高等輸送二 — 質傳 lecture 1 diffusion in dilute solutions 郭修伯 助理教授
TRANSCRIPT
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.