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Diffusion processes by 04W5iVd5

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									                         Diffusion processes

                             Introduction
               Typical scenario of catastrophic pollution
                  Accidental release of contaminant




Transport of conatminant by underground water, streams, rivers, lake or
                     ocean currents, atmosphere




           Diliution of contaminant by water or air (diffusion)




     Mechanism of contaminant transport and dilution crucial for
         understanding and control of polution phenomena
             Management of environmental pollution



 resources                   industry                goods


    Wastes:
                                             Accidental realease of
Transformation                                   contaminants:
 Accumulation                                    Dispersion in
                                                 environment
   Transport




                    Environmental Quality


Effects on                  Effects on               Effects on
  health                     climate                 ecosystem
Possible strategies for pollution control
        Control sources of contaminants
      Prevention – the most recommended

                Waste treatment
       Less recommended choice, only if
                 unavoidable

          Dispersion in environment
           Impact of waste storage
      Accidental release of contaminants
Technical necessity e.g. heat, flue gas emission
              Related problems:
            Conservation of mass
       Point sources/distributed sources
        Continuous/accidental release
                           Basic definitions
                             Concentration:
          Mass M of contaminant in volume V gives concentration
                                       M          kg 
                            c  lim               m3                  (1)
                                   V 0
                                       V          
                                      
          in general:          c  c x , t                            (2)

Relative concentration:
                                  c
                                                          %;  ppm   (3)
                            of solution

        Averages:
                                       
      Ensemble average:             c x , t                           (4)
                                                 t 0 T
     Time average:               
                              c x , t0  
                                            1
                                            T       c dt
                                                  t0
                                                                        (5)

                                   
                              cV  x , t  
                                             1
   Space average:
                                             V    c dV
                                                  V
                                                                        (6)
                         Basic definitions



                             1           1
                             Q 
 Flux average:        cf         cU dA   q dA    (7)
                                A
                                         Q A

  Where:    Q   U dA - volumeflux over surfaceA
                  A

               kg 
            q  2   specific mass flux
              m s

Dilution:       1                                     (8)
             S
                p
                                               ion
             where: p relative volume concentrat
                volume of contaminant
             p                                       (9)
                      total volume
            Mass conservation equation for contaminant




Conservation of mass of conatminant in volume V with no sources within the
volume:

                                         
                          c dV
                       t V
                                      q  n dA  0
                                        A
                                                                   (10)
                      chaneg of mass mass flux
                      contained within
                      volume
Mass conservation equation for contaminant

   Due to Gauss-Ostrogradski transformation

                   c  
                t  xi qi  dV  0
               V               
                                                      (11)


        Since V is an arbitrary volume:

                 c                                  (12)
                       qi  0
                 t xi
              Flux of contaminant:

               qi  U i c  qDi                       (13)
               advection       diffusion

     Substitution of Eq.(13) into Eq. (12) leads to

              c 
                    U i c     qDi             (14)
              t xi             xi
      Mass conservation equation for contaminant

   If fluid is at rest:     Ui  0
                           c     
                                   qDi                                     (15)
                           t    xi
                                   U i
If fluid is incompressible:             0
                                   xi
                          c     c      
                             Ui          qDi                              (16)
                          t     xi    xi
           If flow is turbulent (Reynolds decomposition):
                                U i  U i  ui
                                C C c                                        (17)
                                U i c  U i C  ui c
         Then mass conservation equation becomes:
                          C
                           t
                               Ui
                                   C
                                   xi
                                       
                                           
                                          xi
                                                  
                                              qDi  ui c                      (18)

          Molecular diffusion                          Turbulent flux – closure problem
         Recommended strategies for turbulent transport analysis

Basic mechanisms involved into transport phenomena:
• Advection by flow
• Naturl convection
• Evaporation/condensation
• Entrainment/deposition
• Molecular/turbulent diffusion
• Disperision in shear flows
                    Separation of dominant mechanisms




                Splitting of the area of analysis into subdomians
            Recommended strategies for turbulent transport analysis
                   Experimental versus computer modeling
     Type of analysis          accuracy         Time required           Cost


   In-situ ivestigations           +                 --                   --
        (full scale)

  Experimental (model)             +-                --                   -
     investigations
   Computer modeling               +-                ++                   +

   Dimensional analysis    Order of magnitude       +++                  +++




Conclusions:
•Numerical modeling is nowadays the most recommended, however keep in mind
that: garbage             computer           garbage
•Accuracy depends on physical features assumptions but their validity is often
doubtful e.g. diffusivity in ocean varies in the range 10-9 (molecular) to 105 m2/s
(turbulent)
•Dimensional analysis often useful in rough approximations
                                   Molecular diffusion
    Molecular diffuison plays minor role in environmental transport processes but is a basis
    for understanding other types of diffusion
                   According to Fourier’s (1822) law of heat transfer

