3 sediment transport in rivers

7/28/2019 3 Sediment Transport in Rivers

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Last update 18.12.2009. Original material by Tuomo Karvonen, 2002.Department of Civil and Environmental Engineering, Helsinki University of Technology

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3 SEDIMENT TRANSPORT IN RIVERS

3.1 Introduction

Water flow in rivers causes sediment transport and the relationship between flow factors (e.g.

velocity, depth) and amount of material transported is a very interesting but complex phenomenaand it has several important engineering aspects like erosion around structures, backfilling of

dredged channels or reservoirs, erosion below a dam, morphological changes in rivers, etc.

The purpose of this material is to cover the basic phenomena related to sediment transport both asbed load and as suspended load. In order fully to understand the sediment transport processes the

material included covers first some very important basic concepts of open channel hydraulics

(sections 3.2 and 3.3):

shear stress in laminar and turbulent flow (section 3.2.1 and 3.2.2) bottom shear stress and friction velocity (section 3.2.4 and 3.2.5) classification of flow layers (section 3.2.7)

velocity distribution in open channel flow (sections 3.3.2 and 3.3.3) bed roughness and bottom friction coefficient (sections 3.3.4 and 3.3.7) drag force (section 3.3.6)

The sediment transport theory and equations for calculating bed load and suspended load are given

in section 3.4.

The text is aimed for students who are not very familiar with the subject but basic knowledge of the

theory of open channel hydraulics is needed:

flow types (laminar, turbulent, steady, unsteady) uniform flow, non-uniform flow basic flow factors like hydraulic radius, wetted perimeter, Reynolds number etc. Froudes number, subcritical and supercritical flow friction losses, Manning, Chezy and Darcy-Weissbach friction coefficients continuity equation energy principle (specific energy, Bernoulli equation) momentum principle

The basic theory of the topics described above can be examined e.g. from the web-pages of theYhd-12.2010 Hydraulics.

Most of the equations shown in this material are definitely well-known, and referenced in numerouspapers. In this case the notations and equations collected by Liu (2001) are used. The material

included is not intended to replace the text books. On the contrary, this should encourage to get

acquainted the theory more thoroughly.

Throughout the text the following symbols have been used for the most often used variables.

The symbols for water and sediment properties:

: density of water depending on temperature and e.g. salt concentration; usually around 1000kg m-3

s: density of natural sediments; usually a value 2650 kg m-3 is used s: relative density is denoted as s=s/ : kinematic viscosity of water [10-6 m2 s-1]
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d, d50: grain diameter is denoted as d or d50, which is the 50 % point in the grain size distributioncurve [m]

Flow factors, velocity distribution parameters:

h: water depth [m]

u: the symbol used for time averaged horizontal velocity of water is u [m s-1] w: the symbol used for time averaged vertical velocity of water is w [m s-1] u*: friction velocity [m s-1] lm: mixing length [m] z0: the point of zero velocity in hydraulically rough flow [m] : theoretical thickness of the viscous sublayer [m]

Friction and shear stress parameters:

: von Karman constant (=0.40) ks : bed roughness

Hr: ripple height [m] f : Darcy-Weissbach friction coefficient C: Chezy coefficient S*: sediment-fluid parameter c: critical Shields parameter : shear stress in laminar flow (viscous shear stress) [N m-2] t: shear stress in turbulent flow [N m-2] b: bottom shear stress [N m-2] b': effective shear stress acting on a grains [N m-2]

Sediment transport variables:

qs,b: bed load transport [m2 s-1] qs,s: suspended load transport [m2 s-1] qs,T: total sediment load transport [m2 s-1]

3.2 Laminar, turbulent and bottom shear stresses

3.2.1 Shear stress in laminar flow

In laminar flow the fluid particles move horizontally without macroscopic mixing, i.e. the laminarflow can be visualised as layers which slide smoothly over each other as shown in Fig. 3-1. The

layers close to the bottom have slower velocity and therefore there is a shear stress between the

different "layers" since the upper layer moves faster than the layer below. This shear stress can also

be called viscous shear stress since it is caused by the viscosity of water.

The shear stress can be given by Newtons law of viscosity

dz

dv=(3-1)


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where is density of water and is kinematic viscosity, v is velocity and z is vertical coordinate(positive upwards). The velocity difference dv/dz determines the magnitude of the shear stress. It

has to be remembered that also in the turbulent flow there is narrow laminar flow layer near the

bottom.

Fig. 3-1. Shear stress in laminar flow (u(z) refers to v(z)).

3.2.2 Shear stress in turbulent flow

In turbulent flow the particles move both horizontally and vertically and there is a continuous

"mixing of particles". Some move upwards and some downwards according to a stochastic process.In turbulent flow the instantaneous velocity components in horizontal (x) direction, V, and in

vertical (z) direction, W can be defined as

'wwW

'vvV

+=+=

(3-2)

where v and w are the time averaged velocities in x and z direction, respectively and v and w arethe instantaneous velocity fluctuations. Time averaged values v and w are usually used in the

computations.

In turbulent flow the particles move irregularly causing continuous exchange of momentum

(remember: momentum is mass multiplied by velocity) from one portion of fluid to another and this

momentum exchange is the reason for the turbulent shear stress, which is also called the Reynolds

stress. The turbulent shear stress, given by time-averaging of the Navier-Stokes equation, is

'w'vt = (3-3)

Eq. (3-3) can be used if measurements of velocities in horizontal and vertical direction are availableover a period (e.g. few minutes) and the readings are stored e.g. several times during one second.

Then it is possible to calculate the time-averaged values v and w from the measurement series andthereafter v' and w' can be calculated for each individual measurement. The Reynolds stress can


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then be calculated by calculating the time-averaged value of v'w' (2nd

momentum, i.e. variance) and

by multiplying the value with density as shown in (3-3). Without real measurements equation (3-3)

is not very useful in calculations and therefore further development is needed for those cases.

Prandtls mixing length theory

Prandtl introduced the mixing length theory in order to calculate the turbulent shear stress. As we

remember from above, turbulent shear stress is caused by particle movement from one layer to

another causing momentum exchange. Prandtl's mixing length theory is based on the assumption

that a fluid parcel has to travel over a length lm before its momentum is transferred. Basically it iseasy to understand the concept of the mixing length: since turbulent stress is caused by momentum

transfer, it is useful to have a physical length that defines when the momentum has been transferred.

