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dimensionality of particulate matter dimensionality N of compact particles: ε=1 ε→∞ ε→0 aspect ratio ε = L/d N=3 N=2 N=1 ε>1 Objects with aspect ε<1 ratios 0 < ε < 1 and 1 < ε < ∞ may be described by non- integer dimensionalities. N = 2 + 1/ε N = 1 + 2·ε treatment of irregular shapes: N=3 N=2 N=1 • N = 2 and N = 1 shapes may be bended in the L or A = L×L dimension, respectively. • Irregular cross sections of N = 2 shapes are described by the diameter of the inscribed circle. • Irregular ε > 1 or ε < 1 particles are described by the diameter of the inscribed circle like before, however, taking N = 2 + 1/ε or N = 1 + 2·ε, respectively. • N = 3 shapes are described by the diameter of the inscribes sphere. soda ash particles with N ≈ 3 soda ash particles with ε ≈ 1/6, hence N ≈ 1.33 200 µm 200 µm glass fibres with N = 2 12 µm natural mineral fibres (erionite) with N = 2 2 µm rice husk with N ≈ 1 10 mm sphere, N = 3 fibre, N = 2 chip, N = 1 4π 3 V0 = ⋅ r0 π ⋅ L ⋅ r02 2 A⋅ r0 3 A0 = 4π ⋅ r02 2π ⋅ L ⋅ r 2A A 4π ⋅ r02 3 2π ⋅ L ⋅ r 2 2A 1 S= 0 = = = = V0 4π 3 r0 π ⋅ L ⋅ r0 r0 2 2 A ⋅ r0 r0 ⋅ r0 3 unified law for the surface area to volume ratio of particles A0 N S= = with dimensionality N: V0 r0 sphere, N = 3 fibre, N = 2 chip, N = 1 4π 3 V0 = ⋅ r0 π ⋅ L ⋅ r02 2 A⋅ r0 3 A0 = 4π ⋅ r02 2π ⋅ L ⋅ r 2A A 4π ⋅ r02 3 2π ⋅ L ⋅ r 2 2A 1 S= 0 = = = = V0 4π 3 r0 π ⋅ L ⋅ r0 r0 2 2 A ⋅ r0 r0 ⋅ r0 3 unified law for the surface area to volume ratio of particles A0 N S= = with dimensionality N: V0 r0 A torus has S = 2 / r like a fibre r 2 R V0 = 4π 2 ⋅ R ⋅ r ; A0 = 2π 2 ⋅ R ⋅ r 2 A0 2 S= = independent of R V0 r The regular Platonian bodies have a surface to volume ratio of 3 / ri , where ri is the radius of the inscribed sphere. They thus resemble a body with N = 3 and r = ri. regular Platonian A V ri A 3 = = = S= = = body: a 2 a 3 a V ri 1 1 tetra- 3 12 2 12 6 14.697 hedron = 1.732 = 0.1179 = 0.2041 a octa- 2 3 1 3 2 1 6 6 7.348 hedron = 3.464 = 0.4714 = 0.4082 a 6 1 1 hexa- 2 6.000 hedron = 6.000 = 1.000 = 0.50000 a isoca- 5 3 5 12 (3 + 5 ) 1 12 ( 3⋅ 3 + 5 ) 3.970 hedron = 8.660 = 2.1817 = 0.7558 a dodeca- 15 ⋅ 1 + 5 5 2 1 4 (15 + 7 5 ) 1 2 ⋅ 5 2 + 10 5 11 2.694 hedron = 20.646 = 7.6631 = 1.1135 a sphere, N = 3 fibre, N = 2 chip, N = 1 4π 3 V0 = ⋅ r0 π ⋅ L ⋅ r02 2 A ⋅ r0 3 A0 = 4π ⋅ r02 2π ⋅ L ⋅ r 2A A 4π ⋅ r02 3 2π ⋅ L ⋅ r 2 2A 1 S= 0 = = = = V0 4π 3 r0 π ⋅ L ⋅ r0 r0 2 2 A ⋅ r0 r0 ⋅ r0 3 unified law for the surface area to volume ratio of particles A0 N S= = with dimensionality N: V0 r0 Shapes may be designed which reach N >3. Example: hollow cylinder with outer radius ra and outer radius ri : V0 = π ⋅ L ⋅ (ra2 − ri2 ); A0 = 2π ⋅ L ⋅ (ra + ri ) + 2π ⋅ (ra2 − ri2 ) ⇒ A0 2 2 2 S= = + → for L >> ra . V0 ra − ri L ra − ri L For L = 2 ra and d = ra - ri = ½ ra , 2 2 5 S= + = ⇒ N = 5. ra / 2 2 ⋅ ra ra different bodies used in chemical technology, N > 3 Fractal structures may reach very large N and S values 2· N N N wt. = 1200 g/m 670 g/m 370 g/m S= = = a a/2 r Let n = 12, *) ρ = 3 g/cm2 a = 2 cm Then Sm = 35 m2/g n=0 n=1 n=2 *) a / 312 ≈ 4 nm; wt. = 1 g/m after n steps: 2 2 2 a a a V = a2 ⋅ L V = 5⋅ ⋅ L V = 25 ⋅ ⋅ L V = 5 ⋅ n ⋅ L n 3 9 3 A = 4a ⋅ L a a a 4 A = 5⋅4⋅ ⋅ L A = 25 ⋅ 4 ⋅ ⋅ L A = 5n ⋅ 4 ⋅ n ⋅ L S= , N =2 3 9 3 a 12 36 4 ⋅ 3n S = , N =6 S = , N = 18 S= , N = 2 ⋅ 3n a a a standard reactions of particulate matter „R-D-A-F“ in chemical technology sphere, N = 3 fibre, N = 2 chip, N = 1 4π m0 = ⋅ ρ ⋅ r03 π ⋅ L ⋅ ρ ⋅ r02 2 A ⋅ ρ ⋅ r0 3 4π m(t ) = ⋅ ρ ⋅ r 3 (t ) π ⋅ L ⋅ ρ ⋅ r 2 (t ) 2 A ⋅ ρ ⋅ r (t ) 3 expressing r(t) by a constant dissolution velocity v yields r (t ) = r0 − v ⋅ t 4π m(t ) = ⋅ ρ ⋅ (r0 − v ⋅ t )3 π ⋅ L ⋅ ρ ⋅ (r0 − v ⋅ t ) 2 2 A ⋅ ρ ⋅ ( r0 − v ⋅ t ) 3 m0 − m(t ) definition of a dimensionless reaction α (t ) = ; α (t ) ∈ {0,1} turnover: m0 r03 − (r0 − v ⋅ t )3 r02 − (r0 − v ⋅ t ) 2 r0 − (r0 − v ⋅ t ) α (t ) = r03 r02 r0 3 2 1 v ⋅t v ⋅t v ⋅t α (t ) = 1 − 1 − 1 − 1 − 1 − 1 − r0 r0 r0 N unified law for the reaction turnover of particles v ⋅t α N (t ) = 1 − 1 − with dimensionality N: r0 m0 − m(t ) dimensionless mass turnover α(t): α (t ) = t = time, m0 = initial mass, m0 m = mass at time t m0 ≥ m(t ) ≥ 0 ⇒ 0 ≥ α (t ) ≥ 1 special case: spherical particles: m = 4 π ⋅ r3 ⋅ ρ 3 r = particle radius, ρ = density 4π ⋅ r03 ⋅ ρ − 4 π ⋅ r 3 (t ) ⋅ ρ r03 − r 3 (t ) α (t ) = 3 4π 3 = 3 ⋅ r03 ⋅ ρ r03 local mass turnover law: r (t ) = r0 (t ) − v ⋅ t (here: constant rate v in cm/s) r03 − (r0 − v ⋅ t )3 3 v ⋅t α (t ) = = 1 − 1 − r0 explicite form of the time law: 3 r0 r0 definition of the times t* and t50 a(t ) = 1 for t = = t100 = t * v ( ) of total and 50 % turnover, re- r r spectively a(t ) = 0.5 for t = 0 ⋅ 1 − 3 1 = 0.2063 ⋅ 0 = t50 2 v v layer fluid film (Nernst film δN) core core layer film fluid 1 δN fluid film diffusion Sh ⋅ D fluid c ⋅ t ⋅ 1- c l s 0 radius 1 pore diffusion 4 ⋅ D pore ⋅ t 0 radius surface controlled 1 reaction v⋅t 0 radius ∂c Fick‘s 2nd law: Fick‘s 1st law: J = −D + v c ∂x ∂c ∂ 2c ∂c ∂v = D 2 − v − c Jin Jout ∂c ∂J ∂t ∂x ∂x ∂x continuity: ∂c / ∂t =− ∂t ∂x δ Pr = ƒ(Re) l fluid velocity v δ Pr v∞ v l v l ∂v v ∞ Re = = = η /ρ ν ∂x δ Pr ∂c cs − c∞ ∂ 2c cs =− D concentration c ∂x δN since 2 = 0 ⇒ J = ·(c s − c ∞ ) ∂x δN δN c∞ ν δ N = δ Prƒ(Sc); Sc = D distance from surface surface l= Shsum = Shmin + Shlam + Shturb 2 2 circumference *) *) projection in flow direction v fluid ⋅ l Re = ⇒ ν 4π ⋅ r 2 vfluid v fluid ⋅ 2r l= = 2r Re = , Shmin = 2 2π ⋅ r 2π·r ν 2π ⋅ r ⋅ L vfluid v fluid ⋅ π ⋅ r l= =π ⋅r Re = , Shmin = 0.3 2 L + 2r ν 2r req = A / π vfluid v fluid ⋅ A / π Re = , Shmin = 0 2⋅ A A ν l= = A/π 2π ⋅ req Reynold‘s numbers for individual particles exposed to a lateral flow b π v fluid ⋅ l v fluid ⋅ b ⋅ r Re = = ν ν N 0.037·Re 0.8 ·Sc Shlam = 0.664·Re 1/ 2 1/3 ·Sc ; Shturb = 1 + 2,443·Re −1 ·(Sc 2/3 - 1) Sc = 109 106 105 104 103 107 6 102 10 5 101 10 4 100 10 Sh 103 102 101 Sheff = Shlam + Shturb 2 2 0 10 100 101 102 103 104 105 106 107 Re Sc = 109 108 107 106 105 3 104 10 103 10 2 102 Sh 101 1 5 10 2 100 100 0 1 2 10 10 10 Re α α RN(α), fluid film N=3 2 X = 1.5 laminar X = 1.2 turbulent, low Sc X = 0.1 turbulent , high Sc 1 RN (α ) : t t* ( ) = 1 − 2 −1 / N ⋅ t t50 = 1 − (1 − α )1 / N N t α = 1 − 1 − , t* ∝ r0X t * t / t* α α RN(α) N=3 2 1 RN (α ) : t t* ( ) = 1 − 2 −1 / N ⋅ t t50 = 1 − (1 − α )1 / N N t α = 1 − 1 − , t* = r0 / v t * t / t* very long hollow cylinder with length L, outer radius ra and inner radius ri: ( m0 = π ⋅ ρ ⋅ L ⋅ ra2 − ri2 ) [ m(t ) = π ⋅ ρ ⋅ L ⋅ (ra − v ⋅ t )2 − (ri + v ⋅ t )2 ] ra2 − ri2 + (ra − v ⋅ t )2 − (ri + v ⋅ t )2 α (t ) = ra2 − ri2 1 2⋅v ⋅t 2⋅v ⋅t r −r = 1 − = 1− ; t* = a i ra − ri ra − ri 2⋅v hollow cylinder like before, but with lenth L = ra; let ri → ½ra: 2ra ( m0 = 2π ⋅ ρ ⋅ ra ⋅ ra2 − ri2 ) [ m(t ) = 2π ⋅ ρ ⋅ (ra − v ⋅ t ) ⋅ (ra − v ⋅ t )2 − (ri + v ⋅ t )2 ] α (t ) = ( ) [ ra ra2 − ri2 + (ra − v ⋅ t ) ⋅ (ra − v ⋅ t )2 − (ri + v ⋅ t )2 ] ( ra ra2 − ri2 ) v ⋅t 2⋅v ⋅t v ⋅t 4⋅v ⋅t r −r r = 1 − 1 − ⋅ 1 − → 1 − 1 − ⋅ 1 − ; t* = a i → a ra ra − ri ra ra 2⋅v 4⋅v α α R2(α) short hollow cylinder, ri = ½ra; plotted vs. 2 t / t* long full cylinder, t* = vt / r long hollow cylinder, t* = 2vt / (ra - ri) short hollow cylinder, t* = 2vt / (ra - ri) t / t* α α DN(α) N=3 2 1 t t D1 (α ) : = 0.2500 ⋅ =α2 t* t50 t t D2 (α ) : = 0.1534 ⋅ = (1 − α ) ⋅ ln(1 − α ) + α t* t50 D3 (α ) : t t* = 0.0426 ⋅ t t50 ( = 1 − (1 − α )1 / 3 ) 2 r02 t* = 4⋅ D t / t* α implicit and explicit forms of DN(α): t t D1 (α ) : = 0.2500 ⋅ =α2 t* t50 1 t t α ≈ 1 − 1 − = t* t* t t D2 (α ) : = 0.1534 ⋅ = (1 − α ) ⋅ ln(1 − α ) + α t* t50 1.3793 t α ≈ 1 − 1 − → P2 (α ) t* D3 (α ) : t t* = 0.0426 ⋅ t t50 ( = 1 − (1 − α )1 / 3 ) 2 3 t α = 1 − 1 − t* δα error of the explicit form of D2(α): D2(α) - P2(α) t / t* α D3 R3 D1 R1 t / t* α B. α D3(α) C. A. ∞ 6 1 t α= ⋅ ∑ 2 ⋅ exp − 6k 2 ⋅ π 2 k =1 k t * Jander: t t* [ = 1 − (1 − α)1 / 3 ] 2 = 0.0426 ⋅ t t 50 Ginstling: t t = 3 − 2 ⋅ α − 3 ⋅ (1 − α) 2 / 3 = 0.1101 ⋅ t* t 50 Carter: t t* [ ] = 2.424 ⋅ 2 − (1 + α ) 2 / 3 − (1 − α ) 2 / 3 = 0.1446 ⋅ t t 50 t / t50 α α D3(α) B. ∞ 6 1 t α= ⋅ ∑ 2 ⋅ exp − 6k 2 ⋅ π 2 k =1 k t * A. Jander: C. t t* [ = 1 − (1 − α)1 / 3 ] 2 = 0.0426 ⋅ t t 50 Ginstling: t t = 3 − 2 ⋅ α − 3 ⋅ (1 − α) 2 / 3 = 0.1101 ⋅ t* t 50 Carter: t t* [ ] = 2.424 ⋅ 2 − (1 + α ) 2 / 3 − (1 − α ) 2 / 3 = 0.1446 ⋅ t t 50 t / t50 geometry : V * (t) = fgeox(t)y(t)z(t); x(t) = vxt or 4Dt , y(t), z(t) analogously x(t) = y(t) = d0 , 4π z(t) = vt or 4Dt fgeo = fgeo = 1 x 3 y z fgeo = π x(t) = d0, x(t) = y(t) = z(t) y(t) = z(t) = vt or 4Dt = vt or 4Dt 4π 3 3 V * (t) = v t or = πd vt or 2 0 = d0 v t or 2 2 3 3 3 1 1 = 4d Dt 2 32π 2 2 = 2π d D t 2 2 2 0 = D t 0 3 x y z 3 1 2 n = 3; n = 1; n = 2; 2 2 2 n 1 2 3 V * (t) = Bt , n = 1, 2, 3; , , 2 2 2 nucleusde nsity : m 1 C NV = Ct , m = 0, , 1 ⇒ dNV = 0, dt, Cdt 2 2 t massbalan ce : dVcryst = V * (t)(V0 − Vcryst)dNV ⇒ dVcryst dα ∫ V0 − Vcryst = ∫ 1- α = ∫ V * (t)dNV = n Bt NV for m = 0 n+m = ⇒ −ln(1 − α) = kt ∫ BCtn+m-1dt for m > 0 ( ) α = 1 − exp − kt X , X = n + m, k = k(T) α α Am,n(α) t m +n α = 1 − exp − τ crystal crystal growth nucleation kinetics geometry kinetics const.: diff.: linear: m=0 m=0.5 m=1 N=2 linear: n=1.0 1.0 1.5 2.0 diff.: n=0.5 0.5 1.0 1.5 N=1 linear: n=2.0 2.0 2.5 3.0 diff.: n=1.0 1.0 1.5 2.0 N=3 linear: n=3.0 3.0 3.5 4.0 diff.: n=1.5 1.5 2.0 2.5 t/τ add DH as nuclei → m = 0; gypsum add methyl cellulosis → diffusion kinetics; use β-HH or α-HH as main component in the product N = 2 ⇒ n + m = 0.5 educt educt set product 10 m 10 m 10 m β hemihydrate (β-HH) α hemihydrate (α-HH) set dihydrate (DH) β-CaSO4½H2O α -CaSO4½H2O CaSO42H2O α α Am,n(α) n + m = 0.5 n+m=4 t/τ α volume reactions of nth order, Fn(α): 1 t 1−n 1 − 1 − for n < 1 t * 1 − exp(−t / τ ) for n = 1 Fn (α ) : α = 1 1− 1 for n > 1 t n −1 1 + (2 n−1 − 1) ⋅ t50 diffusion control: napparent = ½·(nintrinsic + 1) α n=½ α Fn(α) 1 2 2 t F½ (α ) = 1 − 1 − , t50 = t * ⋅ (1 − ½ ) t * F1 (α ) = 1 − exp(−t / τ ), t50 = τ ⋅ ln 2 1 F2 (α ) = 1 − t 1− t50 t / t50 special mathematical functions special mathematical functions dy dx 1 ln x definition: x⋅ = =1 dx dy x 1 y = ∫ ⋅ dx = ln x + const. x main values and limits: x>0 x = 1 : ln x = 0; x → 0 : ln x → −∞; x → ∞ : ln x → ∞; ∂ 1 derivatives and integrals: ∂x ln x = ; x ∫ ln x ⋅ dx = x ⋅ ln x − x + const. algorithm („button“): commonly available special mathematical functions sin x, cos x definition: ∂2 y 2 ∂ y 2 λ ω ⋅ 2 = k ⋅ 2 ; u = (2π ⋅ν ) ⋅ 2 =ω /k ∂t ∂x 2π y = A ⋅ sin(ω ⋅ t ± k ⋅ x) + B ⋅ cos(ω ⋅ t ± k ⋅ x) π π sin 0 = 0; sin = 1; sin( z + π ) = − sin z; cos z = sin( z + ) main values and limits: 2 2 sin( − z ) = − sin z; cos(− z ) = cos z; sin 2 z + cos 2 z = 1 ∂ ∂ derivatives and integrals: sin z = cos z; cos z = − sin z ∂z ∂z algorithm („button“): commonly available special mathematical functions erf x, erfc x definition: ∂u ∂ 2u x ∂u dy = a⋅ 2 ; z = and y ≡ ⇒ = −2 z ⋅ dz ∂t ∂x 4⋅ a ⋅t ∂z y ln y = − z 2 + const. ⇒ u = ∫ exp(− z 2 ) ⋅ dz + C1 ⋅ z + C2 ∞ x ∞ 2 2 ∫ exp(− z 2 ) ⋅ dz = π ⇒ π ⋅ ∫ exp(− z 2 ) ⋅ dz + π ⋅ ∫ exp(− z 2 ) ⋅ dz = 1 −∞ 0 x erf ( x ) erfc ( x) main values and limits: erf ( x) + erfc( x) = 1; erf (− x) = −erf ( x); erfc(− x) = 2 − erfc( x); erf (0) = 0; x → ∞ ⇒ erf ( x) → 1 ∂ ∂ 2 derivatives and integrals: erf ( x) = − erfc( x) = ⋅ exp(− x2 ) ∂x ∂x π ∞ 1 ∫ erfc( x) ⋅ dx = ierfc( x) = π ⋅ exp(− x 2 ) − x ⋅ erfc( x) x 4 ⋅ x2 algorithm („button“): erf ( x) ≈ 1 − exp − π erfc(x) erf(x) x = 0.304588 x ½·erfc(x) 1- ½·erfc(x) = ½ + ½·erf(x) x repeated integrals of erfc(x): ∞ 1 ierfc( x) = ∫ erfc( x)dx = ⋅ exp(− x 2 ) − x ⋅ erfc( x) x π ∞ i erfc( x) = ∫ ierfc( x)dx = 2 1 (erfc( x) − 2 x ⋅ ierfc( x) ) = 1 (1 + 2 x 2 ) ⋅ erfc( x) − 4 2 ⋅ x ⋅ exp(− x 2 ) x 4 π i n erfc( x) = 2n ( 1 n−2 ⋅ i erfc( x) − 2 x ⋅ i n −1erfc( x) ) ∞ 2 ∫ x ⋅ erfc( x)dx = 1 erf ( x) + 2 x 2 ⋅ erfc( x) − 