another chapter in the search for the holy grail: a mechanistic basis for hydraulic relations for...
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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
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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.
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NON-DIMENSIONALIZATION
2bf,bT
T2bf
5/2bf
bf5/1
5/2bf
bf5/1
DgDQ
Q,DgD
QQ,Q
BgB~,Q
HgH~
These forms supersede two previous forms, namely
which appear in reference 3 of the previous slide. Note:
DBB,
DHH bfbf
5/25/2 QB~B,QH~H
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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
Btil
de, H
tilde
, 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
Bdi
mtil
de, H
dim
tilde
, 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
10
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
rnbf
rbfbfbf
bf
bf,
bf
DH
SgHHBQ
uUCz
Manning-Strickler form: where Ubf = Qbf/(Bbf Hbf) denotes bankfull flow velocity,
Here we leave r and nr as parameters to be evaluated.
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BEDLOAD TRANSPORT RELATION
Use Parker (1979) approximation of Einstein (1950) relation applied to bankfull flow:
2.11,RD
SHwhere
1DRgDB
Gbf
bf
5.4
bf
c2/3bfG
bf
bf,bTbf
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RELATION FOR CHANNEL-FORMING SHIELDS NUMBER
Base the form of the relation on Parker (1978):
constrc
bf
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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
15
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
2bf,bT
T2bf
5/2bf
bf5/1
5/2bf
bf5/1
DgDQ
Q,DgD
QQ,Q
BgB~,Q
HgH~
DBB,
DHH bfbf
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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
rn
r
nbf
rbfbfbf
bf
bf,
bf HD
HSgHHB
QuUCz
o5/25/2bf H~QH~Q
DHH
]nn)2/1)[(2/5(
o2/1
SB2/3
c
]nn)2/1[(2/1
SB2/3
c2/12/3
BS
BS
H~H
H~1Q
H~1
SH~B~1Cz
]nn)2/1[(2/1
SB2/3
o2/12/3
bfbfbf
bf BSQH~
1SH~B~
1SgHHB
QCz
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RELATION FOR BANKFULL SHIELDS NUMBER
RDSHbf
bf
]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
RH~
RSH~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,1DRgDB
Qq bf
bf
5.4
bf
c2/3bfG
bf
bf,bTbf
]n)5/2)[(2/3(
2/3
So5.4
2/3G
2/3c
5.42/3
GbfSQ
rRH~
r11r
r11rq
5/2T
bf
bf,bTbf QB~R
QDRgDB
]}n)5/2[(]n)5/2)[(2/3{(
2/3
So5.4
2/3GBT
BSQrR
H~
r11rRQ
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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
taus
bf
20
THE RESULTING RELATIONS
rnrHCz
nc Q
ynyT QQ
73.3H~ ]n)2/5(n)4/5()2/3[(o
2/1S
1Br
BS
264.0nn21
25n BSr
0141.0rR
H~ So
0562.0n52n s
00318.0R
H~r11 2/3
S2/3
oB
5.4
G
y
550.0n23n1n SBy
21
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
taus
bf tausbfFitQ
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
DH73.3
SgHHBQ
uUCz
5.4
bf
c2/3bf
bf
bf,bTbf 12.11
DRgDBQ
q
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)
Pred
icte
d B
bf (m
)
predicted Albertapredicted Britain Ipredicted Idahopredicted Coloradoequality1/22
Test against the original data set
0661.0Q70.4B~
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0.1
1
10
0.1 1 10
Reported Hbf (m)
Pred
icte
d H
bf (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
Pred
icte
d 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)
Pred
icte
d Q
bf (m
3 /s)
predictedequality1/22
264.0bf
bfbfbf
bf
bf,
bf
DH73.3
SgHHBQ
uUCz
Test against the original data set
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1
10
100
1 10 100
Reported Bbf (m)
Pred
icte
d B
bf (m
)
predicted Marylandpredicted Britain IIpredicted Tuscanyequality1/22predicted Colo Andr
Test against four new data sets
0661.0Q70.4B~
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0.1
1
10
0.1 1 10
Reported Hbf (m)
Pred
icte
d H
bf
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
Mea
sure
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)
Pred
icte
d Q
bf (m
3 /s)
predicted Marylandpredicted Britain IIequality1/22predicted ColoAndr
Test against three new data sets
264.0bf
bfbfbf
bf
bf,
bf
DH73.3
SgHHBQ
uUCz
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1
10
100
1 10 100
Reported Bbf (m)
Pred
icte
d B
bf (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(r11R
rB n52n
21
51n
Rn11
ry
5.4
G
o
rr11
H~
Rn11
ry
5.4
G
S
rr11
R n
52nS
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 widthAlberta widthIdaho widthColorado widthr = 1.1r = 1r = 0.9Britain I depthAlberta depthIdaho depthColorado depthr = 1.1r = 1r = 0.9Britain I slopeAlberta slopeIdaho slopeColorado sloper = 1.1r = 1r = 0.9
S,H~ ,
B~
Q
B~
H~
S
VARIATION IN r (BANK STRENGTH)
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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 widthAlberta widthIdaho widthColorado widthar = 1.2ar = 1ar = 0.8Britain I depthAlberta depthIdaho depthColorado depthar = 1.2ar = 1ar = 0.8Britain I slopeAlberta slopeIdaho slopeColorado slopear = 1.2ar = 1ar = 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 widthAlberta widthIdaho widthColorado widthay = 1.5ay = 1ay = 0.5Britain I depthAlberta depthIdaho depthColorado depthay = 1.5ay = 1ay = 0.5Britain I slopeAlberta slopeIdaho slopeColorado slopeay = 1.5ay = 1ay = 0.5
Q
S,H~ ,
B~
y
y
y
y
yy
y
y
y
B~
H~
S
VARIATION IN r (CHANNEL RESISTANCE)
38
QUESTIONS?
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