Intro New Scheme Smooth Rough Examples Conclusion New scheme for dry deposition of aerosols Rostislav D. Kouznetsov and Mikhail A. Sofiev
Intro New Scheme Smooth Rough Examples Conclusion The problem Given concentration at some height above ground find the steady-state flux. Deposition velocity: v d ( z 1 ) = J / C ( z 1 ) Approaches: ◮ Fixed or size-dependant deposition velocity ◮ Resistance analogy (aerodynamic + quasi-laminar sub-layers etc. . . ) ◮ Something more fancy. . .
Intro New Scheme Smooth Rough Examples Conclusion On the resistance analogy For gases – straightforward resistance analogy: steady-state flux over any layer is proportional to the difference of concentrations. For particles: J ( z ) = − K ( z ) ∂ C ∂ z + v ( z ) C . Resistance analogy does not apply for finite layers.
Intro New Scheme Smooth Rough Examples Conclusion Evolution of Resistance approaches Slinn 1980: ◮ Concepts of r a , r b and correction. 1 v d = + v s r a + r b + v s r a r b ◮ Suggested for water surfaces. ◮ Rcommended by SP1998 for all surfaces. ◮ Very easy to implement, widely accepted. Petroff and Zhang 2010: Zhang 2001: ◮ “Exponential” form for ◮ Surface dependent r b aerodynamic layer ◮ 4 parameters, 15 LUC, 5 seasons ◮ 10 parameters, 15 LUC, 5 seasons ◮ Rcommended by SP2006 ◮ Finally fits the data
Intro New Scheme Smooth Rough Examples Conclusion The scheme ◮ “Exponential” scheme for finite layers Z2 C2 ◮ Separate treatment of J1 smooth and rough surfaces Z1 C1 Smooth Rough V1 ◮ Rigorously derived V1 Z obs Z obs scheme for smooth surfaces Za Vimp Va Vint Vdif Vs ◮ Small amount of 0 parameters for rough surfaces
Intro New Scheme Smooth Rough Examples Conclusion “Exponential” scheme Steady-state particle flux equation below z 1 : J ( z ) = − K ( z ) ∂ C ∂ z + v ( z ) C = const if v ( z ) = v s = const and C (0) = 0: � z 1 C ( z 1 ) dz J ( z 1 ) = 1 − exp( − v s r ) v s , r = K ( z ) 0 r is the resistance of the layer below z 1 . Can be also solved if v ( z ) � = const and for C (0) � = 0. The layer can be split at any point to evaluate corresponding concentration.
Intro New Scheme Smooth Rough Examples Conclusion Smooth surfaces 10 2 10 1 10 0 ◮ 1D problem ν T + 10 -1 DNS 10 -2 ◮ Universal Zhao&Wu (2006) 10 -3 This study turbulence 1 10 100 profiles 10 0 (normalized by 10 -1 ν , u ∗ ) w + ◮ Well studied 10 -2 ◮ Can be fit with 1 10 100 10 2 rational + / w +2 functions + = ν T 10 1 ◮ Of interest: τ L ν t , w 2 , τ L 1 10 100 z +
Intro New Scheme Smooth Rough Examples Conclusion Smooth surfaces Steady-flux equation above smooth surface J = − ( D + ν p ( z )) ∂ C ( z ) + ( V t ( z ) − v s ) C ( z ) , ∂ z where D and ν p are Brownian and (vertical) eddy diffusivity of particles, and V t is turbophoretic velocity: dw 2 p V t = − τ p dz , where w 2 p is the mean square vertical velocity of a particle due to turbulence. The profiles of turbulence over smooth surfaces are universal and can be approximated with rational functions.
