Aeolian sand transport rates and mid-air collisions

Transcription

1 Kavli Institute of Theoretical Physics Aeolian sand transport rates and mid-air collisions Michael Sørensen University of Copenhagen michael. p.1/37

2 The Aarhus sand gang. p.2/37

3 Transport rate TRANSPORT RATE: Q =Φ MEAN JUMP LENGTH =Φ x(t i ) Φ =MASSFLUXFROMTHEBEDINTOTHEAIR t i time of impact x(t i ) jump length. p.3/37

4 Grain borne shear stress Owen (1964) ρu 2 = AIR BORNE SHEAR STRESS + GRAIN BORNE SHEAR STRESS = T a (y)+t g (y) U friction speed in the grain free wind, ρ density of air Grain borne shear stress at height y: T g (y) =Φv(y) v(y) =theaverageincreaseofthehorizontalvelocitycomponentofa saltating grain while it is above the level y v(y) =[ẋ(t 2 y) ẋ(t 1 y)]1 {y top >y} t 1 y 1st time at height y y top height of grain trajectory t 2 y 2nd time at height y. p.4/37

5 . p.5/37

6 Grain borne shear stress AIR BORNE SHEAR STRESS AT THE BED: T a (0) = πρu 2 0 <π<1 It follows from T a (0) + Φv(0) = ρu 2 that Φ=ρU 2 (1 π) /v(0), and T g (y) =ρu 2 (1 π) a(y) where a(y) =v(y)/v(0).. p.6/37

7 Friction speed Anderson (1986) FRICTION SPEED AT HEIGHT y: U (y) = U 2 T g (y)/ρ = U 1 (1 π)a(y) Sørensen (1991, 2004): 1 x 1 x U (y) =U (1 (1 π)a(y)) Durán and Herrmann (2006): U (0) = πu ρu (0) 2 = π 2 ρu 2 πρu 2. p.7/37

8 Wind profile We only need a good approximation to 1 x in the interval [0, 1 π]: 1 x 1 1 π 1 π x U (y) =U (1 (1 π)a(y)) Eddy viscosity/prandtl turbulence closure (κ = von Kármán s constant): du dy = U κy (1 ( 1 π ) a(y)) U(y) =κ 1 U [ ln(y/y0 ) ( 1 π ) b(y) ] where b(y) = y y 0 z 1 a(z)dz.. p.8/37

9 Assumption: The function b is independent of U. p.9/37

10 . p.10/37

11 . p.11/37

12 Asimplesaltationmodel:thegrainmotion ẍ = H(v)(U(y) ẋ) ÿ + g + H(v)ẏ =0 (x(0),y(0)) = (0, 0) (ẋ(0), ẏ(0)) = (v1,v 0 2) 0 is random H(v) =D(v)/(mv). p.12/37

13 Asimplesaltationmodel:thegrainmotion ẍ = H(v)(U(y) ẋ) ÿ + g + H(v)ẏ =0 (x(0),y(0)) = (0, 0) (ẋ(0), ẏ(0)) = (v1,v 0 2) 0 is random H(v) =D(v)/(mv) Owen (1964): D(v) =δv H(v) =t 1 t = m/δ is the response time of a grain to changes in the wind speed. p.12/37

14 Asimplesaltationmodel:thegrainmotion ẍ = H(v)(U(y) ẋ) ÿ + g + H(v)ẏ =0 (x(0),y(0)) = (0, 0) (ẋ(0), ẏ(0)) = (v1,v 0 2) 0 is random H(v) =D(v)/(mv) Owen (1964): D(v) =δv H(v) =t 1 t = m/δ is the response time of a grain to changes in the wind speed x(t) = t 0 ( ) 1 e (t s)/t U(y(s))ds + t v1 0 ( ) 1 e t/t y(t) = t (v f + v 0 2)(1 e t/t ) v f t v f = gt. p.12/37

15 A simple saltation model: the transport rate Q =Φ x(t i ) t i impact time x(t i ) = κ 1 U [ ti 0 ( 1 e (t i s)/t ) log(y(s)/y0 )ds ( 1 π ) ] t i ( ) 1 e (t i s)/t b(y(s))ds 0 + t v 0 1 ( 1 e t i /t ). Φ=ρU 2 (1 π) /v(0). p.13/37

16 A simple saltation model: the transport rate Q =Φ x(t i ) t i impact time x(t i ) = κ 1 U [ ti 0 ( 1 e (t i s)/t ) log(y(s)/y0 )ds ( 1 π ) ] t i ( ) 1 e (t i s)/t b(y(s))ds 0 + t v 0 1 ( 1 e t i /t ). Φ=ρU 2 (1 π) /v(0) Qg ρu 3 =(1 π) [ α + β π + γ/u ]. p.13/37

17 Saturated saltation Owen (1964): AIR BORNE SHEAR STRESS AT THE BED IS EQUAL TO ρu 2 c FOR ALL U U c U c friction speed at the impact threshold π = U 2 c U 2 = V 2 V = U U c dimensionless friction speed. p.14/37

