高等輸送二 — 質傳

Lecture 1Diffusion in dilute solutions

郭修伯 助理教授

Mass transfer

• Convection– Free convection and forced convection

• Diffusion– diffusion is caused by random molecular motion that

leads to complete mixing.• in gases, diffusion progresses at a rate of about 10 cm/min;

• in liquid, its rate is about 0.05 cm/min;

• in solids, its rate may be only about 0.00001 cm/min

– less sensitive to temperature than other phenomena

Diffusion

• When it is the slowest step in the sequence, it limits the overall rate of the process:– commercial distillations

– rate of reactions using porous catalysts

– speed with which the human intestine absorbs nutrients

– the growth of microorganisms producing penicillin

– rate of the corrosion of steel

– the release of flavor from food

• Dispersion (different from diffusion)– the dispersal of pollutants

Understand diffusion?

• What is Diffusion?– process by which molecules, ions, or other small

particles spontaneously mix, moving from regions of relatively high concentration into regions of lower concentration

• How to study diffusion?– Scientific description: By Fick’s law and a diffusion

coefficient

– Engineering description: By a mass transfer coefficient

Models for diffusion

• Mass transfer: define the flux

• Two models (from assumptions!)– Fick’s law

– Mass transfer coefficient model

))(( areaunittime

removedgasofamountfluxdioxidecarbon

differenceionconcentratdioxidecarbonkfluxdioxidecarbon

lengthcapillary

differenceionconcentratdioxidecarbonDfluxdioxidecarbon

Models

• The choice between the mass transfer and diffusion models is often a question of taste rather than precision.

• The diffusion model– more fundamental and is appropriate when concentrations

are measured or needed versus both position and time

• The mass transfer model– simpler and more approximate and is especially useful

when only average concentrations are involved.

Diffusion in dilute solutions

Diffusion in dilute solutions

• Diffusion in dilute solutions are frequently encountered– diffusion in living tissue almost always involves the transport

of small amounts of solutes like salts, antibodies, enzymes, or steroids.

• Two cases are studied– steady-state diffusion across a thin film

• basic to membrane transport

– unsteady-state diffusion into a infinite slab• the strength of welds

• the decay of teeth

Early work in diffusion

• Thomas Graham (University of Glasgow)– diffusion of gases (1828 ~ 1833); constant pressure– The flux by diffusion is proportional to the

concentration difference of the salt

water

stucco （灰泥） plugglass tube

diffusion gas

airhydrogen

Apparatus for liquids

• Adolf Eugen Fick (~1855)– Diffusion can be described on the same mathematical basis as

Fourier’s law for heat conduction or Ohm’s law for electrical conduction

– One dimensional flux:

– Paralleled Fourier’s conservation equation

z

cADAjJ

111

area across which diffusion occurs

the flux per unit area

concentrationdistance

diffusion coefficient

z

c

z

A

Az

cD

t

c 121

21 1 Fick’s second law:

one-dimensional unsteady-state diffusion

Steady diffusion across a thin film

• On each side of the film is a well-mixed solution of one solute, c10 > c1l

c10

c1lz

l

z

Mass balance in the layer z

rate of diffusion out of the layer at z + z

rate of diffusion into the layer at z

Solute accumulation =

s.s. zzz jjA 110

zzz jjA 110

Dividing A z

zzz

jj zzz

)(0 11

z 0

10 jdz

d

z

cDj

11

21

2

0dz

cdD

21

2

0dz

cdD

B.C.z = 0, c1 = c10

z = l, c1 = c1l

l

zcccc l )( 101101

z

cDj

11

lccl

Dj 1101

c10

c1lz

l

z

linear concentration profile

Since the system is in s.s., the flux is a constant.

Derive the concentration profile and the flux for a single solute diffusing across a thin membrane. The membrane is chemically different from the solutions.

Similar to the previous slide, a steady-state mass balance gives:

21

2

0dz

cdD

B.C.z = 0, c1 = HC10

z = l, c1 = HC1l

l

zCCHHCc l )( 101101

Different boundary conditions are used;where H is a partition coefficient.This implies that equilibrium exists across the membrane surface.Solute diffuses from the solution into the membrane.

c10

c1l

c10

c1l

H H

10

1l

chemical potential: driving force

)(][

1101 lCCl

DHj

[DH] is called the permeability. The partition coefficient H is found to vary more widely than the diffusion coefficient D, so differences in diffusion tend to be less important than the difference in solubility.

l

zCCHHCc l )( 101101

z

cDj

11

Derive the concentration profile and the flux for a single solute diffusing across a micro-porous layer.

