Diffusion
051333 Unit operation in Agro-Industry III
Department of Biotechnology, Faculty of Agro-Industry
Kasetsart University
Lecturer: Kittipong Rattanaporn
1
Outline
Š Definition and mechanism
Š Theory of diffusion
Š Molecular diffusion in gases
Š Molecular diffusion in liquid
Š Mass transfer
2
Definition and mechanism
Š Diffusion is the movement of an individual
component through a mixture.
(the influence of a physical stimulus)
Š Driving force is a concentration gradient of
the diffusing component.
Š A concentration gradient tends to move the
component in the direction to equalize
concentrations and destroy the gradient.
3
Š Consider 2 gases B and E in a chamber, initially
separated by a partition. At some instant in time, the
partition is removed, and B and E diffuse in opposite
direction as a result of the concentration gradients.
4
Š Diffusion is the characteristic of many mass transfer
operations.
Š Diffusion can be causes by an activity gradient such
as by a pressure gradient, by a temperature gradient,
or by the application of an external force field.
Š Role of diffusion in mass transfer.
„
„
„
„
„
Distillation
Leaching
Crystallization
Humidification
Membrane separation
5
Theory of Diffusion
Š Assumption
„
„
„
Diffusion occurs in a direction perpendicular to
the interface between the phases and at a definite
location.
Steady state. (The concentrations at any point do
not change with time)
Binary mixtures.
6
Comparison of diffusion and heat
transfer
Diffusion
Driving force Concentration
gradient
Net flux
Heat transfer
(conduction)
Temperature gradient
The rate of change in The rate of temperature
concentration
change.
From the similarity, the equation of heat conduction
can be adapted to problems of diffusion in solids or fluids.
7
Comparison of diffusion and heat
transfer
Š Differences between heat transfer and mass
transfer
„
„
„
Heat transfer is an energy transition but diffusion
is the physical flow of material.
Heat transfer in a given direction is based on one
temperature gradient and the average thermal
conductivity.
Mass transfer, there are different concentration
gradients for each component and often different
diffusivities.
8
Diffusion quantities
Š Velocity, u (length/time)
Š Flux across a plane, N (mol/area time)
Š Flux relative to a plane of zero velocity, J (mol/area
time)
Š Concentration, c and molar density ρM
(mol/volume)
Š Concentration gradient, dc/dx
9
Molal flow rate, velocity, and flux
The total molal flux (mole/time area)
N = ρMu0
(1)
Where ρM is the molar density of the mixture and u0 is
the volumetric average velocity.
For component A and B, the molal fluxes are
(2)
NA = u A c A
(3)
NB= uBcB
10
The molar flux of component A and B
(4)
JA = cAuA – cAu0 = cA(uA – u0)
JB = cBuB – cBu0 = cB(uB – u0)
(5)
The diffusion flux J is assumed to be
proportional to the concentration gradient
and the diffusivity of component (D)
dc A
J A = − D AB
dx
dc B
J B = − D BA
dx
(6)
(7)
11
Š Equation 6 and 7 are statements of Fick’s
first law of diffusion for a binary mixture.
Š The law is based on:
„
„
„
The flux is in moles per unit time.
The diffusion velocity is relative to the volumeaverage velocity.
The driving potential is in the molar
concentration.
12
Š The Fick’s law is similar to Fourier ‘s law
of heat conduction (8) and Newton’s
equation for shear-stress-strain relationship
(9).
q
∂T
= −k
A
∂x
∂u
σ = −µ
∂y
(8)
(9)
13
Diffusivity
Š DAB is mass diffusivity of component A on
component B through the mixture.
Š The dimensions of mass diffusivity are
length squared divided by time, usually given
as m2/s or cm2/s
14
Š The magnitude of mass diffusivities for liquids or
gases in solid are less than the mass diffusivities for
gases in liquids.