Fig. 3-2. Prandtl's mixing length theory (adapted from Liu 2001) (u refers to v).

Fig. 3-2 shows the time-averaged profile. The fluid particle locating in layer 1 and having the

velocity u1, moves to layer 2 due to eddy motion. There is no momentum transfer during the

movement, i.e. the velocity of the fluid parcel is still u1 when it just arrives at layer 2, and decreases

to u2 some time later by momentum exchange with other fluid in layer 2. This action will speed up

the fluid in layer 2, which can be seen as turbulent shear stress t acting on layer 2 trying toaccelerate layer 2 (Liu 2001).

Note: The momentum transfer discussed above happens in the vertical direction. It is useful to point

out now that later on in this material it is shown that there is momentum transfer between the main

channel and floodplain in the horizontal direction which causes an additional resistance to the main

channel flow. That is, there is a shear stress in the imaginary wall between the main channel flowand the floodplain flow. In this case the momentum is transferred in horizontal direction

(perpendicular to the flow direction) and particle moving from the main channel to floodplain will

speed up the flow velocity in the floodplain, which can be seen as additional stress acting on the

boundary.

The horizontal instantaneous velocity fluctuation of the fluid parcel in layer 2 is (see Fig. 3-2).

dz

dvlvv'v m21 ==

(3-4)

and assuming the vertical velocity fluctuation having the same magnitude


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dz

dvl'w m=

(3-5)

where the negative sign is due to downward movement in this case. The turbulent shear stress isnow (see Eq. (3-3)):

22mt

dz

dvl'w'v

==

(3-6)

The very useful definition called kinematic eddy viscosity [m2 s-1] is:

dz

dvl2m=

(3-7)

which has the same unit than the kinematic viscosity used in the shear stress in laminar flow in Eq.(3-1). The turbulent shear stress can now be defined in a similar way compared to laminar (viscous)

shear stress:

dz

dvt =

(3-8)

3.2.3 Total shear stress

The total shear stress is the sum of viscous shear stress (it is acting also in turbulent flow) and theturbulent shear stress:

dz

dv

dz

dvt +=+=

(3-9)


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3.2.4 Bottom shear stress

The derivation of the bottom shear stress is completely analogous to the derivation of the Chezy

coefficient in the "Hydraulics"-course.

Fig. 3-3. Fluid force and shear stress (adapted from Liu 2001)

Assuming uniform flow and balancing the forces acting on the darker area in Fig. 3-3 we get

oz xS)zh(gsinx)zh(gx == (3-10)

which leads to

oz

S)zh(g = (3-11)

where So is the bottom slope. The bottom shear stress (z=0) in uniform flow is thus

ob ghS= (3-12)

Bottom shear stress b for an irregularly shaped cross-section can be easily derived by taking intoaccount that the shear stress is acting on the whole wetted perimeter P over the whole length x.

= sinxgAxPb (3-13)

and by remembering that hydraulic radius R=A/P the bottom shear stress equation will be

ob gRS= (3-14)

For wide channels R is around h and then (3-12) and (3-14) are almost equal.


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3.2.5 Friction velocity

The bottom shear stress is often converted to so called friction velocity u*, which is defined by

ob

* gRSu == (3-15)

The unit of u* is the unit of velocity [m s-1] and it denotes the fluid velocity very close to the bottom

(see Eq. (3-24) in section 3.3.2).

3.2.6 Viscous sublayer

On the bottom of an unvegetated channel there is no turbulence and the turbulent shear stress t =0,and therefore in a very thin layer above the bottom, viscous stress is dominant, and hence flow is

laminar in that layer called viscous sublayer.

Above the viscous sublayer the turbulent stress dominates and the total stress can be calculatedfrom Eq. (3-16), i.e. as the sum of viscous and turbulent shear stresses.

otvz S)zh(g =+= (3-16)

The total stress decreases linearly towards zero when approaching the water surface (z approacheswater depth h). The distribution of the shear stress as a function of z is completely analogous with

the hydrostatic pressure p=g(h-z) with the exception of the viscous sublayer close to the bottom.

3.2.7 Classification of flow layers

According to Liu (2001) the so called scientific classification of flow layers is the one shown in Fig.

3-4.

Fig. 3-4. Scientific classification of flow regions in unvegetated channels (not to scale). Adaptedfrom Liu (2001).


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Near the bottom there is the thin viscous sublayer where there is almost no turbulence.

Measurements show that the viscous shear stress in this layer is constant and equal to the bottom

shear stress b. Flow in this layer is laminar. In the transition layer viscosity and turbulence areequally important. In the turbulent logarithmic layer viscous shear stress is negligible and the

turbulent shear stress is equal to the bottom shear stress. The Prandtl's mixing length theory wasdeveloped for this layer and it leads to the logarithmic velocity profile as shown later on. The

turbulent outer region consists of about 80-90 % of the total region and velocity is relatively

constant due to the strong mixing of the flow (Liu 2001).

3.3 Velocity distribution

It was described in section 3.2.7 that the measurements show that the turbulent shear stress is

constant in the turbulent logarithmic layer and it equals the bottom shear stress. By assuming that

the mixing length is proportional to the distance to the bottom, lm=z (is von Karman constant),

Prandtl obtained the logarithmic velocity profile. By the modifications done to the original theory,the logarithmic velocity profile applies also in the transitional layer and in the turbulent outer layer.

Measured and computed velocities show reasonable agreement. This means that from anengineering point of view, two different velocity profiles need to be considered (Liu 2001):

logarithmic velocity distribution, which covers the transition layer, turbulent logarithmic layerand turbulent outer layer from Fig. 3-4.

velocity profile in the viscous sublayer

3.3.1 Characterisation of smooth and rough flow

It is necessary to characterise the flow as hydraulically smooth or rough since it influences e.g. the

thickness of the viscous sublayer etc.

A very big series of experiments were carried out by Nikuradse for pipe flows. He introduced the

concept of equivalent grain roughness ks, which is usually called bed roughness for open channel

flow. Based on the experiments it was found that the following criteria can be used to characterise if

flow is hydraulically smooth, rough or in the transitional zone:

flowaltransitionllyhydraulica;70ku

5

flowroughllyhydraulica;70ku

flowsmoothllyhydraulica;5ku

s*

s*

s*

(3-18)

where u* is the friction velocity calculated using Eq. (3-15) and is the kinematic viscosity.


