A Kolmogoroff-type Scaling for the Fine Structure of Drainage Basins

Gary Parker, University of MinnesotaPeter K. Haff and A. Brad Murray, Duke University

Kolmogoroff scaling in turbulent flows:Energy Cascade

•Turbulent energy is produced in eddies scaling with the size of the "box" (e.g. river depth).•The nonlinear terms in the equations of momentum balance grind larger eddies into ever smaller eddies.•The grinding continues until the eddies are so small that the turbulent energy can be effectively dissipated into heat.•There are no smaller eddies.

production of turbulent energy at large scales

dissipation of turbulent energy to heat at small scales

nonlinear transfer of energy from large to small scales

Balance of energy in a steady turbulent flow

Letu = characteristic turbulent velocity (L/T)L = size of the container (e.g. river depth) (L)ν = viscosity of flow (L2/T)P = production rate/mass of turbulent energy (L2/T3)ε = dissipation rate/mass of turbulent energy (L2/T3)uk = Kolmogoroff velocity scale (L/T)ηk = Kolmogoroff length scale

ratetransferspatialratendissipatiorateproduction

)turbulenceofenergykineticmean(t

+−

=∂∂

Lu3

j

iji ~x

uuuP∂∂′′−= 2

k

2k

j

i

j

i ~xu

xu

ην

∂′∂

∂′∂

ν=εu

Make Kolmogoroff scales from the viscosity ν(L2/T) and the dissipation rate ε (L2/T3)

1)( kk4/1k4/13

k =νη

νε=⎟⎟⎠

⎞⎜⎜⎝

⎛εν

=ηuu

Recalling that

Lu3~P εε≅

it is found that4/3

k ~ ⎟⎠⎞

⎜⎝⎛ νηuLL

That is, the higher the Reynolds number of the flow the finer is the turbulence

From Tennekes and Lumley

Does the idea of a Kolmogoroffscale have any application to

drainage basins?

How dense does a drainage network

have to be in order to "cover" the catchment?

Well, how dense?

Consider a (statistically) steady state landscape undergoing uplift at constant rate vu (L/T).

The channels are undergoing incision at rate vI (L/T).

The hillslopes contain a well-developed regolith which moves downslope with a kinematic diffusivity k (L2/T)

HYPOTHESIS 1: Channel incision creates elevation fluctuations.

HYPOTHESIS 2: Hillslope diffusion obliterates elevation fluctuations.

HYPOTHESIS 3: The drainage network must be just sufficiently fine so that rate of creation of elevation fluctuations balances rate of obliteration.

Channel incision in the near-absence of hillslope

diffusion: the SLOT CANYON:

creates amplitude fluctuation

Hillslope diffusion with weak channel incision: MELTED ICE-CREAM

MORPHOLOGY:destroys amplitude

fluctuation

Image courtesy Bill Dietrich

ridge

headwater tributary

Some scales

Lu = length of headwater tributaryAu = αuLu2 = area of headwater catchment: αu ~o(1)Sh = characteristic slope of hillslope in headwater catchment

Some parametersη = bed elevationt = time(x1, x2) = spatial coordinateζ = incision rate(q1, q2) = vector of volume hillslope transport/width k = hillslope diffusivity

η∇−=rr

kq

Exner equation of sediment balance(neglecting porosity)

u2 vk

t+η∇+ς−=

∂η∂

ridge

headwater tributary

Now for simplicity we assume that•All the hillslope transport occurs on the hillslope and•All the incision occurs in the channel

Integrate Exner on hillslope

∫∫ ∫∫ ∫−=⋅∇−=∂η∂

u uA Anuuuu dsqAvdAqAvdAt

rr

steady state

gain due to uplift

loss due to transport from

hillslope to channel

hn kS~nkq∂η∂

−=

Here n is normal to red perimeter of

path integral:

ridge

headwater tributary

Continue integration on hillslope:path integral is around red line

∫ hunun kSL2~qL2~dsq

hn kS~nkq∂η∂

−= ridge

headwater tributary

The following scale relation results:

huh2uuuuu kSL2LvAv α=α=

where αh is an o(1) parameter

That is, the rate of hillslope denudation must just balance with rate of uplift

Incision rate: we assume

⎩⎨⎧

=ςhillslopeon0channelinvI

Thus within the channel of the headwater tributary

uI vvt+−=

∂η∂

steady state

or thus uI vv =

Channel incision must just keep pace with uplift

ηη

η−η=η′

Mean and fluctuating bed elevation

Hasbargen and PaolaVideo clip

Equation of evolution of amplitude of elevation fluctuation

Decompose into mean and fluctuating parts:

ς′+ς=ςη′+η=η

where the overbar denotes an average over an appropriate spatial window and the prime denotes a fluctuation about that average.

Local Exner:

u2 vk

t+η∇+ς−=

∂η∂

Local Exner: u2 vk

t+η∇+ς−=

∂η∂

Spatially averaged Exner: u2 vk

t+η∇+ς−=

∂η∂

Multiply by and reduce:

( ) ( ) ( ) η+η⋅∇⋅η∇−η∇η⋅∇+ης−=η+η∇η+ης−=⎟

⎠⎞

⎜⎝⎛ η

∂∂

u

u22

vkk

vk21

trrrr

η

Multiply local Exner by η, average and

reduce:( ) ( ) ( )( ) ( ) ( )η′∇⋅η′∇−η′∇η′⋅∇+

η+η⋅∇⋅η∇−η∇η⋅∇+

η′ς′−ης−=⎟⎠⎞

⎜⎝⎛ η′+η

∂∂

rrrr

rrrr

kk

vkk21

21

t

u

22

A

B

Subtract B from A to get equation of evolution of amplitude of elevation

fluctuation:

( ) ( ) ( )η′∇⋅η′∇−η′∇η′⋅∇+η′ς′−=⎟⎠⎞

⎜⎝⎛ η′

∂∂ rrrr kk

21

t2

I II III IV

I. Time rate of change of amplitude of elevation fluctuationII. Rate of creation of amplitude fluctuation due to incisionIII. Rate of transport of amplitude fluctuation due to diffusionIV. Rate of destruction of amplitude fluctuation by hillslope

diffusion

Steady state: approximate balance between creation and destruction of

amplitude fluctuation

( ) ( )η′∇⋅η′∇≅η′ς′− rrkNow if most of the destruction is biased toward the finest scale of the drainage basin, i.e. the

headwater catchment, then

( ) ( ) 2hdkSk α=η′∇⋅η′∇rr

where αd is an o(1) constant

Approximate form for the rate of creation of amplitude fluctuation by incision

ηc(t)ηc(t+∆t)

ηpvI∆t

In the headwater catchment, appoximate the channel as incising into a plain with constant elevation ηp.

Channel elevation ηc decreases with speed vI.

Thus:uu

ucup

BLABLA

+η+η

=η where B denotes channel width

ridge

headwater tributary

( )⎪⎩

⎪⎨

⎧

η−η

η−η=η′

channelin)(21

plainon)(21

21

2c

2p2

uIc vv −=−=η&

( )( ) ( )

u

uuc

uu

uchannel

2u

plain

2

incision

2

ABLv)(

BLA

BL21

tA

21

t

21

t

η−η

≅+

η′∂∂

+η′∂∂

=η′∂∂

=ς′η′−

Now

Since

it follows that

Now scaling

Hcc α=η−η

where H is an "effective" flow depth and αc is an o(1) constant.

Thus the rate of creation of fluctuation amplitude by incision is expressed as

u

uuc A

BLHvα=ς′η′−

Balancing this against the rate of destruction of fluctuation amplitude,

2hd

u

uuc kSA

BLHv α=α

Scale relations for headwater catchment:2uuu LA α=Geometric scaling:

huh2uuu kSL2Lv α=α

2hd

u

uuc kSA

BLHv α=α

Rate of hillslope denudation must balance with rate of uplift:

Rates of creation and destruction of elevation fluctuation amplitude must balance:

3/1

2/1eu

12/1e

u

Avk

AL

⎟⎟⎠

⎞⎜⎜⎝

⎛α=

3/22/1eu

2h kAvS ⎟⎟

⎠

⎞⎜⎜⎝

⎛α=

where Ae = BH = effective channel area and

FROM THESE WE OBTAIN

3/1

hd

c2

3/12h

d

c

u1 42

141 ⎟⎟⎠

⎞⎜⎜⎝

⎛ααα

=α⎟⎟⎠

⎞⎜⎜⎝

⎛α

αα

α=α

Let AT = the total area of the drainage basin and LT = (AT)1/2 denote a length scale for the total basin. It then follows that

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛α

αα

α=

T

2/1e

3/1

2/1eu

3/12h

d

c

uT

u

LA

Avk41

LL

Now suppose that Ae is set. Then the ratio Lu/LT becomes smaller (the drainage basin becomes more intricate) asa) the size of the basin L increases,b) the rate of uplift vu increases andc) the hillslope diffusivity k decreases.In addition, Sh becomes larger (headwater hillslopesbecome steeper) as vu increases and k decreases.

"Kolmogoroff" Scaling for Drainage Basins

α1 10.86 αc 5 channel incision parameterα2 1.357 αd 0.5 amplitude dissipation parameterB 5 m αh 2 hillslope diffusion parameterH 2 m αu 0.5 geometric parameter for headwater basinvu 0.1 cm/yrk 300 cm^2/yr α1 10.9

α2 1.36Ae 3.162Rek 0.105 m

Lu 72.68 mSh 0.303θ 16.85 deg

3/1

hd

c2

3/12h

d

c

u1 42

141 ⎟⎟⎠

⎞⎜⎜⎝

⎛ααα

=α⎟⎟⎠

⎞⎜⎜⎝

⎛α

αα

α=α

3/1

2/1eu

12/1e

u

Avk

AL

⎟⎟⎠

⎞⎜⎜⎝

⎛α=

3/22/1eu

2h kAvS ⎟⎟

⎠

⎞⎜⎜⎝

⎛α=

Sample calculation (numbers for k, vu courtesy Bill Dietrich, Oregon Pacific Coast Range)

"Kolmogoroff" Scaling for Drainage Basins

α1 10.86 αc 5 channel incision parameterα2 1.357 αd 0.5 amplitude dissipation parameterB 5 m αh 2 hillslope diffusion parameterH 2 m αu 0.5 geometric parameter for headwater basinvu 1 cm/yrk 300 cm^2/yr α1 10.9

α2 1.36Ae 3.162Rek 1.054 m

Lu 33.74 mSh 1.406θ 54.57 deg

3/1

hd

c2

3/12h

d

c

u1 42

141 ⎟⎟⎠

⎞⎜⎜⎝

⎛ααα

=α⎟⎟⎠

⎞⎜⎜⎝

⎛α

αα

α=α

3/1

2/1eu

12/1e

u

Avk

AL

⎟⎟⎠

⎞⎜⎜⎝

⎛α=

3/22/1eu

2h kAvS ⎟⎟

⎠

⎞⎜⎜⎝

⎛α=

Sample calculation (numbers for k, vu courtesy Bill Dietrich, Oregon Pacific Coast Range)

THE END!!Or maybe not!

But wait! It's not over yet! Can we

explain how submarine drainage

basins form?

Trinity Canyon and associated

drainage network, Eel Margin,

Northern California

And what about the fine scale of tidal drainage

networks?

Barnstable Salt Marsh, Cape Cod,

Massachusetts

Image courtesy Tao Sun, Sergio Fagherazzi and David Furbish

Sample calculationConsider two rivers:a "prototype" with mean velocity U = 4 m/s and depth H = 2 m, anda "model" with mean velocity U = 1 m/s and depth H = 0.125 m.

One is a perfect Froude model of the other. Note L ~ H, u~U/10, ν = 1x10-6 m2/s.

Prototype ηk ~ 0.075 mm Eddies range from ~ 2 m to 0.075 mm

Model ηk ~ 0.11 mm Eddies range from ~ 0.125 m to 0.11 mm

Scale model up to prototype:Eddies range from ~ 2 m to 1.76 mm