Intro New Scheme Smooth Rough Examples Conclusion Smooth surfaces: verification Floor: u*=0.14 m/s u*=0.4 m/s u*=1 m/s 10 0 10 0 10 0 This study, z = 30 cm Z01, inland water 10 −1 10 −1 10 −1 S80 Sippola & Nazaroff (2004), smooth floor 10 −2 10 −2 10 −2 Sehmel & Sutter (1974), water surface + v d 10 −3 10 −3 10 −3 Möller & Schumann (1970), water surface Caffrey et al (1998), 10 −4 10 −4 10 −4 natural lake v d = 5 ⋅ 10 −4 m/s 10 −5 10 −5 10 −5 10 −8 10 −7 10 −6 10 −5 10 −4 10 −8 10 −7 10 −6 10 −5 10 −4 10 −8 10 −7 10 −6 10 −5 10 −4 Particle diameter, m Particle diameter, m Particle diameter, m Wall: u*=0.14 m/s u*=0.4 m/s u*=1 m/s 10 0 10 0 10 0 This study, z = 30 cm This study, z = 1 cm 10 −1 10 −1 10 −1 Sippola & Nazaroff (2004), smooth wall same, smooth ceiling 10 −2 10 −2 10 −2 Wells & Chamberlain (1967) + v d 10 −3 10 −3 10 −3 Liu & Agarwal (1974) El−Shobokshy (1983) 10 −4 10 −4 10 −4 10 −5 10 −5 10 −5 10 −7 10 −6 10 −5 10 −4 10 −7 10 −6 10 −5 10 −4 10 −7 10 −6 10 −5 10 −4
Intro New Scheme Smooth Rough Examples Conclusion Rough surfaces Simple thoughts: ◮ Air moves in a canopy consisting of collectors ◮ Same collectors absorb momentum and matter ◮ Momentum flux is (more or less) well studied ◮ Ratio of corresponding cross-sections gives ratio of deposition velocities Flow-collector interaction: ◮ Re c = Ud c /ν Particle-collector interaction: ◮ Diffusion Sc = ν/ D ◮ Interception d p / d c ◮ Impaction St = 2 τ p U top d c
Intro New Scheme Smooth Rough Examples Conclusion Rough surfaces (starting points) ◮ Correlations for deposition on rough surfaces (FdlM&F, 1982): Π = d p Re 1 / 2 Sc 1 / 3 c d c Y = d p ( V d − v s ) D Parameters: d p / d c , Re c = Ud c /ν , Sc = ν/ D . Y ∼ Π in the diffusion range (small Π) and Y ∼ Π 3 in the interception range (large Π). ◮ Single-element efficiency for spheres and cylinders (P&F, 1984): d p Re c Sc · η = d p U D η = A Π + B Π 3 , d c A ≃ 2 and B ≃ 1 slightly depend on the collector shape. ◮ If u ∗ and z 0 are used as velocity and size, the correlation holds with different coefficients for each surface (Schack, 1985).
Intro New Scheme Smooth Rough Examples Conclusion Rough surfaces (continued) ◮ Relevant velocity scale U top ≃ 3 u ∗ . ◮ Collection scale u ∗ a = d c U top ◮ Ratio u ∗ / U t op does not appear in Π = d p a Re 1 / 2 Sc 1 / 3 ∗ ◮ Reynolds number Re ∗ = u ∗ a ν ◮ Once fitted for the dataset of Chamberlain (grass in wind tunnel, d c = 5 mm), the correlation Y = 2Π + 80Π 3 fits other datasets with a single fitting parameter a .
