1
Chapter 9 MASS TRANSFER
9.1 Introduction
2
9.1.2
Convective Mass Transfer:
This phenomenon occurs at a macroscopic level where bulk of mass transfer takes place due to difference in
We have been dealing with Heat Transfer that occurs whenever there is a temperature difference in a medium. But,
if there exists a difference in the concentration or density of some chemical species in a mixture, there is a natural
tendency for mass to be transferred, minimizing the concentration difference within the mixture. Therefore, mass
transfer is defined as mass in transient as a result of species concentration difference in a mixture.
The phenomenon of heat and mass transfer are quite analogous to each other and may be differentiated as follows:
S. No.
1.
2.
3.
4.
Heat Transfer
Temperature gradient constitutes the
driving force
Rate depends on the driving force
and thermal resistance.
Takes place from higher temperature
to lower temperature.
Ceases when temperature gradient or
temperature difference reduces to zero.
Mass Transfer
Concentration gradient in a mixture
provides the driving force
Rate depends on the driving force
and diffusion resistance.
From higher concentration to lower
concentration.
Ceases when concentration gradient or
concentration difference reduces to zero.
concentration. Mass transfer requires the presence of two regions at different chemical compositions in which there
is movement of species from higher concentration region to lower concentration region. Convective mass transfer is
analogous to convective heat transfer. However, the bulk fluid motion from one location to the other in case of fluid
flow and convective heat transfer, are due to pressure difference and temperature difference, respectively. Due to
concentration difference, a thin film of concentration boundary layer gets formed, like that of the thermal boundary
layer, adjacent to a surface where concentration gradient exists. As in convective heat transfer, the flow of fluid
during mass transfer may be laminar or turbulent, over a flat surface (external flow) or inside pipes or ducts (internal
flow), free or forced, single or multiphase.
9.2 Fick’s law of diffusion
To explain the mass diffusion process, let us consider the system, shown in Figs. 9.1(a) and 9.1(b). Here a thin
partition separates the two gases A and B. When the partition is removed, the two gases diffuse into each other, until
Mass transfer can be broadly classified as:
equilibrium is established and concentration of the gases become uniform throughout the box. The diffusion rate of
9.1.1
mass from the region at high concentration to the region at low concentration is given by Fick’s law which states
Diffusion Mass transfer:
This takes place at a microscopic level as a result of diffusion from regions of high concentration to regions of low
that the mass flux of a constituent, per unit area, is proportional to the concentration gradient. Mathematically it is
concentration. This type is governed by Fick’s law of Mass diffusion and is analogous to Fourier’s law of Heat
expressed as:
.
Conduction.
Mass diffusion due to concentration gradient is known as ordinary diffusion. Mass diffusion occurring due
to temperature gradient in the system is called thermal diffusion. Similarly pressure gradient in the system may
Thus,
C
md,A
 DAB A
A
x
(9.1)
An expression similar to equation (9.1) can also be written for the diffusion of constituent B,
.
C
m d,B
 D BA B
A
x
result in pressure diffusion. However, the thermal and pressure diffusions are usually negligible. But, in centrifuges,
the pressure gradient generated by the centrifugal effect is used to separate liquid solutions and gaseous isotopes. If
(9.2)
.
from the mixture, it is called forced diffusion. In case, the voids (pores) of a porous material like silica gel are
Gas A
m d,A
.
Gas B
m d,B
smaller than the mean free path of the gas molecules, the molecular collisions may be negligible and the free
molecule flow may be initiated, then it is known as Knudsen diffusion. And if the size of the gas molecules is close
to the void-size, then the absorbed molecules move along the walls of the void. This is known as surface diffusion.
The particles whose diameter is under 0.1 µm, as of mist or soot, act like large molecules then their diffusion due to
Concentration
electric or magnetic forced field is applied externally to separate the electrically charged or magnetized molecules
Imaginary plane
CA
 .

