Diffusion Chapter 9 (old book)General process of flow

Heat as the example

Fick’s First law

Use of diffusion Coefficientspersonal passive samplers,SO2 accommodation coef.Particles

Probability distribution using random movementDiffusion distances

Fick’s Second LawDiffusion in a GC columnDiffusion in a sphere

Estimating Diffusion Coefficientsgaseswater; liquids

Turbulent DiffusionLake systemAtmospheric System

1

Heat

Templow2 Temphigh1

x

The heat that flows thru a slab of material is proportional to the cross sectional area, A, of the slab and the time, t, for a given Temp

Heat flow is also to Temp/x for a given A and time if Temp/x is small

if we think about really small thickness of x

dqdt

kA dTdx

dq/dt = the rate of heat transfer with respect to time

dT/dx = the temperature gradient

k = thermal conductivity

2

dqdt

kA dTdx

k has the units of

material k

Al 4.9x10-2

Steel 1.1x10-2

Pb 8.3x10-3

air 5.7x10-6

glass 2.0x10-4

Temphigh1Templow2

L

at steady state for a const. temp gradient across the rod

3

Diffusion

1b. cool 1a. hot

2b. low voltage 2a. high voltage

3b. low mass 3a. high mass

4b. low pressure 4a. high hydrostatic pressure

flux = flow area-1 time-1

a gradient drives the flow

4

page 184 table 9.1

Page 184 Table 9.1

5

Let us consider a gas diffusing in into a zone where it is constantly collected or removed

O3

O3 O3

O3 [O3]O3

O3

x inlet x

= ;

( )( )

[ ]moles Otime area

D Ox

3 3

If we measure the # of moles of O3 collected over a period of time; know the diffusion coef. for O3 in air, [O3] can be calculated

6

Diffusion and sticking coefficients

the average speed of gas molecules is given by

cRTMw

8

The rate of collisions per unit time with a wall of surface A in a given volume is

rate = 1/4 c x area x concgas

rate RTMw

A Cgas 2

rate/A = # molecules time-1 area-1 = flux

if we think about the # of effective collisions, i.e. the ones that actually stick to the wall, a factor is introduced called

“sticking” factor

surface recombination

accommodation coefficient

removal rate RTMw

A Cgas 2

Judeikis et al. were interested in the effectiveness of coal surfaces in the uptake of SO2 gas.

7

rate/A = # molecules time-1 area-1 = flux

flux = radial velocity Cgas D C/r

r SO2

coal soot coatingmeasure SO2 on surface

-D d [SO2]/d r = rad. vel x [SO2]

ln {[SO2]/[SO2,o]} = rad. vel x r /D

ln {[SO2]/SO2,o]} = krate t

8

Removal of SO2 along a tube reactor coated withfly ash. The accommodation coef. = 4.4x10-4. Thetotal pressure was 55 torr, with O2 = 6 torr andSO2 = 9m torr; % RH= 0

9

The Stokes-Einstein Equation (particles)

Let us now think of diffusion in terms of chemical potential

We can think of the driving force of diffusion as the negative gradient of the chemical potential, i.e.

d/dx = free energy/mol /dx

The frictional force resisting the flow, due to an imbalance in chemical potential, is the frictional coef. f (force/velocity) on each molecule x the velocity, v, of the flow,

for a mole this force is f vNo

f v No= -d/dx

Flux has the units of moles, molecules or mass per area per time

Flux = moles/(cm2 time)

Conc x velocity = moles/cm3- x cm/time =

moles/(cm2 time)

We can define a diffusion velocity caused by a driving force or chemical potential, or the concentration gradient such that:

C v = Flux

recall

10

d RTP

dp

substituting for P from PV=nRT, and n/V = C

P= CRT

d RTC

dC

f Nov = -d/dxv RT

N CdC dx

f 0/

C v = Flux = RTN

dC dxf 0

/

11

Flux RTN

dC dxf 0

/

diffusion coef D= RT/(f No)