Intro New Scheme Smooth Rough Examples Conclusion Rough surfaces (continued) Π- Y correlation. “Learning” and “Control” datasets. 10 8 Aerosol, grass a=2mm Aerosol, plastic grass a=2mm 10 6 Thorium, grass a=2mm Thorium, short grass a=3mm Thorium, plastic grass a=2mm 10 4 Thorium, toweling a=4mm Thorium, glass a=10mm Water, grass a=2mm 10 2 Water, toweling a=4mm Sehmel & Sutter (1974), This study d p v d / D gravel, a=0.5mm 10 6 Single-element Clough (1975), moss, a=0.5mm 10 0 Sippola & Nazaroff (2003), 10 4 insulated floor, a=0.5mm d p v d / D same, insulated wall, a=0.5mm 10 -2 This study 10 2 Single-element 10 -4 10 0 10 -6 10 -2 10 -8 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 10 1 10 2 10 -3 10 -2 10 -1 10 0 10 1 10 2 d p /d c Re 1/2 Sc 1/3 d p /d c Re 1/2 Sc 1/3 ◮ “Learning” – given a , adjusted coefficients ◮ “Control” – fixed coefficients, adjusted a
Intro New Scheme Smooth Rough Examples Conclusion Impaction 10 0 (a) Stokes number can be expressed 10 -1 Efficiency 10 -2 through the same scale: 10 -3 Re=20 Re=100 10 -4 St = 2 τ p U top = 2 τ p u ∗ Re=421 Re=1685 10 -5 d c a 0.01 0.1 1 10 Stokes number St = 2 τ p U/d c Effective stokes number (accounting 10 0 (b) 10 -1 Efficiency 10 -2 for viscous layer): 10 -3 − 1 = St − u ∗ − 1 10 -4 St e = St − Re 2 Re ∗ , 2 Approximation c 10 -5 U top 0.01 0.1 1 10 St - Re -1/2 The efficiency of impaction is approximated: � � � − 0 . 1 1 exp if St e > 0 . 15, St e − 0 . 15 − √ St e − 0 . 15 η imp (St e ) = 0 if St e ≤ 0 . 15.
Intro New Scheme Smooth Rough Examples Conclusion Rough surfaces summary Deposition velocity within in-canopy layer: V d ( z 0 ) = v dif + v int + v imp + v s , v dif = u ∗ · 2Re − 1 / 2 Sc − 2 / 3 , ∗ � 2 � d p Re 1 / 2 v int = u ∗ · 80 , ∗ a 2 u ∗ v imp = u ∗ · η imp (St e ) . U top Aerodynamic layer is accounted with exponential scheme: 1 V d ( z 0 ) exp( − v s r a ) + 1 1 V d ( z 1 ) = (1 − exp( − v s r a )) . v s
Intro New Scheme Smooth Rough Examples Conclusion Overall scheme Z2 C2 J1 Z1 C1 Smooth Rough V1 V1 Z obs Z obs Za Vimp Va Vint Vdif Vs 0 Smooth or rough is decided from roughness Reynolds number. Transition occurs: 2 < u ∗ z 0 /ν < 4
Intro New Scheme Smooth Rough Examples Conclusion Illustration: “grass” and “snow” u*=0.35 m/s u*=0.7 m/s u*=1.4 m/s This study, smooth 10 -1 10 -1 10 -1 Z01, grass S80 This study, a=2mm Same, no impaction 10 -2 10 -2 10 -2 Grass + Sticky grass v d 10 -3 10 -3 10 -3 10 -4 10 -4 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 Particle diameter, m Particle diameter, m Particle diameter, m u*=0.16 m/s u*=0.37 m/s u*=0.63 m/s This study, smooth 10 -1 10 -1 10 -1 Z01, snow S80 This study, a=0.5mm Same, no impaction 10 -2 10 -2 10 -2 Sippola & Nazaroff (2003), insulated floor + v d 10 -3 10 -3 10 -3 10 -4 10 -4 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 10 -8 10 -7 10 -6 10 -5 10 -4 Particle diameter, m Particle diameter, m Particle diameter, m
Intro New Scheme Smooth Rough Examples Conclusion Account for the thermophoresis ◮ Thermophoresis: particles move to cooler regions v TF = α (Kn) · ν dT / dz = α (Kn) · Pr F T T T ◮ The order of magnitude: 1 mm/s per kW/m2. ◮ Is important in the laminar layer for sub-micro range. ◮ Easy to estimate when TF enhances deposition (heat flux towards the surface) and/or when the surface is homogeneous ◮ Can block deposition to water surfaces for sub-micro particles
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