m
d,A


 A 


CB
 . 
 m d,B 
 A 


concentration gradient is known as Brownian motion. However, particles whose diameter is greater than 1 µm, are
Partition of thin membrane
not affected by diffusion as their motion is governed by Newton’s law. Mass diffusion may also take place in
Fig. 9.1(a) Diffusion of gas A into gas B
turbulent flow systems where accelerated diffusion rates will occur as a result of the rapid eddy mixing process. This
x
Fig. 9.1(b) Concentration profile and
flow direction of gases A and B
is quite similar to the mixing process that increases the heat transfer and viscous action in the turbulent flow.
Similarly, concentration gradient can give rise to a temperature gradient and a consequent heat transfer. These two
effects are termed as coupled phenomena and may be treated by the methods of irreversible thermodynamics.
Where, DAB and DBA are the proportionality constants known as diffusion coefficients from A towards B, and B
towards A respectively in m2/s. This is the measure of how fast a species diffuses into the medium.
.
.
m d,A and m d,B = mass fluxes per unit time in kg/s
3
4
CA and CB = mass concentration of component A and B per unit volume in kg/m3. The concentration gradient
The net heat flux at any isotherm will be,
C x is negative since the concentration decreases in the flow direction.
Referring to the Figs. 9.1(a) and 9.1(b), let us understand the physical mechanism of diffusion of gas A into
the gas B through the imaginary plane shown by the dashed line. The concentration of component A is greater on the
left side of the imaginary plane than on the right side. The higher concentration means that there are more molecules
per unit volume in that region. If the system is a gas or a liquid, the molecules move about in random fashion, and
higher the concentration, the more molecules will cross a given plane per unit time. Thus, on the average, more
molecules are moving from left to right across the plane than in the opposite direction. This results in a net mass
transfer from the region of high concentration to the region of low concentration. Also, the fact that the molecules
collide with each other, influences the diffusion process strongly. Thus in a mixture of gases, there is a decided
difference between the collision of like molecules and the collision of unlike molecules. That is, the collision
q q
q
q



A  A  x  A  y  A  z
(9.3)
Where q x , q y , q z are given by Fourier’s law of heat conduction as:
q x  kA x
T
T
T
, q y   kA y
and q z   kA z
y
x
Z
And the net mass flux at any constant concentration line is,
.
 .
md,A  md,A

 A
A


.
  .
   md,A
  A
x 

  .
   md,A
  A
y 

.



z
(9.4)


.

C
m d,A 
C A ;  md,A 
m
C
Where,  d,A    D
 D A
 D A ; 
 A 

x
between two like molecules will not appreciably alter the basic molecular movement, because the two molecules are
x
 A 

y
y
 A 

z
Z
identical and it does not make any difference whether one or the other of the two molecules crosses a certain plane.
It can be seen that Fourier’s law of heat conduction given by equation (9.3) is quite similar to Fick’s law of Mass
The collision of two unlike molecules, say molecules A and B, might result in molecule B crossing some particular
Diffusion.
plane instead of molecule A. This can be understood by considering two teams playing football. If one or two
Diffusion Mass transfer is also analogous to Newton’s law of viscosity and Ohm’s law of electrical conductance.
players of team A take the ball, the ball will go towards the opposite team B, that is, only in one direction. But, if
Newton’s law of viscosity for shear stress between fluid layers is,
the ball is stopped or kicked by a player of the team B, the ball may go in any direction, not essentially towards the