                         Fick’s law-thermal analogy
                                                                                     (19)
                                              T
                           heat flux   k
                                              x
                           where : k  thermal diffusivit y coeffient

                          Fick’s (1855) analogy for mass flux
                                       c
                            qi   D                                                 (20)
                                       xi
                            where:
                            q  mass flux per unit surfaceand unit time
                            D- diffusion constant[L2 /T]

                                 kg     m2   kg 
Note on the units:            q  2 ; D  ; c  3 
                                m s     s   m 
                                    Fick’s law
  General form of diffusion equation:


                     gradient of transported 
        flux                                
                      quantity                 
       i.e. it is gradient/flux relationsh ip for mass ( D), momentum ( )
       or heat (k ) diffusion process

D - physical constant depending on the properties of fluid and contaminant

     Example:
                                                                             
     For water:          D~10-9   [m2/s]          Schmidt number Sc 
                                                                           D
                         ν ~ 10-6 [m2/s]          for water          Sc  10 3
                         k ~ 10-5 [m2/s]          for air            Sc  1
     For air:
                         D ~ ν ~k ~10-5 [m2/s]
            A simple model for gradient/flux relationship

  Suppose a 1D pipe with concentration gradient in x-direction, two boxes
  created in a pipe i.e. left (L) and right (R) containing different numbers of
                          molecules NL=10 and NR=20

     L                R                             L               R




                                                        ΔNL
    Δx               Δx                        Δx                       Δx

                                                        ΔNR




             (A)                                              (B)
Concentration of molecules in the initial (A) and consecutive moment (B) for
                          1D diffusion process
           A simple model for gradient/flux relationship

  If each molecule has propability p to pass from one box to the another,
  then if (say) p=1/5 after time t+Δt, because

                           1              1
                     N R  20  4; N L  10  2
                           5              5

                     in a new time step
                                             
                     N L  10  2  4  12; N R  20  4  2  18

                                After each time step

                      N Ln 1)  N Ln ) ; N Rn 1)  N Rn )
                        (          (        (          (


                                       We obtain

                         N 
                            R
                             n 1       
                                                          
                                      N Ln 1  N Rn   N Ln    

  And the process continues until the uniform distribution of the molecules is
approached – the concentration difference dimnishes with time and the process
                    tends toward uniform concentration
            Analytical description of gradient/flux relationship
Each molecule has mass m

                             M L  N L m; M R  N R m
   The mass flux during time Δt in positive x-direction is

       mass
               m p N L  m pN R  p M L  M R   negative mass flux
        flux 

                           Concentration in each box:

                             ML              MR
                       cL        ;      cR
                            1 x           1 x
                     Mass flux per unit area and unit time

                          x                x 2  c x  2c    
                    qx  p cL  cR    p                  
                          t                t  x 2 x 2
                                                 
                                                                 
                                                                 
        Analytical description of gradient/flux relationship
       Mass transfer should not be dependent on the size of the box
                               x 2 
                        lim  p        const  D diffusivity 
                        x 0  t 
                        t 0       
                   And hence neglecting higher order terms
                                             c
                                  qx   D
                                             x
               We obtained a simple 1D model of Fick’s law
Conclusions:
•Boxes must be large enough, so that we can apply probability analysis

                                N L ; N R  1
•Boxes must be small enough to apply first order approximation in Taylor series
                                   2c  c
                                x 2 
                                  x    x
•Both the above conditions are always satisfied in normal conditions for
molecular diffusion because the mass of a single molecule is small an their
number is very large even in small volumes
  Similarity solutions and properties of the 1D diffusion equation

Assume 1D diffusion in fluid at rest:
                              c    2c
                                 D 2
                              t   x                                (21)

    And uniform concentration along y and z axes
                             c c
                                  0                                (22)
                             y z




                  Coordinates for 1D diffusion problem
      Similarity solutions and properties of the 1D diffusion equation

                        Total mass of the contaminant:
                                                
                               M   c dV  A  c dx
                                     V          
                                                                            (23)
                      That means c will be sought in the form

                                  M   1     M 1
                             c                                          (24)
                                  A Length A L

                                  Let us introduce:
                             L  referencelength scale
                             T - referencetime scale
                           Dimensional analysis yields:

                             c    c
                               D 2        L2  DT
                             T   L                                          (25)


