Transcript Slide 1

Another Chapter in
THE 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
1
What is the physical basis relations for bankfull geometry of
gravel-bed streams?
2
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
3
THE GOAL:
A Mechanistic Description of the Rules Governing Hydraulic
Relations at Bankfull Flow in Alluvial Gravel-bed Rivers
The Parameters:
Qbf
QbT,bf
=
=
Bbf
Hbf
S
D
g
R
=
=
=
=
=
=
bankfull discharge (m3/s)
volume bedload transport rate at bankfull
discharge (m3/s)
bankfull width (m)
bankfull depth (m)
bed slope (1)
surface geometric mean or median grain size (m)
gravitational acceleration (m/s2)
submerged specific gravity of sediment ~ 1.65 (1)
The Forms Sought:
Hbf ~ Qnbfh
, Bbf ~ Qnbfb
, S ~ Qnbfs
, QbT ,bf ~ QnbfbT
4
DATA SETS
1.
2.
3.
4.
Alberta streams, Canada1
Britain streams (mostly Wales)2
Idaho streams, USA3
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 morphology
of mountain streams, Journal of Hydraulic Engineering, 129(11),
2003.
4 Pitlick, J. and Cress, R., 2002, Downstream changes in the channel of a
large gravel bed river, Water Resources Research 38(10), 1216,
doi:10.1029/2001WR000898, 2002.
5
NON-DIMENSIONALIZATION
~ g1/ 5Hbf
H  2/ 5
Qbf
~ g1/ 5Bbf
, B  2/ 5
Qbf
ˆ 
, Q
Qbf
gDD2
QbT ,bf
ˆ
, QT 
gDD2
These forms supersede two previous forms, namely
Hbf
ˆ
H
D
Bbf
ˆ
, B
D
which appear in reference 3 of the previous slide. Note:
~ ˆ 2 / 5
ˆ
H  HQ
~ ˆ 2 / 5
ˆ
, B  BQ
6
WHAT THE DATA SAY
The four independent sets of data form a coherent set!
100
~
B
Britain width
Alberta width
Idaho width
Colorado width
Britain depth
Alberta depth
Idaho depth
Colorado depth
Britain slope
Alberta slope
Idaho slope
Colorado slope
Btilde, Htilde, S
10
1
~
H
0.1
S
0.01
0.001
0.0001
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
1.0E+07
Qhat
7
REGRESSION RELATIONS BASED ON THE DATA
~
ˆ 0.00004
H  0.379Q
~
ˆ 0.0661
, B  4.70Q
ˆ 0.344
, S  0.100Q
To a high degree of approximation,
~ ~
H  Hc  0.379
Remarkable, no?
1.E+02
Bdimtilde, Hdimtilde, S
y = 4.6977x0.0661
1.E+01
y = 0.3785x4E-05
1.E+00
Bdimtilde
Hdimtilde
S
Power (Hdimtilde)
Power (Bdimtilde)
Power (S)
1.E-01
y = 0.1003x-0.3438
1.E-02
1.E-03
1.E-04
1.E+02
1.E+03
1.E+04
1.E+05
Qdim
1.E+06
1.E+07
8
WHAT DOES THIS MEAN?
Hbf ~ Qbf0.4 or
0.379 0.4
Hbf 
Qbf
g
Bbf ~ Qbf0.461 or

/2
Bbf  4.70 g D5s50


0.0661
S ~ Qbf0.344 or

5/2

S  0.100 g Ds50


0.344
g1/ 2  Qbf0.461

 Q0.344
 bf
9
THE PHYSICAL RELATIONS NECESSARY TO
CHARACTERIZE THE PROBLEM
Required: four relations in the four unknowns
Hbf, 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).
10
RESISTANCE RELATION
Manning-Strickler form: where Ubf = Qbf/(Bbf Hbf) denotes
bankfull flow velocity,
nr
Ubf
Qbf
 Hbf 
Cz 

 r 

u,bf Bbf Hbf gHbf S
 D 
Here we leave r and nr as parameters to be evaluated.
11
BEDLOAD TRANSPORT RELATION
Use Parker (1979) approximation of Einstein (1950) relation
applied to bankfull flow:

bf
q 
QbT ,bf
Bbf RgD D
w here
H S
bf  bf
RD
,
 
 G 
 3/2
bf


1 
 

c

bf



4 .5
 G  11.2
12
RELATION FOR CHANNEL-FORMING SHIELDS NUMBER
Base the form of the relation on Parker (1978):

bf

c

 r  const

13
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
nbT
bf
QbT ,bf ~ Q
where nbT must be evaluated.
14
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.
~ ~
H  Ho
~
nB
ˆ
, B  B Q
nS
ˆ
, S  S Q
~
Ho  0.379 , B  4.70 , nB  0.0661 , S  0.100 , nS  0.344
Recall that
~ g1/ 5Hbf
H  2/ 5
Qbf
~ g1/ 5Bbf
, B  2/ 5
Qbf
Hbf
ˆ
H
D
ˆ 
, Q
Qbf
gDD2
Bbf
ˆ
, B
D
QbT ,bf
ˆ
, QT 
gDD2
15
RESISTANCE RELATION
The desired form is
nr
Cz 
Ubf
Qbf
H 
ˆ nr