Micro-porous layer

No longer one-dimensional

Effective diffusion coefficient is used

)(][

1101 lCCl

DHj )(

][1101 l

eff CCl

HDj

Homogeneous membrane Micro-porous layer

Deff = f (solute, solvent, local geometry)

Membrane diffusion with fast reactionA solute is diffusing steadily across a thin membrane, it can rapidly and reversibly react with other immobile solutes fixed with the membrane. Derive the solute’s flux.A mass balance for reactant 1 gives:

rate of diffusion out of the layer at z + z

rate of diffusion into the layer at z

Solute accumulation =

rate of consumption by reaction

s.s

zArjjA zzz 1110 110 rjdz

d

A mass balance for (immobile) product 2 gives:

zAr 10 10 r

10 jdz

d

The reaction has no effect.

Well-stirred solutions

Diaphragm （隔板） cell

• Two well-stirred volumes separated by a thin porous barrier or diaphragm.

• The diaphragm is often a sintered glass frit/ a piece of filter paper.

• Calculate the diffusion coefficient when the concentrations of the two volumes as a function of time are known.

Assuming the flux across the diaphragm quickly reaches its steady-state value, although the concentrations in the upper and lower compartments are changing with time:

)(][

,1,11 upperlower CCl

DHj

Pseudo steady-state for membrane diffusion

H includes the fraction of the diaphragm’s area that is available for diffusion.

Overall mass balance on the adjacent compartments

1,1 Aj

dt

dCV lower

lower

1,1 Aj

dt

dCV upper

upper

A is the diaphragm’s area

upperlowerupperlower VV

AjCCdt

d 111,1,1

)(][

,1,11 upperlower CCl

DHj

upperlowerupperlower VV

AjCCdt

d 111,1,1

upperlowerupperlowerupperlower VV

CCl

DHACC

dt

d 11)(

][,1,1,1,1

)( ,1,1,1,1 upperlowerupperlower CCDCCdt

d

upperlower VVl

HA

11][

upperlower

upperlower

CC

CCC

,10

,10

,1,1*1

*1

*1 CDC

dt

d *1ln

1C

tD

Find the flux across a thin film in which diffusion varies sharply (i.e., the diffusion coefficient is not a constant). Assume that below some critical concentration c1c, diffusion is fast, but above this concentration it is suddenly much slower.

c10

c1lzc

l

small diffusion coefficient

large diffusion coefficient

left

c1c

10 jdz

d

dz

dcDj L

11

cc c

cL

zdcDdzj

1

1010 1 c

c

L ccz

Dj 1101

right10 j

dz

d

dz

dcDj R

11

l

cc

c

cR

l

zdcDdzj

1

111

lcc

R cczl

Dj 111

The flux is the same across both films:

l

ccDccDj lcRcL 111101

B

BB

A

AA

tissueB

BBgas

A

AA

i

lHD

lHD

plHD

plHD

p,1,1

1

Skin diffusionSkin behaves as if it consists of two layers, each of which has a different gas permeability. Explain how these two layers can lead to the rashes observed.

concentration gas pressure

Assuming that the gas pressure is in equilibrium with the local concentration:

For layer A, gasiA

gas ppl

zpp ,11,11

For layer B, itissueB

Ai pp

l

lzpp 1,111

p1,gaslA+lB

p2,gas

p2,tissue

p1,tissueA Bgas outside

the body

p1i

The flux through layer A equals that through layer B :

)(][

)(][

,11

1,11

tissueiB

BB

igasA

AA

ppl

HD

ppl

HDj

p1+p2

Unsteady diffusion in a semiinfinite slab - free diffusion

• Any diffusion problem will behave as if the slab is infinitely thick at short enough times.