team A. The molecules would, in general, have different masses; thus the mass transfer would be influenced by the
u
y
(9.5)
collision. Using the kinetic theory of gases it is possible to predict analytically the diffusion rates for some systems
The transport of energy is given by the heat conduction equation, while the viscous shear equation describes the
by taking into account the collision mechanism and molecular weights of the constituent gases. In gases the
transport of momentum across fluid layers, and the diffusion law gives the transport of mass.
diffusion rates are clearly dependent on the molecular speed, and consequently on temperature since the temperature
9.2.2 Fick’s law for perfect gases (low pressure gases)
indicate the average molecular speed.
The diffusion process described through the Figs. 9.1(a) and 9.1(b), is basically occurring in two ways. While the
9.2.1 Analogy between conduction heat transfer and diffusion mass transfer
gas A is diffusing into gas B, at the same time the gas B is diffusing in gas A. In using Fick’s law, one may consider
It is known that the diffusion mass transfer is analogous to Fourier’s law of heat conduction. Let us examine their
mass flux per unit area and mass concentration as it is in the equations (9.1) and (9.2) or these equations may be
similarity through the Figs. (9.2 and 9.3). Figure (9.2) shows heat transfer rates on a constant temperature line which
expressed in terms of molar concentration and fluxes. There is no general rule to say which type of expression will
are quite similar to the mass transfer rates on a constant concentration line, shown in the Fig. (9.3).
be most convenient. The problem under consideration will itself determine which one should be used. For gases,
q
A
 y
 .

 m d,A 
 A 

y
e)
lin
Fig. 9.2 Heat flux on an Isotherm
e
nt in
ta n l
ns tio
Co ntra
e
nc
q
 A 
x
Co
nt
e
rm tur
he ra
ot pe
Is Tem
sta
on
(C
q
A
 z
Fick’s law may be expressed conveniently in terms of partial pressures by making use of the perfect gas equation of
 .

 m d,A 
 A 

z
 .

 m d,A 
 A 

x
state. However, they will be valid only for gases at low pressures or at a state where the perfect gas equation of state
applies.
The perfect gas equation is generally expressed as: p  RT or pV  mRT 
2
mR o T
 NR o T
M
(9.6)
3
Where, p is pressure in N/m , ρ is density in kg/m , R is Gas constant in J/kgK, Ro is Universal Gas constant (=8315
J/kmol K), M is molecular weight in kg/kmol, T is temperature in Kelvin, V is volume in m3, m is mass in kg, N
Fig. 9.3 Mass flux on a constant
concentration line
=m/M is number of moles in kmol. The gas constant R for a particular gas may be expressed in terms of the
universal gas constant Ro and the molecular weight of the gas as: R 
Ro
.
M
5
6
Here, density (ρ) represents the mass concentration used in Fick’s law. In case, the concentration (C) is used on the
MA and MB : Molecular weights of constituents A and B.
molar basis, then its unit will be kmol/m3. In such a case, the density will be ρ=CM, where M is the molecular
Equation (9.10) offers a convenient expression for calculating the diffusion coefficient for various compounds and
weight.
mixtures. However, the experimental values of the diffusion coefficient if available should be preferred. Molecular
Since Fick’s law uses concentration equivalent to the density, having the unit kg/ m3, for convenience C= ρ has been
volume and molecular Weight of some constituents are listed in Table 9.1.
used throughout the text. And, if necessary, the concentration given in kmol/m3 may be converted to kg/m3 by
multiplying with the molecular weight of the constituent, being used.
Using,

