1

Another Chapter inTHE SEARCH FOR THE HOLY GRAIL:

A MECHANISTIC BASIS FOR HYDRAULIC RELATIONS FOR BANKFULL FLOW IN GRAVEL-BED RIVERS

Gary Parker

With help from François Metivier and John Pitlick

2

What is the physical basis relations for bankfull geometry of gravel-bed streams?

3

Where do the following relations come from?

• Bankfull Depth Hbf ~ (Qbf)0.4

• Bankfull Width Bbf ~ (Qbf)0.5

• Bed Slope S ~ (Qbf)-0.3

where Qbf = bankfull discharge

4

THE GOAL:

A Mechanistic Description of the Rules Governing Hydraulic Relations at Bankfull Flow in Alluvial Gravel-bed Rivers

The Parameters:

Qbf = bankfull discharge (m3/s)QbT,bf = volume bedload transport rate at bankfull

discharge (m3/s)Bbf = bankfull width (m)Hbf = bankfull depth (m)S = bed slope (1)D = surface geometric mean or median grain size (m)g = gravitational acceleration (m/s2)R = submerged specific gravity of sediment ~ 1.65 (1)

The Forms Sought:bTsbh n

bfbf,bTnbf

nbfbf

nbfbf Q~Q,Q~S,Q~B,Q~H

5

DATA SETS1. Alberta streams, Canada1

2. Britain streams (mostly Wales)2

3. Idaho streams, USA3

4. Colorado River, USA (reach averages)

1 Kellerhals, R., Neill, C. R. and Bray, D. I., 1972, Hydraulic and geomorphic characteristics of rivers in Alberta, River Engineering and Surface Hydrology Report, Research Council of Alberta, Canada,No. 72-1.2 Charlton, F. G., Brown, P. M. and Benson, R. W., 1978, The hydraulic geometry of some gravel rivers in Britain, Report INT 180, Hydraulics Research Station, Wallingford, England, 48 p. 3 Parker, G., Toro-Escobar, C. M., Ramey, M. and Beck S., 2003,The effect of floodwater extraction on the morphologyof mountain streams, Journal of Hydraulic Engineering, 129(11), 2003.4 Pitlick, J. and Cress, R., 2002, Downstream changes in the channel of alarge gravel bed river, Water Resources Research 38(10), 1216,doi:10.1029/2001WR000898, 2002.

6

NON-DIMENSIONALIZATION

2

bf,bTT2

bf5/2

bf

bf5/1

5/2bf

bf5/1

DgD

QQ,

DgD

QQ,

Q

BgB~

,Q

HgH~

These forms supersede two previous forms, namely

which appear in reference 3 of the previous slide. Note:

D

BB,

D

HH bfbf

5/25/2 QB~

B,QH~

H

7

WHAT THE DATA SAY

0.0001

0.001

0.01

0.1

1

10

100

1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07

Qhat

Bti

lde,

Hti

lde,

S

Britain widthAlberta widthIdaho widthColorado widthBritain depthAlberta depthIdaho depthColorado depthBritain slopeAlberta slopeIdaho slopeColorado slope

H~

B~

S

The four independent sets of data form a coherent set!

8

REGRESSION RELATIONS BASED ON THE DATA

y = 0.3785x4E-05

y = 4.6977x0.0661

y = 0.1003x-0.3438

1.E-04

1.E-03

1.E-02

1.E-01

1.E+00

1.E+01

1.E+02

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Qdim

Bd

imti

lde,

Hd

imti

lde,

S

BdimtildeHdimtildeSPower (Hdimtilde)Power (Bdimtilde)Power (S)

344.00661.000004.0 Q100.0S,Q70.4B~

,Q379.0H~

To a high degree of approximation,

379.0H~

H~

c Remarkable, no?

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WHAT DOES THIS MEAN?

4.0bfbf

4.0bfbf

Qg

379.0H

orQ~H

461.0bf

2/10661.02/550sbf

461.0bfbf

QgDg70.4B

orQ~B

344.0bf

344.02/550s

344.0bf

QDg100.0S

orQ~S

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THE PHYSICAL RELATIONS NECESSARY TO CHARACTERIZE THE PROBLEM

Required: four relations in the four unknownsHbf, Bbf, S, QbT,bf.

1. Resistance relation (Manning-Strickler):

2. Gravel bedload transport relation (Parker 1979 approximation of Einstein 1950):

3. Relation for channel-forming Shields number bf*

(Parker 1978): and

4. Relation for gravel yield from basin (not determined solely by channel mechanics).

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RESISTANCE RELATION

rn

bfr

bfbfbf

bf

bf,

bf

D

H

SgHHB

Q

u

UCz

Manning-Strickler form: where Ubf = Qbf/(Bbf Hbf) denotes bankfull flow velocity,

Here we leave r and nr as parameters to be evaluated.