Stokes (including Cunningham’s slip factor) showed that for unit spheres and nonturbulent viscous flow that the resisting force on a particle flowing through a fluid is

f = 6r/Cc

where = is the viscosity of the medium (poise)air(20

oC) = 1.83x10-4 g/(cm sec) r is the

radius of the particles and

Cc = 1+/r(A+ Qe-rb/) where is the mean free path of air = 0.067 m

particle Cc

size (m)0.01 22.20.05 4.970.1 2.870.25 1.690.5 1.331 1.165 1.03

12

The Randomness of Diffusion

Consider 17 boxes arranged in a row PAGE 185 FIGURE 9.2

13

page 186 figure 9.3

fit to random walk distribution Normal Gaussian Distribution p m

nmnn( ) exp//

/

2 22

1 2 2

p xx

( ) exp/

12 21 2

2

2

14

Where we would like to go with this is relate the sigma, , which is a basic feature of the normal distribution to the diffusion coefficient D

p xx

( ) exp/

12 21 2

2

2

p mn

mnn( ) exp//

/

2 22

1 2 2

22= n/2

if we multiply this by x, the distance across a box, can be related to an actual distance

We will then calculate the flux across from one box to another

The concentration gradient which caused the flux

Substitute into Flux = -D dC/dt

15

BOX # 3 4 5 6x

40 0 8 5t

60 0 24 0 4 6t

0 42 0 14 7t

0 28 0 7 8t

If we look at 5th box in seventh time step (7t), 12 particles are entering box #5 from the left and

2 from the right

at the eighth (8t) step, 7 particles leave box #5 and go back to box #4

this gives a net flux of 5 particles between the 7th and 8th steps or Flux = 5/2t

We now define an average change in concentration per length between adjacent boxes because for every step one box is emptied and the adjacent one filled

C = N/x

the spatial conc. gradient dC/dx is

dC/dx = C/x = N/x2

16

If we look at time step 6, the gradient dC/dx driving the diffusion between boxes for time step 7 is

dC/dx = -N/x2 = -(24-4)/x2

recalling that F = -D dC/dx and F= 5/(2t)

t= nt and x = n1/2/2 x

x = (2Dt)1/2

for three dimensional movement

s = (4/Dt)1/2

17

Example How long does it take for a gas molecule of biphenyl and 0.25 m particle to diffuse from the center of a 5 cm sphere to the a walls of the sphere? Assume a diffusion coef of 0.06 cm2/sec for biphenyl and 1.6x10-6

cm2/sec for the particle.

2.5 cm

for biphenyl s = (4/Dt)1/2

t = 82 seconds

for the aerosol, D= 1.62x10-6 cm2/sec(d= 0.25 m) t = 35.5 days

in water biphenyl diffusion is much slower than in air

D10-6 so

t = tens of days

18

Fick’s second law

Fick’s second law attempts to express the change in concentration with respect to time with the change in Flux

dC/dt = f (flux)

Consider an elemental volume (box) with a flux of material in and out

FX

FO x

A mass balance on the elemental volume per unit time (both in and out)

mass= conc x vol); flux = mass/(area time)

flux x area= mass/time

19

mass= conc x vol); flux = mass/(area time)

mass/time = (conc x vol)/time

d massd time

d conc vold time

( )( )

( )( )

= -area flux

V dC/dt = -area flux

V = area x

division by V

dC/dt = -flux/x;

as x --> zero

x o xflux dflux

dxlim

1

it appears that

dC/dt = - dflux/dx

Ct

Fxx cons t

x

t const

tan

Ct

Fxx cons t

x

t const

tan

20

Ct x

D Cx

D Cx

( )2

2

If we think of three dimensions

Ct

Fx

Fy

Fz

x y z

Ct x

Cx y

Cy z

Czx y z (D ) (D ) (D )

Ct

D Cx

Cy

Cz

2 2 2

2 2 2

21

1. diffusion in the x direction

A solution in the x direction for a long tube, where diffusion in the y and z direction is insignificant, is

C x t MDt

xDt

( , )( )

exp/

2 41 2

2

2

where = (2Dt)1/2

Figure 9.5 Page 193

22

What kind of diffusion can we expect for a compound traveling down a 30 m fused silica column (0.25mm id); assume a flow of 1 cm3/min?

id vol of the col = (0.025cm/2)2x 30m x100cm/m;

vol =1.47cm3 ; Flow = ??