Table 9.1: Molecular volume and molecular Weight of some constituents
Gas
p ,
p
p M
p
p M
C A   A  A  A A (for species A) and C B  B  B  B B (for species B)
R AT
R oT
RT
R BT
R oT
(9.7)
Consequently, Fick’s law of diffusion (equation 9.1) for component A into component B, if isothermal diffusion is
considered, can be written as:
Air
Oxygen
Hydrogen
Sulphur
Phosphorus
Iodine
Molecular
Volume
29.9
7.4
14.3
25.6
27
37
Molecular
Weight
28.9
32.0
2.0
256.5
123.89
253.81
Gas
Nitrogen
Carbon dioxide
Water
Ammonia
Fluorine
Bromine
Molecular
Volume
15.6
34
18.8
25.8
8.7
27.0
Molecular
Weight
28
44.0
18.0
17.0
37.99
159.81
.
C
M dp A
m d,A
  D AB A   D AB A
A
x
R o T dx
(9.8)
Diffusion in liquids and solids
Fick’s law of diffusion can be used for problems involving liquid and solid diffusion. There may be diffusion of
Similarly for the diffusion of component B into component A under isothermal condition equation (5.2) becomes:
liquid into gas or air. Drying of clothes, timber, soil are examples where water diffuses into the air. Diffusion of
.
perfume, petrol, diesel, etc. are other examples of liquid to gas diffusion. Disappearing of dry ice placed in the
M dp B
m d,B
  D BA B
A
R o T dx
(9.9)
atmosphere represent solid to gas diffusion while salt and sugar dissolving in water or milk are solid to liquid
diffusion. Similarly, diffusion of carbon into iron during case-hardening, doping of semiconductors for transistors
and migration of doped molecules in semiconductors are examples of solid to solid diffusion.
For a binary mixture of two species A and B, by definition, we have:
.
m  mA  mB ; C  CA  CB ;   A  B ; N  N A  N B where, N= m /M=Number of moles passing a plane (kmol/s)
Diffusion in solids is complex because of the strong influence of the molecular force fields on the process. For these
systems Fick’s law of equation (9.1) is often used, along with experimentally determined diffusion coefficient,
For gases at low pressures, Dalton’s law of partial pressure gives: p  p A  p B
although this relation may not adequately describe the physical process. The numerical value of the diffusion
Concentration or density can also be expressed in a non dimensional form in terms of mass fractions as:
coefficients for liquids and solids is much smaller than for gases, primarily because of the larger molecular force
m
m V C A A p A N A
Mass of species A
wA 
 A  A




Total Mass of the mixture
m
mV
C

p
N
fields, the increased number of collision and the consequent reduction in the freedom of movement of molecules.
wB 
and
m
m V C B B p B N B
Mass of species B
 B  B




; V=volume in m 3
Total Mass of the mixture
m
mV
C

p
N
9.3 General mass diffusion equation in a stationary Medium
We can derive generalized form of the mass diffusion equation by the same procedure used for obtaining the heat
9.2.3 Diffusion coefficients
Diffusion in gases
Gilliland has proposed the following semi-empirical equation for estimating the diffusion coefficient in gases:
D  0.04357

T
1
3
2
1
p VA  VB
3
3

2
1
1

MA Mb
Figure 9.4 shows mass transfer on each face of the element in the cartesian co-ordinate system. Let, RA = rate of
increase of mass of species A due to chemical reaction per unit volume of the mixture, kg/s.m3 and mx, my and mz
be diffusion mass flow rates in the three directions. The species generation (RA) is a volumetric phenomenon and
such reactions that occur within the medium are called homogeneous reactions. This is analogous to the internal heat
generation in case of the heat conduction.
2
D – diffusion coefficient, m /s
However, in some cases the chemical reaction may occur at the surface due to contact between the medium and the
T – Kelvin
p – total system pressure in N/m
(9.10)
conduction equation. For this purpose, let us consider an element of sides dx, dy and dz having volume dv = dxdydz.
2
VA and VB : molecular volumes of constituents A and B as calculated from the atomic volumes
surroundings. This type of species generation is called heterogeneous reaction and is analogous to the specified
surface heat flux condition.
7
 .

 m d,A 

 y  dy
y
the medium, the concentration of species A are maintained at the wall surfaces but vary within it along the flow
 .

 m d,A 

z
RA


 m d,A 

 z  dz
.
z
dx
direction.
Plane Surface
 .

 m d,A 

 x  dx
dy
 .

 m d,A 

x
8
Referring to the [Fig. 9.5(a)] and applying Fick’s law of diffusion for the species A through the wall of thickness ∆x,
after integration from x=x1 to x=x2, we have
dz
 .