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3.3.2 Velocity profile in the turbulent layer

In the turbulent layer the total shear stress is assumed to contain only the turbulent shear stress. The

total shear stress increases linearly with depth

)h

z1()z( bt = (3-19)

According to Prandtl's mixing length theory shown in section 3.2.2 and Eq. (3-6)

22mt

dz

dul

= (3-6)

Now comes the modification of the original theory. Instead of assuming that the mixing lengthlm=z like Prandtl assumed, lm is replaced by equation

5.0m )

h

z1(zl = (3-20)

it is possible to combine (3-20) and (3-6) to get

z

u

z

/

dz

du *b

=

= (3-21)

Eq. (3-21) can be integrated from z0 to h to get the logarithmic velocity profile.

=0

*

z

zln

u)z(u (3-22)

The integration constant z0 is the elevation corresponding to zero velocity (see Fig. 3-5).

Note the similarity to the calculation of aerodynamic resistance ra needed in the estimation of

potential evapotranspiration rate using the Penman-Monteith equation! In these calculations theroughness length is usually around 10-13 % of the crop height!


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Fig. 3-5. Velocity distribution in hydraulically smooth and rough flow (graphs not to scale).

The integration constant z0 (m) is here based on the study conducted by Nikuradse for pipe flows.

+

=

flowaltransitionllyhydraulica;70ku

5k033.0u

11.0

flowroughllyhydraulica;70ku

k033.0

flowsmoothllyhydraulica;5ku

u11.0

z

s*s

*

s*s

s**

0

(3-23)

The friction velocity u* is the fluid velocity very close to the bottom and it is the flow velocity at

elevation z=z0e, i.e.

*e0zzuu ==

(3-24)

3.3.3 Velocity profile in the viscous sublayer

In the case of hydraulically smooth flow there is a viscous sublayer whereas in the rough flow there

is a height z0 with zero flow (Figs. 3-5 and 3-6). Viscous shear stress is constant in the viscous

sublayer and it is equal to the bottom shear stress as shown in Fig. 3-4.

bdz

dv== (3-25)

Eq. (3-25) can be integrated assuming that velocity v=0 when z=0. Integration gives

zv

z)z(v2*

b

=

= (3-26)


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giving a linear velocity distribution in the viscous sublayer. The thickness of the viscous sublayer

can be obtained by finding the z value where the logarithmic velocity distribution intersects the

linear distribution giving a theoretical thickness of the viscous sublayer (Liu 2001).

*v6.11

= (3-27)

3.3.4 Bed roughness

One important question is how to determine the bed roughness ks? Nikuradse made his experiments

by glueing grains of uniform size to pipe surfaces. In completely flat bed consisted of uniformspheres the bed roughness ks would be the diameter of the grains. This cannot be found in nature,

where the bed material is composed of grains with different size and bottom itself is not flat but itincludes ripples or dunes (see section 3.4.3).

According to Liu (2001), the following ks values have been suggested based on different type of

experiments:

concrete bottom k s=0.001 - 0.01 m flat sand bed k s= (1 - 10)d50 bed with sand ripples ks=(0.5 - 1.0)Hr: Hr= ripple height [m]

Fig. 3-6. Summary of the velocity profiles in hydraulically smooth and hydraulically rough flow.


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3.3.5 Connection between bottom shear stress and Chzy coefficient

If one compares the well-known Chzy equation

RSCU = (3-28)

and the equation for friction velocity u*,

gRSu* = (3-15)

it is possible to derive a connection between C, u* and average velocity U.

g

u

UC

*

= (3-29)

In the next step it is necessary to calculate the average velocity from the logarithmic velocity profile

by integrating it from z0 to h (Liu 2001).

)ez

hln(

u

h

z1)

z

hln(

u

dz)z

zln(

h

udz)z(u

h

1U

0

*0

0

*

h

0z0

*h

0z

+

=

=

==

(3-30)

In this way it is possible to relate the Chezy coefficient, water depth h and roughness height/rough

flow and critical velocity u* as follows (Liu 2001):

=

70ku

;flowroughllyHydraulicak

h12log18

5ku

;flowsmoothllyHydraulica3.3

hu12log18

)ez

hln(

gC

s*

s

10

s**10

0

(3-31)

where the equations for z0, (3-23) have been utilized.

3.3.6 Drag force

Flowing water moving past an object will exert a force called drag force

2DD AUC

2

1F = (3-32)


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where CD is the coefficient and A is the projected area of the object to the flow direction. Values of

CD are usually around 1.0..2.0 but experiments are needed for more accurate determination of the

values. Drag force comes partly from the skin friction when water moves around the object and

partly from the pressure the moving water exerts on the object.

Correspondingly to the drag force another force called lift force FL can be defined. It has the sameshape than the drag force:

2LL AUC

2

1F = (3-33)

where CL is the lift coefficient which also need to be determined experimentally.

3.3.7 Bottom friction coefficient f

Liu (2001) shows derivation of a dimensionless friction coefficient f which corresponds to the

Darcy-Weissbach friction coefficient originally derived for pipe flow. The derivation is based on

examining the forces acting on a grain resting on the bed. The drag force is slightly modified from

(3-32) by multiplying the depth averaged velocity U by an empirical coefficient to take intoaccount the fact that the true velocity near the grain on the bottom is somehow related to U. Then it

is possible to examine the shear stress b acting on the grain by saying that the horizontal force isthe drag force acting on A' which is the projected area of the grain to the horizontal plane:

2DD )U(AC

2

1F = (3-34)

222D

Db fU

2

1U)

'A

AC(

2

1

'A

F=== (3-35)

where f is empirical friction coefficient corresponding the Darcy-Weissbach coefficient in pipeflow.

)'A

AC(f 2D= (3-36)

Eq. (3-36) is not useful and therefore the following derivation is needed by utilising Eqs (3-35), (3-

15) and (3-28):

2o

2o

2b

C

g2

RSC

gRS2

U

2f =

=

= (3-37)

where it has been assumed that flow is uniform. Eq. (3-37) can finally be converted to hydraulically

smooth and rough flow conditions by utilising (3-31).


