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Reaction Kinetics

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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 Pr—ƒ(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 → ½—ra:
                                                                              2—ra
                       (
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) = fgeo—x(t)—y(t)—z(t); x(t) = vx—t or 4—D—t ,
                                  y(t), z(t) analogously

                 x(t) = y(t) = d0 ,
       4π        z(t) = v—t or 4—D—t
fgeo =                                     fgeo = 1         x
        3

                                                                y
                                                        z
                     fgeo = π
                                      x(t) = d0,
x(t) = y(t) = z(t)
                                      y(t) = z(t) = v—t or 4—D—t
= v—t or 4—D—t
         4π 3 3
V * (t) = —v —t or       = π—d —v—t or
                              2
                              0                  = d0 —v —t or
                                                             2   2

          3
              3 3
                                       1   1
                                                 = 4—d —D—t
                                                         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) = B—t , n = 1, 2, 3; , ,
                                    2 2 2
nucleus€de nsity :
           m     1                C
NV = C—t , m = 0, , 1 ⇒ dNV = 0,      dt, C—dt
                 2               2— t
mass€balan ce :
dVcryst = V * (t)—(V0 − Vcryst)—dNV ⇒
   dVcryst          dα
∫ V0 − Vcryst = ∫ 1- α = ∫ V * (t)—dNV =
       n
  B—t —NV for m = 0                                       n+m
=                                    ⇒ −ln(1 − α) = k—t
  ∫ B—C—tn+m-1—dt for m > 0

               (   )
α = 1 − exp − k—t 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-CaSO4—½H2O                 α -CaSO4—½H2O               CaSO4—2H2O
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 = q—s



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 
                                      − ⋅ ln1 − ⋅ 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 +     ⋅ ln1 −
                                                                            + 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
language:English
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