What?
River response to base level rise
and other boundary conditions
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Flow
Sediment transport
Mass conservation and equilibrium profile
Effects of changing boundary conditions
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Play with the models of Gary Parker
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Dr. Maarten Kleinhans
Summer course climate change and fluvial systems
Course materials of Prof. Gary Parker
– http://www.ce.umn.edu/~parker/
– …/~parker/morphodynamics_e-book.htm
and have fun!
1D SEDIMENT TRANSPORT MORPHODYNAMICS
with applications to
RIVERS AND TURBIDITY CURRENTS
© Gary Parker November, 2004
Why?
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We need basic (trained) intuition of the
effects of conservation of mass
The morphodynamic system
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flow
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– Input = output - storage
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Equilibrium river profile is helpful concept to
test effects of changing boundary conditions
– Upstream: discharge, sediment feed
– Downstream: base level
– Along the river: initial conditions of slope,
sediment composition, entrenchment…
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Sediment tr.
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morphology
n
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Flow
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Flow from old laws:
u =
(R
2/ 3
S 1/ 2
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u = C RS
What’s in a name…
Manning law
Chezy law
8gRS
u =
f
n
)
Darcy-Weisbach law
Here: Manning-Strickler for friction
Introduction
River flood waves
Hydraulic roughness
Bedforms
Sediment transport
Mixture effects
Channel patterns
Downstream fining
Bars, bends, islands
Overbank sedimentation
Hydraulic geometry
example
connections
Channel-forming discharge
• common frequency 1-2.33 years
• definitions based on:
• sediment transport frequency
• channel dimensions
• Herein: assume simple channel dimensions!
• because channel margins (levees, banks) formed
at discharge which just floods the banks ~ bankful
discharge
• equilibrium assumed!
1 /6
H
C = α r  
 kc 
1
Bankfull (channel forming) flow discharge: the flux
Sediment moved over time
Q = Au
A ≈ Wh →
Q = Whu
S=u3 or u5
Q flow discharge m3/s
u flow velocity
S
H depth
bankfull
S avg
Q
Q
Area (bankfull)
Qchannel forming
Width
X probability
Intermittency of bankfull discharge
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Approach to full discharge regime
Intermittency of bankfull discharge
Backwater curve
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Subcritical flow, decrease to S=0 (basin)
water surface elevation (base
level) is raised at t = 0 by e.g.
installation of a dam
– Typically 0.03-0.1 for small flashy to large rivers
n
wetted
Perimeter
Compute sediment transport for I
flood
sediment supply
remains constant
at qsa
Q
low flow
η
antecedent equilibrium bed
profile established with load qsa
before raising base level
t
Shear stress
About Froude: subcritical and supercritical flow
bed shear stress (and sediment mobility)
• slow
• downstream control
• Fr < 1
• fast
• no downstr. control
• Fr > 1
τ = ρ gR sin (S ) ≈ ρ gRS
τ* =
τ
[(ρ s − ρ )gD50 ]
Flow shear stress on the bed (Newton)
Shields number:
Sediment-entraining ‘force’ vs.
sediment-detraining ‘force’
2
About sediment transport:
Sediment transport
10
Meyer-Peter and Mueller type:
(
qt∗ = α t τ ∗ − τ c∗
n
n
)
nt
Suspended transport
Shields mobility number
n
, τ ∗ > τ ∗c
α=8, n=1.5
Einstein parameter:
qt
q =
[( ρ s − ρ ) / ρ s ]gD D
∗
t
RIJN
ALLIER
0,1
2.
Braiding, often Fr~1
3.
4.
♣
Stream
power
1.
♦
2.
Meandering, often Fr<<1
5.
6.
7.
(Van den Berg, 1995)
1.
2.
3.
4.
5.
6.
gravel
100
either as fixed boundary condition (mountains, large disequilibrium rivers) or from sediment transport and input!