 r  bf   rH
u,bf Bbf Hbf gHbf S
 D 
Now using the definition of Cz, the non-dimensionalizations and the
relations
~ ~
H  Ho
~
ˆ nB
, B  B Q
ˆ nS
, S  S Q
it is found that
Cz 
Qbf
1
1
ˆ [(1/ 2)nS nB ]
 ~ ~3 / 2 1/ 2  ~3 / 2
Q
Bbf Hbf gHbf S BH S
Ho B1S/ 2
H
~ ˆ 2 / 5~
ˆ 2 / 5H
But Hˆ  bf  Q
 Q Ho so that
D
1
Cz  ~ ~3 / 2 1/ 2
BH S
( 5 / 2)[(1/ 2)nS nB ]
ˆ 
H
1
[(1/ 2)nS nB ]
ˆ


 ~3 / 2
Q

~
~
Hc B1S/ 2
H3c / 2B1S/ 2  Ho 
1
16
RELATION FOR BANKFULL SHIELDS NUMBER
By definition
Hbf S
 
RD

bf
Using the relations
~ ~
H  Ho
it is found that
~
ˆ nB
, B  B Q
ˆ nS
, S  S Q
~
2 / 5~
ˆ
Q
H
S
H

o S ˆ [(2 / 5 )nS ]
bf 

Q
R
R
This can be rewritten as
bf  rc
~
H

o S ˆ [(2 / 5 )nS ]
, c 
Q
rR
17
RELATION FOR GRAVEL TRANSPORT AT BANKFULL
FLOW
Recall that

bf
q 
QbT ,bf
Bbf RgD D
 
 G 

bf
3/ 2

c 
 1   
bf 

4.5
,
Hbf S
 
RD

bf
Now from the last relation of the previous slide,

bf
q  Gr
3/ 2
 1
1  
 r
4.5
 
~

1
H
  oS 
3/ 2
 Gr 1  
 r   rR 
4.5
 3/ 2
c
3/ 2
ˆ (3 / 2)[(2 / 5)nS ]
Q
Using the previously-introduced non-dimensionalizations,

bf
q 
ˆ
Q
T

~ ˆ 2/5
RgD D
RBQ
QbT ,bf
Bbf
Thus
3/ 2
~
1  HoS  ˆ {(3 / 2)[(2 / 5)nS ][(2 / 5)nB ]}
3/ 2
ˆ
QT  RBGr 1  
Q


 r   rR 
4.5
18
EVALUATION OF THE CONSTANTS
From the regression relations,
~
Ho  0.379 , B  4.70 , nB  0.0661 , S  0.100 , nS  0.344
In addition, for natural sediment it is reasonable to assume
R  1.65
In the Parker approximation of the Einstein relation,
c  0.03
r  1.63
0.1
tausbf
The data of the four sets
indicate an average value
of bf* of 0.04870, or thus
1
0.01
0.001
100
1000
10000
100000
Qhat
1000000
10000000
19
THE RESULTING RELATIONS
ˆ nr
Cz  rH
r   
nr 
ˆ n
c   Q
~ [(3 / 2)(5 / 4)nS (5 / 2)nB ]
Ho
 3.73
1 1/ 2
B S
51

n

n
 S B   0.264
22

~
HoS
 
 0.0141
rR
n 
2
 ns  0.0562
5
4.5
ny
ˆ
ˆ
QT   y Q
~
 1
G 1   BH3o / 23S/ 2
 r
y 
 0.00318
R
3
ny  1  nB  nS  0.550
2
20
TEST OF RELATION FOR Cz
using all four data sets
100
Cz
ˆ 0.263
Cz  3.43H
Cz
Fit
10
1
1
10
100
1000
Hhat
21
TEST OF RELATION FOR bf*
using all four data seta
1
ˆ 0.0562
bf  rc  0.0230Q
tausbf
0.1
tausbf
FitQ
0.01
0.001
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
Qhat
22
FINAL RESULTS
If we assume mechanistic relations of the following form:
Cz 
qbf 
Ubf
Qbf
H 

 3.73  bf 
u,bf Bbf Hbf gHbf S
 D 
QbT ,bf
Bbf RgD D
 
 11.2 bf
Hbf S
 
 1.63 c
RD

bf
3/2
0.264
resistance

 
1 

 


c

bf
4.5
bedload transport
channel-forming Shields number
ˆ  0.00318Q
ˆ 0.550
Q
T
sediment yield relation
then we obtain the results
~
~
ˆ 0.0661
H  0.379 , B  4.70Q
ˆ 0.344
, S  0.100Q
ˆ 0.0562
c  0.0141Q
The first three of these correspond precisely to the data!
23
~
ˆ 0.0661
B  4.70 Q
Test against the
original data set
Predicted Bbf (m)
1000
predicted Alberta
predicted Britain I
predicted Idaho
predicted Colorado
equality
1/2
2
100
10
1
1
10
100
1000
Reported Bbf (m)
24
~
H  0.379
Test against the
original data set
Predicted Hbf (m)
10
predicted Alberta
predicted Britain I
predicted Idaho
predicted Colorado
equality
1/2
2
1
0.1
0.1
1
10
Reported Hbf (m)
25
ˆ 0.344
S  0.100Q
Test against the
original data set
Predicted S
0.1
predicted Alberta
predicted Britain I
predicted Idaho
predicted Colorado
equality
1/2
2
0.01
0.001
0.0001
0.0001
0.001
0.01
0.1
Reported S
26
Ubf
Qbf
 Hbf 
Cz 

