A Kolmogoroff-type Scaling for the Fine Structure of Drainage Basins Gary Parker, University of Minnesota Peter 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 nonlinear transfer of dissipation of turbulent energy energy from large to turbulent energy to at large scales small scales heat at small scales Balance of energy in a steady turbulent flow ∂ (mean kinetic energy of turbulence) = ∂t production rate − dissipation rate + spatial transfer rate Let u = 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 ∂u u3 ∂u′ ∂u′ u 2 P = −u′u′ i ~ ε = ν i i ~ ν k i j ∂x ∂x η2 ∂x j L j j k Make Kolmogoroff scales from the viscosity ν (L2/T) and the dissipation rate ε (L2/T3) 1/ 4 ⎛ ν3 ⎞ u η ⎜ ⎟ 1/ 4 k k ηk = ⎜ ⎟ uk = (νε) = 1 ⎝ ε ⎠ ν Recalling that u3 P ≅ ε ε ~ L it is found that 3 / 4 η ⎛ ν ⎞ k ~ ⎜ ⎟ L ⎝ uL ⎠ That is, the higher the Reynolds number of the flow the finer is the turbulence From Tennekes and Lumley Does the idea of a Kolmogoroff scale 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 Some scales Lu = length of headwater tributary 2 Au = αuLu = area of headwater catchment: αu ~o(1) Sh = characteristic slope of hillslope in headwater catchment ridge headwater tributary Some parameters η = bed elevation t = time (x1, x2) = spatial coordinate ζ = incision rate (q1, q2) = vector of volume hillslope transport/width k = hillslope diffusivity r qr = −k∇η Exner equation of sediment balance (neglecting porosity) ridge ∂η = −ς + k∇2η + v headwater ∂t u 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 r dA = v A − ∇ ⋅ qrdA = v A − q ds ∫∫∂t u u ∫∫u u ∫ n AAuu gain loss due to steady due to transport from state uplift hillslope to channel ∂η q = −k ~ kS ridge n ∂n h Here n is normal headwater to red perimeter of tributary path integral: Continue integration on hillslope: path integral is around red line ∂η q = −k ~ kS ridge n ∂n h qnds ~ 2Luqn ~ 2LukSh ∫ headwater tributary The following scale relation results: 2 vuAu = αuvuLu = 2αhLukSh 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 ⎧vI in channel ς = ⎨ ⎩0 on hillslope Thus within the channel of the headwater tributary ∂η = −v + v or thus vI = vu ∂t I u steady Channel incision must just keep pace state with uplift Mean and fluctuating bed elevation η′ = η − η η η Video clip Hasbargen and Paola Equation of evolution of amplitude of elevation fluctuation Decompose into mean and fluctuating parts: η = η + η′ ς = ς + ς′ where the overbar denotes an average over an appropriate spatial window and the prime denotes a fluctuation about that average. Local Exner: ∂η = −ς + k∇2 η + v ∂t u ∂η 2 Local Exner: = −ς + k∇ η + v ∂t u ∂η 2 Spatially averaged Exner: = −ς + k∇ η + v ∂t u ∂ ⎛ 1 2 ⎞ 2 Multiply by η ⎜ η ⎟ = −ςη + kη∇ η + vu η = ∂t 2 and reduce: ⎝ ⎠ r r r r − ςη + k∇ ⋅ ()η∇η − k()∇η ⋅ ()∇ ⋅ η + vu η A ∂ ⎛ 1 1 ⎞ Multiply local ⎜ η2 + η′2 ⎟ = −ςη − ς′η′ Exner by η, ∂t ⎝ 2 2 ⎠ average and r r r r + k∇ ⋅ ()η∇η − k()∇η ⋅ ()∇ ⋅ η + vu η reduce: r r r r + k∇ ⋅ η′∇η′ − k ∇η′ ⋅ ∇η′ B ()()() Subtract B from A to get equation of evolution of amplitude of elevation fluctuation: ∂ ⎛ 1 ⎞ r r r r ⎜ η′2 ⎟ = −ς′η′ + k∇ ⋅ (η′∇η′)− k(∇η′)⋅ (∇η′) ∂t ⎝ 2 ⎠ IIIIII IV I. Time rate of change of amplitude of elevation fluctuation II. Rate of creation of amplitude fluctuation due to incision III. Rate of transport of amplitude fluctuation due to diffusion IV. Rate of destruction of amplitude