12.163/12.463 Surface Processes and Landscape Evolution
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September, 2004
VI. Erosional Channel Networks
A. Plan-form Network Properties
Dendritic channel network most common, scale-invariant.
Other forms (trellis, radial patterns, etc.) ! lithologic and structural control.
Dd = drainage density
Dd =
!l "1
[m ]
A
Ac
xc 2
1
!
1
2xc
! = spacing between channels
Dd = f (xc ) ! lithology, climate, tectonics
Scaling Laws Drainage Networks
Hack’s Law (1957):
Length of a channel is related to drainage area.
l ! A0.52"0.67
l ! A0.6
A = ka x h x = along-stream distance
h 1.67
Horton’s Laws:
Numbers of channels, lengths, areas all grow in steady geometric progression.
w = order of channel
Second order channel ! two or more first order channel join
Third order channel ! two or more second order channel join
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
Stream numbers (Bifurcation Ratio):
n
Rn = Rb = w = constant (3-5) 4
nw+1
Stream areas:
a
Ra = w+1 = constant
aw
–
September, 2004
notation: (range) mode
(3-6) 4
Stream segment lengths:
l
Rl = w+1 = constant (1.5-3) 2
lw
B. Channel Longitudinal Profiles (Empirical):
1960’s, 1970’s Statistical empirical studies ask what form of equation describes long profile?
Various researchers argue for: Power law
z ! x b , logarithmic, semi-log, exponential
forms best-fit channel profiles in their study.
Idea behind this work: downstream increase in Qw ! greater transport
capacity/efficiency of erosion
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
September, 2004
! gentler channel slope
Problem: basin shape governs relation between Qw and x .
Qw = kq Ac , 0.7 ! c ! 1 ,
c < 1 due to: infiltration losses, non-uniform rainfall, flood plain storage
Combine with Hack’s law:
A = ka x h ; ka = 6.69 , h = 1.667
Q w = k q k a x hc
c
Thus otherwise identical but different shape basins could have very different river
profiles.
Flint (1974) removes problem of basin shape and resolves this debate (more or less)
Flint’s law, power-law between S and A.
S = k s A "!
ks = steepness index, θ = concavity index
Derive estimates from regression of logS vs. logA plots:
log S = log(k s A "! )= log k s + log A "! = log k s " ! log A
Commonly observed:
0.4 ! " ! 0.6 ; but can vary downstream (spatial variation in geology, tectonics,
dis-equilibrium conditions, etc).
Problem: ks , θ covary … small difference in θ causes huge change in apparent ks .
Need to normalize somehow for useful intercomparison. Two methods:
S r = k s Aref
"!
; S = k sn A "!ref
Discussed at length later and in lab.
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
September, 2004
C. Revisit the Channel Width Problem
Recall empirical observation (robust, all data on both alluvial and bedrock channels):
W ! Qw
12
Lets consider where this comes from in terms of hydrology and Parker’s channel width
closure.
Combine conservation of mass of water and Drainage Basin Hydrology and solve
for channel width as a function of drainage area and flow velocity and depth:
Q w = u hW ; Qw = kq Ac
W=
k q Ac
uh
Write Parker’s Channel Width Closure in terms of dimensional shear stress:
! * = (1 + " )! *c ; " = 0.2 ! 0.4 ; τ*c = 0.06 {for gravel}
! * = 2 {for sand}
# b = (1 + " )0.06(! s $ ! )gD (gravel)
" b = 2(! s # ! )gD (sand)
# b = " (! s $ ! )gD ; where α depends on gravel vs. sand bed
Use Conservation of Momentum
" b = !ghS ; " b = !C f u 2
Use first and second relations for boundary shear stress to substitute in for h and
velocity, respectively, in the relation for W, then substitute in the channel closure
condition for boundary shear stress:
W=
k q ! 3 2 g C f Ac S
[" (!
# ! )gD ]
32
s
Substitute observed relation for river profile concavity:
S = k s A "!
W=
k q k s ! 3 2 g C f A c $#
[" (!
$ ! )gD ]
32
s
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
September, 2004
So if c ~ 1 (typical) and θ ~ ½, then W ! A1 2 ! Qw 1 2
Empirical relations are internally consistent, which is a nice check, but this does
mean that the channel width problem and river profile problem are tightly coupled
– solving one requires solving both, ultimately.