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=

70ku;flowroughllyHydraulica

k

h12log

06.0

5ku

;flowsmoothllyHydraulica

3.3

hu12log

06.0

C

g2f

s*2

s

10

s*2

*10

2

(3-38)

Friction coefficient equation (3-38) together with Eq. (3-35) provide in some cases a useful way to

calculate the bottom shear stress b.

3.4 Calculation of sediment transport in open channels

3.4.1 Sediment transport types

The total transport of sediments, qs,T can be divided to (e.g. Liu 2001)

Bed load transport, qs,b (see section 3.4.4) suspended load, qs,s (see section 3.4.5) wash load, qs,w

Bed load transport is the part of the total load which is more or less continuously in contact with the

bed. The bed load is in close relation to the effective shear stress (section 3.4.3) which acts directly

to the grain surface.

Suspended load is the part of total load which is moving without continuous contact with the bed.

The appearance of ripples will increase bed shear stress and thus the suspended load is related to the

total bed shear stress.

Wash load is composed of very fine particles transported by water but they are not originated from

the bed. The calculation of wash load is not discussed here.

The sum bed load transport, qs,b and suspended load, qs,s is called bed-material load. They are

moving in different layers in the water. If the bottom is completely flat, which usually is not the

case, the thickness of the layer is typically some grain diameters. E.g. Einstein (1950) suggested

that the thickness of the bed load layer is 2d50. When the bed is rippled, the thickness of the bed

load transport layer is often suggested to be the on the order of the ripple height Hr or the bedroughness ks (Bijker 1971).


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Fig. 3-7. Ripples and dunes (adapted from Liu 2001).

The transport formulas have been developed assuming that the lateral transport in the river is very

difficult to forecast and therefore the unit is m3

(m s)-1

, i.e. cubic meters per second per meter width.Sediment transport is calculated in terms of volume of sediment and if it is necessary to take into

account river aggradation/degradation, then sand porosity must be taken into account (see Section

3.5).

Before giving the equations used in the calculation, a brief theory of settling velocity and threshold

of sediments is needed.

3.4.2 Settling velocity and threshold of sediments, Shields diagram

The first step is to develop equation for calculation ofsettling velocity of a grain in still water. The

forces acting on a grain in still water moving downwards at constant velocity v s are the downward

submerged weight of the grain and drag force acting upwards. Drag force acting upwards

2s

2

D2

DD v4

dC

2

1AUC

2

1F

== (3-39)

and submerged weight acting downwards

6

dg)(

3

s

(3-40)

and by equating these

6

dg)(v

4

dC

2

1 3

s2s

2

D

=

(3-41)

the settling velocity can be solved from Eq. (3-42)

== sD

s s;C3

gd)1s(4v (3-42)

The drag coefficient CD of a grain can be calculated based on the Reynolds number


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Re

364.1CD += (3-43a)

=

dvRe s (3-43b)

If Eq. (3-43a) is substituted to (3-42), the settling velocity vs can be calculated from

8.2

d

36gd)1s(5.7

d

36

v

2

s

+

= (3-44)

The next step is to consider threshold of sediments in steady flow. The forces acting on a grain

lying on the bottom are given in Fig. 3-8. The driving force is the drag force exerted on the grain the

resistant force is the friction force.

Fig. 3-8. Vertical and horizontal forces acting on a grain resting on the bed (adapted from Liu

2001).

The drag force is calculated from (3-45) assuming that the velocity used in the drag force equation

is the friction velocity u* close to the bottom multiplied by an empirical coefficient .

2*

2

D2

DD )u(4

dC

2

1AUC

2

1F

== (3-45)

The resisting force is the friction force=friction coefficient f multiplied by the difference between

the downward acting submerged weight W' and upward acting lift force FL (see Fig. 3-8).

2*

2

LL )u(

4

dC

2

1F

= (3-46a)


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6

dg)('W);F'W(f

3

sL

= (3-46b)

The next step is to define a critical friction velocity u*,c, i.e. the situation when the forces are in

balance and the grain is just starting to move.

=

2c*,

2

L

3

s2

c*,

2

D )u(4

dC

2

1

6

dg)(f)u(

4

dC

2

1(3-47)

By re-arranging (3-48) it is possible to obtain a form

2L

2D

2

2c*,

3

4

CfC

f

gd)1s(

u

+

=

(3-48)

where the parameter is defined as the Shields parameter

gd)1s(

u2*

= (3-49)

Now it is possible to define three different conditions when sediments start to move:

gd)1s(

uparameterShieldscriticalor

ustressshearbottomcriticalor

uvelocityfrictioncriticaluu

2c*,

cc

2c*,c,bc,bb

c*,c*,*

=>

=>>

(3-50)

How to determine the critical Shields parameters c? Originally it was taken from the Shieldsdiagram giving c as a function of the grain Reynolds number Re=dgu*/, where dg is acharacteristic grain diameter. The Shields diagram (not shown here) is inconvenient because friction

velocity u* is both in the x-axis in Re and y-axis which is calculated from Eq. (3-49). Madsen et al.(1976) converted the Shields diagram in to the diagram shown in Fig. 3-9 giving the relationship

between the critical Shields parameterc and a so called sediment-fluid parameter S*

=4

gd)1s(dS* (3-51)

The modified Shields diagram is shown below in Eq. (3-52) in the form that can be used e.g. in

Excel-calculations without the need to look the critical Shields parameterc from the graph.


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==

++=

4

gd)1s(dSand)Slog(xwhere

03167.1x636397.0x054252.0x20307.0x06043.0x002235.0)log(

**10

2345c

10

(3-52)

How to use Eq. (3-52) or the diagram? Calculate first the sediment-fluid parameter S* and then from

Eq. (3-52) (or from the graph 3-9) the critical Shields parameterc, which in turn can be used in Eq.(3-50) to calculate the critical friction velocity u*,c and the critical bottom shear stress b,c.

Fig. 3-9. The modified Shields diagram giving critical Shields parameter c as a function of thesediment-fluid parameter S*. The diagram is also given as a fifth order polynomial.