Sediment continuity:
transport
S ∂η
∂q
(1− λ p )
= - b gradient
∂t
∂x
Roughness predictor C
Sediment transport predictor
Width of the channel W
Grain size
λ=poros
η=bed level
qb=transp
t=time
x=location
Seven equations needed (3)
Channel change
General rate of change:
sand
0.1
1
10
Grain size/diameter (mm)
Boundary conditions: Q, H (downstream for Fr<1)
Water continuity: Q=uwh (= mass conservation)
Chezy (or Darcy-Weisbach or Manning) u=C(hS) 1/2
Slope S
♥
♠
silt
Seven equations needed (1)
1.
♣ Allier
♦ Meuse
♥ Rhine
♠ Volga
GRENSMAAS
Shields criterion
0,01
About channel pattern:
Bed load transport
1
∂η
∂q
(1 − λp )
=- b
∂t
∂x
Exner: ∂η/∂ t~∂q b /∂ x
So after a sudden change the gradient (and thus ∂qb/∂x) is
large
Therefore morph change fast
But then gradient decreases and morph change less fast
y(t) = yequil +/- α e-β t
Exponential decrease or increase with representative T:
Note:
1.
Slope of equilibrium channel:
Slope
react
s slo
wly
More water, less sediment input: smaller slope
Less water, more sediment input: larger slope
1.
2.
3. Sea-level rise/fall: specify as boundary condition H
4. Climate change: specify as boundary condition Q (qb)
5. Tectonics: specify as raising/falling bed level
2.
parameter
Equilibrium slope diffusive character:
1.
2.
~63 % of change accomplished at T
Bump ->local flow acceleration ->increase sediment
transport ->bump removed!
BUT: bedforms and bars! Other extra mechanisms
involved
time
3
Laboratory
Seven equations needed (3)
discharge
and
sediment feeder
Note:
3.
Roughness predictor:
slope i
water depth h1
water
depth
h0
grain flow
thickness
hg
Very
unce
rtain
1. Grain size
2. Bedforms ! (no bedforms in large grains, large bedforms
in small grains)
3. Bars (braid bars, meanders, etc.)
sediment bed
4.
Sediment transport predictor (bedload, suspended
load)
Unce
rtain
5.
Depends on
Width of the channel
channel pa
ttern
1. NO PHYSICAL PREDICTOR AVAILABLE
2. Bank erosion and sediment uptake
3. Bank stability: soil type, antecedent deposits,
vegetation…
River models in practice
Note: a river may react in various ways to
changing Q,Qs, and how is not well known
1. Morphological change
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2. River pattern change
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3. Meander/bar wavelength change
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4. Sediment composition (e.g. coarse top-layer
or fine deposit)
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2-4 are all ignored in the computer exercises.
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Long profiles of rivers
Quasi-equilibrium long profiles
Often concave! But straight slope expected?
Long Profile of the Amazon River
•
•
•
•
•
•
•
2500
2000
1500
1000
500
0
-7000
• “quasi” implies not equilibrium where sediment
output equals input over each reach. That would
nearly always give a straight slope.
Causes of concavity:
3000
η (m)
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Upstream specification Q and q b
Downstream specification H (or h)
Along river specification of D grain size and W
width
Along river specification of initial S
Empirical roughness predictor is calibrated (check
H)
Empirical sediment transport predictor is calibrated
(check rate of bedlevel change)
So no bank erosion; assume fixed banks (Dutch
canals…)
Examples: Sobek, Wendy
-6000
-5000
-4000
-3000
-2000
-1000
Subsidence
Sea level rise -> downstr. slope decreases
Delta progradation -> downstr. slope decreases
Downstream sorting of sediment -> fining
Abrasion of sediment -> fining
Effect of tributaries: increase of discharge!
Antecedent relief: drop from mountains to the plain
0
x (km)
4
Effect concavity on width
The Kosi River flows into a zone of rapid
subsidence. Subsidence forces a
streamwise decline in the sediment load in
a similar way to sea level rise. Note how
the river width decreases noticeably in the
downstream direction.
Kosi River and Fan, India (and adjacent
countries).