4 π ⋅ x ⋅ exp(− x 2 ) x repeated derivatives of erf(x): ∂ 2 erf ( x) = ⋅ exp(− x 2 ) ∂x π ∂2 4 erf ( x) = − ⋅ x ⋅ exp(− x 2 ) ∂x 2 π i²erfc(x) ierfc(x) erfc(x) x ∂erf(x) / ∂x erf(x) ∂²erf(x) / ∂x² x special mathematical functions Γ(x) Let us consider the faculty function n! : f (n) = n! with n ∈ . It can be defined by a recoursive definition, namely, by: f (n) = n ⋅ f (n − 1) . This allows one to extend the faculty to real numbers: f ( x) = x ⋅ f ( x − 1) ⇒ f ( x) =" x!" with x ∈ . A look at the solution of the standard integral, solved by partial itegration, ∫ t n e −t dt = −t n e −t + n ⋅ ∫ t n −1e −t dt = −t n e −t − n ⋅ t n −1e −t + n ⋅ (n − 1) ⋅ ∫ t n − 2 e −t dt = ... Note that all these terms leads to the following possibility to represent n! : ∞ become zero at the bound- f (n) = n!= ∫ t n ⋅ exp(−t ) ⋅ dt . aries t → 0 and t → ∞. 0 EULER chose the following definition for his Gamma function for x : ∞ f ( x − 1) = ∫ t x −1 ⋅ exp( −t ) ⋅ dt ≡ Γ( x) , thus Γ(n + 1) = n! 0 special mathematical functions ∞ Γ(x) definition: Γ( x, z ) = ∫ t x −1 ⋅ exp(−t ) ⋅ dt (incomplete) z ∞ Γ( x) = ∫ t x −1 ⋅ exp(−t ) ⋅ dt (complete) 0 π x → 0 ⇒ Γ( x) → ; Γ( 1 ) = π ; Γ(1) = Γ(2 ) = 1; Γ( 3 ) = 1 main values and limits: x 2 2 2 x ⋅ Γ( x) = Γ( x + 1) ⇒ Γ( N + 1) = N ! algorithm („button“): 1 1 1 Γ( x) ≈ 2π ⋅ exp ( x − 1 ) ⋅ ln x − x + − + 2 12 ⋅ x 360 ⋅ x 3 1260 ⋅ x 5 Γ( x + 5) Γ( x) = x ⋅ ( x + 1) ⋅ ( x + 2) ⋅ ( x + 3) ⋅ ( x + 4) Γ(x) x 1/Γ(x) is an oscillating function; note that: π = sin(π ⋅ x) Γ( x) ⋅ Γ(1 − x) 1/ Γ(x) x 1 lim Γ( x) = x →0 x Γ(x) 1/x Γ(x) x Γ(x) main values of Γ(x) 3! Γ(1/2) = √π Γ(3/2) = ½ ·√π Γ(1) = 1 Γ(2) = 1 Γ( N + 1) = N ! 2! 0! 1! x particle size distributions http://en.wikipedia.org/wiki/Log-normal_distribution http://www.owlnet.rice.edu/~msci301/Log-normal%20graph%20paper.jpg general distribution range of r: r (rmin, rmax); rmin may → 0 distribution parameters: rmax may → ∞ fineness parameters: e.g. rnn = radius at which Q(r) = nn % ; width of distribution: e.g ±δr around rnn comprising 90 % differential distribution: q(r) = a function of particle size r and the distribution parameters r cumulative distribution: Q(r) = ∫ q(r ) ⋅ dr lower limit of r general convolution integral r for ƒ = β·rx : F (r ) = ∫ β ⋅ r x ⋅ q(r ) ⋅ dr lower limit of r average of ƒ: f = ∫ β ⋅ r x ⋅ q(r ) ⋅ dr entire distibution Note: Any property of an individual particle may be presented in terms of a series of expressions ƒ = β·rx. Example: β = 4π, x = 2 ⇒ 4π·r2 the linear distribution - a simple example: 1.0 r − rmin Q(r) = rmax − rmin abundance Q(r) ∂Q (r ) 1 0.5 q(r ) = = ∂r rmax − rmin rmax + rmin r50 = 0.0 2 rmin r50 rmax particle radius r rmax rmax 1 r2 1 rmax − rmin 2 2 r = ∫ r ⋅ q(r ) ⋅ dr = ⋅ rmax − rmin 2 rmin = ⋅ 2 rmax − rmin rmin 1 (rmax + rmin ) ⋅ (rmax − rmin ) rmax + rmin = ⋅ = = r50 2 rmax − rmin 2 the linear distribution - a simple example: 1.0 r − rmin Q(r) = rmax − rmin abundance Q(r) ∂Q (r ) 1 0.5 q(r ) = = ∂r rmax − rmin rmax + rmin r50 = 0.0 2 rmin r50 rmax particle radius r x+ r x +1 − rmin1 r β F (r ) = ∫ β ⋅ r x ⋅ q(r ) ⋅ dr = ⋅ x + 1 rmax − rmin rmin x +1 x+ β rmax − rmin1 ⋅ , x ≠ −1 rmax x + 1 rmax − rmin ƒ(r ) = β ⋅ r x ⇒ ƒ = ∫ β ⋅ r x ⋅ q (r ) ⋅ dr = rmin ln(rmax / rmin ) β⋅ , x = −1 rmax − rmin the linear distribution - a simple example: 1.0 ƒ(r ) = β ⋅ r x ⇒ abundance Q(r) x +1 x+ 0.5 β rmax − rmin1 ⋅ , x ≠ −1 x + 1 rmax − rmin ƒ = ln(rmax / rmin ) 0.0 β⋅ , x = −1 rmin r50 rmax rmax − rmin particle radius r 4π 4π A = 4π ⋅ r , β = 4π , x = 2; m = 2 ⋅ ρ ⋅r , β = 3 ⋅ ρ , x = 3; ⇒ 3 3 A 4π rmax − rmin π 3 3 rmax − rmin 4 rmax − rmin 4 4 3 3 Sm = = 3 ⋅ r − r 3 ρ ⋅ r − r = ρ ⋅ r4 − r4 m max min max min max min A 3 3 3 ln(rmax / rmin ) Sn = = , β = , x = −1 ⇒ S n = ⋅ m ρ ⋅r ρ ρ rmax − rmin the linear distribution - a simple example: 1.0 abundance Q(r) Let rmax = 500 µm, 0.5 rmin = 100 µm, ρ = 2.65 g / cm3 0.0 rmin r50 rmax particle radius r 4 rmax − rmin 3 