kc
qehc, 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 twounknown constants (P and cmax). Our rescaled hindered equation becomes
∂c
∂τ− P ∂
∂η
[c
(1− c
cmax
)n]=∂2c
∂η2, (126)
subject to boundary/initial conditions
(i) P c(
1− ccmax
)n= − ∂c
∂ηat η = 1,
(ii) ∂c∂η
= −1 at η = 0,
(iii) c = ch,0 at τ = 0,
where ch,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, providingthe upward flux remains in the interval 0 ≤ φu ≤ φu,max. We can write thesolutions to (126) as a linear combination of the steady, time independentsolution cs = cs(η), and a perturbed, time dependent solution c = c(η, τ).(i.e c = cs + c). We would like to determine whether the final solutions to(126) yields a stable (c→ cs as τ →∞) or an unstable solution (c 6−→ cs asτ →∞). 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
)]=∂2cs∂η2
, (127)
Subject to boundary conditions
52
(i) Pcs
(1− cs
cmax
)= −∂cs
∂ηat η = 1,
(ii) ∂cs∂η
= −1 at η = 0.
By integrating up and applying the boundary condition at η = 1, we are leftwith a non-linear, first order ODE given by
−Pcs(
1− cscmax
)=∂cs∂η
. (128)
Separation of variables yields
−P∫
dη =
∫dcscs
+
∫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 mustchoose 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
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 steadysolution has been introduced. This occurs at much higher concentrations andcorresponds 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 (lowconcentration) 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, weset a lower limit for P in (134), and if P drops below this limit, we do notobtain any real solutions. If P = 4/cmax, the two steady solutions collideand 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 perturbedconcentration (c) and check whether it tends to 0 as time tends to infinity.
54
5.3.2 Unsteady Solution (n = 1)
By substituting c = 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
)=∂2c
∂η2+∂2cs∂η2
,
(135)subject to boundary/initial conditions
(i) −P c− Pcs + Pcmax
c2s + 2P
cmaxcsc+ P
cmaxc2 = ∂c
∂η+ ∂cs
∂ηat η = 1,
(ii) ∂c∂η
+ ∂cs∂η
= −1 at η = 0,
(iii) c+ cs = ch,0 at τ = 0.
We can simplify both the PDE and the boundary conditions using what weknow 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
)=∂2c
∂η2, (136)
subject to boundary/initial conditions
(i) −P c+ Pcmax
(2csc+ c2) = ∂c∂η
at η = 1,
(ii) ∂c∂η
= 0 at η = 0,
(iii) c = ch,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
∂ηterm in the PDE, and the c2
term in boundary condition (i). These non-linear terms make it impossibleto solve the differential equation analytically, therefore a MATLAB solvercalled pdepe has been used to produce graphical results.
The solver pdepe solves initial-boundary value problems for systems of parabolicand elliptic PDEs in the one space variable x and time t. The ordinary dif-ferential equations resulting from discretisation in space are integrated toobtain 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
distribution or the Rouse number, we will see how the solutions behave overtime. To simplify the solution, we will normalise the vertical distance (z)and rescaled concentration (c) by dividing them by η and cmax respectively.The results will provide us with a better understanding of the stability of thetwo 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 thedistribution of sediment in suspension varies as we move with the flow. Wefound that two steady states exist: one for low concentrations (cs,1), and onefor high concentrations (cs,2). At time 0, we set up the suspended sedimentsuch that it is distributed uniformly throughout the river. However, we takeinto 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) willevolve 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 Rousenumber is set as P = 2, and the initial concentration is ch,0/cmax = 0.5.
We begin with fixing the Rouse number (P ) to be 2, and beginning the initialconcentration distribution as ch,0/cmax = 0.5. From (123), this happens to
56
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 moresuspended sediment settles onto the riverbed and we move closer to the steadystate 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 takenas P = 2. The initial concentration is set as ch,0/cmax = 0.01.
Figure 21 shows how the solution behaves when we set the initial concentra-tion as ch,0/cmax = 0.01. Once again, over time the concentration distributiontends to the steady concentration state cs,1. This confirms that we pick upa stable solution for low concentrations. This solution yields the same resultthat was seen when we took the settling velocity to be constant.
