diffusion processes introduction typical scenario of catastrophic pollution accidental release of...

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:

Transformation

Accumulation

Transport

Accidental realease of contaminants:

Dispersion in environment

Environmental Quality

Effects on health

Effects on climate

Effects on 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

3

0

limm

kg

V

Mc

V

txcc ,

ppmsolutionof

c;%

Concentration:

Mass M of contaminant in volume V gives concentration

(1)

in general: (2)

Relative concentration: (3)

Averages:

Ensemble average: txc ,

Time average:

Tt

t

dtcT

txc0

0

1, 0

Space average: V

V dVcV

txc1

,

(4)

(5)

(6)

Basic definitions

AA

f dAqQ

dAcUQ

c11

AdAUQA

surfaceover flux volume-

flux mass specific 2

sm

kgq

Flux average: (7)

Where:

Dilution:

volumetotal

tcontaminan of volume

ionconcentrat volumerelative :where

1

p

p

pS (8)

(9)

Mass conservation equation for contaminant

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

volume

withincontained

flux mass mass of chaneg

0

dAnqdVct AV

(10)


Due to Gauss-Ostrogradski transformation

0

dVqxt

c

V

ii

0

ii

qxt

c

diffusion advection Diii qcUq

(11)

Since V is an arbitrary volume:

(12)

Flux of contaminant:

(13)

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

Dii

ii

qx

cUxt

c

(14)


0iU

Dii

qxt

c

0

i

i

x

U

If fluid is at rest:

(15)

If fluid is incompressible:

Diii

i qxx

cU

t

c

If flow is turbulent (Reynolds decomposition):

cuCUcU

cCC

uUU

iii

iii

Then mass conservation equation becomes:

cuqxx

CU

t

CiDi

iii

(16)

Molecular diffusion Turbulent flux – closure problem

(17)

(18)

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

coeffienty diffusivit thermal :where

flux heat

kx

Tk

Fick’s law-thermal analogy

According to Fourier’s (1822) law of heat transfer

(19)

Fick’s (1855) analogy for mass flux

/T][Lconstant diffusion

unit time and surfaceunit per flux mass

:where

2D-

q

x

cDq

ii

(20)

Note on the units:

3

2

2;;

m

kgc

s

mD

sm

kgq

Fick’s law

General form of diffusion equation:

processdiffusion )(heat or

)( momentum ),( massfor iprelationshlux gradient/f isit i.e.

quantity

ed transportofgradient

k

D

flux

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

Example:

For water: D~10-9 [m2/s]

ν ~ 10-6 [m2/s]

k ~ 10-5 [m2/s]

For air:

D ~ ν ~k ~10-5 [m2/s]

1 air for

10 for water

number Schmidt

3

Sc

Sc

DSc

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

Δx Δx Δx Δx

LR

(A) (B)

Concentration of molecules in the initial (A) and consecutive moment (B) for 1D diffusion process

ΔNL

ΔNR

A simple model for gradient/flux relationship

182420;124210

step timenew ain

2105

1;420

5

1

RL

LR

NN

NN

)()1()()1( ; nR

nR

nL

nL NNNN

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

After each time step

We obtain

nL

nR

nL

nR NNNN 11

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

mNMmNM RRLL ;

The mass flux during time Δt in positive x-direction is

flux mass negative

RLRL MMppNmNpmflux

mass

Concentration in each box:

x

Mc

x

Mc R

RL

L

1;

1

Mass flux per unit area and unit time

2

22

2 x

cx

x

c

t

xpcc

t

xpq RLx

Analytical description of gradient/flux relationship

ydiffusivitlim2

00

Dconstt

xp

tx

x

cDqx

1; RL NN

Mass transfer should not be dependent on the size of the box

And hence neglecting higher order terms

We obtained a simple 1D model of Fick’s law

Conclusions:

•Boxes must be large enough, so that we can apply probability analysis

•Boxes must be small enough to apply first order approximation in Taylor series

x

c

x

cx

2

2

•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

2

2

x

cD

t

c

Assume 1D diffusion in fluid at rest:

(21)

And uniform concentration along y and z axes

0

z

c

y

c(22)

Coordinates for 1D diffusion problem


Total mass of the contaminant:

V

dxcAdVcM

LA

M

LengthA

Mc

11

scale timereference -

scalelength reference

T

L

(23)

That means c will be sought in the form

(24)

Let us introduce:

Dimensional analysis yields:

DTLL

cD

T

c 2

2 (25)

Time scale and the length scale in the diffusion process are naturally related

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:

DtA

Mc

1


(26)

i.e. concentration may be obtained for a similarity solution

txfDtA

Mc ,

1

The only possibility for non-dimensional function f is:

(27)

212

ss

m

m

Dt

xff

(28)

That finally yields:

Dt

xf

DtA

Mc ;

1 (29)

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

Introducing Eq.(29) into the r.h.s of diffusion equation:

f

DtA

M

x

c

d

fdf

d

dff

232

2

2

2 1;

(30)

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

ftDt

ftDA

M

t

c

2

11

2

13

(31)

Hence the diffusion equation takes the following form

fff 2

1 (32)

And finally:

ff 2 (33)

