This document discusses various types of molecular diffusion and flux. It begins by defining Fick's law of diffusion and describing how molecular flux occurs from regions of high concentration to low concentration. It then provides several examples of evaluating molar flux in different systems and conditions, including: equimolar countercurrent diffusion, dilute concentrations, diffusion through a stagnant gas, axial diffusion neglected with forced convection, and Knudsen diffusion. The document also discusses relationships between mass, molar, and volumetric fluxes and how fluxes change based on the reference frame. It concludes by introducing macroscopic and microscopic approaches to formulating mass transfer models.

3. drying
Industrial
distillation
Evaporation
Absorption

5. In the absence of other
gradients such as
-Temperature
-Electrical
-Gravitational potentials
Molecules of a given species
( e.g. A) within a single phase
will always diffuse from region
oh high to low concentrations.

6. This will result in a Molar Flux of
species A (NA) in the direction of
decreasing concentration.
Flux is the flow rate FA per unit
area normal to the direction
of flow
(NA) is a Vector quantity ( )
moles
2 .s
m
moles
2 .s
m
A
NA
FA
=

7. NH3
1 2
NH3
Steady state fluxes
Determine NA for ammonia and mass flux for
ammonia nA through stationary frame at steady
state? There are 12.4x10 molecules of NH3 that
pass through an area A (2m ) over time interval 10 s
23
2
High
Concentration
Low
Concentration

8. Number of NH3 molecules
Avogadro's No.
No. of Moles =
12.04 x 10
23
6.02 x 1023
No. of Moles = =2 Mol
Two moles pass through an area of 2m in 10 s. therefore
the No. of moles passing through 1 m per second ( Molar
flux NA)
2
2
2 mol
(2 m ) ( 10
s)
NA= = 0.1 mol/m .s
2
2
The mass Flux = nA = NA x MWA
nA = 0.1 mol/m .s x 17g/mol
nA = 1.7 g/m .s
2
2

9. The Molar Flux NA is the result of two
contributions:
1)The molecular diffusion flux
produced by a concentration
gradient JA
2) Flux resulting from the bulk motion of
the fluid BA
NA JA BA
+
=

10. What is the difference between
the two contributions?
Gas mixture A &B
Well mixed
(no concentration
gradient)
A
B
A
B
B
B
A
B
A
A
A
B
A
B
B
B
A
B
A
A
A
B
A
B
B
B
A
B
A
A
UA
Velocity UA
The flux of A
resulting from Bulk
motion
BA = XA (NA+ NB)
BA = XA (NT)
The flux of A is
only
JA = 0
BA
Well mixed
(no concentration
gradient)

11. The Bulk flow term can be expressed in
terms of concnteration CA and average
molar velocity U
M
U
M
CA
=
BA
U
M
=xi Ui
Xi = Ci
C
Particles velocity
Mole fraction

12. B
A
A
A
B
A
B
B
B
A
A
A A
A
A
UA
A
A
A
A
A
B B
B
B
B
JA
Let see if the mixture is not spatially uniform
( Concentration gradient exist) and moving with
molar average velocity U
M
In this case there will
be both diffusion flux
of A,(JA) relative to the
motion of the mixture
+ Bulk flow of A (BA)
Thus total molar Flux of
A is given by
NA JA BA
+
=
can be expressed either in
terms of concentration of A
BA
NA JA CA U
+
=
M
XA (NA+ NB)
NA JA+
=
M
Or in terms of mole fraction

13. Remember :
The Flux of A Relative to Fixed Frame ( e.g. Your
Desk) Is the product of concentration of A (CA) and
the particle velocity ( UA)
NA = CA X UA 1
The Flux JA is the flux of A Relative to the molar average
velocity U
The molar average velocity for a binary system is
M
U
M
=xi Ui U
M
= xAUA + xB
UB

14. (UA –U)
M
What this difference tells us?
It tells us the velocity at which
species A is moving relative to
average velocity of all the species

15. Flux MJA Relative to U
M
MJA = CA ( UA –U )
M
MJA = CAUA – CA U
M

NA
CA ( )
xAUA + xB UB

Factoring the reciprocal of the total concentration (1/C )
from the terms in parentheses m we have
C
(CA UA + CA UB )
CA
CA U
M
= xA
= NA + NB
( )

16. MJA = CAUA – CA U
M
NA
CA ( )
xAUA + xB UB


xA (NA + NB)
NA= MJA + xA ( NA + NB )
From all these equations we
get
NA=-C DAB  xA + xA ( NA + NB )

17. Evaluating the Molar Flux
1- Equimolar Counter Current
Diffusion
EMCD A
B
A B
A B
Fluxes of A & B are Equal
in magnitute and flow
counter to each other
Mathematically
NA = - NB A diffuse from
the Bulk to
Catalyst Surface
where it
isomerizes to
form B
A diffuse from
the Bulk to
Catalyst Surface
where it
isomerizes to
form B

18. NA=-C DAB  xA + xA ( NA + NB )
Mathematically
NA = - NB
substitute into general equation
NA=-C DAB  xA + xA ( NA + {-NA} )
0
NA=-C DAB  xA =MJA
EMCD Equation
For constant total concentration
NA=-DAB  CA =MJA

19. Evaluating the Molar Flux
1-Dilute Concentrations
Such case occur when : The mole
fraction of the diffusing Solute & the
Bulk motion in the direction of diffusion
are small.
The Flux is similar to constant total
concentration
NA=-DAB  CA =MJA

20. NA=-C DAB  xA + xA ( NA + {-NA} )
0
The Equation above can also be reduced to that
similar to Constant total concentration and
dilute concentration
For porous catalyst systems when the pore
radii are very small. Diffusion under these
condition is known as Knudsen diffusion (Dk)
and occur when the mean free path
 >> diamter of the catalyst . Here the reacting
molecules collide more often with pore wall
than with other. In this case we neglect the
Bulk terms and the flux of species A for
Knudsen diffusion is expressed as follows:
NA=-Dk CA =MJA