 m d,A 

y
Mass diffusion of species A:
.
 x  x  x
 C  CA2   CA1  CA2 
md,A  DA  A1
; where R Mass diff  2 1 

x
R Mass diff .
DA
DA


x
Considering the generalized equation (9.13) for concentration variation during the diffusion mass transfer, the 1D
Fig. 9.4 Mass diffusion rates on the faces of an
element dxdydz in the cartisian co-ordinate system
steady state concentration equation is,
Making a mass flow rate balance on the element of sides dx, dy and dz, we have
 D AB
C A
C
C A 
C A
 
dxdz 
.dydz  D AB dydz  A 
D AB dydz 
dx  D AB
x
x
x 
x 
y
C
C A 
C A
 
dxdy   D AB dxdy  C A
 D AB dxdz  A 
D AB dxdz 
dy  D AB
y
y 
y 
z
z
C A 
C A
 
 D AB dxdy 
dxdydz
dz  R A dxdydz 
z 
z 
t
This will simplify to give,
If DAB is constant, then
C A   
C A
 
 D AB
 D AB

x 
x  y 
y
 2CA
x 2

 2CA
y 2

 2CA
z 2
  
C A 
C A
   D AB
  RA 
z 
t
 z 
R
1 C A
 A 
D AB D AB t
R
1 C A
or,  C A  A 
D AB D AB t
2
(9.16)
 2CA
 0.
x 2
Twice integration of this equation, gives a non dimensional form of the concentration variation of species A in the
(9.11)
medium B as: C A  C A1  x  x1 (This is a linear variation along the plane surface)
C A 2  C A1 x 2  x1
(9.17)
Equations (9.16) and (9.17) are similar to the following 1D steady heat conduction equations for a plane wall
obtained using Fourier’s law:
.
q
(9.12)
(9.13)
kA  T1  T2 
x
 T1  T2  ;

R Heat cond.
where, R Heat cond. 
x
kA
and
T  T1
x  x1
 2T

(a linear temperature variation) using
0
T2  T1 x 2  x1
x 2
The mass diffusion of species A (equation 9.16) can be represented by electrical circuit as shown in Fig. 9.5(b)
.
Equation (9.13) is analogous to the general heat conduction equation:  2 T  q  1 T
k  t
which is similar to the circuits for heat transfer and current flow.
m
Similarly we can derive mass diffusion equations for the cylindrical and the spherical co-ordinate systems. They will
Species B
be,
CA1
For a cylindrical system
R
1   C A  1   C A   2 C A
1 C A


 A 
r

r r  r  r 2    
D AB D AB t
z 2
(9.14)
Species A
R Mass diff.
CA 2
1D diffusion mass transfer
CA
.
m d,A
m
CA1
Area
A
q
q
C. A2
m d,A
T1
R Heat cond.
T2
1D conduction heat transfer
For a Spherical system
 CA
CA  R A
1   2 CA 
1
1
 
1 CA


r

 sin 

r 2 r 
r  r 2 sin 2  2 r 2 sin   
  D AB D AB t
2
I
(9.15)
9. 3.1 One dimensional Steady Mass diffusion in a stationary medium
Let us consider a stationary permeable body of certain thickness, area and density (Species B) that can be a liquid or
solid through which there is mass diffusion of species A. Under steady mass diffusion, with no chemical reaction in
x1
dx
x
x2
Fig. 9.5 (a) Mass diffusion of
species A in a plane surface
I
V1
R Electrical
V2
current (I) flow
Fig.9.5(b) Electrical circuit for
mass transfer, heat transfer
and current flow
9
Cylindrical Surface
Considering a cylindrical medium of species B and diffusion of species A into it as shown in Fig. 9.5(c), we can
md,A
C  CA2  CA1  CA2 
 2lD A1