Note that similarity solution allowed to transform the p.d.e into the ordinary differential equation


After integration of Eq.(33) we obtain:

Cff 2

oddx

(34)

Determination of the constant C

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

C(x,t) C(x,t)

x xInitial injection at t=0 At t >0

(a) (b)

evenc


oddevenoddfoddfevenffc

0C

ff2

And due to symmetry requirements

(35)

And finally:

(36)

That yields the solution:

Dt

x

DtA

MCtxcCf

4exp

1,

4exp

2

1

2

1

(37)

Note that C1 the only unknown quantity

14

exp2

1

Dt

dx

Dt

xC

a

dxxa

22exp

41

1 C


From the contaminant mass conservation equation Eq. (23)

(38)

From tables of definite integrals we find

(39)

Substitution into eqaution (38) leads to:

(40)

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

Dt

x

DtA

Mtxc

4exp

4

1,

2

(41)

Statistical measures of concentration distribution establidhed due to diffusion processs

First order moment – expected value:

0

:..,,,

][

dxxcxE

eioddcxevencoddx

dxxcxE

Second order moment – variance:

dx

Dt

x

tDdxxcx

M

A

4exp

4

1 222

Having known that the following definite integral is:

Dt

aanda

dxxax2

1;

2exp

3222

Variance can be expressed as:

Dt22 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:

2

1p

+λ-λ

∞ ∞Δ – Dirac delta

function

ξ

p(ξ)

(43)


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

ntX 21 (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:

2

2

2exp

2

1

x

xp (45)


Variance of the radnom walk process:

jinini 22222

21

221

2

0ji

22 n

(46)

Since the steps are independent:

(47)

Then for random walk process:

(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:

nt

Adn the variance of random walk process:

(50)

tVn

tV

n

tVnVnn

22222

(51)


tV 2 (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:

DV (53)

Example: N molecules of mass m injected into infinitely thin layer


Total mass of injected molecules:

0MmN

dxxpN

dxxpmN

(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

(55)

Mass of the contaminant in volume Adx:

Concentration of the contaminant:

xpA

M

dxA

dxxpNm

volume

masstxc 0,

Since the pdf is Gaussian then the concentration reads:

(56)

(57)

(58) tVwherex

A

Mtxc

2

2

20 ,

2exp

2

1,


If diffusion process is described by Fick’s law:

2

2

x

cD

t

c

tDwherex

A

Mtxc 2,

2exp

2

1, 2

2

20

the solution for initial mass M0 injected at x=0:

From comparison of Fick’s law and random walk:

VD 2

1

Conclusion:

Any random walk process produces the diffusion with diffusivity

V2

1


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


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.:

2

2

0 2exp

2

1

x

NN

0

2

;5.6

1,1

Nn

nfor

13exp2.1605.6,

2xnXN

(59)

(60)

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

Random walk

4 8 28 48 84 128 140 160 140 128

Gaussian

process3.7 10.1 23.4 46.8 80.1 117.8 148.3 160.2 148.3 117.8

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 moveCreation of impulse in brownian motion of macroscopic particles

Brownian motion

Suppose that the particle is spherical and obeys Stokes law:

velocityparticle

viscosity

particle of radius:where

6forcedrag

U

r

Ur

(59)

Equation of motion for particle:

FUrdt

Udm

6 (60

Inertia forceRandom impulse (+ or -)

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?

VD 2

1

scale velocity

scale time or

scale velocity

scalelinear

VV

r

m

rU

Um

a

U

66

To evaluate the diffusivity of random walk we need two scales

From dimensional analysis the time scale can be estimated:

a

U

m

Ur

m

Fa 6 -on accelerati

(61)

Brownian motion

Between the impulses the motion equation is homogeneous:

tUtUU

tU

dtdt

m

r

U

dUUr

dt

dUm

explnln

66

00

constant)(Boltzman /1038.1 :where

2

1

23

20

KJk

TkUm

m

kTUV

220

2

(62)

Velocity scale U0 is determined by the equipartition of energy

(63)

Hence the velocity scale is:

(64)

And diffiusivity due to brownian motion can be evaluated as:

r

kT

r

m

m

kTVDbm

3

1

6

22 (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

D

dt

dDt 22

dt

d

Growth rate of the contaminant cloud due to moelcular diffusion:

(66)

Conclusion:

the larger the cloud

The smaller growth rate:

Extemely large time scales needed to achieve efficinet mixung by moldeular diffusion, for example in water:

dayss

sm

m

D49

29

101010

1

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


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

0

2

2

0

n

1 :is scale timediffuison theofreduction theand

is scale time then thepartssmalleer n intosplit is cloud theif ,

t

t

Dn

tD

t

n

n


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


Case 2: L

Cloud of contaminant distorted by turbulent eddy (random)

Speed-up of the molecular diffusion by break-up of the mixing volume


Case 3: L

Small turbulent eddies distributed within the cloud of contaminant

Molecular mixing augmented by turbulent eddies


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

scales

molecular

eddies turbulentof

scalessmallest

•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

diffusion processes introduction typical scenario of catastrophic pollution accidental release of...

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mass flux20note

volume v

transport phenomena

flux of contaminant

d pipe

space average

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