3.4.3 Effective shear stress

Various types of bedforms were shown in Fig. 3-7. According to laboratory experiments the

sequence of bedforms with increasing flow intensity is (Liu 2001):

Flat bed Ripples Dunes High stage flat bed

Antidunes


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Ripples are formed at relatively weak flow intensity and are linked with fine material, with d50 less

than 0.7 mm. The size of ripples is primarily controlled by grain size as follows

50r50r d1000L;d*100H == (3-53)

where Hr is the ripple height (m) and Lr is the length of ripples. At low flow intensity the rippleshave fairly regular form with upstream slope 6 and downstream slope 32.

The dunes have the smaller shape than ripples but they are much larger and they linked with d50bigger than 0.6 mm. Antidunes are formed when flow is supercritical (Froude number greater than

1.0).

Bed roughness ks was briefly discussed in section 3.3.4 for different kind of surfaces.

= bedrippledd100

bedflatd)10..1(

k 50

50

s (3-54)

Bed roughness is thus related to the absence/presence of ripples.

In the presence of ripples, the total shear stress b consists of two parts: Effective shear stress b' originating from the skin friction (grain surface friction) Shear stress due to the form pressure of the ripples b''

''b

'bb

+= (3-55)

Effective shear stress b' is important since it is acting on the single grains and therefore it is crucialfor estimating the bed load transport as will be shown in the section 3.4.4.

In the case of flat bed b'' is zero and bed roughness is taken as ks=2.5d50 and the effective shearstress is calculated as (see Eqs (3-35) and (3-38)):

22

50

10

2'b

U

)

d5.2

h12log(

06.0

2

1fU

2

1

== (3-56)

where h is water depth and U is average velocity. In the case of rippled bed, b' is as given abovebut the total stress is larger due to ripples

22

r

10

2b U

)H

h12log(

06.0

2

1fU

2

1

== (3-57)

where it has been assumed that bed roughness ks equals the height of the ripples Hr.


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3.4.4 Calculation of bed load transport

In bed load transport it is essential to take into account both the bottom shear stress b and theeffective shear stress b' acting on single sediment (skin friction).

In many methods bed load transport is expressed in the form (e.g. Liu 2001).

gd)1s(d

q b,sB

= or gd)1s(dq Bb,s = (3-58)

Therefore, the difference between the methods is in many cases the way to calculate the

dimensionless function B. Unless specified, assume that d=d50 in bedload transport equations.

Meyer-Peter formula

Meyer-Peter (1948) formula is also based on large amount of experimental data and it based on thecritical Shields parameter calculated using Eq. (3-52)

5.1cB )'(8 = (3-59)

where ' is effective Shields parameter calculated as

gd)1s(

/'

'b

= (3-60)

Einstein-Brown formula

According to Liu (2001): The principle of Einstein's analysis is as follows: the number of deposited

grains in a unit area depends on the number of grains in motion and the probability that the

hydrodynamic forces permit the grains to deposit. The number of eroded grains in the same unit

area depends on the number of grains in that area and the probability that the hydrodynamic forces

are strong enough to move them. For equilibrium conditions the number of grains deposited must

be equal to the number of grains eroded, which, together with experimental data fitting gives

350

2

350

2

3B

gd)1s(

36

gd)1s(

36

3

2K

)'(K40

+=

=

(3-61)

where ' is effective Shields parameter calculated as previously.

Kalinske-Frijlink formula

Kalinske-Frijlink (1952) formula is based on curve fitting of all the data available at that time. It

does not use the dimensionless function B.


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='b

50b50b,s

gd)1s(27.0expd2q (3-62)

3.4.5 Suspended load transport

The calculation of suspended load transport is more complicated than that of bed load transport. The

formulation shown in this text is based on steady flow conditions. It is assumed that turbulentfluctuations keep the sediment in suspension. Both the vertical sediment concentration distribution

c(z) and velocity distribution u(z) have to be estimated.

The estimation of the velocity distribution has been given in sections 3.3.2 and 3.3.3. In turbulentlayer the velocity distribution is calculated from Eqs (3-22) and (3-23) and in the viscous sublayer

from Eq. (3-26).

The derivation of the concentration distribution is based on the following assumptions:

concentration c(h) = 0, i.e. at water surface concentration is zero the bed load transport is assumed to take place in a layer with thickness a in Fig. 3-10 and

suspended load takes place in the layer from z = a to z = h

the reference concentration ca needed is calculated assuming constant concentration in the bedload transport layer

the total suspended load can be calculated by integrating the c(z) multiplied by u(z) from z = ato z = h (see Fig. 3-10.)

Based on the assumptions given above it is possible to derive equation for the concentration

distribution c(z) (Liu 2001).

)*u

sv(

aah

a

z

zhc)z(c

= (3-63)

It is assumed that the thickness of the bed load transport layer is equal to the bed roughness ks and

in this layer there is a constant sediment concentration ca.


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Fig. 3-10. Illustration of vertical distribution of c(z) and u(z) (adapted from Liu 2001).

The depth averaged velocity in the bed load transport layer Ub is calculated assuming that close to

the bottom there is a viscous sublayer with linear velocity distribution and above that a turbulent

logarithmic velocity profile. This leads to an equation which can be used to calculate averagevelocity in the bed load transport layer.

*

sk

e0z 0

*0

*

sb u34.6dz)

z

zln(

uez

u

2

1

k

1U

+

= (3-64)

and in this way the bed load transport can be given as

asbb,s ckUq = (3-65)

which finally gives the average sediment concentration in the bed load transport layer.

s*

b,s

sb

b,sa

ku34.6

q

kU

qc == (3-66)

Now the concentration profile c(z) in Eq. (3-63) can be calculated and the suspended load can be

calculated by two different methods.

Method 1

Method 1 is based on direct numerical integration of c(z) multiplied with u(z)


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=h

as,s dz)z(c)z(uq (3-67)

The idea is to calculate both c(z) and u(z) from z=a to z=h and use e.g. discretisation small enough

(e.g. 0.005..0.01 m) to calculate the integral numerically.

Method 2

Method 2 given by Liu (2001) is based on quasi-analytical solution of the integral as shown in Eq.

(3-68). The Einstein integrals I1 and I2 needed in (3-68) have to be evaluated by integrating

numerically Eqs (3-69). Since numerical integration is needed also in the method 2 it seems

preferable to use method 1 and integrate numerically only one equation.