18 Saturated saltation Φ=ρU 2 ( 1 V 2 ) /v(0), T g (y) =ρu 2 ( 1 V 2 ) a(y) U(y) =κ 1 U [ ln(y/y0 ) ( 1 V 1) b(y) ] Qg ρu 3 = ( 1 V 2)[ α + βv 1] = ( 1 V 1)( 1+V 1)[ α + βv 1]. p.15/37

19 Saturated saltation Φ=ρU 2 ( 1 V 2 ) /v(0), T g (y) =ρu 2 ( 1 V 2 ) a(y) U(y) =κ 1 U [ ln(y/y0 ) ( 1 V 1) b(y) ] Qg ρu 3 = ( 1 V 2)[ α + βv 1] = ( 1 V 1)( 1+V 1)[ α + βv 1] Sørensen (2004): Sørensen (1991): Qg ρu 3 Qg ρu 3 = ( 1 V 2)[ α + βv 1 + γv 2] = ( 1 V 1)[ α + βv 1]. p.15/37

20 Sørensen (2004) formula Iversen and Rasmussen (1999): Homogeneous sand of size 170 µm α =0,β =2.1,γ =3.0 Dimensionless transport rate Dimensionless friction speed. p.16/37

21 Sørensen (2004) formula Iversen and Rasmussen (1999): Natural sand of typical size 230 µm Råbjerg Mile α = 1.8,β =0,γ =15.9 Dimensionless transport rate Dimensionless friction speed. p.17/37

22 Bagnold s focal point U(y) = κ 1 U [ ln(y/y0 ) ( 1 V 1) b(y) ] = κ 1 U [ln(y/y 0 ) b(y)] + κ 1 U c b(y) Suppose a height ȳ exists where U(ȳ) does not depend on U U(ȳ) =κ 1 U [ln(ȳ/y 0 ) b(ȳ)] + κ 1 U c b(ȳ) Only possible if ln(ȳ/y 0 )=b(ȳ), but ln(ȳ/y 0 ) b(ȳ) = ȳ y 0 z 1 (1 a(z))dz 0 (a(z) 1) ln(ȳ/y 0 )=b(ȳ) only possible, if ȳ is sufficiently close to the bed that a(ȳ) 1 In that case ln(y/y 0 )=b(y) for all y ȳ. p.18/37

23 Bagnold s focal point U(y) = κ 1 [ U ln(y/y0 ) ( 1 π ) b(y) ] = κ 1 U [ln(y/y 0 ) b(y)] + κ 1 U (0)b(y) πu = U (0) U (0) = f(u ) Suppose κ 1 U [ln(ȳ/y 0 ) b(ȳ)] + κ 1 f(u )b(ȳ) does not depend on U Then where f (U )= (µ 1) µ =ln(ȳ/y 0 )/b(ȳ) 1. p.19/37

24 Bagnold s focal point f(u )= (µ 1)U + κ U c = f(u c )= (µ 1)U c + κ So κ = µu c and U (0) = (µ 1)U + µu c Owen (1964): µ =1 Since necessarily U (0) 0, thisformulacanonlyholdforsufficiently small values of U : 1 V 1 µ 1. p.20/37

25 Wind profile πu = U (0) = (µ 1)U + µu c implies π =1 µ(1 V 1 ) π =(µ 1) 2 2µ(1 µ)v 1 + µ 2 V 2 1 π = µ(1 V 1 )(2 µ + µv 1 ) U(y) =κ 1 U [ ln(y/y0 ) µ ( 1 V 1) b(y) ] U(ȳ) =κ 1 U [ ln(ȳ/y0 ) µ ( 1 V 1) b(ȳ) ] = κ 1 U c ln(ȳ/y 0 ). p.21/37

26 Transport rate Φ=ρU 2 µ(1 V 1 )(2 µ + µv 1 )/v(0) Qg ρu 3 = µ(1 V 1 )(2 µ + µv 1 ) [ α + βv 1] Durán and Herrmann (2006). p.22/37

27 Transport rate Φ=ρU 2 µ(1 V 1 )(2 µ + µv 1 )/v(0) Qg ρu 3 = µ(1 V 1 )(2 µ + µv 1 ) [ α + βv 1] Durán and Herrmann (2006) µ =ln(ȳ/y 0 )/b(ȳ). p.22/37

28 Transport rate Φ=ρU 2 µ(1 V 1 )(2 µ + µv 1 )/v(0) Qg ρu 3 = µ(1 V 1 )(2 µ + µv 1 ) [ α + βv 1] Durán and Herrmann (2006) µ =ln(ȳ/y 0 )/b(ȳ) U(y) =κ 1 U [ ln(y/y0 ) µ ( 1 V 1) b(t ) ] = κ 1 U ln(y/ỹ 0 ) for y T T = top of saltation layer ln(ỹ 0 )=ln(y 0 )+µ(1 V 1 )b(t ) b(y) = y y 0 z 1 a(z)dz. p.22/37