Š In solids, the mass diffusivities range from 10-9 to
10-1 cm2/s, in liquids, the mass diffusivities range
from 10-6 to 10-5 cm2/s and for gases, the mass
diffusivities range from 5x10-1 to 10-1 cm2/s
Š The mass diffusivity magnitudes are a function of
temperature and concentrations; in the case of gases
the mass diffusivity is substantially influenced by
pressure.
15
Molecular diffusion in gas
16
Š Using Ideal gas law
p B = ρ B RB T
(10)
Š the gas constant RB for gas B can be written in terms of the
Universal gas constant Ru
Ru
RB =
MB
(11)
Š Ru is the universal gas constant 8.314 (m3 Pa)/(g-mol K) and
MB is molecular weight of gas B.
Š Thus,
p
(12)
ρ = B
B
RBT
or
pB M B
ρB =
RuT
(13)
17
Š Since ρB is mass concentration, substitute equation 13
in equation 6
∂  pB M B 

J B = − D BE 
∂x  Ru T 
D BE M B ∂p B
JB =−
Ru T ∂x
(14)
(15)
Š The mass diffusivity DBE refers to diffusivity of gas B in gas
E. Similarly, we can express diffusion of gas E in gas B
D EB M E ∂p E
JE =−
Ru T ∂x
(16)
18
Š Steady-State Diffusion of gases
Assume the mass diffusivity does not depend on concentration,
from equation 6, we obtain
dc A
(17)
J A = − D AB
dx
By separating variables and integrating:
x2
cA2
x1
c A1
J A ∫ dx = − D AB ∫ dc A
D AB ( c A1 − c A 2 )
JA =
( x 2 − x1 )
(18)
(19)
19
Š For the condition of steady state diffusion, the
concentration at the boundaries must be constant
with time and diffusion is limited to molecular
motion with in the solid being described.
Š The mass diffusivities are not influenced by
magnitude of concentration and no temperature
gradient.
20
Example 1 Molecular diffusion of Helium
in Nitrogen
Š A mixture of He and H2 gas is contained in a pipe
at 298 K and 1 atm total pressure which is constant
throughout. At one end of the pipe at point 1 the
partial pressure pA1 of He is 0.60 atm and at the
other end 0.2 m pA2 of is 0.20 atm. Calculate the
flux of He at steady state if DAB of He-N2 mixture
is 0.687 x 10-4 m2/s
21
Equimolar Counter diffusion in Gases
Š Molecules of A diffuse to the right and B to
the left.
Š The total pressure is constant, the net moles
of A diffusing to the right must equal the net
moles of B to the left.
(20)
JAz = -JBz
22
Š Fick’s law for B for constant c,
dc B
J B = − D BA
dz
Š Now P = pA + pB = constant then
c = c A + cB
Š Differentiating both sides
dcA = - dcB
(21)
(22)
(23)
23
Š Equating Fick’s law to (21)
dc B
dc A
J A = − D AB
= − J B = −( − D BA )
dz
dz
(24)
Š Substitute (23) into (24) and canceling like term
(25)
DAB = DBA
Š For a binary mixture of A and B the diffusivity
coeff. DAB for A diffusing in B in the same as DBA
for B diffusing into A.
24
Example 2 Equimolar Counterdiffusion
Š Ammonia gas (A) is diffusing through a uniform
tube 0.10 m long containing N2 gas (B) at 1.0132 x
105 Pa and 298 K. At one end of the pipe at point 1
the partial pressure pA1 is 1.013 x 104 Pa and at the
other end pA2 of is 0.507 x 104 Pa. The diffusivity
DAB = 0.230 x 10-4 m2/s. Calculate the flux JA and
JB at steady state.
25
General case for Gases A and B
Plus Convection
Š When the whole fluid is moving in bulk or
convective flow to the right.
Š The molar average velocity of the whole fluid
relative to a stationary point is uM (m/s)
Š Component A is still diffusing to the right, its
diffusion velocity uAd is measured relative to the
moving fluid.