+=

=

=

2s

1a*

h

a

)*u

sv(

a0

*

h

as,s

I)k033.0

hln(Iacu6.11

dzah

a

z

zhc)

z

zln(

u

dz)z(c)z(uq

(3-68)

=

=

1

A

*z

*z

)1*

z(

2

1

A

*z

*z

)1*

z(

1

dB)Bln(B

B1

)A1(

A216.0I

dBB

B1

)A1(

A216.0I

(3-69)

*

s*

su

vz;hzB;

hkA

===(3-70)

By combining Eqs (3-68) and (3-66) the final equation used in the method 2 is obtained:

+= 2

s1b,ss,s I)

k033.0

hln(Iq83.1q (3-71)


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3.4.6 Total sediment transport

The total sediment transport is the sum of the bed load transport and suspended load transport.

Bijker's method is based on summing the bed load transport from Eq. (3-58) and suspended load

transport either from numerical integration of (3-67) or from (3-71).

++=+= 2

s1b,ss,sb,sT,s I)

k033.0

hln(I83.11qqqq (3-72)

Engelund's method is based on one single equation

g)1s(

d

gd)(

U*05.0q 505.1

50s

b2T,s

= (3-73)

3.4.7 Examples of sediment transport equations

[sedim_bed_suspended.xls]

3.4.7.1 Numerical integration

In suspended sediment transport it is necessary to integrate numerically either the integral

= ha

s,s dz)z(c)z(uq (3-67)

or the Einstein integrals I1 and I2.

=

=

1

A

*z

*z

)1*

z(

2

1

A

*z

*z

)1*

z(

1

dB)Bln(B

B1)A1(

A216.0I

dBB

B1

)A1(

A216.0I

(3-69)

*

s*

s

u

vz;

h

zB;

h

kA

=== (3-70)

Integration means simply calculation of area between the integration lower and upper limit

restricted by the function to be integrated. Numerical integration is based on dividing the integration

interval into small trapezoids and integral is the sum of area of these trapezoids.


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A convenient way is to use Excel to numerical integration and the best way to illustrate the method

is to take an integral with known analytical solution and then compare the numerical solution with

the analytical one.

Example 3-1. Calculate numerically the integral

+ =2

0

22

0

dx)2x(dx)x(f (3-74)

using interval length x=0.01, i.e. N=200 intervals. Here f(x)=x2+2. The exact solution can beeasily found in this case to be 20/36.666667. The numerical integration is done in such a way thatthe function value f(x) in Eq. (3-74) is calculated for all values starting from integration lower limit

(x1=0 here) up to and including the integration upper limit (xN+1=2 here). The numerical

approximation of the integral Inum can then be calculated as follows:

[ ] x)x(f5.0..)x(f)x(f)x(f5.0I 1N321num ++++= + (3-75)

where the first and last function values f(x1) and f(xN+1) have the weight 0.5 since these values are

included only in one trapezoid whereas all the other function values are included in the calculation

of the area of two small trapezoids.

The results of the integral are shown in Fig. 3-11. The result is a very good approximation as shownbelow, i.e. numerical value is 6.6667 compared to exact=20/3. The Sum shown in the graph is

calculated by taking the first and last value with weight 0.5 as shown in Eq. (3-75).

Fig. 3-11. Numerical integration example.

Example S-1.: Numerical integrationdelx 0.01

i x f(x)

1 0 2

2 0.01 2.0001

3 0.02 2.0004

4 0.03 2.0009

. . .

. . .

197 1.96 5.8416

198 1.97 5.8809

199 1.98 5.9204200 1.99 5.9601

201 2 6

Sum 666.67

Integral 6.6667


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3.4.7.2 Bed load transport

Example 3-2. Calculate bed load transport when input data is given in Fig. 3-12. The calculation of

bed load transport is relatively straightforward. The following input data are needed (Fig. 3-12).

The relative density s=s/ is also calculated.

Fig. 3-12. The input data needed for calculation of bed load transport.

The algorithm for calculating bed load transport can be summarised as follows.1) Estimate the bed roughness ks and the ripple height Hr. In the case that ripple height is not

measured, it can be assumed that Hr=100d50. If the bottom is assumed to be flat, bed roughnessvalue ks is around 2.5d50. In the case of ripples it can be assumed that ks=(0.75..1.0)Hr. In the

examples calculated below it is assumed that ks=Hr.

2) Calculate bottom friction factor f needed in the calculation of bottom shear stress b in Eq. (3-35). Eq. (3-38) can be used to calculate f. The checking of the form of flow (rough or smooth)

has to be done when friction velocity u* is known.

3) The critical Shields parameterc needed in many bed load transport equations is calculated from

Eq. (3-52) or taken from Fig. 3-8 as a function of the sediment-fluid parameter S*.4) Bottom shear stress b and effective shear stress b' have to be calculated in the next step fromEqs (3-35) and (3-56).

5) Effective Shields parameter' can also be calculated at this stage from Eq. (3-60).

The summary of the variables calculated using the input data given in Fig. 3-12 is given in Fig. 3-13.

Fig. 3-13. Calculated variables needed in bed load transport calculations.

All the necessary variables are now available for calculating the bed load transport using the Eqs

given in (3-58)..(3-62). The results are shown for Meyer-Peter, Einstein-Brown and Kalinske-

Frijlink equations in Fig. 3-14. It can be seen that there is a quite big difference between the results

given by different methods indicating how difficult (and stochastic as nature) the problem is.

Input data

visc 1.00E-06 Viscosity (m^2/s)

d50_g 2.50E-04 d50 (m)

hhh 1.5 Water depth (h)

UUU 1 Average velocity (m/s)

Karman 4.00E-01 Von Karman constant

roow 1010.0 Density of water (kg/m^3)

roos 2650.0 Density of sediment (kg/m^3)

sss 2.624 s=roos/roow

Calculated variables needed in Bed load transport Eqs.

Hr 0.0250 Hr=Ripple height =Hr=100*d50 (here)

ks 0.0250 Bed roughness =ks = Hr (here)

fff 0.007349024 friction coefficient f Sstar 3.944080428 Sediment-fluid parameter Sx

Shields 0.044041829 Critical Shields parametercTau 3.711257324 Bottom shear stress b(N/m^2)Tau_eff 1.523671109 Effective shear stress b'Sh_eff 0.37882477 Effective Shields parameter'


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Fig. 3-14. Bed load transport in units m3

(m s)-1

.