Image from NASA;
https://zulu.ssc.nasa.gov/mrsid/mrsid.pl
Response to base level rise
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Response to change in sediment supply
Backwater curve and sea level rise
Together generate ‘accomodation space’
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Ultimate water surface
Increase in load (but Q unchanged):
aggradation
Decrease: degradation
final equilibrium bed profile in
balance with load qt > qta
Initial water surface
transient aggradational profile
Ultimate bed
Initial bed
sediment supply
increases from qta
to qt at t = 0
η
η
antecedent equilibrium bed profile
established with load qta
transient bed profile
(prograding delta)
Examples aggradation/degradation
Delta progradation
Bed evolution
160
140
100
80
25
20
40
degradation
20
0
0
2000
4000
6000
8000
10000
Bed evolution
Distance in m
90
80
aggradation
0 yr
20 yr
40 yr
60 yr
80 yr
100 yr
Ultimate
70
60
50
40
Elevation in m
60
Elevation in m
Elevation in m
Bed evolution (+ Water Surface at End of Run)
0 yr
5 yr
10 yr
15 yr
20 yr
25 yr
Ultimate
120
bed 0 yr
bed 20 yr
bed 40 yr
bed 60 yr
bed 80 yr
bed 100 yr
bed 120 yr
ws 120 yr
15
10
5
0
30
20
-5
10
0
10000
20000
30000
40000
50000
0
0
2000
4000
6000
Distance in m
8000
10000
Distance in m
5
Example delta progradation
Response to sudden faulting
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Back to equilibrium
Time scale depends on transport rate and
fault height
Missouri River prograding
into Lake Sakakawea,
North Dakota.
Image from NASA
website:
https://zulu.ssc.nasa.gov/mrsid/mrsid.pl
η
Computer exercises
1.
Response to upstream Q and qs
n RTe-bookAgDegNormal.xls
2.
Response to downstream base level
n RTe-book1DRiverwFPRisingBaseLevelNormal.xls
Optional
3. Gilbert-type delta building
n RTe-bookAgDegBW.xls
4.
Computer exercises - sample
Response to faulting
n RTe-bookAgDegNormalFault.xls
Calculation of River Bed Elevation Variation with Normal Flow Assumption
(Qf)
(Inter)
(B)
(D)
Calculation of ambient river conditions (before imposed change)
Assumed parameters
Q
70 m^3/s
Flood discharge
If
0.03
Intermittency
The colored boxes:
B
25 m
Channel Width
indicate the parameters you must specify.
D
30 mm
Grain Size
The rest are computed for you.
(lamp)
(kc)
λp
kc
(S)
S
0.35
75 mm
0.008
Bed Porosity
Roughness Height
If bedforms are absent, set kc = ks, where ks = nk D and nk is an order-one factor (e.g. 3).
Ambient Bed Slope
Otherwise set kc = an appropriate value including the effects of bedforms.
Computed parameters at ambient conditions
H
0.875553 m
Flow depth (at flood)
0.141503
Shields number (at flood)
τ*
q*
0.232414
Einstein number (at flood)
qt
0.004859 m^2/s
Volume sediment transport rate per unit width (at flood)
Gt
3.05E+05 tons/a
Ambient annual sediment transport rate in tons per annum (averaged over entire year)
Calculation of ultimate conditions imposed by a modified rate of sediment input
Gt f
7.00E+05 tons/a
Imposed annual sediment transport rate fed in from upstream (which must all be carried during floods)
qtf
0.011161 m^2/s
Upstream imposed volume sediment transport rate per unit width (at flood)
τ ult ∗
Sult
0.211523
Ultimate equilibrium Shields number (at flood)
0.014207
Ultimate slope to which the bed must aggrade
Hult
0.736984 m
Ultimate flow depth (at flood)
Click the button to perform a calculation
Calculation of time evolution toward this ultimate state
L
qt,g
∆x
∆t
10000 m
0.011161 m^2/s
1.67E+02 m
0.01 year
length of reach
sediment feed rate (during floods) at ghost node
spatial step
time step
Ntoprint
Nprint
M
αu
Duration of calculation
200 Number of time steps to printout
5 Number of printouts
60 Intervals
0.5 Here 1 = full upwind, 0.5 = central difference
10 y e a r s
Computer exercises - sample
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