D. Transport-Limited Incising Channels (Alluvial but Erosional):
Conservation of Mass (sediment) with Uplift:
!z
1 !qs
=U "
!t
1 " # p !x
Generalized Sediment Transport Rule
Qc = Wqs = K f A f S
m
nf
For gravel bedload, m f = 1, n f = 1
For sand total load, m f = 4 3, n f = 5 3
Details of the formulation depend on the channel closure rule, sediment transport
equation, treatment of downstream fining, etc {Derivations below}
! b = (1+ " )! c or W = kwQ b
Bedload transport: qs ! " b 3/2 , n f =
Rapid downstream fining: m f
2 3
! =1
3 2
1.5
Steady-State Profile
By definition for transport-limited channels:
Qc = Qs
Steady-state sediment flux for uniform rock uplift is:
Q s = ! g AU ; where βg denotes bedload fraction of total sediment flux
Setting these equal and substituting the general sediment transport rule:
m
KfA f S
nf
' ( gU
S =%
% K
& f
= !g AU
1
$n
" A
"
#
f
1! m f
nf
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
IF m f = 2, n f = 2 ! "
September, 2004
1
; values given above will be inconsistent.
2
Suspended sediment transport on hillslopes ( qs ! " b 3 , no change in width) is
1
consistent with m f = 2, n f = 2 ! "
– shown by Willgoose, 1991.
2
But the relevant question is whether sediment transport relations and channel
closure rules for alluvial rivers (not hillslopes) are consistent with observed river
profile concavities. To address this, we need to look at how the physics influences
the exponents in the generalized transport relation, mf and nf.
Gravel Transport (MPM) with Parker Channel Closure
Conservation Mass (Water)
Q = u hW = k q A c
Conservation Momentum
" b = !ghS ; " b = !C f u 2
combining these gives:
"b
32
=
! 3 2 g C f SQ
W
=
! 3 2 gk q C f SA c
W
Channel closure (Parker, gravel)
3
(! b " ! c )
32
(" b # " c )3 2
& ' #2
; cw = $
= c w! b
!
%1 + ' "
! 3 2 gc w k q C f SA c
=
W
32
MPM Gravel Transport (from Alluvial Profile Lectures)
Qc = Wq s =
"
32
8W
(! # ! cr )3 2
((" s # " ) " )g
Substitute in relation for (! b " ! c )3 2 :
Qc =
Qc =
! 3 2 gc w k q C f SA c
8W
! 3 2 ((! s " ! ) ! )g
W
8c w k q C f
((! s " ! ) ! )
Ac S
Thus we find:
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
Qc = Wqs = K f A f S
m
Kf =
September, 2004
nf
8c w k q C f
; mf = c ~ 1 ; nf = 1
((! s " ! ) ! )
Note: initiation of motion threshold has not been ignored, it has been subsumed
into the channel closure rule.
Implication of Steady-State Profile Concavity Index:
1
1! m f
' ( gU $ n
" A n
S =%
% K "
& f #
1! mf mf !1
" =!
=
= 0 !! PROBLEM
nf
nf
f
f
Sand Transport (Engelund and Hansen) with Parker Channel Closure
Conservation Mass (Water)
Q = u hW = k q A c
Conservation Momentum
" b = !ghS ; " b = !C f u 2
as before combining these gives:
"b
32
=
! 3 2 gk q C f
W
Ac S
Channel closure (Parker, sand)
! * = 2 {for sand}
" b = 2(! s # ! )gD (sand)
Combine above relations to give:
"b
52
"b
52
= " b" b
=
32
= 2(! s # ! )gD
! 3 2 gk q C f
W
2(! s # ! )! 3 2 g 2 k q D C f
Ac S
Ac S
W
Engelund and Hansen Sand Total Load Equation (from Alluvial Profile Lectures)
Qc = Wq s
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
q s* =
September, 2004
0.05 5 2
!*
Cf
qs
q s* =
; "* =
((! s " ! ) ! )gD D
q s = q s*
((" s # " ) " )gD D =
"b
(! s # ! )gD
((" s # " ) " )gD D 5 2
!b
2.5
C f [(" s # " )gD ]
0.05
Substitute the relation for ! b 5 2
Qc = Wq s =
0.1k q
(! s " ! )!
Ac S
Cf
Thus we find (again):
Qc = Wqs = K f A f S
m
Kf =
nf
0.1k q
(! s " ! )!