Time scale and the length scale in the diffusion process are naturally related
Similarity solutions and properties of the 1D diffusion equation
                                 Example:
Consider a diffusion process after initial injection of the the contaminant.
             The time scale is time t elapsed after injection.
 After substitution into Eq.(25) the concentration can be evaluated as:

                                      M 1
                                 c                                       (26)
                                      A   Dt
    i.e. concentration may be obtained for a similarity solution

                                        f  x, t 
                                 M 1
                            c                                            (27)
                                 A   Dt
      The only possibility for non-dimensional function f is:
                                                     
                                                     
                                   x         m     
                             f  f               12
                                   Dt    m 2                             (28)
                                                s
                                          s  
                                                  

                             That finally yields:
                                                                               (29)
                                   f  ;  
                            M 1                           x
                       c     
                            A   Dt                        Dt
Analytical solution of 1D diffusion equation for initial impulse injection of
                                contaminant
                 Introducing Eq.(29) into the r.h.s of diffusion equation:

                     df                d2 f        2c M 1
                f     ;       f                             f         (30)
                     d                d 2       x 2
                                                         A Dt 32


           and introducing Eq.(29) into the l.h.s. of the diffusion equation

                       c M  1    1        1  
                                      f      f                           (31)
                       t   A  2 D t 3
                                            Dt 2t 

            Hence the diffusion equation takes the following form

                                  
                                      1
                                         f   f   f                     (32)
                                      2
                                        And finally:


                                       f   2 f                         (33)
     Note that similarity solution allowed to transform the p.d.e into the ordinary
                                  differential equation
 Analytical solution of 1D diffusion equation for initial impulse injection of
                                 contaminant

                             After integration of Eq.(33) we obtain:

                                       f  2 f   C                               (34)

                             Determination of the constant C

                                      x    odd
                       C(x,t)                                               C(x,t)

       (a)                                            (b)




         Initial injection at t=0      x                          At t >0            x

Expected evolution of the concentration. Concentration of contaminant at the
initial injection (a) and after time elapsed t (b)

                                           c  even
Analytical solution of 1D diffusion equation for initial impulse injection of
                                contaminant

c f        f  even        f   odd    f  odd   even  odd
                   And due to symmetry requirements

                                C 0                                  (35)

                             And finally:

                                                                     (36)
                            f         f
                                     2
                         That yields the solution:

                     2                                   x2 
          f  C1 exp    cx, t   C1
                                          M          1
                     4                                    4 Dt 
                                                        exp     
                                                                        (37)
                                        A          Dt          


                  Note that C1 the only unknown quantity
Analytical solution of 1D diffusion equation for initial impulse injection of
                                contaminant
        From the contaminant mass conservation equation Eq. (23)
                                    
                                         x 2  dx
                                C1  exp 
                                         4 Dt  Dt  1
                                               
                                                                          (38)

                    From tables of definite integrals we find
                               
                                                             
                                        
                                  exp  a 2 x 2 dx         a
                                                                             (39)

                   Substitution into eqaution (38) leads to:

                                                        1
                                             C1 
                                                        4                      (40)


   And finally we obtain the equation for concentration evolution due to molecular
                                       diffusion

                                                      x2 
                        c  x, t  
                                     M        1
                                                      4 Dt 
                                                  exp     
                                     A       4Dt                             (41)
Statistical measures of concentration distribution establidhed due
                      to diffusion processs
                 First order moment – expected value:
                           
                [E]        x cx dx
                           

                 x  odd , c  even,              x  c   odd ,   i.e. :
                           
                E    x cx dx  0
                           

                          Second order moment – variance:
                                                         
                                                               x2 
                                     x cxdx  4 Dt exp  4Dt dx
                      A                                1
                    2                  2
                                                                     
                      M                                             

          Having known that the following definite integral is:
                 
                                                  
                     2       2 2
                                
                   x exp  a x dx             2a   3
                                                        ; and a 
                                                                       1
                                                                   2 Dt

                      Variance can be expressed as:

     2  2Dt      Variance is a measure of the contaminant displacement from the
                   initial positiion i.e. is a measure of teh size of the cloud
                 Diffusion and random walk model
                    Definition of radnom walk process:
Step ξ of the length λ randomly to the right or to the left (drunkard’s walk

                               t                                       (42)

      Probablity density function (pdf) of the random walk is:

                            p       
                                   1
                                                                                (43)
                                   2
       ∞                                                        ∞
                                                                      Δ – Dirac delta
                                        p(ξ)                              function




                                                                                 ξ
       -λ                                                        +λ
                      Diffusion and random walk model

 Consider n successive and independent (i.e. no memory of previous step) ξi steps;
 i=1…n and the random walk variable X(t):

                           X t   1   2     n                        (44)

                Where: X(t) is the particle position after n steps

                  Problem: what is the pdf of the process X(t)?

