A Two-Phase Continuum Theory for Windblown Sand

Transcription

1 A Two-Phase Continuum Theory for Windblown Sand Jim Jenkins Civil and Environmental Engineering Cornell University Alexandre Valance Institut de Physique de Rennes Université de Rennes 1 Suspension and transport of sand by saltation and gas velocity fluctuations in a turbulent shearing flow

2 Desertification Nouakchott, Mauritania Dune formation and migration associated with wind-blown sand

3 Saltation (sauter: to jump) A grain lifted from the bed by a strong turbulent eddy is accelerated by the mean turbulent shear flow and, upon colliding with the bed, rebounds and ejects other particles. The drag of the particles on the wind eventually limits the number of particles that can participate. At higher wind speeds, turbulent suspension, associated with the velocity fluctuations of the wind, becomes increasingly important, and collisions between particles above the bed may play a role in their suspension and transport.

4 Aeolian Transport Two-phase, turbulent flow Ŝ H û Û Ĥ Friction velocity / Shields parameter ( f ˆ ˆ ) H ρ S* S ˆ ( ˆˆˆ s ) H / ρ gd u* S / ˆ Air drag: νd(u ˆ ˆ u) ˆ f ˆ 3 ρ ˆ ˆ 2 ˆ 18µ ˆ D (U u) ˆ 3T ˆ ˆ2 1 d + + d f

5 Experiments Creyssels, Dupont, Valance (Rennes) Ould El Moctar (Nantes), Rasmussen (Aarhus) Wind-Sand Tunnel, Aarhus, Denmark

6 Volume Fraction ν = ν exp ŷ ˆl ˆl 4ˆd

7 Continuum Theory Particle horizontal momentum ds νd = + (U u) dy σ Particle vertical momentum: dp = ν dy Particle fluctuation energy dq = + su γ dy Particle pressure p =ν T Particle shear stress? Particle energy flux?

8 Continuum Theory Single particle trajectories without vertical drag Upward: Downward: dξ x ξ y = D(U ξ x) dy dξx ξ y = D(U ξ x) dy Multiply by ν, sum, and average 2 d νξ y ( ξ x + ξ x) = νd ξ y( ξ x ξ x) dy 2u ( ξ +ξ ) x x p νξ =ν T 2s νξ y( ξ x ξ x) 2 y du p Ds dy =α q νξ y ( ξ x u) +ξ y + νξy ( ξx u) + ξ y B ν ξ y( ξ x ξx)( ξ x +ξx)/2 ξ y( ξ x ξx) u q=

9 Particle Shear Stress Continuum versus discrete simulation 15 u = 17.5, T = 2, S* =.35, α = 2 Discrete Simulation Contniuum model 1 y/d s

10 Splash Beladjine, et al. Phys. Rev. E 75, 6135 (27) Incident sphere Rebounding sphere Ejected spheres Ejected spheres Ejected spheres

11 Momentum of rebounding particles ξ= e( ξ ) ξ= (.87.72sin θ) ξ ξ =.3 ε( ξ ) ξ =.15 ξ sin θ y y y Total number N of particles N(ξ) = 1+13(1 e 2 ) ξ 4 1, if ξ > 4 1, if 1 ξ 4, if ξ < 1 Velocity distribution function 2 2 n ξ ( x u) ξ y f( ξ ) = exp 2πT 2T ( ν =π n /6)

12 Mass flux!m = (N 1)ξ y f(ξ)dξ ξy = 13 2π 74 2 π n T 2 u (4 u ) u n 2T e T πt 2u e (4 u ) 2 2T Momentum flux!m = π 6 ( ξ ξ)ξ y f(ξ)dξ ξ y!m x = ν T u T.33T!M y = ν T.12 u T + T u.8 u + ν T 2

13 u With!M y = p = ν T, 4.6 T = Boundary-Value Problem Steady, fully-developed flow:!m = u = 2, T = 2 and s!m x =.6ν T dν dy = ν T du dy = 2D s p D = (U u) + 3T + σ σr ds D(U u) dy = ν ρ d(gd) = µ / ρ s σ= R ρ f f f du dy = (S* s)σ (S* s)σ κ ( y + y )