3 cm 2 Sm = ⋅ 4 = 30.0 ρ rmax − rmin 4 g 3 ln(rmax / rmin ) cm 2 Sn = ⋅ = 45.6 ρ rmax − rmin g log-normal 0<r <∞ distribution parameters: r84 r s = 2 ⋅ ln = 2 ⋅ ln 50 = 2 ⋅ σ ; r50 r50 r16 2 1 1 r differential distribution: q(r ) = ⋅ ⋅ exp− 1 ⋅ ln s s⋅ π r r50 r cumulative distribution: Q(r) = 1 − 1 erfc 1 ⋅ ln s 2 r50 general convolution integral β x2 ⋅ s2 r x⋅s F (r ) = ⋅ r50 x ⋅ exp ⋅ erfc 1 ⋅ ln s − for ƒ = β·rx : 2 4 r50 2 x2 ⋅ s2 average of ƒ: f =β ⋅ r50 x ⋅ exp 4 N 1 N mass-related specific surface area: S mass = ⋅ exp − ⋅ s 2 4 2 ρ ⋅ r50 log-normal distribution s = 0.5 1 1.5 2 4 abundance Q 1 r Q(r) = 1 − erfc s ⋅ ln 1 2 r50 r / r50 log-normal distribution s = 0.5 1 1.5 2 4 abundance Q 1 r Q(r) = 1 − erfc s ⋅ ln 1 2 r50 log r / r50 ( ) ( exp σ 2 / 2 = exp s 2 / 4 ) log-normal grid σ = s/ 2 parallel through pole abundance particle size as radius or diameter pole ( ) ( exp σ 2 / 2 = exp s 2 / 4 ) log-normal grid σ = s/ 2 0. 2 the ordinate: 5. 1 Q = 1 − ½ ⋅ erfc(x) = ½ + ½ ⋅ erf(x) ⇒ 0. 1 4 ⋅ x2 y = 2 ⋅ Q − 1 = erf(x) ≈ 1 − exp − ⇒ π 5. 0 abundance Q π 0. 0 x = inverf(y) ≈ ⋅ − ln(1 − y 2 ) 2 procedure: 5. 0 - • draw an ordinate with a linear scale x; • select a Q value (e.g.: Q = 0.98); 0. 1 - • calculate 2·Q - 1 (= 0.96 in the example); • calculate x (= 1.414 in the example), 5. 1 - for Q < 0.5, take -x instead of x; • mark position on the x scale (blue line); 0. 2 - • transfer Q = 0.98 on the Q (scale red line) particle size as radius or diameter pole ( ) ( exp σ 2 / 2 = exp s 2 / 4 ) log-normal grid σ = s/ 2 parallel through pole abundance particle size as radius or diameter pole The distribution Qi(r) can be related to number (i=0), length (i=1), area (i=2), volume or mass (i=3). The median r50,i is converted to r50,j, j ≠ i, by j −i 2 r50, j = r50,i ⋅ exp ⋅s 2 Example: Conversion from fibre number (i=0) to fibre mass (j=2) yields r50, 2 2−0 2 = r50,0 ⋅ exp ( ) ⋅ s = r50,0 ⋅ exp s 2 2 • Sieve analysis yields r50,3. • Counting particle from a micrograph yields r50,0. • Evaluating images by an image analyzer yields r50,2. ( ) ( exp σ 2 / 2 = exp s 2 / 4 ) log-normal grid σ = s/ 2 glass sand 90 % < 500 µm parallel through pole abundance s = 0.63·√2 10 % < 100 µm = 0.89 r50 = 220 µm Smass = 51 cm²/g particle size as radius or diameter pole 3 2 3 v ⋅t v ⋅t v ⋅t v ⋅t local R3(α) kinetics: α ( r , t ) = 1 − 1 − = 3⋅ − 3⋅ + r r r r v⋅t ∞ convolution integral A(t): A(t ) = ∫ q (r ) ⋅ dr + ∫ α (r , t ) ⋅ q (r ) ⋅ dr 0 v⋅t 1 1 v ⋅t solution: A(t ) = 1 − ⋅ erfc ⋅ ln s 2 r50 3 v ⋅t s2 1 v ⋅t s + ⋅ ⋅ exp ⋅ erfc ⋅ ln s + 2 r 4 r50 2 2 1 3 v ⋅t − ⋅ 2 r 2 ( ) ⋅ exp s ⋅ erfc ⋅ ln s v ⋅t r50 + s 1 v ⋅t 3 9 ⋅ s2 1 v ⋅t 3⋅ s + ⋅ ⋅ exp ⋅ erfc ⋅ ln s + 2 r 4 r50 2 4⋅ z2 approximation for erfc(z): erfc( z ) ≈ 1 − sign( z ) ⋅ 1 − exp − π local kinetics: α ( y ) = 1 − (1 − y )3 = 3 ⋅ y − 3 ⋅ y 2 + y 3 v ⋅t 4⋅ D ⋅t R3(α) or D3(α) : y= or = r r solution: A( y ) = 1 − 1 ⋅ erfc(1 ⋅ ln y ) 2 s + 3y 2 ⋅ exp ( )⋅ erfc( ⋅ ln y + ) s2 4 1 s s 2 − 3 y2 2 ⋅ exp(s )⋅ erfc( ⋅ ln y + s ) 2 1 s + y3 2 ⋅ exp( )⋅ erfc( ⋅ ln y + ) 9s2 4 1 s 3s 2 • Set criterion for dissolution progress, e.g., A(y) = 0.995; vary y in the formula until it matches this value; from known v or D value, calculate the time demand t0.995. • Alternatively, observe the time demand to reach A(t) = 0.995; vary v or D until the corresponding y value yields this value. r50 = 220 µm, s = 0.89 r50 = 220 µm, s = 1.40 reaction progress A(t) r50 = 350 µm, s = 0.89 v = 12 nm/s = 43.2 µm/h t in hours r50 = 220 µm, s = 0.89 r50 = 220 µm, s = 1.40 reaction progress A(t) r50 = 350 µm, s = 0.89 v = 12 nm/s = 43.2 µm/h tcrit = 8 h t in hours power rmin < r < rmax distribution parameters: m m ⋅ r m −1 differential distribution: q(r ) = rmax − rmin r m − rmin m cumulative distribution: Q(r ) = m rmax − ln(r / rmax ) β ⋅m⋅ for x = − m rmax − rmin general convolution integral F (r ) = x+ x+ for ƒ = β·rx : β ⋅ m rmaxm − rminm ⋅ othertwise x+m m rmax − rmin