29 Grain speeds Horizontal component of grain speeds measured by laser-doppler technology, Rasmussen and Sørensen (2005).. p.23/37

30 The function v b(y) = y y 0 z 1 a(z)dz a(y) =v(y)/v(0) v(y) =[ẋ(t 2 y) ẋ(t 1 y)]1 {y top >y} By approximations in Rasmussen and Sørensen (2005): [ e v(y) = 2νλ 1 a y ỹ a 2 y +1 K 2 +2 ỹ (νu 2 (y) U) ( ) ( 2 ỹ(a 2 y +1) K 2 2 ) ] ỹ 1 a 2 y +1 ) ( K 1 (2 ỹ) ỹ(a 2 y +1) K 1 2. p.24/37

31 [ e v(y) = 2νλ 1 a y ỹ a 2 y +1 K 2 +2 ỹ (νu 2 (y) U) The function v ( ) ( 2 ỹ(a 2 y +1) K 2 2 ) ] ỹ 1 a 2 y +1 ) ( K 1 (2 ỹ) ỹ(a 2 y +1) K 1 2 U(y) =U u 20 Exponential(λ) E(u 10 u 20 )=νu 20 y = t [u 2 (y) v f ln(1 + u 2 (y)/v f )] v f = gt a y = 1 λ(v f + u 2 (y)) ỹ = y/(a y t v f ). p.25/37

32 Other transport rate formulae Qg ρu 3 = ( 1 V 1) f(v ) Forms of the function f(v ) proposed by other authors Owen (1964) (1 + V 1 )(α + βv 1 ) Kind (1976) C(1 + V 1 ) Owen (1980) C(1 + V 1 ) Ungar and Haff (1987) βv 1 (1 + V 1 ) Sørensen (1991) α + βv 1 Sauermann et al. (2001) (1 + V 1 )(α 1+CV 2 + γv 1 ) Sørensen (2004) (1 + V 1 )(α + βv 2 + γv 1 ) Durán and Herrmann (2006) and present µ(2 µ + µv 1 ) [ α + βv 1]. p.26/37

33 Mid-air collisions Sørensen (1988, 1991): Soft-bed hypothesis Sørensen and McEwan (1996) Probability that a saltator hits a reptator below height 1 cm Grains are spheres with radius r Saltator moves along a straight line with speed u and angle ϕ Time of mid-air collision: T [0, 1/u 2 ] c(y) concentration (by number) of grains at height y P (T [t, t + dt] T>t)=c(t u 2 t)4πr 2 udt. p.27/37

34 Probability of mid-air collision Probability density function of T : t f T (t) =au 2 c(1 u 2 t)exp ( au 2 0 ) c(1 u 2 s)ds, 0 t 1/u 2 Probability of mid-air collision: 1 P (T = u 1 2 )=1 exp ( a 1 0 ) c(s)ds Probability density of hight at which collision occurs (Y =1 u 2 T ): ( 1 ) f Y (y) =ac(y)exp a c(s)ds 0 y 1 y. p.28/37

35 Concentration profile ( 1 c(y) =m 1 Φ ẏ(t 1 y) 1 ) ẏ(t 1 y) Fitted an exponential curve to concentration found by McEwan and Willetts (1991) for spherical grains of radius 150 µm c(y) =c 0 exp( νu). p.29/37

36 Probability of mid-air collisions Probability of mid-air collisions (p) U angle c 0 ν p p.30/37

37 Height of collision. p.31/37

38 Velocities after the collision Coefficient of restitution for quartz grains of sand size: 0.7 Anderson and Haff (1991), experiments at University of Aberdeen Velocity of descending saltator after collision: v(u, α, β) =u cos(ϕ) sin(ϕ) 0 h(α, β) Velocity of target grain after collision: u h(α, β) h(α, β) =0.85 cos(ϕ)cos 2 (β)+sin(ϕ)cos(β)sin(β)sin(α) sin(ϕ)cos 2 (β)+cos(ϕ)cos(β)sin(β)sin(α) cos(β)sin(β)cos(α), sin(β) = L 2r 7-8 % of kinetic energy of saltators lost. p.32/37

39 Grain trajectories after collision The impact velocity u i of a grain that at height y 0 has velocity v 0 : u i1 = v 01 +(U v 01 )(1 exp( t i /t )) u i2 = v 02 t i g + y 0 /t u i3 = v 03 exp( t i /t ) exp( t i /t )=1 v f t i y 0 t (v 02 + v f ) Previous model with U(y) =U. p.33/37

40 . p.34/37

41 Mean horizontal displacement Mean horizontal displacement (cm) after collision (or potential collision) U No Collision Collision (cm/s) Saltator Target Saltator Target p.35/37

42 Mean number of grains ejected Mean number of grains ejected when the grains hit the sand surface Anderson and Haff (1991): n =1.75v i U No Collision Collision (cm/s) Saltator Both grains Both grains p.36/37

43 Momentum extracted from wind Mean momentum extracted from the wind by the saltator and the target grain after collision (or potential collision) U (cm/s) No Collision Collision p.37/37