Now, to understand more about the behaviour of the solution at higherconcentrations, we start the initial concentration close to the steady solu-tion cs,2. Figure 22 shows an example for ch,0/cmax starting at 0.79. As timeincreases, the solution begins to tend towards the steady state cs,2, howeverif we follow the behaviour for long enough, we begin to move away from thesteady state. This is an interesting observation as it suggests that the steadysolution for higher concentrations could be unstable.
57
Figure 22: A plot of the steady solution cs,2 with the Rouse number takenas P = 2. (The solution cs,1 is not shown). The initial concentration is setas ch,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 awaybefore hitting the solution. If the initial concentration is low enough, thestable 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 ofsediment in suspension being deposited onto the bed. However, if we choosea sufficiently high initial concentration, we have a situation in which thesolution will not approach the stable state cs,1. Instead, the erosion fluxseems to dominate the downward flux, allowing more sediment to be pickedup off the bed into suspension. We will define the point at which the initialconcentration no longer tends to a steady state as
ch,0cmax
= 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 candetermine the value of Clim for a given Rouse number. Figure 23 shows thisrelationship.
58
Figure 23: A plot showing a relationship between Clim and P , where we havetaken cmax = 6.
For smaller magnitudes of the Rouse number, the size of Clim decreasesrapidly. 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 initialconcentration such that we reach a stable solution. On the other hand, aswe increase the Rouse number, Clim seems to plateau and approach an upperlimit given by
C∗lim = limP→∞
Clim(P ) (138)
Given the maximum concentration cmax, and choosing P to be sufficientlylarge, one can determine an accurate approximation of this limit.
We would now like to investigate what happens if we do not approach asteady solution over time. From earlier calculations, if we choose the initialconcentration such that
ch,0cmax
> Clim(P ), (139)
then we do not approach the steady state cs,1. This situation is shown inFigure 22.
59
As we move away from the steady solutions, the concentration of sedimentin suspension increases. However, the concentration can never exceed cmaxsince this is the maximum concentration which is physically possible. Forthis reason, we must enforce that the (rescaled) up flux (φu) vanishes forc ≥ cmax. So the up flux becomes
φu = c
(1− c
cmax
)H
(1− c
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 ≥ cmaxas
∂c
∂τ=∂2c
∂η2, (141)
subject to boundary/initial conditions
(i) ∂c∂η
= 0 at η = 1,
(ii) ∂c∂η
= 0 at η = 0,
(iii) c = ch,0 at τ = 0,
where the boundary condition at the free surface has altered due to φu = 0for c ≥ cmax, and furthermore, the boundary condition at the bed has beenchanged so the erosion flux is also 0. This is because no more sediment canbe eroded into suspension if the concentration at the bed is already at amaximum.
The result is simply the well known heat equation. At the point whenc = cmax, no more mass of sediment can be added to the concentrationdistribution. So as a result of the heat equation, we would expect the linerepresenting 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 caseany 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.
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These eigenvalues should give us a good understanding of the stability ofeach steady state.
Going back the rescaled hindered problem in (126), and taking n = 1, wehave
∂c
∂τ− P ∂
∂η
[c
(1− c
cmax
)]=∂2c
∂η2, (142)
Subject to boundary/initial conditions
(i) P c(
1− ccmax
)= − ∂c
∂ηat η = 1,
(ii) ∂c∂η
= −1 at η = 0,
(iii) c = ch,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 techniqueknown as linearisation. Linearisation is used to find a linear approximationof a function at a given point, hence, we can linearise around our steadystates to yield a linear approximation of (126). We begin by taking a smallperturbation around cs, given by
c = cs(η) + εc(η, τ), (143)
where epsilon is defined such that ε� 0.
Now, substituting (143) into our differential equation, and only consideringthe terms of O(ε), we have
∂c
∂τ− P ∂
∂η
[c
(1− 2cs
cmax
)]=∂2c
∂η2, (144)
subject to boundary/initial conditions:
(i) P c(
1− 2cscmax
)= − ∂c
∂ηat 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 methodcalled separation of variables (as was used to solve the unsteady concentrationequation). We begin by rewriting c as
c = F (η)T (τ), (145)
61
where F and T are functions of η and τ respectively.