14 y= : u = 2, T = 2, s =.6ν T, U= y= 21: s= Parameter: ν Include suspension by the turbulent velocity fluctuations Particle vertical momentum = dˆp dŷ ˆρs ν ĝ + ˆD ν Δ ˆV Turbulent suspension force ν Δ ˆV = µ ˆ ρˆ T f dν dˆ y Turbulent viscosity.9 ˆµ T = ˆρ f.165 κ ( ŷ + ŷ ) ˆk Velocity fluctuations

15 ˆk (.9) (Sˆ s) ˆ = f.165 ρˆ H Boundary-Value Problem Steady, fully-developed flow:!m = u = 2, T = 2 and s!m x =.6ν T dν ν = dy T + Dµ µ T T 1 σ k.41( y + y ) (.9) k (S* s) σ.165 [ ] du dy = 2D s p D = (U u) + 3T + σ σr ds D(U u) dy = ν ρ d(gd) = µ / ρ s σ= R ρ f f f du dy = (S* s)σ µ T

16 y= : u = 2, T = 2, s =.6ν T, U= y= 21: s= Parameter: ν Influence of Suspension S*=.35 and.98 Grain Velocities u = 2, T = 2, s =.6ν T, α = Dimensionless Height Dimensionless Grain Velocity

17 Suspension moves the grain velocities in the right direction.

18 Influence of Suspension Air Velocities Suspension moves the air velocities in the right directions.

19 Influence of Suspension Particle Concentration Dimensionless Height Concentration Above 5 particle diameters, suspension increases the dimensionless decay length.

20 Uniform, Unsteady (saturation time) c u t = cv u y + s y + c D σ (U u) s = σp αd u y c v t v = cv y + y ( p + b) c D σ v c b = σp D v y c t = c v y v c y + ε y c y c T p = ct U t = σ S y cd(u u) S = 1 σ κ y + y ( ) U y 2 σ = ρs ρ, D =.3 f ( ) 2 + v 2 U u Re, Re = d(gd) µ / ρ f y = : s =.6cT, U =, v = β ( u u ), c y = c T T = ( u / 4.6) 2 β = y = H: s =, S = S * + S *, b =, cu =.1 t = : steady solution c, u, v =, U.

21 Linearize about the steady solution: S * / S * << 1 u c t = c v u y + s y + c D σ (U u ) D D +c σ (U u ) + c σ ( U u ) s = T D c u αd D y + c u y + u c y D =.3 U u R D =.3 U u v c t ( ) c D = y p + b σ v c b = σc T v D y c t = v c y v c y + ε c y c T p = c T U t = σ S y c D (U u ) c D (U u ) c D ( U u ) ( ) S = 2 κ y + y σ 2 U y U y

22 Take S * =.5. Boundary conditions y = : s =.6 c T, U =, v = β u, c y = c T y = H: s =, S* =.5, b =, c u =.1 Initial conditions t = : u, v, U, c, S * = S * Dimensional parameters cgs d =.25, g = 98, µ f / ρ f =.15 Dimensionless parameters u = 21.7, T = 22, σ = 22, α = 2 κ =.41, ε =.1, β = Solve the system using Matlab pdepe.

23 u'(y,t): S* =.5, u = 21.7, S*' = Time t Distance y 15 c' (y,t): S* =.5, u = 21.7, S*' = Time t Distance y 15

24 v' (y,t): S* =.5, u = 21.7, S*' = Time t Distance y 15 B. Andreotti, P. Claudin, O. Pouliquen, Geomorphology 123, (21)

25 What have we done? We ve formulated a two-phase continuum theory for saltation that also accommodates turbulent and collisional suspension. The theory permits boundary-value problems to be phrased and solved for steady, uniform flows and initial-boundary-value problems to be phrased and solved both for unsteady, uniform flows and steady, developing flows. The solutions reproduce the distributions and feature of flows seen in laboratory flows, including the characteristic times for the adjustment of the different mechanisms of suspension to changes in conditions. At least one of these adjustments occurs in a nonmonotone way.