; where R Mass diff 
r
R Mass diff .
ln  o 
 ri 
(9.18)
But, when there is concentration gradient, there will be a simultaneous flow of species in the direction of decreasing
(9.19)
concentration at a diffusion velocity of vd. The average velocity of each species can then be determined as:
vA  v  vd,A and vB  v  vd,B
and the total mass flow rates will be,
Similarly for a spherical shaped medium as the species B and diffusion of species A into it, shown in Fig. 9.5(d), the
equations for mass diffusion and concentration variation can be written as:
and
gradient, there will be no molecular mass diffusion and the velocity of all species will be equal to the mass averaged
velocity of the flow, that is , vA = vB = v.
Spherical Surface
md,A  4Dri ro
(9.22)
medium where mass transfer is by diffusion only, as discussed earlier. However, if there is no concentration
r
ln  o 
 ri 
2lD
1   C A 
ln  r r1 
CA  CA1
and

(a lograthmic variation)
r
 0;
r r  r 
CA2  CA1 ln  r2 r1 
.
 v  B v B C A v A  C B v B
m mA mB


or v  A v A  B v B . Thus, v= A A

 w A vA  w BvB
A
A
A

C
Here, v is the mass averaged velocity of the flow. The mass averaged velocity v=0 will correspond to a stationary
write the equations for mass diffusion and concentration variation as follows:
.
10
CA1  CA2
 ro  ri 
 C  CA2  ;
 A1
R Mass diff .
where R Mass diff
r  r 
 o i
(9.20)
4Dri ro
mA  A vA A  A  v  vd,A  A  A vA  A vd,A A  mc,A  md,A
(9.23)
mB  B vBA  B  v  vd,B  A  B vA  B vd,BA  mc,B  md,B
(9.24)
.
.
.
.
.
1   2 C A 
1 r1   1 r 
CA  CA1

r
  0;
r 2 r 
r 
CA2  CA1 1 r1   1 r 2 
(9.21)
.
.
.
.
.
Where, mc,A and mc,B are convective mass transfer rates and m d,A and m d,B are diffusion mass transfer rates
for species A and B, respectively.
Introducing Fick’s law of diffusion (equation 9.1), m d,A  D AB A CA and m d,B  D BA A CB
x
x
.
Species A
Species
B
ln(ro / ri )
 R Mass diff.
2lD
C A1
CA1
CA 2
Fig.9.5(c) 1 D Mass diffusion in cylindrical medium
.
.
the total mass flow rates become: m A  mc,A  D AB A CA or m A  CA v  D AB CA (for species A)
x
A
x
(9.26)
.
Concentration
profile C A
ro  ri
4ri ro D
ri
Sp
ec
ies
A
ri
Species
B
ro
m d,A
Concentration
profile C A
Sp
ec
ies
A
Species A
.
ro
m d,A
(9.25)
.
, l
length
.
.
CA 2
 R Mass diff.
Fig.9.5(d) 1 D Mass diffusion in a spherical medium
9. 3.2 One dimensional Steady Mass diffusion in a moving medium
.
.
and m B  mc,B  D BA A CB or m B  CB v  D BA CB
x
A
x
Using the equation for total mass flowrate,
m mA mB
,


A
A
A
C  
C  
C  
C B 

(9.28)
v  Cv  C A v  D AB A   C B v  D BA B   Cw A v  D AB A   Cw B v  D BA
x  
x  
x  
x 

