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1
River response to base level riseand other boundary conditions
Dr. Maarten KleinhansSummer course climate change and fluvial systems
Course materials of Prof. Gary Parker
What?n Flown Sediment transportn Mass conservation and equilibrium profilen Effects of changing boundary conditions
n Play with the models of Gary Parker– http://www.ce.umn.edu/~parker/– …/~parker/morphodynamics_e-book.htm
and have fun!
1D SEDIMENT TRANSPORT MORPHODYNAMICSwith applications to
RIVERS AND TURBIDITY CURRENTS© Gary Parker November, 2004
Why?
n We need basic (trained) intuition of the effects of conservation of mass– Input = output - storage
n 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…
flow
Sediment tr.
morphology
The morphodynamic system
n Introductionn River flood wavesn Hydraulic roughnessn Bedforms
n Sediment transportn Mixture effects
n Channel patternsn Downstream finingn Bars, bends, islandsn Overbank sedimentationn Hydraulic geometry
exampleconnections
Flown Flow from old laws:
n Here: Manning-Strickler for friction6/1
=
cr k
HC α
( )
fgRSu
RSCunSR
u
8
2/13/2
=
=
= What’s in a name…Manning law
Chezy law
Darcy-Weisbach law
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!
2
Sediment moved over time
Q
S
S=u3 or u5
X probability
Q
Savg
Qchannel forming
uWhQWhA
uAQ
=→≈
=
Bankfull (channel forming) flow discharge: the flux
Area (bankfull)H depthbankfull
Width
u flow velocity
wettedPerimeter
Q flow discharge m3/s
Intermittency of bankfull discharge
n Approach to full discharge regimen Intermittency of bankfull discharge
– Typically 0.03-0.1 for small flashy to large rivers
n Compute sediment transport for I
t
Qlow flow
flood
Backwater curve
n Subcritical flow, decrease to S=0 (basin)
η
antecedent equilibrium bed profile established with load qsa
before raising base level
water surface elevation (base level) is raised at t = 0 by e.g.
installation of a dam
sediment supply remains constant
at qsa
About Froude: subcritical and supercritical flow
• slow
• downstream control• Fr < 1
• fast• no downstr. control
• Fr > 1
bed shear stress (and sediment mobility)
Flow shear stress on the bed (Newton)
Shields number:Sediment-entraining ‘force’ vs.sediment-detraining ‘force’
( )
( )[ ]50
*
sin
gD
gRSSgR
s ρρττ
ρρτ
−=
≈=
Shear stress
3
Sediment transport
n Meyer-Peter and Mueller type:
n α=8, n=1.5n Einstein parameter:
( ) ∗∗∗∗∗ >−= cn
ctttq ττττα ,
DgDq
qss
tt ]/)[( ρρρ −
=∗
About sediment transport:
0,01
0,1
1
10
Shi
elds
mob
ility
num
ber
Shields criterion
Bed load transport
Suspended transport
sand gravelsilt0.1 1 10 100
Grain size/diameter (mm)
RIJN
GRENSMAAS
ALLIER
(Van den Berg, 1995)
Streampower
Grain size
♣ Allier♦ Meuse♥ Rhine♠ Volga ♣
Braiding, often Fr~1
Meandering, often Fr<<1
♦
♥
♠
About channel pattern: Seven equations needed (1)1. Boundary conditions: Q, H (downstream for Fr<1)2. Water continuity: Q=uwh (= mass conservation)3. Chezy (or Darcy-Weisbach or Manning) u=C(hS)1/2
4. Slope S1. either as fixed boundary condition (mountains, large dis-
equilibrium rivers) or from sediment transport and input! 2. Sediment continuity:
5. Roughness predictor C6. Sediment transport predictor7. Width of the channel W
xq
t)1( b
p ∂∂
=∂η∂
λ− -
λ=porosη=bed levelqb=transpt=timex=location
S transportgradient
Channel changeGeneral rate of change:
1. Exner: ∂η/∂t~∂qb/∂x2. So after a sudden change the gradient (and thus ∂qb/∂x) is
large3. Therefore morph change fast4. But then gradient decreases and morph change less fast5. y(t) = yequil +/- αe-β t
6. Exponential decrease or increase with representative T:
xq
t)1( b
p ∂∂
=∂η∂
λ− -
time
parameter
~63 % of change accomplished at T
Seven equations needed (3)Note:1. Slope of equilibrium channel:
1. More water, less sediment input: smaller slope2. Less water, more sediment input: larger slope3. Sea-level rise/fall: specify as boundary condition H4. Climate change: specify as boundary condition Q (qb)5. Tectonics: specify as raising/falling bed level
2. Equilibrium slope diffusive character:1. Bump ->local flow acceleration ->increase sediment
transport ->bump removed!2. BUT: bedforms and bars! Other extra mechanisms
involved
Slope reacts slowly
4
dischargeand
sediment feeder
sediment bed
waterdepth
h0
water depth h1
slope i
grain flow thickness
hg
Laboratory Seven equations needed (3)Note:3. Roughness predictor:
1. Grain size2. Bedforms! (no bedforms in large grains, large bedforms
in small grains)3. Bars (braid bars, meanders, etc.)