; mf = c ~ 1 ; nf = 1
Cf
Same prediction (and problem) for steady-state river profile concavity index.
Sand Transport (Engelund and Hansen) with Channel Width ~ A
Conservation Mass (Water)
Q = u hW = k q A c
Conservation Momentum
" b = !ghS ; " b = !C f u 2
as before combining these gives:
"b
32
"b
52
=
! 3 2 gk q C f
W
( )
= "b
32 53
=
Ac S
(
! 5 2 gk q C f
W
)
53
53
A 5c 3 S 5 3
Engelund and Hansen Sand Total Load Equation (from Alluvial Profile Lectures)
Qc = Wq s
q s* =
0.05 5 2
!*
Cf
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12.163/12.463 Surface Processes and Landscape Evolution
K. Whipple
q s* =
qs
; "* =
((! s " ! ) ! )gD D
q s = q s*
((" s # " ) " )gD D =
September, 2004
"b
(! s # ! )gD
((" s # " ) " )gD D 5 2
!b
2.5
C f [(" s # " )gD ]
0.05
Substitute the relation for ! b 5 2
Qc = Wq s =
Qc =
((! s " ! ) ! )gD D ! 5 2 (gk q C f
2.5
W53
C f [(! s " ! )gD ]
W 0.05
0.05k q
53
2
16
((! s " ! ) ! ) C f
13
23
g DW
)
53
A 5c 3 S 5 3
A 5c 3 S 5 3
Empirical channel closure (Hydraulic Geometry)
W = k w Q b = k w k q A bc
b
Combine above relations to give:
Qc =
0.05k q
(5 ! 2 b ) 3
((" s ! " ) " ) k w
2
23
Cf
16
13
g D
A (5c ! 2bc ) 3 S 5 3
Thus we find:
Qc = Wqs = K f A f S
m
Kf =
0.05k q
nf
(5 " 2 b ) 3
((! s " ! ) ! )2 k w 2 3 C f 1 6 g 1 3 D
;
for c ~ 1 and b ~ ½ ; mf ~ 4/3 ; nf = 5/3
Prediction for steady-state river profile concavity index:
" =!
1! mf
nf
=
mf !1
nf
=
1
; Perhaps reasonable for sandy, suspension5
dominated alluvial rivers.
D. Transitions from Alluvial to Bedrock (Mixed) Channels
Non-dimensional Bedrock Channel Number , N br
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12.163/12.463 Surface Processes and Landscape Evolution
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September, 2004
Under what conditions are channels detachment-limited (DL) vs. transport limited (TL)?
N br =
TL: Qc ! Qs
DL: Qc > Qs ,
Definition:
Qs
Qc
N br < 1 ! DL (mixed) , N br ! 1 " TL
Steady state:
Qs = ! g AU, Qc = K f A f S
m
nf
1
! U $n ' m
Assume channel is DL, S = # & A n
"K %
1
(
m+
'
n
!
$
U
mf *
Qc = K f A # & A n *" K %
)
,
nf
Mountain channels ! gravel bedload ! n f = 1
Qs ! g nf 1" nf 1"m f +
=
K U
A
Qc K f
n
N br =
n
mn f
n
1. If K goes up (wetter/stormier; weaker rock) or K f goes down (coarse gravel) or
! g goes up (lots of gravel) ! TL
2. If and only if n < n f (n < 1), does U goes up ! DL
3. If n = n f (n = 1), N br ! f (U )
4. If concavity of TL system is less than concavity of DL ! expect transition to TL
downstream (vice versa)
1
! U $ n 'm m
DL: S = # & A n ,
! concavity (derived in next lectures)
"K %
n
1
f
" !gU % n f 1(m
1! mf
nf
'' A
TL: S = $$
, " concavity
K
n
# f &
f
Whipple and Tucker (2002) give predictions for the critical drainage area
for this transition, and for a given drainage area, the critical rock uplift rate
for the transition.
5. If concavities are same N br ! f (A)
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September, 2004
Note: in this case, equations for steady-state channel slope above indicate that DL
and TL channels can have identical longitudinal profiles.
Next Lecture: Idealized Model for Bedrock, Detachment-limited Channel
Transport capacity: Qc
Sediment Supply (Flux): Qs
Qs / Qc ! very small
Erosion is governed by ability to “detach” or incision into bedrock.
!q
Not limited by s .
!x
11