Diffusion Chapter 9 (Old Book)

Diffusion Chapter 9 (old book)

General process of flow

Heat as the example

Fick’s First law

Use of diffusion Coefficients

personal passive samplers,

SO2 accommodation coef.

Particles

Probability distribution using random movement

Diffusion distances

Fick’s Second Law

Diffusion in a GC column

Diffusion in a sphere

Estimating Diffusion Coefficients

gases

water; liquids

Turbulent Diffusion

Lake system

Atmospheric System

Heat

Templow2 Temphigh1

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 DTemp

Heat flow is also µ to DTemp/Dx for a given A and time if DTemp/Dx is small

if we think about really small thickness of Dx

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

· dT/dx = the temperature gradient

· k = thermal conductivity

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

Temphigh1

Templow2

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

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

page 184 table 9.1

Page 184 Table 9.1

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

O3 O3

O3 [O3]

x inlet x

= ;

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

Diffusion and sticking coefficients

the average speed of gas molecules is given by

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/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

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

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

flux = radial velocity ·Cgas· a = -D DC/D r

Dr SO2

coal soot coating
measure SO2 on surface

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

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

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

Removal of SO2 along a tube reactor coated with

fly ash. The accommodation coef. = 4.4x10-4. The

total pressure was 55 torr, with O2 = 6 torr and

SO2 = 9m torr; % RH= 0

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.

dm/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 ·v·No

f ·v· No= -dm/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

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

P= CRT

f · No·v = -dm/dx

C v = Flux =

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 = 6phr/Cc

where h= is the viscosity of the medium (poise)
hair(20oC) = 1.83x10-4 g/(cm sec) r is the
radius of the particles and

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

particle Cc

size (mm)

0.01 22.2

0.05 4.97

0.1 2.87

0.25 1.69

0.5 1.33

1 1.16

5 1.03

The Randomness of Diffusion

Consider 17 boxes arranged in a row PAGE 185 FIGURE 9.2

page 186 figure 9.3

fit to random walk distribution Normal Gaussian Distribution

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

2s2= n/2

if we multiply this by Dx, the distance across a box, s 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

BOX # 3 4 5 6 Dx

40 0 8 5Dt

60 0 24 0 4 6Dt

0 42 0 14 7Dt

0 28 0 7 8Dt

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

2 from the right

at the eighth (8Dt) 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 = 5/2Dt

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

DC = DN/Dx

the spatial conc. gradient dC/dx is

dC/dx = DC/Dx = DN/Dx2

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

dC/dx = -DN/Dx2 = -(24-4)/ Dx2

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

t= nDt and sx = n1/2/2 Dx

sx = (2Dt)1/2

for three dimensional movement

s = (4/p Dt)1/2

Example How long does it take for a gas molecule of biphenyl and 0.25 mm 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/pDt)1/2

t = 82 seconds

for the aerosol, D= 1.62x10-6 cm2/sec

(d= 0.25 mm) t = 35.5 days

in water biphenyl diffusion is much slower than in air

D»10-6 so

t = tens of days

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

FDX

FO Dx

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

Dmass= D(conc x vol); Dflux = Dmass/(area Dtime)

Dflux x area= Dmass/Dtime

Dmass= D(conc x vol); Dflux = Dmass/(area Dtime)

Dmass/Dtime = D (conc x vol)/Dtime

= -area · Dflux

DV dC/dt = -area · Dflux

DV = area · Dx

division by DV

dC/dt = -Dflux/Dx;

as Dx --> zero

it appears that

dC/dt = - dflux/dx

If we think of three dimensions

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

where s= (2Dt)1/2

Figure 9.5 Page 193

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 = p(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= s

· so our peak would broaden by

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

· 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

2. Diffusion in the Radial Direction

PAH

converting to polar (radial) coordinates

x= r sinq cosf, y = r sinq sin f , z = r cos q

if diffusion is only in the radial direction

PAH

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

Diffusion between two parallel plates

Cout

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

C/Co= 1 - 1.5265f2/3 +1.5 f +0.0342 f4/3

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

Estimating Diffusion Coefficients

Factors that influence diffusion

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

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

organic
air

where N= # air molecules/vol

d= collision diameter of air and organic

z= 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 µ radius2 µ cross sect. area of molecule

V µ r3or rµ V1/3; so r2 or area µ V2/3

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

and molar volume =

Figure 9.6 top page 195

See Figure 9.6 page 195

Molecular weight vs Diffusion Coefficients

Page 195 Fig 9.6 bottom

Diffusivities in water page 196 Figure 9.7

Estimating Gas Phase Diffusion Coefficients (page 197)

equation on page 197 with definitions

Estimating molar volumes

1. molar volume, V =

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

Calculating liquid Diffusion Coefficients

V = molar volume

h = 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 h = A / T+B

For other liquids Wilke-Chang (1955) give

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

water 2.6

CH3-OH 1.9

ethanol 1.5

Heptane 1

Benzene 1

page 409 In Barrow Physical Chemistry 1962 Table 12.5

Reynolds Numbers

A 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 the

viscosity of the medium

Inertial forces/viscous forces = Re#

Re# = rm x vel x d/h

h/rm = n = (kinomatic for air) =

Re# = vel x d/n

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

for flow in a pipe

laminar region 1-2000

intermediate region 2100 - 4000

turbulent flow >4000

A 1 mm 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/n = 3500x12x2.54/60 x 15x2.54/.151= 479000
Turbulent Diffusion

for molecular diffusion

sx = (2Dt)1/2

If we look a molecular diffusion times

(page 201 Table 9.5)

For transport by advection

Flux = C v

Fad = C*v (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.

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, Da, at a flow of Qex the velocity vx = Qex/ Da

Since Flux = D C · vx= Qex/ Da ( - 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 s2 about its center of mass. The patch after time t will grow and be displaced

s2 = 2Et and ¶s2/¶t = 2E

Figure 9.9 page 206

The growth of s2 becomes faster with increasing s2.

Figure 9.10 page 207

Table 9.6 page 206

How do we determine turbulent diffusion coefs in vertical mixing?

1) from the change in heat content D

; from Fickian flux Cp = heat capacity

substituting and solving for Ez