Substituting this into (144) yields
T ′
T=F ′′
F+ P
(F(
1− 2cscmax
))′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 whichare given by
dT
dτ+ λT = 0︸ ︷︷ ︸
(∗)
d2F
dη2+ P
d
dη
[F
(1− 2cs
cmax
)]+ λF = 0︸ ︷︷ ︸
(∗∗)
. (147)
We can solve (∗) easily since it is simply a first order, linear differentialequation. The solution is
T (τ) = e−λτ , (148)
where the integration constant has been left out as we can absorb it into theF 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 byapplying the boundary conditions. The idea is to substitute our steady so-lutions (cs,1 and cs,2) into (∗∗), and use the eigenvalues to determine thestability 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 increasethe time parameter (τ). This implies the steady concentration solution mustbe stable.
Case 2: At least one of the λn’s is negative.
If one or more of our eigenvalues is negative, then the perturbed solutionwill not vanish as we increase the rescaled time (τ). Due to the exponentialin (148), the perturbed concentration will exponentially increase with time.This implies the steady concentration solution must be unstable.
62
Since (∗∗) is a linear, second order ODE, we can assume the general solutionwill 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 Mapleto solve (∗∗) we find that
f1,n(η) =B±(2∆n + P )e−
12
(2∆n−3P )η + (2∆n − P )e−12
(2∆n−P )η
(1 +B±ePη)2, (150)
f2,n(η) =−B±(2∆n − P )e
12
(2∆n+3P )η − (2∆n + P )e12
(2∆n+P )η
(1 +B±ePη)2, (151)
where B± is defined in (131), and the transformation λn = P 2
4−∆2
n has beenmade to simplify the expression.
Now, the boundary conditions for (∗∗) can be determined from boundaryconditions (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 enableus to determine α and β in terms of ∆n, B± and P . Using Maple, we obtaintwo 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 =
(αβ
)the
corresponding eigenvector.
It is a fundamental result of linear algebra that an equation has a non-zero
63
solution if, and only if, the determinant of the matrix is zero. So, to havenon-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 (indoing so determining B±). We can do this since the hindered concentrationequation 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 Rousenumber is fixed to P = 2. Since (24) implies cmax > 2, we always find twosolutions as expected. The blue line corresponds to B+, and the red linecorresponds to B−. (155) for ∆n.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
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 Rousenumber 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 thestability of each solution depends on the eigenvalues λn. Since we made the
64
transformation λn = P 2
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 P2
or purelyimaginary.
• The steady solution is unstable if there exists a ∆n such that ∆n >P2
.
Since P , cmax and B± are arbitrary constants, we can fix them in order tofind 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 givenabove, 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 toB+ must be stable.
Furthermore, using Maple to solve (155) for ∆n with P , cmax and B− fixedas 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 toB− must be unstable.
Figure 25 shows a graphical result of the solutions to (155) for B− by definingthe 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 thetrivial solution λ = 0, and the solution at 0 corresponds to λ = 1. But wesee there are two more solutions at ±1.25, these correspond to λ < 0 andhence confirm the fact that cs,2 is unstable.
65
Figure 25: A plot showing the relationship between FB−(∆) := M11M22 −M12M21, and ∆. The Rouse number is taken as fixed at 2 and cmax is fixedat 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 cmaxand B± vary. We have the following cases:
Case 1:
If cmax > 2, we obtain 2 solutions. The solution cs,2 corresponds to B− whichranges between 0 and 1. In this case the ∆n’s are strictly greater than 1, thisimplies cs,2 is unstable. The solution cs,1 corresponds to B+ which rangesbetween 1 and ∞. For 1 < B+ < 2 the ∆n’s are strictly less than 1, and forB+ ≥ 2 the ∆n’s are purely imaginary. This implies that the solution cs,1 isstable.
Case 2:
If cmax < 2, we do not obtain any solutions, so there are no stabilities todetermine. This corresponds to the case where the up flux is 0 which wasredefined in (140) in terms of the Heaviside step function.