fluctuation by hillslope diffusion Steady state: approximate balance between creation and destruction of amplitude fluctuation r r − ς′η′ ≅ k(∇η′)(⋅ ∇η′ ) Now if most of the destruction is biased toward the finest scale of the drainage basin, i.e. the headwater catchment, then r r 2 k(∇η′)⋅ (∇η′) = αdkSh where αd is an o(1) constant Approximate form for the rate of creation of amplitude fluctuation by incision ridge ηp vI∆t headwater ηc(t) ηc(t+∆t) tributary In the headwater catchment, appoximate the channel as incising into a plain with constant elevation ηp. Channel elevation ηc decreases with speed vI. η A + η BL Thus: η = p u c u where B denotes channel width Au + BLu Now ⎧ 1 2 (ηp − η) on plain 1 2 ⎪ ()η′ = 2 ⎨1 2 ⎪ (η − η)2 in channel ⎩2 c Since η& c = −vI = −vu it follows that ∂ 1 2 ∂ 1 2 ()η′ A + ()η′ BL ∂t 2 u ∂t 2 u ∂ 1 2 − η′ς′ = ()η′ = plain channel ≅ ∂t 2 A + BL incision u u BLu (η − ηc )vu Au Now scaling η − ηc = α cH 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 BLu − η′ς′ = αcHvu Au Balancing this against the rate of destruction of fluctuation amplitude, BLu 2 αcHvu = αdkSh Au Scale relations for headwater catchment: 2 Geometric scaling: Au = αuLu Rate of hillslope denudation must α v L2 = 2α L kS balance with rate of uplift: u u u h u h Rates of creation and destruction of BL α Hv u = α kS2 elevation fluctuation amplitude must c u d h Au balance: FROM THESE WE OBTAIN 1/ 3 2 / 3 L ⎛ k ⎞ ⎛ v A1/ 2 ⎞ u = α ⎜ ⎟ S = α ⎜ u e ⎟ 1/ 2 1⎜ 1/ 2 ⎟ h 2 ⎜ ⎟ A e ⎝ vuA e ⎠ ⎝ k ⎠ 1/ 3 1/ 3 1 ⎛ α ⎞ 1 ⎛ α ⎞ where A = BH = effective ⎜ c 2 ⎟ ⎜ c ⎟ e α1 = ⎜4 αh ⎟ α2 = ⎜4 ⎟ channel area and αu ⎝ αd ⎠ 2 ⎝ αdαh ⎠ Let AT = the total area of the drainage basin and LT = 1/2 (AT) denote a length scale for the total basin. It then follows that 1/ 3 1/ 3 L 1 ⎛ α ⎞ ⎛ k ⎞ ⎛ A1/ 2 ⎞ u = ⎜4 c α2 ⎟ ⎜ ⎟ ⎜ e ⎟ ⎜ h ⎟ ⎜ 1/ 2 ⎟ ⎜ ⎟ LT αu ⎝ αd ⎠ ⎝ vuA e ⎠ ⎝ LT ⎠ Now suppose that Ae is set. Then the ratio Lu/LT becomes smaller (the drainage basin becomes more intricate) as a) the size of the basin L increases, b) the rate of uplift vu increases and c) the hillslope diffusivity k decreases. In addition, Sh becomes larger (headwater hillslopes become steeper) as vu increases and k decreases. 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 parameter α2 1.357 αd 0.5 amplitude dissipation parameter B5m αh 2 hillslope diffusion parameter H2m αu 0.5 geometric parameter for headwater basin vu 0.1 cm/yr k300cm^2/yr α1 10.9 α2 1.36 1/ 3 2 / 3 Ae 3.162 1/ 2 Lu ⎛ k ⎞ ⎛ vuA e ⎞ Rek 0.105 m = α ⎜ ⎟ S = α ⎜ ⎟ 1/ 2 1⎜ 1/ 2 ⎟ h 2 ⎜ ⎟ A e ⎝ vuA e ⎠ ⎝ k ⎠ 1/ 3 1/ 3 Lu 72.68 m 1 ⎛ α ⎞ 1 ⎛ α ⎞ α = ⎜4 c α2 ⎟ α = ⎜4 c ⎟ Sh 0.303 1 ⎜ h ⎟ 2 ⎜ ⎟ αu ⎝ αd ⎠ 2 ⎝ αdαh ⎠ θ 16.85 deg 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 parameter α2 1.357 αd 0.5 amplitude dissipation parameter B5m αh 2 hillslope diffusion parameter H2m αu 0.5 geometric parameter for headwater basin vu 1cm/yr k300cm^2/yr α1 10.9 α2 1.36 1/ 3 2 / 3 Ae 3.162 1/ 2 Lu ⎛ k ⎞ ⎛ vuA e ⎞ Rek 1.054 m = α ⎜ ⎟ S = α ⎜ ⎟ 1/ 2 1⎜ 1/ 2 ⎟ h 2 ⎜ ⎟ A e ⎝ vuA e ⎠ ⎝ k ⎠ 1/ 3 1/ 3 Lu 33.74 m 1 ⎛ α ⎞ 1 ⎛ α ⎞ α = ⎜4 c α2 ⎟ α = ⎜4 c ⎟ Sh 1.406 1 ⎜ h ⎟ 2 ⎜ ⎟ αu ⎝ αd ⎠ 2 ⎝ αdαh ⎠ θ 54.57 deg 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 calculation Consider two rivers: a "prototype" with mean velocity U = 4 m/s and depth H = 2 m, and a "model" with mean velocity U = 1 m/s and depth H = 0.125 m.