21. This type is found in systems where
two phases are present.
Evaporation and gas absorption are
typical process
Evaluating the Molar Flux
Diffusion through a Stagnant Gas
If there is a stagnant gas, there will be
no net flux of B w.r.t. a fixed coordinate
i.e.
NB = 0

22. NA=-C DAB  xA + xA ( NA + NB )
NB = 0
Substituting
We get
NA=-C DAB  xA + xA NA
Rearranging
yields
NA= C DAB  xA
1 - xA
-1

23. NA= C DAB  xA
1 - xA
-1
NA=-C DAB  xA + xA NA
Take this term to LHS
NA- xA NA = -C DAB  xA
NA (1- xA) = -C DAB  xA
This will lead us to this equation
NA= C DAB ln ( ) = C DAB ln
1 - xA xB

24. JAz
JAx
Z
x
Evaluating the Molar Flux
Axial Diffusion effect is nglected
and flux result from forced
convection
When the flux of A results from forced convection. We
assume that the diffusion in the direction of flow (
which in this case z direction), Jaz , which is very
small in comparison with the Bulk flow.

25. Evaluating the Molar Flux
Axial Diffusion effect is nglected
and flux result from forced
convection
When the flux of A results from forced convection. We
assume that the diffusion in the direction of flow (
which in this case z direction), Jaz , which is very
small in comparison with the Bulk flow.
One contribution in the z direction ,
BAz (Uz  /Ac {m /m .s}) where  volumetric flow
rate in the direction of flow and Ac is the cross
sectional area. JAz is neglected, while JAx normal
to the direction of the flow, may not be nglected.
NA = Baz = CA Uz  CA /Ac
We can express the molar flow rate FA =  CA when not
accounting for diffusional effects, However when accounting
for diffusional effects The molar flow rate FAz is = NA Ac
3 2

26. FA = NAz Ac = - CDAB + xA (Nz) Ac
dxA
dz


Where Nz is the total molar average flux in z
direction of all n species , i.e
Nz =  Niz
i=1
m

27. Mass Flux
Molar Flux
Reference system
nA = A UA
NA = CA UA
Fixed Coordinates
jA = A(UA- U )
JA = CA (UA- U )
Molar average Velocity
jA = A (UA- U )
JA = CA (UA- U )
Mass average Velocity
jA = A (UA- U )
vJA = CA (UA- U )
Volume average Velocity
Fluxes for Binary Systems
M
M
M
M
m
m m
m
v v
v

28. Relationships between Fluxes
Relationship between
mass and molar fluxes
nA = NA MA MjA = mjA
M
MA
MjA = MJA
MA
mJA = MJA
MB
M

29. Relationships between Fluxes
Molar flux Relative to U
M
NA +NB = C U
M
MJA + MJB = 0
NA = MJA + xA ( NA +NB
NA = MJA + CA U
M

30. Relationships between Fluxes
Mass flux Relative to U
m
nA +nB =  U
m
mjA + mjB = 0
nA = mjA + A ( nA +nB)
nA = mjA +  A U
m

31. Changing from one reference to
another Hooyman et al (1953) derived
a general diffusion coefficient
expressed as
JA =-DAB xA
1-A
1-xA
1
V
Molar
volume of
the mixture
Weighing
factor relating
to the
appropriate
velocity

32. A
Reference velocity
Fick`s Law

A
mJA
xA
MJA
CA VA
vJA
U
m
U
M
U
v
Weighing factors Values

33. Formulation of Mass Transfer Models
Macroscopic models
Describe industrial Process
but do not provide a detailed
description of the process.
Because the properties of the
system are averaged over
position.
The only independent
variable is time , so its easy
to solve the derived
differential equation of
macroscopic models

34. Macroscopic material balance
Inlet (i)
Applying the law of conservation of matter to a
stream flowing into a volume element fixed in
space
Outlet (i)
Volume
element
General Material Balance in the
RATE FORM
Rate of
Accumulation
Rate of
Transport
into
Rate of
Transport
Out
Rate of
generation
Rate of
Transport
Out

35. General Material Balance for total flow of mass
V V U S U S
( )
Mass of
material
within the
system
Average
velocity of
the fluid
Dividing both sides by and shrinking the
time to zero we get
dt
d
(  V ) ( U S) ( U S) ( U S)

36. Note:
1- Since total mass can not be neither created
nor destroyed both the generation and
consumption terms are not present. (no
chemical reaction).
2- the mixing within the volume element is
perfectly mixed concentration leaving or
entering the volume is the same.
3- The subscript on the exit term (Outlet) will
be dropped from the equation because the
exiting stream is a variable.
4- The S in and out are the same.
THE MASS BALANCE BECOMS
dt
d
S
1
(  V ) ( U) ( U) Absence of
Coordinates

37. General Microscopic Material
Balance

38. Derivation of continuity equation
Eulerian approach
In this approach the mass of component
A moves at a velocity, UA through a
fixed volume element
And Can be extended to cases where the
volume element moves with some
reference velocity ( UM, Um, U )
M m v

39. Start:
considering a differential element of fixed shape
 Flow entering and leaving the element at all direction.
 Substitute mass flux terms for species A into the
general balance equation
Rate of
Accumulation
Rate of
Transport
into
Rate of
Transport
Out
Rate of
generation
Rate of
Transport
Out

40. (nAx - nAx ) Y Z
x X +x
(nA - nAx ) X Z
Y Y +Y
(nA - nAx ) X Y
Y Y +Y
rA X Y Z
*
X Y Z A


t