66
Case 3:
If cmax = 2, we obtain one (repeated) solution corresponding to B± = 1. Inthis 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 theparameter n is typically in the range (2.4, 4.65). Let us consider how thebehaviour of our solution differs if we take n = 3. The problem we wish tosolve is
∂c
∂τ− P ∂
∂η
[c
(1− c
cmax
)3]
=∂2c
∂η2, (163)
subject to boundary/initial conditions
(i) P c(
1− ccmax
)3
= − ∂c∂η
at η = 1,
(ii) ∂c∂η
= −1 at η = 0,
(iii) c = ch,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 solverpdepe to show the how concentration distribution changes through time, we
67
are left with an accurate description of where the two solutions lie. Figure 26shows 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 concentrationequation for n = 3. The arrows show the direction in which the solutionevolves over time, where solutions for t close to 0 haven’t been included. TheRouse number is taken as P = 2.
If we compare the result to the case where n = 1, we notice that the positionof 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 andcloser together. We see once again that if we choose the initial concentrationto be high enough, then we reach a point, denoted Clim, where we do notapproach a stable solution. Using an iterative method in MATLAB we canonce again determine the size of Clim for different magnitudes of the Rousenumber (P ). This relationship is shown in Figure 27 (1).
The behaviour of Clim follows a very similar pattern to the case where wetook n = 1. However, if we hold the value of cmax constant, the size of Pat which Clim hits 0 will increase. This is due to the up flux (φu) having adifferent distribution over c depending on the parameter n.
68
1.5 2 2.5 3 3.5 4 4.5 50
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 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
concentrat ion ( c/cmax)height(z
/h)
(2)
Figure 27: (1) A plot showing a relationship between Clim and P , where wehave taken cmax = 6. (2) A plot showing how the solution behaves when westart the initial concentration at ch,0/cmax = 0, along with the Rouse numberP = 1.4, and cmax = 6.
In Figure 27 (2) we see an example of how the solution behaves if we takeP low enough such that Clim drops below 0. The initial concentration is setto 0, and as time increases we notice that the steady solution is completelybypassed.
6 Conclusion
In this project, we constructed a differential equation which describes thevertical distribution of sediment in a uniform flow. The turbulent nature ofthe flow was modelled using an empirically found relationship with the eddydiffusivity (Dyer and Soulsby 1988). We first considered fully-developed flowprofiles with different representations of the eddy diffusivity.
We then went on to consider how the concentration develops over time. Givenan initial reference concentration, we could determine how the distribution ofsediment evolves. This was achieved by solving a Sturm-Liouville eigenvalueproblem where asymptotic representations of the eigenvalues were found forsufficiently small Reynold’s numbers. This provided an accurate solutionwhich compared reasonably well with experimental results. Further investi-gations were taken to determine what affect the Rouse number had on therate of evolution and accuracy.
69
Finally, a concept known as hindered settling was imposed which takes thesettling velocity’s dependence on the concentration into account. Hinderedsettling tends to occur at higher concentrations since the interactions betweenparticles would be more likely. We found that there were two solutions to thehindered settling equation and using Maple we could analytically determinetheir stability. We discovered that the solution for lower concentrations wasstable and the solution at higher concentrations was unstable, as a result ofhindered settling. Once again, the Rouse number played a vital role in howthe distribution of concentration evolves and whether it approaches a steadystate.
There were many parts of this project that could have undergone furtherinvestigations. For example, we could have modelled the evolution of theconcentration given a non-uniform initial concentration. Or linearisationcould have been trialled for larger n, since n typically takes values in therange (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 cohesivesediment in a uniform flow. This is because of the difficulty in mathe-matically modelling turbulence, with additional unpredictabilities arising forhigher concentrated flows.
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Batchelor, G. K. (2000). “An introduction to fluid dynamics.” CambridgeUniversity Press.
Camenen, B.a, and van Bang, D. P.b (1983). “Modelling the settling of sus-pended sediments for concentrations close to the gelling concentration.”aCemagref, France. bSaint Venant Laboratory, France.
Dankers, P. J. T. (2006). “On the hindered settling of suspensions of mudand mud-sand mixtures.” The Netherlands.
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Dorrell, R. M.,1 and Hogg, A. J.2 (2012). “Length and time scales of re-sponse of sediment suspensions to changing flow conditions.” 1Geographyand Environment, Univ. of Southampton, UK. 2Centre for Environ-mental and Geophysical Flows, School of Mathematics, Univ. of Bris-tol, UK.
Dyer, K. L., Soulsby, R. L. (1988). “Sand transport on the continentalshelf” Annu. Rev. Fluid Mech., 20, 295-324.
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