Š The velocity of A relative to the stationary point is
the sum of the diffusion velocity and the average or
convective velocity
26
uA
uAd
uM
uA = uAd + uM
Mutiply by cA
cAuA = cA uAd + cAuM
(26)
(27)
Rewrite (27) in terms of flux
NA = JA + cAuA
(28)
27
Let N is the total convective flux of the whole stream
relative to the stationary point
N = cuM = NA + NB
NA + NB
uM =
c
Substitute (30) to (28)
cA
N A = J A + (N A + N B )
c
(29)
(30)
(31)
28
When JA is Fick’s law.
dc A c A
N A = − D AB + (N A + N B )
dz c
(32)
Š Equation 32 is the final generation for diffusion
plus convection .
Š For equimolar counter diffusion , NA = NB and the
convective term in (32) become zero.
Š Then NA = JA = -NB = -JB
29
Special case for A diffusing
through stagnant, Nondiffusing B
Š In this case one boundary at the end of the
diffusion path is impermeable to component
B, so it cannot pass through.
Š For example, the evaporation of organic
solvent into air, Gas adsorption into liquid.
30
Š To derive the case for A diffusing in stagnant,
nondiffusing B, NB = 0 is substitute into the general
equation (32)
dc A c A
N A = − D AB + N A
dz c
(33)
Š Keeping the total pressure P constant,
D AB dp A p A
NA =−
+ NA
RT dz P
(34)
31
Š Rearranging and integrating,
pA 
D AB dp A

N A 1 −  = −
RT dz
 P
(34)
D AB pA 2 dp A
N A ∫ dz = − ∫
z1
RT pA1  1 − p A 


 P
(35)
D AB P
P − pA2
NA =
ln
RT (z 2 − z1 ) P − p A1
(36)
z2
32
Š Eq. 36 can write in a form of an inert B log mean
value.
p B 2 − p B1
p A1 − p A 2
(37)
PBM =
=
ln( p B 2 / p B1 ) ln( P − p A 2 / P − p A1 )
Š Substitute 37 into 36
D AB P
( p A1 − p A 2 )
NA =
RT (z 2 − z1 )PBM
(38)
33
Example 3 Diffusion of water through
stagnant, Nondiffusing Air
Š Water in the bottom of a narrow tube is held at a
constant temperature of 293 K. The total pressure of
air is 1.01325x105 Pa (assumed dry air) and the
temperature is 293 K. Water evaporates and
diffuses through the air in tube and diffusion path is
0.1524 m long.
„
„
„
Calculate the rate of evaporation at steady state
The diffusivity of water vapor at 293 K and 1 atm is
0.25 x 10-4 m2/s
The vapor pressure of water at 293 K is 0.0231 atm
34
Molecular diffusion in liquids
35
Š Diffusion of solutes in liquids is very important in
many industrial process, especially in separation
process.
Š The rate of molecular diffusion in liquids is slower
than in gases because the molecules are very close
together compared to gas.
Š In generally, the diffusion coefficient in a gas will
be greater than in a liquid but the flux in a gas is not
that much greater because the concentrations in
liquids are considerably higher than in gases.
36
Š In diffusion in liquids an important
difference from diffusion from gases is that
the diffusivities are often quite dependent on
the concentration of the diffusing component
37
Calculation of diffusion in liquids
Š Equimolar counterdiffusion
N A = −N B
(39)
D AB (c A1 − c A 2 ) D AB c av (x A1 − x A 2 ) (40)
=
NA =
z 2 − z1
z 2 − z1
ρ   ρ1 ρ 2 

c av =   =  +  2
 M  av  M 1 M 2 
(41)
38
Š Diffusion of A through nondiffusing B
„
An example is a dilute solution of propionic acid
in a water solution being contacted with toluene
D AB c av
( x A1 − x A 2 )
NA =
(z 2 − z1 )x BM
(42)
39
Š Where
x B 2 − x B1
x BM =
ln( x B 2 / x B1 )
(43)
For dilute solution xBM is close to 1.0 and c is
constant
D AB (x A1 − x A 2 )
(44)
NA =
( z 2 − z1 )
40
Example4 : Diffusion of Ethanol
through Water
An ethanol-water solution in the form of stagnant film
2.0 mm thick at 293 K is in contact at one surface
with an organic solvent in which ethanol is soluble
and water is insoluble. Hence, NB= 0. At point 1 the
concentration of ethanol is 16.8 wt% and the solution
density is 978.2 kg/m3. The diffusivity of ethanol is
0.740 x 10-9 m2/s.