Warning! With "suitable" input data (e.g. d50=50 m, h=2 m and U=0.3 m s-1) it is possible that thecalculated value for the effective Shields parameter ' will be smaller than the critical Shields

parameterc leading to a case that the power in Meyer-Peter equation (3-59) cannot be calculated('-c will be negative). This type of situation can be found when d50 and U are both quite small. Inthese cases e.g. the Einstein-Brown formula works properly.

3.4.7.3 Calculation of suspended load transport

Example 3-3. Calculate suspended load transport when input data is given in Fig. 3-12.

In the calculation of the suspended load transport it is necessary to follow the procedure:

1) Calculate the settling velocity vs from Eq. (3-42)

2) Calculate the friction velocity u* from bottom shear stress using Eq. (3-15).

3) The thickness of the viscous sublayer is calculated, but for rough flow this value is not

necessarily needed in the calculation of the velocity distribution u(z).4) Check that flow is hydraulically rough from (3-18).

5) In the calculation of the velocity distribution u(z) the elevation z0 corresponding to zerovelocity has to be calculated from Eq. (3-23) (see Fig. 3-6).

6) The average concentration ca in the bed load transport layer is needed in the concentration

distribution function c(z). In calculating ca from Eq. (3-66) it is necessary to choose somebed load equation. Meyer-Peter was chosen in the results given in Fig. 3-15. Einstein-Brown

might be better to ensure that the problem with Meyer-Peter (warning in 3.4.7.2) does not

prevent the calculation.

7) In the Method 1 given in Eq. (3-67) it is necessary to numerically integrate c(z)u(z) from

z=a to z=h. The integration interval was in this case divided to 200 trapezoids and z canthen be calculated as z = (h-a)/200 where a is assumed to be equal to the bed roughness ks(see Fig. 3-10).

8) In Method 2 it is necessary to calculate numerically the Einstein integrals I1 and I2, which

are needed in Eq. (3-71). Therefore it is necessary to calculate the auxiliary variables A and

z* from Eq. (3-70).

Fig. 3-15. Variables needed in the calculation of suspended load transport.

Bed load transport (m^3/(m*s)

Meyer 2.44479E-05 Meyer-Peter bed load

K 0.520409322 coeff K in Einstein formula

Einst 1.78536E-05 Einstein-Brown bed load

Kalinske 1.48606E-05 Kalinske-Frijlink

Variables needed in suspended load calculations

wss 0.028911172 Settling velocity (m/s)

uss 0.060617755 Friction velocity ux (m/s)

Thick_v 0.000191363 Thickness of the viscous sublayer (m)

u*ks/Visc 1515.443871 Test if flow hydraulically rough>70

z0 0.000825 Elevation corresponding to zero velocity

ca 0.002544554 Max. concentration at z=a=ks

delz 0.007375 Interval in numerical integration (200 points)

AAA 0.016666667 A in Einstein integrals

Zstar 1.192355778 zx in Einstein integrals


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Finally, it is possible to calculate the suspended load transport either by numerically integrating (3-

67) (Method 1) or by calculating numerically the Einstein integrals and then applying (3-71) by

selecting some bed load transport equation (Method 2, Meyer-Peter chosen for qs,b in the results

given below).

Fig. 3-16. Results from the two methods to calculate suspended load transport.

As shown in Fig. 3-16, the results of methods 1 and 2 differ slightly. It is likely that Method 1 is

more accurate if the integration interval is small enough. In Method 2 the problems are related to

several approximations done in deriving Eq. (3-71) and in the Einstein integrals which also need to

be estimated numerically. Moreover, in Method 2 qs,b is needed and the choice of the formula

influences the suspended load result. In Method 1, qs,b is also needed when ca is calculated butotherwise qs,s from Method 1 is independent from the choice of the equation used to calculate qs,b.

3.4.7.4 Calculation of total sediment transport

Example 3-4. Calculate total sediment transport, i.e. the sum of the bed load transport and

suspended load transport.

The two methods, Bijker's method (using Meyer-Peter) and Engelund's method give the results

shown in Fig. 3-17.

Fig. 3-17. Total sediment transport and the fraction of suspended load transport from total sediment

transport.

The results show quite big difference between the two methods. Moreover, a very useful result isthe fraction of suspended load transport from the total transport. In this case about 80 % from the

total transport is in suspended form. In this example d50 = 0.25 mm (2.5x10

-4

m). As shown in thefollowing example, the results are quite different when d50 is smaller.

Example 3-5. Calculate bed load transport, suspended load transport and total load transport when

input data is shown in Fig. 3-18.


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Fig. 3-18. Input data for sediment transport case when d50 = 60 m and velocity U=0.4 m s-1.Calculated variables needed in the equations are shown also.

The results for bed load transport, suspended load transport and total sediment transport are shown

in Fig. 3-19. Due to the fact that d50 is much smaller than in Examples 3-2..3-4, the bed load

transport is very small compared to suspended load transport. In this case 99 % of the total load istransported in suspended form.

Note that in this case the Meyer-Peter equation does not work if velocity would be e.g. only 0.3 m s-

1

. It would lead to a situation that the calculated value for the effective Shields parameter ' wouldbe smaller than the critical Shields parameterc. In this case Einstein-Brown should be used.

Input data

visc 1.00E-06 Viscosity (m^2/s)

d50_g 6.00E-05 d50 (m)

hhh 2 Water depth (h)

UUU 0.4 Average velocity (m/s)

Karman 4.00E-01 Von Karman constant

roow 1010.0 Density of water (kg/m^3)

roos 2650.0 Density of sediment (kg/m^3)

sss 2.624 s=roos/roow

Calculated variables needed in Bed load transport Eqs.

Hr 0.0060 Hr=Ripple height =Hr=100*d50 (here)

ks 0.0060 Bed roughness =ks = Hr (here)

fff 0.004624336 friction coefficient f

Sstar 0.463727259 Sediment-fluid parameter Sx

Shields 0.150818517 Critical Shields parametercTau 0.373646336 Bottom shear stress b(N/m^2)Tau_eff 0.179006174 Effective shear stress b'Sh_eff 0.185440207 Effective Shields parameter'


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Fig. 3-19. Bed load transport, suspended load transport and total sediment transport for the casewhen d50 = 60 m and velocity U=0.4 m s-1.