 3.73 

u,bf Bbf Hbf gHbf S
 D 
0.264
Test against the
original data set
Predicted Qbf (m3/s)
10000
1000
predicted
equality
1/2
2
100
10
1
1
10
100
1000
Reported Qbf (m3/s)
10000
27
~
ˆ 0.0661
B  4.70 Q
Predicted Bbf (m)
100
Test against four new
data sets
predicted Maryland
predicted Britain II
predicted Tuscany
equality
1/2
2
predicted Colo Andr
10
1
1
10
100
Reported Bbf (m)
28
~
H  0.379
Predicted Hbf
10
Test against four new
data sets
predicted Maryland
predicted Britain II
equality
1/2
2
predicted Tuscany
predicted Colo Andr
1
0.1
0.1
1
10
Reported Hbf (m)
29
ˆ 0.344
S  0.100Q
Measured S
0.1
Test against four new
data sets
predicted Maryland
predicted Britain II
equality
1/2
2
predicted Tuscany
predicted Colo Andr
0.01
0.001
0.0001
0.0001
0.001
0.01
0.1
Reported S
30
Ubf
Qbf
 Hbf 
Cz 

 3.73 

u,bf Bbf Hbf gHbf S
 D 
0.264
Test against three
new data sets
Predicted Qbf (m3/s)
1000
100
predicted Maryland
predicted Britain II
equality
1/2
2
predicted ColoAndr
10
1
0.1
0.1
1
10
100
Measured Qbf (m3/s)
1000
31
BRITAIN II STREAMS: ROLE OF BANK STRENGTH
Class 1 has least vegetation, Class 4 has most vegetation
Predicted Bbf (m)
100
Class 1
Class 2
Class 3
Class 4
equality
1/2
2
10
1
1
10
Reported Bbf (m)
100
32
RELATION BETWEEN VEGETATION DENSITY AND BANK
STRENGTH, BRITAIN II STREAMS
3
r
2.5
2
1.5
1
1
2
bf  rc
3
Vegetation Class
4
33
HOW WOULD VARIED BANK STRENGTH (r), SEDIMENT
SUPPLY (Y) AND RESISTANCE (r) AFFECT HYDRAULIC
GEOMETRY?
ˆ
Cz  rH
nr
ˆ n
c   Q
ˆ  Q
ˆ ny
Q
T
y

~
nB
ˆ
B  B Q
  1

  G  1   r  
~
r

Ho   


 y r




4.5
~ ~
H  Ho
B 
y
4.5
 1
R G 1   (r )3 / 2
 r
  1

  G  1   r  
r

S  R   


 y r




4.5
ˆ nS
S  S Q
 1
 
 1nR
nB 
1
1nR
1 1
2
 n  nr
5 2
5



nS 
2
 n
5
34
VARIATION IN r (BANK STRENGTH)
100
~
B
10
~ ~ S
Bt,
B, HHt,
,S
1
~
H
0.1
0.01
S
0.001
0.0001
1.E+02
1.E+03
1.E+04
1.E+05
ˆ
Qhat
Q
1.E+06
1.E+07
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
35
VARIATION IN y (GRAVEL SUPPLY)
100
10
B, H, S
~ ~Ht, S
Bt,
1
~
B
~
H
0.1
0.01
S
0.001
0.0001
1.E+02
1.E+03
1.E+04
1.E+05
ˆ
Qhat
Q
1.E+06
1.E+07
Britain I width
Alberta width
Idaho width
Colorado width
ar
r = 1.2
r = 1
ar
r = 0.8
ar
Britain I depth
Alberta depth
Idaho depth
Colorado depth
r = 1.2
ar
r = 1
ar
r = 0.8
ar
Britain I slope
Alberta slope
Idaho slope
Colorado slope
r = 1.2
ar

arr = 1

arr = 0.8 36
VARIATION IN r (CHANNEL RESISTANCE)
100
~
B
10
~ ~Ht, S
Bt,
B, H, S
1
~
H
0.1
0.01
S
0.001
0.0001
1.E+02
1.E+03
1.E+04
1.E+05
ˆ
Qhat
Q
1.E+06
1.E+07
Britain I width
Alberta width
Idaho width
Colorado width
y = 1.5
ay
ay
y = 1
ay
y = 0.5
Britain I depth
Alberta depth
Idaho depth
Colorado depth
y = 1.5
ay
y = 1
ay
y = 0.5
ay
Britain I slope
Alberta slope
Idaho slope
Colorado slope

ayy = 1.5

ayy = 1

ayy = 0.5
37
QUESTIONS?
38