The flow time = 1.47cm3/ 1cm3 per min = 1.47 min

A typical diffusion coef= 0.07 cm2/sec,

(2Dt)1/2=

so our peak would broaden by

4 x = 4(2x0.07x1.47*60)1/2 = 14.06 cm; why 4

the carrier travels 30 m or 30x100 cm in 1.47 min

this equals 3,000 cm/1.47 min = 34 cm/sec

our peak would broaden in this time

14.06 cm /34cm/sec = ~0.4 sec

23

2. Diffusion in the Radial Direction

PAH

Ct

D Cx

Cy

C

z

2 2 2

2 2 2)

converting to polar (radial) coordinates

x= r sincos, y = r sinsin , z = r cos

Ct

D Cr

Cr r

Cr r

Cr

C

2

2

2

2 2 2

2

22 1 1

sincot

if diffusion is only in the radial direction

Ct

D Cr

r Cr

2

22

( ) ( )C rt

D C rr

2

2

PAH

dr24

These types of systems can be solved with numerical techniques to calculate the C at successive depths of dr into the particle over time

U= Cr

U p U p D tr

U p U p U pt t t( ) ( ) ( ) ( ) ( )

2 1 2 1

25

Diffusion between two parallel plates

Cout

W

L Cin H

Let’s say that we wanted to strip a gas and not particles

Solutions to the partial differential equations take on the form

CC

Aout

inn

n

n

1

exp

C/Co= 1 - 1.52652/3 +1.5 +0.0342 4/3

where = 8 x D x L x W/(H x flow)

26

0

20

40

60

80

100

% T

rans

mis

sion

0 10 20 30 40 50 60 Flow Rate (cfm)

Organic Gases (D=0.05cm2/s) Particles (D=0.0005cm2/s)

Large Stripper System Transmission Properties

organic gas D=.05

particles D=.00005

27

Estimating Diffusion Coefficients

Factors that influence diffusion

average distance traveled between collisions,i.e. mean free path,

more collisions for a given distance translates into a lower mean free path

organic air

org

airz N org air,air /( ) ,

11 1 2 2

where N= # air molecules/vol= collision diameter of air and organicz= molecular wt ratio of organic/air

We would predict that diffusion coefficients would decrease with increasing (molecular weight)1/2 and effective collision diameters squared.

Diameter2 radius2 cross sect. area of molecule

V r3or r V1/3; so r2 or area V2/3

If we assume the volume of a molecule is molar vol V

28

and molar volume = molecular weightliquiddensity

in cm mol/

Figure 9.6 top page 195

See Figure 9.6 page 195

29

Molecular weight vs Diffusion Coefficients

Page 195 Fig 9.6 bottom

30

Diffusivities in water page 196 Figure 9.7

31

Estimating Gas Phase Diffusion Coefficients (page 197)equation on page 197 with definitions

32

Estimating molar volumes

1. molar volume, V = molecular weightliquiddensity

incm mol3 /

for benzene

Mw= 78, density =0.88gcm-3=89 cm3/mole

2. sum of atom size---> diffusion

for benzene V= 6(C) +6(H) +ring

V= 6x16.5+6x2.0-20.2 = 90.8 cm3mol-1

33

35

Calculating liquid Diffusion Coefficients

D xx V

cmw

13 26 10 5

114 0 5892 1. ( sec ). .