                          Central Limit Theorem (CLT):
   The probability density function of the sum of n independent random
 variables tends towards the normal distribution with variance equal to the
variance of the sum whatever the individual distributions of these variables
                     provided n tends towards infinity



                 Pdf of normal (Gaussian) distribution process is:
                                             x2 
                            px  
                                      1
                                         exp  2 
                                             2                               (45)
                                     2         
                  Diffusion and random walk model
                    Variance of the radnom walk process:

 2  1   2  i     n 2  12   22   i2    n2  2i j   (46)

                      Since the steps are independent:
                                                                            (47)
                                    i j  0
                       Then for random walk process:

                                   2  n2                                 (48)

    If τ is teh time required for every step and V is velocity such that:

                                    V                                   (49)

                     Then the time t to make n steps is:
                                  t  n                                   (50)

                   Adn the variance of random walk process:

                                                2
                                            t    t2
               2  n 2  nV  2
                                        n V   V   V  t            (51)
                                            n    n
                        Diffusion and random walk model

                                     2    V   t                             (52)

Close analogy of random walk process to molecular diffusion with λ analogous to mean
free path and V analogous to molecular agitation (≈ temperature), which finally leads to
the expression for diffusivity:

                                       V   D                                   (53)

            Example: N molecules of mass m injected into infinitely thin layer
                   Diffusion and random walk model
                     Total mass of injected molecules:
                                       N m  M0                          (54)

If the molecules move according to random walk due to collisions then
according to Central Limit Theorem probability p(x)dx that one given molecule
is in the slice (x,x+dx) is the Gaussian pdf and the number of molecules in
slice (x,x+dx) yields
                                      N  px dx                         (55)

                     Mass of the contaminant in volume Adx:

                                       N m px dx                        (56)

                             Concentration of the contaminant:

                                           Nm p x dx M 0
                   c  x, t                              px 
                                   mass
                                                                        (57)
                                  volume     Adx        A
                Since the pdf is Gaussian then the concentration reads:

                                      x2 
                cx, t          exp  2 , where  2   V t
                           M0 1
                                                                          (58)
                           A  2     2 
                                          
       Diffusion and random walk model
    If diffusion process is described by Fick’s law:
                        c    2c
                           D 2
                        t   x

     the solution for initial mass M0 injected at x=0:
                           x2 
     cx, t  
                M0 1
                       exp  2 , where  2  2 Dt
                           2 
                A  2         

       From comparison of Fick’s law and random walk:

                             1
                          D  V
                             2
                        Conclusion:
Any random walk process produces the diffusion with
                    diffusivity

                             1
                               V
                             2
             Diffusion and random walk model
Example: Diffusion 1024 particles injected at x=0 and t=0 and then
                 dispersed due to random walk




    Illustration of particles diffusion by random walk process
                      Diffusion and random walk model
    In order to smooth the distribution let us calculate the average distribution for the
      intermediate step n=6.5 and compare it with the Gaussian distribution, i.e.:
                                            1      x2 
                                   N  N0      exp  2 
                                                   2 
                                                                                            (59)
                                           2         

  for   1,   1   2  n                                                        x2 
                                                       N  X , n  6.5  160.2 exp  
                                                                                            (60)
       n  6.5; N 0                                                                 13 
                                                                                       


             -7       -6     -5      -4      -3       -2        -1          0          1           2
   x

Random       4        8      28      48      84      128       140        160        140           128
 walk


Gaussian   3.7     10.1     23.4 46.8 80.1         117.8     148.3 160.2 148.3                117.8
process



  Conclusion: after only 6.5 steps the distribution obtained from random
             walk is already close to the Gaussian process
                                  Brownian motion
R. Brown (1826) – observation of motion of small particles
A. Einstein (1905) – rigorous explanation
practical application: motion of aerosol particles
Starting point for analysis: macroscopic particle suspended in fluid,
radius a ≈ 1 μm, particle subjected to molecular agitation i.e. collisions with molecules of
    fluid (much smaller than suspended particle)


Numbers of collisions do not balance




  Finite probability to have more
       impulses on one side



          Particle will move
                                                 Creation of impulse in brownian
                                                 motion of macroscopic particles
                      Brownian motion
    Suppose that the particle is spherical and obeys Stokes law:
                                     
                drag force  6   r U
                 where: r  radius of particle
                                                                     (59)
                             viscosity
                           
                           U  particle velocity

                  Equation of motion for particle:
                          
                        dU                                         (60
                      m      6   rU  F
                         dt


                                              Random impulse (+ or -)
Inertia force
                                              Form fluid molecules


Due to random impulse the motion of particle as in random walk
                      Brownian motion
   Problem: what is the diffusivity of this process?

