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