Calculate the steady state flux NA.
41
Molecular diffusion in solid
42
Š Rate of diffusion in solids is generally slower
than rates in liquids and gases.
Š For example
„
„
„
„
Leaching of food
drying of foods
Diffusion and catalytic reaction in solid catalyst
Separation of fluids by membrane
43
Š We can classify transport in solids into two
types of diffusion
„
„
Diffusion in solids following Fick’s law.
Diffusion in porous solids that depends on
structure.
44
Diffusion in solids following Fick’s law
Š This type of diffusion does not depend on the
actual structure of solid,
Š The diffusion occurs when the the fluid or
solute diffusing is actually dissolve in the
solid to form a more or less homogenous
solution.
45
Š Using the general equation for binary diffusion
(Eq. 32)
dc A c A
N A = − D AB + (N A + N B )
dz c
Š The bulk flow term is usually small, so it is
neglected.
dc A
N A = − D AB
dz
(45)
46
Š Integration of equation 45 for a solid slab at steady
state
(
c A1 − c A 2 )
(46)
N =D
A
AB
( z 2 − z1 )
Š For the case of diffusion radially through a cylinder
wall inner radius r1 and outer r2 and length L,
NA
dc A
= − D AB
2πrL
dr
(47)
47
2πL
N A = D AB (c A1 − c A 2 )
ln r2 r1
( )
Š
Š
Š
(48)
The diffusion coefficient is not dependent upon the
pressure of gas or liquid on the outside of the solid.
The solubility of gas or liquid is directly
proportional to pA by Henry ‘s law.
For gas,
Sp A kgmolA
(49)
c =
A
3
22.4 m solid
48
Diffusion in porous solid that
depends on structure
Š The porous solid that have pores or interconnected
voids in the solid would effect the diffusion.
Š For the situation where the voids are filled with
liquid or gas, the concentration of solute at
boundary is diffusing through the solvent in the
void volume takes a tortuous path which is greater
than z2 – z1 by a factor, τ, called tortuous.
49
Š For a dilute solution,
εD AB (c A1 − c A 2 )
NA =
τ ( z 2 − z1 )
(50)
Š Where ε is the open void fraction
τ is the correction factor of the path longer
than z.
50
Š Combined into an effective diffusivity
D Aeff
ε
= D AB
τ
(51)
Š For diffusion of gases in porous solids
εD AB ( p A1 − p A 2 )
NA =
τRT (z 2 − z1 )
(52)
51
Mass transfer
52
Š Mass transfer is the phrase commonly used
in engineering for physical processes which
involve molecular and convective transport
of atom and molecule within physical
systems.
ref: http://en.wikipedia.org/wiki/Mass_transfer
53
Š There are many difference of mass transfer
but the most cases of mass transfer is treating
using the same type of equations, which
feature a mass transfer coefficient, km.
Š The mass transfer coefficient km is defined as
the rate of mass transfer per unit area per unit
concentration difference (m3/m2 s, m/s).
54
•
JA
mB
km =
=
(c B1 − c B 2 ) A(c B1 − c B 2 )
(53)
The coefficient represents the volume (m3) of
component B transported across a boundary of one
square meter per second
55
For the relationship of the ideal gas law, the mass
transport due to convection become
k AM B
( pB1 − pB 2 )
mB = m
RuTA
•
(54)
When the specific application of mass transport is
water vapor in air, the equation 54 is used for
computing the convective transport of water vapor in
air.