                           1
                        D   V
                           2
To evaluate the diffusivity of random walk we need two scales

     linear scale                   time scale
                           or 
   V  velocity scale               V  velocity scale

  From dimensional analysis the time scale can be estimated:

                       F 6   rU              U
     acceleration - a                    
                       m     m                 a

                U   m U      m
                       
                a 6   rU 6   r                              (61)
                           Brownian motion
        Between the impulses the motion equation is homogeneous:

               dU                dU        6  r          dt
            m       6  rU                    dt  
                dt               U            m                       (62)
                                                  t
            ln U    ln U 0  U t   U 0 exp   
                     t
                                                 
      Velocity scale U0 is determined by the equipartition of energy

             1
               m U 02  k T                                            (63)
             2
             where : k  1.38 10  23 J / K   (Boltzman constant)

                         Hence the velocity scale is:

                                             2kT
                            V  U
                               2       2
                                       0                              (64)
                                              m
And diffiusivity due to brownian motion can be evaluated as:

                                    2kT   m    1 kT
              Dbm  V 2                  
                                     m 6  r 3   r
                                                                        (65)
                      Brownian motion

       Example: diffisuivity for spherical particles in air


   R [μm]              0.1                1                   10


Dbm [m2/s]       1.3x10-10         1.3x10-11         1.3x10-12



Note: compare these values with kinematic viscosity for air
equal ν= 1.5x10-5 [m2/s]
i.e. diffusiivity of brownian motion is small compared with
diffusivity of air
                     Dispersion by turbulent motion

       Growth rate of the contaminant cloud due to moelcular diffusion:
                                                  d D
                           2  2 Dt                                      (66)
                                                  dt 
                               Conclusion:
                           the larger the cloud                      

                                                                     d
                               The smaller growth rate:                 
                                                                     dt
Extemely large time scales needed to achieve efficinet mixung by moldeular
diffusion, for example in water:
                                        1m
                                           2
                                                 109 s  104 days
                               D            m
                                       109
                                             s

Note: time constant for reaction of human nerves –diffusion of ions through the
cellular membrane of the thickness h=10-6 m, τ=1ms
                     Dispersion by turbulent motion

                               How to intensify diffusion ?
                          A possible way to speed up diffusion




                     Break up the cloud into smaller fragments

         If the size of the cloud is σ2, then the diffusion time scale is: 2
                                                                         
                                                                                   
    2
       , if the cloud is split into n smalleer parts then the time scale is t n   
                                                                                   n
t0 
     D                                                                             D
                                                             2
                                                  t 1
and the reduction of the diffuison time scale is : n   
                                                  t0  n 
                       Dispersion by turbulent motion
             What can be the role of turbulence in mixing enhancement ?
                         Turbulence -> superposition of eddies

Case 1: the size of the turbulent eddy L is much larger than the size of the cloud σ




            Cloud transported by turbulent eddy without any change of shape



                        No speed-up of the molecular diffusion
                      Dispersion by turbulent motion

     Case 2:       L 




               Cloud of contaminant distorted by turbulent eddy (random)



Speed-up of the molecular diffusion by break-up of the mixing volume
               Dispersion by turbulent motion

Case 3:     L  




     Small turbulent eddies distributed within the cloud of contaminant




              Molecular mixing augmented by turbulent eddies
                           Dispersion by turbulent motion

     Possible mechanisms of turbulence interaction with molecular diffusion




                                        Vortex stretching




          Full range of eddies (from largest to smallest) appears in turbulent flow




Even if initially only largest eddies exist (case 1) after a while eddies of sizes corresponding
to cases 2 and 3 will be developed
            Vortex stretching




               Vortex stretching



Steeper contration gradients inside the vortex



    Since {flux} ~ {concentration gradients}


       Enhanced diffusion in radial direction
                Summary of turbulent diffusion

•Turbulence cannot directly enhance mixing and homogenisation at
molecular level because fo disparity of scales

      smallest scales         molecular
                                     
      of turbulent eddies     scales   

•Turbulence can make molecular mixing more efficient by
     Reducing the local size of the contaminant volume cases 2+3
     Making local concentration gradients steeper thus enhamcing
     the diffusive mass flux

								
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