c10

c1

position z

time

At time zero, the concentration at z = 0 suddenly increases to c10

rate of diffusion out of the layer at z + z

rate of diffusion into the layer at z

Solute accumulation =

zzz jjAzcAt

111

Mass balance on the thin layer Az

zzz jjAzcAt

111

Dividing A z

zzz

jjc

tzzz

)(11

1

z 0

11 j

zt

c

z

cDj

11

21

21

z

cD

t

c

Fick’s second law

ztatcc

ztatcc

zallfortatcc

,0

0,0

,0

11

101

11

Boundary conditions

21

21

z

cD

t

c

erf101

101

cc

cc

0

22erf dse s

z

cDj

11

11041

1

2

ccet

D

z

cDj Dt

z

11001 cct

Dj z

Dt

z

4

Free diffusion with fast reactionA solute is diffusing steadily across a semiinfinite slab, it can rapidly and reversibly react with other immobile solutes fixed within the slab. Derive the solute’s flux.

A mass balance for reactant 1 gives:

rate of diffusion out of the layer at z + z

rate of diffusion into the layer at z

Solute accumulation =

rate of generation by reaction

zArjjAzcAt zzz

1111 121

21 r

z

cD

t

c

For a first-order reaction 11 kcr

121

21 kc

z

cD

t

c

21

21

1 z

c

k

D

t

c

Use to replace D k

D

1

The reaction has left the mathematical form of the answer unchanged, but it has altered the diffusion coefficient.

21

21

1 z

c

k

D

t

c

21

21

z

cD

t

c

erf101

101

cc

cc

0

22erf dse s

11001 cct

Dj z

Dt

z

4

If the same B.C.s are used:

erf101

101

cc

cc

0

22erf dse s

tk

D

z

14

11001 1cc

tk

Dj z

Without reaction With first-order fast reaction

A sharp pulse of soluteThe initial sharp concentration gradient relaxes by diffusion in the z direction into the smooth curves. Calculate the shape of these curves.

Position zz

Mass balance on the differential volume Az

rate of diffusion out of this volume

rate of diffusion into this volume

Solute accumulation

in Az

=

zzz AjAjzcAt

111

Dividing A z

z 0

11 j

zt

c

z

cDj

11

21

21

z

cD

t

c

)(,0,0

0,0,0

0,,0

1

1

1

zA

Mczt

z

czt

czt

Boundary conditions

far from the pulse, the solute concentration is zero

at z = 0, the flux has the same magnitude in the positive and negative directions

all the solute is initially located at z = 0A: the cross-sectional area over which diffusion is occurringM: the total amount of solute in the system(z): the Dirac function (length)-1

MAdzzA

MAdzc

)(1

)(11 scst

c

21

2

21

2 )(

dz

scd

z

c

and

Laplace transform2

12

1)(

)(dz

scdDscs

Second order linear O.D.E.

z and c1 are independent variables

21

21

z

cD

t

c

Apply Laplace Transform to solve

21

21

z

cD

t

c

011

cscDz

D

sz

D

s

BeAec

1s regards as constant

The boundary condition:

Laplace transform

02

11 zatDA

M

dz

cd

Laplace transform

zatc 01

inverse transform

)(,0,0 1 zA

Mczt

0,,0 1 czt

zD

sz

D

s

BeAec

10

2

11 zatDA

M

dz

cd

zatc 01

zD

s

es

D

DA

Mc

21Dt

z

eDtA

Mc

4

1

2

4

Gaussian curve

The steady dissolution of a spherical particleThe sphere is of a sparingly soluble material, so that the sphere’s size does not change much. However, the material quickly dissolves in the surrounding solvent, so that solute’s concentration at the sphere’s surface is saturated. The sphere is immersed in a very large fluid volume, the concentration far from the sphere is zero. Find the dissolution rate and the concentration profile around the sphere.

r

Mass balance on a spherical shell of thickness r located at r from the sphere:

rate of diffusion out of the shell

rate of diffusion into the shell

Solute accumulation

within the shell

=

rrr jrjrrcrt

12

12

12 444

s.s

rrr jrjr 12

12 440

Dividing 4r2r

r 0

12

2

10 jr

dr

d

r

r

cDj

11

dr

dcr

dr

d

r

D 1222

0

0,

)(,

1

110

cr

satccRr

Boundary conditions

r

Rsatcc 0

11 )(

r

cDj

11

)(120

1 satcr

RDj

The growth of fog droplets and the dissolution of drugsExample:

The diffusion of a solute into the cylinderThe cylinder initially contains no solute. At time zero, it is suddenly immersed in a well-stirred solution that is of such enormous volume that its solute concentration is constant. The solute diffuses into the cylinder symmetrically. Find the solute’s concentration in this cylinder as a function of time and location.

r

z

Mass balance on a cylindrical shell of thickness r located at r from the central axis:

rate of diffusion out of the shell

rate of diffusion into the shell

Solute accumulation

within the shell

=

rrr rLjrLjrcrLt

111 222

rrr rLjrLjrcrLt

111 222

Dividing 2rLr

r 0

11

1rj

rrc

t

r

cDj

11

r

cr

rr

D

t

c

11

0,0,0

)(,,0

0,,0

1

110

1

r

crt

surfaceccRrt

crallt

Boundary conditions

r

cr

rr

D

t

c

11

0,0,0

)(,,0

0,,0

1

110

1

r

crt

surfaceccRrt

crallt

Dimensionless:

)(1

1

1

surfacec

c

0R

r 2

0R

Dt

1

0,0,0

0,1,0

1,,0

all

1

0,0,0

0,1,0

1,,0

all

Assume: )( fg Using the method of “Separation of variables”…

Please refer to my lecture note number 8 for the “applied mathematics”.

10

1

)()(

2

nn

nn

neJJ

1

01

/

00

1

1

20

2

)(21

)( nnn

RtDn

RrJ

eRrJ

surfacec

cn

Diffusion across a thin, moving liquid filmThe concentrations on both sides of this film are fixed by electrochemical reactions, but the film itself is moving steadily.

c10

x

z

Direction of diffusion

c1l

moving liquid film

Assumptions:• the liquid is dilute• the liquid is the only resistance to mass transfer• diffusion in the z direction• convection in the x direction

control volume

Mass balance on a control volume W x z :

xxxxxzzz zWvczWvcxWjxWjzxWct

11111

rate of diffusion in the x direction

rate of diffusion in the z direction

Solute accumulation

in Wxz

=

xxxxxzzz zWvczWvcxWjxWjzxWct

11111

s.s. Neither c1 nor vx change with xDividing Wx zx 0z 0

dz

dj10

dz

dcDj 1

1

21

2

0dz

cdD

lcclz

ccz

11

101

,

,0

B.C. l

zcccc l )( 101101

lccl

Dj 1101

The flow has no effect!

Diffusion into a falling filmA thin liquid film flows slowly and without ripples down a flat surface. One side of this film wets the surface; the other side is in contact with a gas, which is sparingly soluble in the liquid. Find how much gas dissolve in the liquid.

Assumptions:• the liquid is dilute• the contact between gas and liquid is short• diffusion in the z direction• convection in the x direction

Mass balance on a control volume W x z :

xxxxxzzz zWvczWvcxWjxWjzxWct

11111

rate of diffusion in the x direction

rate of diffusion in the z direction

Solute accumulation

in Wxz

=

x

zsolute gas

Liquid with dissolved solute gas

l control volume

xxxxxzzz zWvczWvcxWjxWjzxWct

11111

s.s. Dividing Wx zx 0z 0

xvcxz

j1

10

dz

dcDj 1

1

21

21

/ z

cD

vx

c

x

0,,0

)(,0,0

0,,0

1

11

1

clzx

satcczx

czallx

B.C.xvDx

zerf

satc

c

/41

)(1

1

)(101 satcx

vDj x

z

vx ~ constant

What we have done are:1. We write a mass balance as a differential equation2. Combine this with Fick’s law3. Integrate this to find the desired result

11 cl

Dj

11 ct

Dj

For thin film

For thick slab

Fourier Number

))((

)( 2

timeD

length Much larger than unity …………. Assume a semiinfinite slab

Much less than unity ..…Assume a steady state or an equilibrium

Approximately unity ……..………. Used to estimate the process

Example: Hydrogen has penetrated about 0.1 cm into nickel, D = 10-8 cm2/sec, estimate the operation time of the process.

1))((

)( 2

timeD

length1

)sec)(/10(

)10(28

21

timecm

cm

Approximately 10 days.