V = molar volume

= solution viscosity in centipoise (10-2g cm-

1sec-1) at the temperature of interest. The units of poise refer to the property of fluids that requires a shearing force of one dyne (g cm/sec2) of two parallel layers of one cm2 at velocity of 1 cm/sec over a gradient of one cm

log = A / T+B

For other liquids Wilke-Chang (1955) give

D x Mw Tx V

cmliq

7 4 10 8 1 2

0 62 1. ( ) ( sec )

/

.

where is the solvent association term, Mw is the molecular weight of the solvent, is the solvent viscosity, T is the temp.

water 2.6CH3-OH 1.9ethanol 1.5Heptane 1Benzene 1

page 409 In Barrow Physical Chemistry 1962 Table 12.5

36

Reynolds NumbersA body moving through a fluid creates

1. Inertial forces, i.e. forces due to the acceleration or deceleration of small fluid masses near the body

2. Viscous frictional forces due to theviscosity of the medium

Inertial forces/viscous forces = Re#

Re# = m x vel x d/

/m = = (kinomatic for air) =

Re# = vel x d/

= 0.00121 g/cm-1sec-1 /1.82x10-4 g/cm3= 0.151cm2sec-1

for flow in a pipe

laminar region 1-2000intermediate region 2100 - 4000turbulent flow >4000

A 1 m aerosol is flowing in a 16” duct with a velocity of 3500 ft/min. What is the Re# in the duct?

Re# = duct diameter x rel. velocity of air to duct/v

Re# = vel x d/ = 3500x12x2.54/60 x 15x2.54/.151= 479000

37

Turbulent Diffusion

for molecular diffusion

x = (2Dt)1/2

If we look a molecular diffusion times

(page 201 Table 9.5)

38

For transport by advection

Flux = C v

Fad = Cv (m L-2 t-1)

(M L-3 T-1)

For advection the time scale is simply

(advection) (diffusion)

For distances larger than L advection is more important than diffusion.

In your homework, you will calculate the critical distances for typical air and water advection and then explain what it means.

An expression for Turbulent Flux

It is possible to develop a turbulent diffusion coef. analogous to a molecular diffusion coef.

Let’s assume that turbulent fluctuations cause a change in concentration along with an associated flow Qex of volume between C1

and C2 over some distance Lx.

Lx

39

C2

C1

x

If Lx is small, the concentration difference C1 -C2 can be described by the product of Lx and the slope of the curve at the point it intercepts the plane between C1 and C2:

C1-C2 = - Lx C/x

Since this moves across some area, a, at a flow of Qex the velocity vx

= Qex/ a

Since Flux = C vx= Qex/ a ( - Lx C/x)

and

This is property of the fluid motion and not the substance

described by C

Effect of eddies on dispersion

Consider a patch of ink on a water surface that is turbulent. It is characterized by the size of its concentration variance 2 about its center of mass. The patch after time t will grow and be displaced

40

= 2Et and 2/t = 2E

Figure 9.9 page 206

41

The growth of 2 becomes faster with increasing 2.

Figure 9.10 page 207

42

Table 9.6 page 206

43

How do we determine turbulent diffusion coefs in vertical mixing?1) from the change in heat content

; from Fickian flux

Cp = heat capacity

substituting and solving for Ez

2) Radioactive tracersRadon-222 gas is released fromRadium-226 in sediments

44

the radon diffuses into the water, whereit is transported upward by turbulence

Broecker sets up a Fick’s second law flux expression for excess radon, i.e. radon beyond the “conc.” of that which should be in equilibrium with existing Ra (hence your book uses the word activity)

if there is little change in Rn/t after some time

= zero

a solution to this at Rn/t = zero

Rn (h) = Rn(h=0) exp[-(/Ez )1/2 h]

so if we plot

ln Rn(h) vs h we get a slope of ........

45

Figure 9.13 page 212

46

Parameters used to characterize vertical mixing in Lakes

In a lake water is usually stratified with the mostdense water at the bottom.

A stability time N, is introduced which describes howhow long it will take to “restore” a water parcel which ismoved vertically to a different density regime.

g= gravity acceleration = density

47

the coef. of thermal expansion is related to thechange in density with temperature by

d/dT = - This gives

N2 = -g dT/dz

Finally N2 is related to the vertical eddy diffusion coef.

Ez = a (N2)-q

48

Turbulent Diffusion in the Atmosphere

49

Fisk’s 2nd law

Solutions in 1,2,3 dimensions

If we just think in two dimensions; ie there is no diffusion in the x direction compared to y and z, and substitute an emissions strength for X (emissions in g/time normalized fro velocity)

50

Historically

Ky and Kz have been related to a Gausian such that

substituting

51