•
k m AM B p
(W1 − W2 )
mB =
(55)
0.622 RuTA
56
The Dimensional analysis for predicting
the mass transfer coefficient.
The variables are grouped in the dimensionless number.
N Sh
km dc
=
DAB
µ
N Sc =
ρDAB
ρud c
µ
k
=
ρc p DAB
N Re =
N Le
(56)
(57)
(58)
(59)
57
Š Consider a fluid flow over a flat plate. for the
boundary layer from the leading edge of the plate, we
can write the equation for concentration.
∂ 2c A
∂c A
∂c A
= DAB
+ uy
ux
∂y 2
∂y
∂x
(60)
58
Š From the equation 60, cA represents concentration of
component A at location within the boundary layer.
ì ìcp
=
Pr antl number = N Pr =
ρá k
(61)
The Prandtl number provides the link between velocity
and temperature profile.
µ
If
(62)
=1
ρD AB
then velocity and concentration profile have the same
shape.
µ
Schmidt number = N Sc =
(63)
ρD
AB
59
Š The concentration and temperature profile will have
the same shape if
α
=1
(64)
DAB
The ratio
α
Lewis number = N Le =
(65)
DAB
The functional relationship that correlated these
dimensional number for forced convection are :
N sh = f (N Re , N Sc )
N Sh =
N Sc =
(66)
total mass transferred
total mass transferred by molecular diffusion
(67)
molecular diffusion of momentum
molecular diffusion of mass
(68)
60
Š These correlations are based on the assumption:
„ constant physical properties
„ no chemical reaction in the fluid
„ small bulk flow at the interface
„ no viscous dissipation
„ no interchange of radiant energy
„ no pressure, thermal, or forced diffusion.
61
Laminar Flow Past a Flat Plate
Š When Nre < 5x105
N sh , x =
k m, x x
DAB
1/ 2
1/ 3
= 0.322 N Re
L N Sc
N Sc ≥ 0.6
(69)
Km,x is the mass transfer coefficient at a fixed location.
The dc used in the Sherwood and Reynolds number is the
distance from the leading edge of the plate.
N sh , L
=
k m, L x
DAB
1/ 2
1/ 3
= 0.664 N Re
N
L Sc
N Sc ≥ 0.6
(70)
when flow is laminar over the entire length of the plate. The
dc is the total length of the plate.The mass transfer
coefficient is the average value for the entire plate .
62
Turbulent Flow Past a Flat Plate
Š When Nre > 5x105
N sh , x =
km, x x
DAB
4/5
1/ 3
= 0.0296 N Re,
x N Sc
0.6 < N Sc < 3000
(71)
The characteristic dimension (dc) used in the
Sherwood and Reynolds number is the distance
from the leading edge of the plate.
N sh , L =
km,L x
DAB
0.8
N Sc0.33
= 0.036 N Re
(72)
. The characteristic dimension (dc) is the total
length of the plate.
63
Laminar Flow in a Pipe
1/ 3
N Sh ,d
N N 
k d
= m c = 1.86 Re,d Sc 
DAB
 L / dc 
N Re < 10,000
(73)
Turbulent Flow in a Pipe
Where the characteristic dimension, dc, is the
diameter of the pipe.
N Sh ,d =
km dc
0.8
1/3
= 0.023 N Re,
d N Sc
DAB
N Re > 10,000
(74)
Where the characteristic dimension, dc, is the
diameter of the pipe.
64
Mass Transfer for Flow over
Spherical objects
1/ 2
2/3
0.4
N Sh ,d = 2.0 + (0.4 N Re,
0
.
06
N
)N
+
d
Re, d
Sc
(75)
For mass transfer from a freely falling liquid
droplet
1/ 2
1/3
(76)
N Sh ,d = 2.0 + 0.6 N Re,
)N
d
Sc
65
The end.
66