4. Sediment transport predictor (bedload, suspended load)
5. Width of the channel1. NO PHYSICAL PREDICTOR AVAILABLE2. Bank erosion and sediment uptake3. Bank stability: soil type, antecedent deposits,
vegetation…
Very uncertain
Uncertain
Depends on channel pattern
Note: a river may react in various ways to changing Q,Qs, and how is not well known
1. Morphological change
2. River pattern change
3. Meander/bar wavelength change
4. Sediment composition (e.g. coarse top-layer or fine deposit)
2-4 are all ignored in the computer exercises.
River models in practicen Upstream specification Q and qb
n Downstream specification H (or h)n Along river specification of D grain size and W
widthn Along river specification of initial Sn Empirical roughness predictor is calibrated (check
H)n Empirical sediment transport predictor is calibrated
(check rate of bedlevel change)n So no bank erosion; assume fixed banks (Dutch
canals…)n Examples: Sobek, Wendy
Long profiles of rivers
n Often concave! But straight slope expected?
Long Profile of the Amazon River
0
500
1000
1500
2000
2500
3000
-7000 -6000 -5000 -4000 -3000 -2000 -1000 0
x (km)
η (m
)
Quasi-equilibrium long profiles• “quasi” implies not equilibrium where sedimentoutput equals input over each reach. That would nearly always give a straight slope.
Causes of concavity:• 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
5
Effect concavity on width
Kosi River and Fan, India (and adjacent countries).
Image from NASA;https://zulu.ssc.nasa.gov/mrsid/mrsid.pl
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.
Response to base level rise
n Backwater curve and sea level risen Together generate ‘accomodation space’
ηInitial bed
transient bed profile (prograding delta)
Ultimate bed
Initial water surface
Ultimate water surface
Response to change in sediment supply
n Increase in load (but Q unchanged): aggradation
n Decrease: degradation
η
antecedent equilibrium bed profile established with load qta
final equilibrium bed profile in balance with load qt > qta
transient aggradational profile
sediment supply increases from qta
to qt at t = 0
Bed evolution
0
10
20
30
40
50
60
70
80
90
0 2000 4000 6000 8000 10000
Distance in m
Ele
vatio
n in
m
0 yr20 yr40 yr60 yr80 yr100 yrUltimate
Examples aggradation/degradationBed evolution
0
20
40
60
80
100
120
140
160
0 2000 4000 6000 8000 10000
Distance in m
Ele
vati
on
in m
0 yr5 yr10 yr15 yr20 yr25 yrUltimate
aggradation
degradation
Delta progradation
Bed evolution (+ Water Surface at End of Run)
-5
0
5
10
15
20
25
0 10000 20000 30000 40000 50000
Distance in m
Ele
vati
on
in m
bed 0 yrbed 20 yrbed 40 yrbed 60 yrbed 80 yrbed 100 yrbed 120 yrws 120 yr
6
Missouri River progradinginto Lake Sakakawea,
North Dakota.Image from NASA
website:https://zulu.ssc.nasa.gov/mrsid/mrsid.pl
Example delta progradation Response to sudden faulting
n Back to equilibriumn Time scale depends on transport rate and
fault height
η
Computer exercises1. Response to upstream Q and qsn RTe-bookAgDegNormal.xls
2. Response to downstream base leveln RTe-book1DRiverwFPRisingBaseLevelNormal.xls
Optional3. Gilbert-type delta buildingn RTe-bookAgDegBW.xls
4. Response to faultingn RTe-bookAgDegNormalFault.xls
Calculation of River Bed Elevation Variation with Normal Flow Assumption
Calculation of ambient river conditions (before imposed change)Assumed parameters
(Qf) Q 70 m^3/s Flood discharge(Inter) If 0.03 Intermittency The colored boxes:(B) B 25 m Channel Width indicate the parameters you must specify.(D) D 30 mm Grain Size The rest are computed for you.
(lamp) λ p 0.35 Bed Porosity(kc) kc 75 mm Roughness Height If bedforms are absent, set kc = ks, where ks = nk D and nk is an order-one factor (e.g. 3).
(S) S 0.008 Ambient Bed Slope Otherwise set kc = an appropriate value including the effects of bedforms.
Computed parameters at ambient conditionsH 0.875553 m Flow depth (at flood)
τ* 0.141503 Shields number (at flood)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)
τult∗ 0.211523 Ultimate equilibrium Shields number (at flood)
Sult 0.014207 Ultimate slope to which the bed must aggrade Click the button to perform a calculation
Hult 0.736984 m Ultimate flow depth (at flood)
Calculation of time evolution toward this ultimate state
L 10000 m length of reach Ntoprint 200 Number of time steps to printoutqt,g 0.011161 m^2/s sediment feed rate (during floods) at ghost node Nprint 5 Number of printouts
∆ x 1.67E+02 m spatial step M 60 Intervals∆ t 0.01 year time step αu 0.5 Here 1 = full upwind, 0.5 = central difference
Duration of calculation 10 years
Computer exercises - sample
Computer exercises - sample