12

BEDLOAD TRANSPORT RELATION

Use Parker (1979) approximation of Einstein (1950) relation applied to bankfull flow:

2.11,RD

SH

where

1DRgDB

Qq

Gbf

bf

5.4

bf

c2/3

bfG

bf

bf,bTbf

13

RELATION FOR CHANNEL-FORMING SHIELDS NUMBER

Base the form of the relation on Parker (1978):

constrc

bf

14

RELATION FOR GRAVEL YIELD FROM BASIN AT BANKFULL FLOW

This relations is external to the channel itself, and instead characterizes how the channels in a watershed interact with the unchannelized hillslopes. The necessary relation should be a dimensionless version of the form

where nbT must be evaluated.

bTnbfbf,bT Q~Q

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WORKING BACKWARD

Rather than working forward from the basic physical relations to the hydraulic relations, let’s work backward and find out what the form the physical relations must be to get the observed hydraulic relations.

SB nS

nBo QS,QB

~,H

~H~

344.0n,100.0,0661.0n,70.4,379.0H~

SSBBo

Recall that

2

bf,bTT2

bf5/2

bf

bf5/1

5/2bf

bf5/1

DgD

QQ,

DgD

QQ,

Q

BgB~

,Q

HgH~

D

BB,

D

HH bfbf

16

Now using the definition of Cz, the non-dimensionalizations and the relations

it is found that

But so that

RESISTANCE RELATION

SB nS

nBo QS,QB

~,H

~H~

The desired form is

r

r

nr

n

bfr

bfbfbf

bf

bf,

bf HD

H

SgHHB

Q

u

UCz

o5/25/2bf H~

QH~

QD

HH

]nn)2/1)[(2/5(

o2/1

SB2/3

c

]nn)2/1[(

2/1SB

2/3c

2/12/3

BS

BS

H~H

H~

1Q

H~

1

SH~

B~

1Cz

]nn)2/1[(

2/1SB

2/3o

2/12/3bfbfbf

bf BSQH~

1

SH~

B~

1

SgHHB

QCz

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RELATION FOR BANKFULL SHIELDS NUMBER

RD

SHbfbf

]n)5/2[(Soccbf

SQrR

H~

,r

By definition

Using the relations

it is found that

This can be rewritten as

SB nS

nBo QS,QB

~,H

~H~

]n)5/2[(So5/2

bfSQ

R

H~

R

SH~

Q

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RELATION FOR GRAVEL TRANSPORT AT BANKFULL FLOW

Recall that

Now from the last relation of the previous slide,

Using the previously-introduced non-dimensionalizations,

Thus

RD

SH,1

DRgDB

Qq bf

bf

5.4

bf

c2/3

bfG

bf

bf,bTbf

]n)5/2)[(2/3(

2/3

So

5.42/3

G

2/3

c

5.42/3

GbfSQ

rR

H~

r

11r

r

11rq

5/2

T

bf

bf,bTbf

QB~

R

Q

DRgDB

Qq

]}n)5/2[(]n)5/2)[(2/3{(

2/3

So

5.42/3

GBTBSQ

rR

H~

r

11rRQ

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From the regression relations,

In addition, for natural sediment it is reasonable to assume

In the Parker approximation of the Einstein relation,

The data of the four setsindicate an average valueof bf

* of 0.04870, or thus

EVALUATION OF THE CONSTANTS

344.0n,100.0,0661.0n,70.4,379.0H~

SSBBo

65.1R

03.0c

63.1r

0.001

0.01

0.1

1

100 1000 10000 100000 1000000 10000000

Qhat

tau

sb

f

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THE RESULTING RELATIONS

rnrHCz

nc Q

ynyT QQ

73.3H~ ]n)2/5(n)4/5()2/3[(

o2/1

S1

BrBS

264.0nn2

1

2

5n BSr

0141.0rR

H~

So

0562.0n5

2n s

00318.0R

H~

r1

1 2/3S

2/3oB

5.4

G

y

550.0n2

3n1n SBy

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TEST OF RELATION FOR Czusing all four data sets

1

10

100

1 10 100 1000

Hhat

Cz Cz

Fit

263.0H43.3Cz

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0.001

0.01

0.1

1

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Qhat

tau

sb

f

tausbf

FitQ

TEST OF RELATION FOR bf*using all four data seta

0562.0cbf Q0230.0r

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FINAL RESULTS

If we assume mechanistic relations of the following form:264.0

bf

bfbfbf

bf

bf,

bf

D

H73.3

SgHHB

Q

u

UCz

5.4

bf

c2/3

bf

bf

bf,bTbf 12.11

DRgDB

Qq

cbf

bf 63.1RD

SH

0562.0c Q0141.0

344.00661.0 Q100.0S,Q70.4B~

,379.0H~

550.0T Q00318.0Q

resistance

bedload transport

channel-forming Shields number

sediment yield relation

The first three of these correspond precisely to the data!

then we obtain the results

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1

10

100

1000

1 10 100 1000

Reported Bbf (m)

Pre

dic

ted

Bb

f (m

)

predicted Albertapredicted Britain Ipredicted Idahopredicted Coloradoequality1/22