3.5 River aggradation and degradation models (movable bed models)

3.5.1 Introduction

Sediment transport in rivers is a very special problem in fluvial hydraulics. Transported sediments

may render nearly any man-made structure useless. By 1973, 33 % of the U.S. reservoirs built

before 1935 had lost from 25 % to 50 % of their original capacity while another 14% of thesereservoirs had had their capacity reduced between 25 % and 50 %. One out of ten reservoirs lost all

usable storage (ASCE, 1975). Other problems due to sediment transport are structural or

morphological deterioration due to degradation below dams, deterioration of water intake entrance

conditions, filling up of irrigation channels, and shifting of stream alignment, not to mention

pollution and ecological consequences. The essential difficulty of dealing with the situation is thepresent human incapacity to obtain a satisfactory quantitative description of sediment transport

phenomena (Cunge 1980).

Existing mathematical models are nearly all based upon the idea that it should be possible tosimulate hydrological flow conditions and the concomitant change in longitudinal profile of a river

over a period of 20-50 years, using a chosen sediment transport formula. The mathematical models


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described in this chapter are one dimensional, representing only longitudinal bed profiles,

longitudinal free surface profiles and sediment transport as a function of time and hydraulic flow

conditions. Nevertheless they can be used to solve numerous problems associated with river bed

evolution in response either to natural conditions or to man-made developments.

The following natural phenomena are examples of what has been or can be successfully simulatedwith such models (Cunge 1980):

Bed variations during floods in the lower reaches of rivers flowing into the sea (Perdreau andCunge, 1971; Bouvard, Chollet and Cunge, 1977; Chollet, 1977).

Delta formation at river mouths, alluvial fan formation below foothills (Chang and Hill, 1977). Bed variations during floods in the vicinity of gorges (Blanchet, 1971) or at river crossings (de

Vries, 1973a).

Bed variations downstream of tributaries or at bifurcations (de Vries, 1973b). Long term natural evolution of a river bed (Chen and Simons, 1975).

Some of the river improvement problems which may be studied with the help of such mathematicalmodels are:

Modifications of water flow and sediment transport due to dam construction (Chollet andCunge, 1979).

Establishment of long term operating rules for dams in order to preserve useful reservoir volumeby flushing operations; determination of necessary reservoir volume (Lugiez, 1976; Chollet and

Cunge, 1979).

Modifications of river morphology in response to river training works, cutoffs or alignmentchanges (Verdet, 1975).

Deposition of materials, and dredging operations (Verdet, 1975). Modifications to river morphology due to the withdrawal of water for irrigation, industrial use,

etc. (de Vries, 1973a).

3.5.2 Modelling river degradation and aggradation

The sediment transport equation developed in section 3.4 can be applied in so called river

aggradation/degradation models. The partial differential equations to be solved are shown in Eqs (3-

76):

3/10

22

f

tot,stot,s

f

22

h

qnS

0x

q

q

hqz)p1(

t

0ghSx

zgh

2

gh

h

q

xt

q

0x

q

t

h

=

=

+

+

=++

++

=+

(3-76)

where t is time [s], x is distance in the horizontal direction [m] h is water depth [m], q is dischargein cubic meters per meter width unit width [m2 s-1], z is the bottom elevation the channel, Sf is


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friction slope calculated e.g. using the Manning equation, p is porosity of the bed material, q s,tot is

the total sediment transport calculated as the sum of bed load transport and suspended load transport

according the methods described in sections 3.3.3-3.3.5.

The first two equations of (3-76) are the "standard" Saint Venant equations applied in a form that

assumes a wide channel and transport of water and sediments are calculated per meter width due tothe fact that in long channel the methods cannot predict transversal transport of sediments and

therefore it is practical to consider a one meter wide "tube".

The third equation is capable of taking into account the fact that sediment transport changes themorphology of the river. E.g. sediment can be eroded (=degradation) from a certain section of the

river and will be settled down when velocity decreases (aggradation).

Examples of river bottom degradation and aggradation

Some illustrative computer simulations are shown as animated images. The simulations have been

carried out with a computer program AGG_DEG.EXE that solves simultaneously the three partialdifferential equations (3-74): continuity equation for water and sediment and momentum equation

for water. The model calculates the changes in the channel bottom level as influenced by changes in

sediment transport in the channel.

The initial situation is that uniform flow conditions are prevailing in a 5000 m long channel:

discharge is 4.83 m3/s per meter width flow is uniform and surface slope equals the bottom slope So=0.0015 and water depth is 1.73 m Manning coefficient n is very small = 0.02 sediment discharge is constant in the whole channel and elevation of the river bottom does not

change

A) The only change is that sediment inflow from upstream is increased by 50 %. The result

of increase in sediment inflow is aggradation of the sediment starting from the upstream

end of the channel. [sed_ani3.gif]

B) A dam is built at downstream end and water depth behind the dam is 5.0 m. i.e. increase

from 1.73 to 5.0 m. The result is decreased velocity above the dam and sediment

aggradation that slowly starts to fill the reservoir decreasing the useful storage volume.[sed_ani2.gif]

C) The dam has been built at downstream end and sediment inflow from upstream isincreased by 50 %. Increase in sediment inflow from upstream and the dam at

downstream end together have the effect that bottom elevation of the channel isincreased along the whole channel section. [sed_ani4.gif]

D) The dam has been built and there is an upstream sedimentation reservoir that decreases

the sediment inflow from upstream by 50 %. The result of decreased sediment inflow is

a gradual degradation at the upstream end and aggradation at downstream end due to the

influence of the dam. [sed_ani5.gif]

E) The dam is at downstream end and upstream sedimentation reservoir cuts the sediment

inflow from upstream. Due to the fact that sediment inflow is negligible from upstream,

the river bed degrades heavily at the upstream end and the degraded material is

aggradated above the dam. [sed_ani1.gif]

In the animated images the brown line is the original bottom elevation and the yellow line showsthe new bottom elevation as influenced by changes in sediment transport in the channel.

3 sediment transport in rivers

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