m − ln(r / rmax ) β ⋅m⋅ for x = −m rmax − rmin average of ƒ: f = x+ x+ β ⋅ m rmaxm − rminm ⋅ othertwise x+m m rmax − rmin m power distribution m = 0.5 1 1.5 2 abundance Q m r r Q(r) = max r / rmax power distribution, final rmin m = 0.5 1 2 abundance Q 1.5 r m − rmin m Q(r) = m rmax − rmin m r / rmax log distribution, final rmin limit of power distribution for m → 0 abundance Q ln r − ln rmin Q(r) = ln rmax − ln rmin r / rmax power and log distribution, final rmin m→ 0 1 2 abundance Q = 0.1 0.5 1.5 r m − rmin m Q(r) = m rmax − rmin m r / rmax exponent m for example no. 1: example no. 2 no.1 parallel through pole abundance power grid pole particle size as radius or diameter RRBS 0<r <∞ distribution parameters: ro = r63 m m ⋅ r m−1 r differential distribution: q(r ) = ⋅ exp− r rom o m r cumulative distribution: Q(r) = 1 − exp− r o general convolution integral x r m F (r ) = β ⋅ rox ⋅ Γ + 1, r for ƒ = β·rx : m o x average of ƒ: f = β ⋅ rox ⋅ Γ + 1 m N −1 Γ + 1 ⋅ N m mass-related specific surface area: S mass = ρ ⋅ ro N Γ + 1 m RRBS distribution 1 - 1/e = 0.6321 abundance Q s = 0.5 m r 1 1.5 2 4 Q(r) = 1 − exp− r o r / ro RRBS distribution 1 - 1/e = 0.6321 abundance Q m r Q(r) = 1 − exp− r o s = 0.5 1 1.5 2 4 log r / ro 1 Γ 1 − RRBS grid m m abundance pole particle size as radius or diameter layer formation at plane surfaces in open and closed systems incongruent dissolution, open system: mass loss per surface area q q = a·t + b· t initial v q∝t surface D exchange zone bulk material q∝ t tim e t incongruent dissolution, open system: q = mass loss per surface q D = L = v ⋅ t + ⋅ erf ( z ) + D ⋅ t ⋅ ierfc( z ) area in g/(cm²) ρ v ρ = density in g/cm³ D D D ⋅t = v ⋅ t + − v ⋅ t + ⋅ erfz ( z ) + ⋅ exp(− z 2 ) L = total dissolution depth v v π in cm L = v ⋅ t + LD LD= depth of depleted layer in cm D ⋅t → ∝ t for z → 0 D = diffusion coefficienct π in cm²/s D v = dissolution velocity → v ⋅t + ∝ t1 for z → ∞ in cm/s v D D D ⋅t LD = − v ⋅ t + ⋅ erfc( z ) + ⋅ exp(− z 2 ) v v π D → for z → ∞ v v ⋅t v z= = ⋅ t 4⋅ D ⋅t 2⋅ D layer growing at outer surface, open system, intrinsic saturation constraint: ( ) t e mi t q = a· 1 + b ⋅ t − 1 q a e r a e c a f r u s r e p s s ol s s a m initial v q∝ t surface bulk material q∝t c0 cin → cs growing layer layer growing at outer surface, open system: cin − c0 r = turnover rate in g/(cm²/s) r = D⋅ 1rst Fick law for r L q L = layer thickness in cm L =ε⋅ q = mass loss per surface ρ area in g/(cm²) r = ∂q / ∂t ρ = density in g/cm³ c r ε = mass share of dissolved matter r = r0 ⋅ 1 − in ⇒ cin = cs ⋅ 1 − c r cs = saturation concentration s 0 in g/cm³ D = diffusion coefficienct in cm²/s ∂q r0 ε ⋅ r0 = ; β= ⇒ dt 1 + β ⋅ q D ⋅ cs ⋅ ρ intrinsic saturation constraint for q= 1 β ( ) ⋅ 1 + 2 ⋅ β ⋅ r0 ⋅ t − 1 the turnover rate: r → 0 if cin → cs → r0 ⋅ t ∝ t1 for short t 2 ⋅ r0 ⋅ t → ∝ t for long t β layer growing at outer surface, open system, no intrinsic saturation constraint: cin − c0 r = turnover rate in g/(cm²/s) r = D⋅ 1rst Fick law for r L q L = layer thickness in cm L =ε⋅ ρ q = mass loss per surface area in g/(cm²) r = ∂q / ∂t ρ = density in g/cm³ ε = mass share of dissolved matter ∂q D ⋅ (cin − c0 ) ⋅ ρ D = diffusion coefficienct = in cm²/s dt ε ⋅q ⇒ D ⋅ (cin − c0 ) ⋅ ρ q ⋅ dq = ⋅ dt ε q = 2⋅ D ⋅t ⋅ (cin − c0 ) ⋅ ρ / ε ∝ t adsorbed layer, closed system, no saturation constraint: ( ) t e mi t q = a· 1 + b ⋅ t − 1 q a e r a e c a f r u s r e p s s ol s s a m initial v q∝ t surface bulk material q∝t c = qs volume con- adsorbed monomolecular layer (resorbed from dissolved matter) centration of dissolved matter adsorbed layer, closed system, no saturation constraint: θ = surface coverage by adsorption, k ⋅c condition for θ= closed system Langmuir law; k = adsorption constant 1+ k ⋅ c s = surface area to solution volume c = q⋅s ratio in cm-1 r = ∂q / ∂t q = mass loss per surface r0 area in g/(cm²) r = r0 ⋅ (1 − θ ) = 1+ k ⋅ c ∂q r0 ~ = ~ ; β = k ⋅s ⇒ dt 1 + β ⋅ q concentration-dependent constraint 1 ( ~ q = ~ ⋅ 1 + 2 ⋅ β ⋅ r0 ⋅ t − 1 β ) for the turnover rate: r → 0 if cin → cs → r0 ⋅ t ∝ t1 for short t 2 ⋅ r0 ⋅ t → ~ ∝ t for long t β no layer growth, no resorption, closed system ⇒ overall saturation constraint: condition for r = turnover rate in g/(cm²/s) r = ∂q / ∂t closed system q = mass loss per surface c = q⋅s area in g/(cm²) s = surface area to solution volume c q⋅s r = r0 ⋅ 1 − = r0 ⋅ 1 − c ratio in cm-1 s cs cs = saturation concentration in g/cm³ dq = r0 ⋅ dt s 1− ⋅ q cs overall saturation constraint cs s − ⋅ ln1 − ⋅ q = r0 ⋅ t c s s c q= s 1 − exp − r0 ⋅ s ⋅ t ⋅ s cs layer growing at outer surface, closed system: cin − c r = turnover rate in g/(cm²/s) r = D⋅ condition for 1rst Fick law for r L closed system q L = layer thickness in cm L = ε ⋅ ; q = c⋅s ρ q = mass loss per surface area in g/(cm²) r = ∂q / ∂t s = surface area to solution volume cin r ratio in cm-1 1 − ⇒ cin = cs ⋅ 1 − r = r0 ⋅ r ρ = density in g/cm³ cs 0 ε = mass share of dissolved matter cs = saturation concentration 1 − ( s / cs ) ⋅ q ε ⋅ r0 ∂q in g/cm³ = r0 ⋅ ; β= ⇒ D = diffusion coefficienct dt 1+ β ⋅ q D ⋅ cs ⋅ ρ in cm²/s 1+ β ⋅ q ⋅ dq = r0 ⋅ dt 1 − ( s / cs ) ⋅ q intrinsic saturation constraint for the turnover rate: r → 0 if cin → cs r0 ⋅ s ⋅ t β ⋅ cs q ⋅ s − = 1 + ⋅ ln1 − + β ⋅q cs s cs layer formation on particles ζ = 1 - [(A(t)/A0 ]1/2 = ∆r(t)/r0 A0 A(t) 10 mm observation of a dissolving particle with N = 3: r (t ) r0 − ∆r (t ) ∆r (t ) A(t ) ∆r (t ) = = 1− = ; define observable ζ = r0 r0 r0 A0 r0 A0 and A(t) is the shadow image (projection) of the observed particle at t = 0 and at t > 0, respectively. linear dissolution: v ⋅t r t ζ = ; t (ζ → 1) = t* = 0 ⇒ =ζ r0 v t* diffusion controlled dissolution: 4D ⋅ t r02 t t ζ = ; t (ζ → 1) = t* = ⇒ =ζ 2 ⇔ζ = r0 4D t* t* kinetics with growing layer at outer surface: ∆r (t ) B = ζ = A ⋅ 1 + ⋅ t − 1 r0 A limit for t → 0 yields: limit for t → ∞ yields: abbreviation: B 2v 4D 2D 2D ζ → ⋅ t with B = ζ → AB ⋅ t with AB = , A= =w 2 r0 r02 v ⋅ r0 v ⋅ r0 this yields the rate equations: 2D 2 2 1 + v ⋅ t − 1 and t = r0 ⋅ ζ 2 + r0 ⋅ ζ r02 r0 ζ = ⋅ for ζ → 1, t → t* ⇒ t* = + v ⋅ r0 D 4D v 4D v reading in the form t/t*: 4D t 1 v ⋅ r0 2D 1 + (4 D / v ⋅ r0 ) t = ⋅ζ 2 + ⋅ ζ and ζ = ⋅ 1+ ⋅ − 1 t * 1 + 4D 4D v ⋅ r0 (2 D / v ⋅ r0 )2 t * 1+ v ⋅ r0 v ⋅ r0 t for w → 0, →ζ 2 t 1 2⋅w 1+ 2 ⋅ w t t* = ⋅ζ 2 + ⋅ ζ and ζ = w ⋅ 1 + ⋅ − 1 t * 1+ 2 ⋅ w 1+ 2 ⋅ w w2 t* t for w → ∞, →ζ t* kinetics with layer formed by incongruent dissolution: ∆r (t ) = ζ = A ⋅ t + B ⋅ t = A ⋅ y + B ⋅ y 2 with y = t r0 limit for t → 0 yields: limit for t → ∞ yields: abbreviation: 4D v 2D ζ → A ⋅ t with A = ζ → B ⋅ t with B = =w r02 r0 v ⋅ r0 resolving the rate equation for y yields: 2 A 4B A2 4B y= ⋅ 1 + 2 ⋅ ζ − 1 = t and y = 2 1+ ⋅ ⋅ ζ − 1 = t 2B A 4B2 A 2 for ζ → 1, t → t*; thus we obtain the for further calculations, following implicit rate equation: we keep in mind: 2 2 2 A2 4B 4B 1 + 2 ⋅ζ − 1 1 + 2 ⋅ζ − 1 t* = 1+ ⋅ − 1 4B 2 A 2 t = A = w t* 4B 2 A 4B 1+ 2 −1 1 + −1 t* = 1+ ⋅ − 1 A w 2B A 2 let us now investigate the limiting time law of the rate equation: 2 2 1 + ⋅ζ − 1 t w = t* 2 1 + −1 w 2 2 ⋅ζ → w =ζ ⇔ζ = t 2D t for w = → 0 we obtain v ⋅ r0 t* 2 t* w 2 2 1 2 1 2D t 1 + ⋅ ⋅ζ − 1 ⋅ζ t for w = → 0 we obtain → 2 w = w =ζ 2 ⇔ζ = v ⋅ r0 t * 1 + 1 ⋅ 2 −1 1 t* 2 w w thus, if D/r0 << v, an R type kinetics is found, while D/r0 >> v yields a D type kinetics. let us now derive the explicit rate law as a function of t/t*; for this purpose, we use the expressions for t* and √t* derived previously: 2 A2 4B A 4B t* = ⋅ 1 + 2 − 1 and t* = ⋅ 1 + 2 − 1 4B 2 A 2B A this is inserted in the rate equation 2 t t A2 4B t A2 4B t ζ = A⋅ t + B ⋅t = A⋅ t * ⋅ + B ⋅ t *⋅ = 1+ ⋅ ⋅ − 1 + 1+ ⋅ − 1 ⋅ t* t * 2B 2 2 t* A t * 4B A 2 t 2 2 t w 2 t w 2 t 2 1 + − 1 ⋅ ζ = w⋅ + ⋅ 1 + − 1 ⋅ = ⋅ 2 ⋅ 1 + − 1 ⋅ + 1 + − 1 ⋅ w t* 2 w t* 2 w t* w t * 2D t for w = → 0 we obtain ζ → v ⋅ r0 t* 2D t for w = → ∞ we obtain ζ → v ⋅ r0 t* like before Summary: ∆r (t ) 2⋅ D abbreviations: ζ = ; w= r0 ν ⋅ r0 kinetics with layer growing t 1 2⋅w = ⋅ζ 2 + ⋅ζ at the outer surface: t * 1+ 2 ⋅ w 1+ 2 ⋅ w →ζ 2 for w → 0; → ζ for w → ∞ 1+ 2 ⋅ w t ζ = w⋅ 1+ ⋅ − 1 w2 t* → t / t * for w → 0; → t / t * for w → ∞ 2 2 kinetics with layer formed t 2 2 = 1 + ⋅ ζ − 1 ÷ 1 + − 1 by incongruent dissolution: t* w w →ζ for w → 0; → ζ 2 for w → ∞ 2 2 t w 2 t 1 + − 1 ⋅ ζ = w⋅ + ⋅ 1 + − 1 ⋅ Note: w t* 2 w t* The term ζ takes the position of t/t* terms in the explicit forms → t /t* for w → 0; → t / t * for w → ∞ α = α(t/t*) of the R(α) kinetics. ζ ζ = ∆r(t)/r0 D1(α): ζ = √(t/t*) leaching & dissolution; w = 2D/(v·r0) = 1; → D1 for w >> 1 → R1 for w << 1 growing reaction layer; w = 2D/(v·r0) = 1; → R1 for w >> 1 → D1 for w << 1 R1(α): ζ = t/t* t / t* application to N = 2: α = 1 - ( 1 - ζ ) 2, dissolving silica fibre with a growing layer of reaction products 1 m reformulation for N = 3: α = 1 - ( 1 - ζ ) 3 1.0 1. congruent R3(α) 2 0.8 3 4 2. incongruent turnover α 0.6 D3(α) d lle ro 1 nt co 0.4 w 3. leaching flo 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 4. reaction normalized time t/t* layer two-phase diffusion 1 phase a phase b diffusion coefficients: Da Db Ja = Jb Da ca (t = 0) cb (t = 0) w= =1 =0 Db c0 c0 c / c0 . . surface equilibrium: ca(x=0) cb(x=0) cb ( x = 0) K= ca ( x = 0) ∂ca ∂ca ∂ 2 ca ∂cb ∂cb ∂ 2 cb J a = − Da ⋅ ; = Da ⋅ 2 J b = − Db ⋅ ; = Db ⋅ 2 ∂x ∂t ∂x ∂x ∂t ∂x 0 0 x / x0 1 phase a phase b diffusion coefficients: Da Db Ja = Jb Da ca (t = 0) cb (t = 0) w= =1 =0 Db c0 c0 c / c0 . . surface equilibrium: ca(x=0) cb(x=0) cb ( x = 0) K= ca ( x = 0) ca K −x cb K ⋅w x = 1− ⋅ erfc = ⋅ erfc c0 K +w 4⋅ D ⋅t c0 K + w 4⋅ D ⋅t a b 0 0 x / x0 verification of the surface equilibrium boundary condition: ca K −x cb K ⋅w x = 1− ⋅ erfc = ⋅ erfc c0 K +w 4⋅ D ⋅t c0 K + w 4⋅ D ⋅t a b ca ( x = 0) K w cb ( x = 0) K ⋅ w = 1− = = c0 K +w K +w c0 K +w K ⋅w cb ( x = 0) K + w = =K q.e.d . ca ( x = 0) w K +w verification of the flow boundary condition: 1 ∂c / c 1 ∂c / c ⋅ J a = − Da ⋅ a 0 = ⋅ J b = − Db ⋅ b 0 = c0 ∂x c0 ∂x ∂ K − x ∂ K ⋅w −x − Da ⋅ 1 − ⋅ erfc = − Db ⋅ ⋅ erfc 4⋅ D ⋅t ∂x K + w 4 ⋅ D ⋅ t ∂x K + w a b K 2 ⋅ Da x2 K ⋅w 2 ⋅ Db x2 ⋅ ⋅ exp − ⋅ ⋅ exp − 4⋅ D ⋅t K + w 4π ⋅ Da ⋅ t 4⋅ D ⋅t K + w 4π ⋅ Db ⋅ t a b Ja = Jb must be fulfilled everywhere, especially at x = 0: K 2 ⋅ Da K ⋅w 2 ⋅ Db ⋅ = ⋅ K + w 4π ⋅ Da ⋅ t K + w 4π ⋅ Db ⋅ t Da = w q.e.d . Db phase a phase b c / c0 K = 1, w = 1 cb 1 x = ⋅ erfc c0 2 4⋅ D ⋅t b Da t in cm2 = 1.25 ca 1 −x 0.002 0.05 0.20 = 1 − ⋅ erfc c0 2 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 1, w = 0.25 Da = 0.25 ⇒ Db = 16 ⋅ Da Db cb w x = ⋅ erfc c0 1 + w 4⋅ D ⋅t b ca 1 −x = 1− ⋅ erfc c0 1+ w 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 1, w = 4 Da 1 = 4 ⇒ Db = ⋅ Da Db 16 cb w x = ⋅ erfc c0 1 + w 4⋅ D ⋅t b ca 1 −x = 1− ⋅ erfc c0 1+ w 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 0.5, w = 1 1 cb ( x = 0) = ⋅ ca ( x = 0) 2 cb K x = ⋅ erfc c0 K + 1 4⋅ D ⋅t b ca K −x = 1− ⋅ erfc c0 K +1 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 2, w = 1 cb ( x = 0) = 2 ⋅ ca ( x = 0) cb K x = ⋅ erfc c0 K + 1 4⋅ D ⋅t b ca K −x = 1− ⋅ erfc c0 K +1 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 0.5, w = 0.25 Da = 0.25 ⇒ Db = 16 ⋅ Da Db 1 cb ( x = 0) = ⋅ ca ( x = 0) 2 cb K ⋅w x = ⋅ erfc c0 K + w 4⋅ D ⋅t b ca K −x = 1− ⋅ erfc c0 K +w 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 2, w = 0.25 Da = 0.25 ⇒ Db = 16 ⋅ Da Db cb ( x = 0) = 2 ⋅ ca ( x = 0) cb K ⋅w x = ⋅ erfc c0 K + w 4⋅ D ⋅t b ca K −x = 1− ⋅ erfc c0 K +w 4⋅ D ⋅t a x / x0 phase a phase b c / c0 K = 0.5, w = 4 Da 1 = 4 ⇒ Db = ⋅ Da Db 16 1 cb ( x = 0) = ⋅ ca ( x = 0) 2 x / x0 phase a phase b c / c0 K = 2, w = 4 Da 1 = 4 ⇒ Db = ⋅ Da Db 16 cb ( x = 0) = 2 ⋅ ca ( x = 0) x / x0

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posted: | 11/26/2011 |

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