Test against the original data set

0661.0Q70.4B~

25

0.1

1

10

0.1 1 10

Reported Hbf (m)

Pre

dic

ted

Hb

f (m

)

predicted Albertapredicted Britain Ipredicted Idahopredicted Coloradoequality1/22

Test against the original data set

379.0H~

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0.0001

0.001

0.01

0.1

0.0001 0.001 0.01 0.1

Reported S

Pre

dic

ted

S

predicted Albertapredicted Britain Ipredicted Idahopredicted Coloradoequality1/22

344.0Q100.0S Test against the original data set

27

1

10

100

1000

10000

1 10 100 1000 10000

Reported Qbf (m3/s)

Pre

dic

ted

Qb

f (m

3 /s)

predictedequality1/22

264.0

bf

bfbfbf

bf

bf,

bf

D

H73.3

SgHHB

Q

u

UCz

Test against the original data set

28

1

10

100

1 10 100

Reported Bbf (m)

Pre

dic

ted

Bb

f (m

)

predicted Marylandpredicted Britain IIpredicted Tuscanyequality1/22predicted Colo Andr

Test against four new data sets

0661.0Q70.4B~

29

0.1

1

10

0.1 1 10

Reported Hbf (m)

Pre

dic

ted

Hb

f predicted Marylandpredicted Britain IIequality1/22predicted Tuscanypredicted Colo Andr

Test against four new data sets

379.0H~

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0.0001

0.001

0.01

0.1

0.0001 0.001 0.01 0.1

Reported S

Me

as

ure

d S

predicted Marylandpredicted Britain IIequality1/22predicted Tuscanypredicted Colo Andr

Test against four new data sets

344.0Q100.0S

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0.1

1

10

100

1000

0.1 1 10 100 1000

Measured Qbf (m3/s)

Pre

dic

ted

Qb

f (m

3 /s)

predicted Marylandpredicted Britain IIequality1/22predicted ColoAndr

Test against three new data sets

264.0

bf

bfbfbf

bf

bf,

bf

D

H73.3

SgHHB

Q

u

UCz

32

1

10

100

1 10 100

Reported Bbf (m)

Pre

dic

ted

Bb

f (m

)

Class 1Class 2Class 3Class 4equality1/22

BRITAIN II STREAMS: ROLE OF BANK STRENGTHClass 1 has least vegetation, Class 4 has most vegetation

33

1

1.5

2

2.5

3

1 2 3 4

Vegetation Class

r

cbf r

RELATION BETWEEN VEGETATION DENSITY AND BANK STRENGTH, BRITAIN II STREAMS

34

HOW WOULD VARIED BANK STRENGTH (r), SEDIMENT SUPPLY (Y) AND RESISTANCE (r) AFFECT HYDRAULIC

GEOMETRY?

rnrHCz

2/35.4

G

yB

)r(r1

1R

rB n5

2n

2

1

5

1n

Rn1

1

ry

5.4

G

o

rr1

1H~

Rn11

ry

5.4

G

S

rr1

1R

n5

2nS

nc Q

ynyT QQ

SnS QS

oH~

H~

BnB QB

~

35

0.0001

0.001

0.01

0.1

1

10

100

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Qhat

Bt,

Ht,

S

Britain I width

Alberta width

Idaho width

Colorado width

r = 1.1

r = 1

r = 0.9

Britain I depth

Alberta depth

Idaho depth

Colorado depth

r = 1.1

r = 1

r = 0.9

Britain I slope

Alberta slope

Idaho slope

Colorado slope

r = 1.1

r = 1

r = 0.9

S,H~ ,

B~

Q

B~

H~

S

VARIATION IN r (BANK STRENGTH)

36

0.0001

0.001

0.01

0.1

1

10

100

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Qhat

Bt,

Ht,

S

Britain I width

Alberta width

Idaho width

Colorado width

ar = 1.2

ar = 1

ar = 0.8

Britain I depth

Alberta depth

Idaho depth

Colorado depth

ar = 1.2

ar = 1

ar = 0.8

Britain I slope

Alberta slope

Idaho slope

Colorado slope

ar = 1.2

ar = 1

ar = 0.8Q

S,H~ ,

B~

rr

r

r

r

r

rrr

B~

H~

S

VARIATION IN y (GRAVEL SUPPLY)

37

0.0001

0.001

0.01

0.1

1

10

100

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Qhat

Bt,

Ht,

SBritain I width

Alberta width

Idaho width

Colorado width

ay = 1.5

ay = 1

ay = 0.5

Britain I depth

Alberta depth

Idaho depth

Colorado depth

ay = 1.5

ay = 1

ay = 0.5

Britain I slope

Alberta slope

Idaho slope

Colorado slope

ay = 1.5

ay = 1

ay = 0.5Q

S,H~ ,

B~

y

y

y

y

yy

y

y

y

B~

H~

S

VARIATION IN r (CHANNEL RESISTANCE)

38

QUESTIONS?