which becomes: Cv  Cv  w A  w B   D AB
Hence,  D AB
caused by some external force. In such cases the chemical species are transported both by diffusion and convection.
Let us consider a mixture of two gases A and B moving at different velocities vA and vB in the x direction. Also let
(9.27)
and substituting the respective values from equations (9.26) and (9.27) we have
Certain practical situations, like evaporation of water from river or lake under the influence of wind velocity or
mixing of two fluids flowing through a channel, represent moving medium where bulk movement of the fluids are
(for species B)
.
CA
CB
 D BA
x
x
CA
CB
 D BA
=0 since  w A  w B   1
x
x
.
.
.
Therefore, m d,A  m d,B  0 or m d,A   m d,B
(9.29)
these gases diffuse into one another due to presence of concentration gradients so that their mass concentrations C A
and CB and mass fractions wA and wB , respectively vary with the direction x.
Using m  mA  mB ; C  CA  CB and   A  B , we can write
This indicates that diffusion of species A must be equal and opposite in direction to the diffusion of species B.
Since, CA  CB  C (is constant),
CA
C
  B , leads to DAB  DBA
x
x
(9.30)
11
Equimolar counter diffusion
This indicates that the rate of diffusion of species A and species B must be equal in magnitude but opposite in
Now consider a physical situation called equimolar counter diffusion as shown in Fig. 9.5(e). The reservoirs contain
direction.
mixtures of two gases at different concentrations and are connected with a long pipe of small diameter to make the
1
mass transfer one dimensional. The reservoirs are assumed to be large enough, so that the concentrations of the
w A1
gases do not change with time and steady state exist in the pipe. Let NA and NB represent the steady state molar
w B1
Reservoir for
Gas A
a molecule of B, and vice versa. This is an example of mass diffusion in a moving medium. At section 1, the mass
concentrations are CA1 and CB1, while at the section 2 they are CA2 and CB2. Also let the reservoirs be at the same
pressure and uniform temperature. If CA1> CB1 and CB2 > CA2 then the Gas A will diffuse from left to the right and
NA
Assuming that the gases obey the perfect gas law, so that
p A1
pA  pB  p (constant), CA  CB  C (constant) and w A  w B  1
w A2
NB
x
the Gas B will diffuse from right to the left.
p B1
Introducing, CA and CB in terms of partial pressures from equation (9.7) into the equations (9.26) and (9.27), we get
1
2
pA  pB  p
p B p B2
pA
Partial
pressures
p A2
Fig. 9.5(e) Equimolar diffusion of gas A and gas B
the molar diffusion rates as:
.
p 
p 
mA
A 
mB
A 
 NA =
p A v  D AB A  and
 NB =
p B v  D BA B 
MA
R o T 
x 
MB
R o T 
x 
At steady state, the total pressure of the system remains constant so that
(9.31)
p A  p B  p and
NA
w A 
NB
w B 
p 
p 
p
p
Or
=
w A v  DAB
and
=
w B v  D BA
where w A  A and w B  B (9.32)
A R o T 
x 
A R oT 
x 
p
p
N NA NB
Now, considering the molar flow rates:
and substituting the respective values from equations (9.32)


A
A
A
we have
dp
dp A
dp
0
  B , since
dx
dx
dx
Similarly we can also write that w A  w B  1 and
(9.37)
dw A
dw
 B
dx
dx
(9.38)
Since each molecule of A is replacing a molecule of B, the molar diffusion rates are equal (NA = -NB) as above, we
again get
DAB =DBA = D. The calculation of D may be made with equation (9.10)
w A 
w B 
pv
p 
p 

w A v  DAB
+
w B v  DBA
R o T R o T 
x  R oT 
x 
Now, considering equations (9.8) and (9.9), and integrating between section 1 and 2 of the Fig. 9.5(e) we have,
w A
w B
or v   w A  w B  v  D AB
 D BA
x
x
Thus,  D AB
Mass
fractions
0
diffusion rates of components A and B, respectively. In the steady state situation, each molecule of A is replaced by
.
wA  wB  1
w B2
wA wB
Reservoir for
Gas B
9.3.3
12
.
w A
w B
 D BA
=0
x
x
(9.33)
m d,A DM A  p A1  p A2  DM A  w A1  w A2 





A
R o T  x  R o T 
x

(9.34)
A similar relation as the equation (9.39) can also written for the species B. Equation (9.39) is similar to the equation
(for species A)
(9.39)
(9.16) which is for a plane surface.
Now the molar diffusion rates of species A and B, using Fick’s law for perfect gases, are:
.
N d,A 
9.3.4
m d,A
A dp A
pA dw A
  D AB
  D AB
and
MA
R o T dx
R o T dx
stagnant air layer, as shown in Fig. 9.6. The free surface of the water is exposed to air in the tank.
m d,B
A dp B
pA dw B
  D BA
  D BA
MB
R o T dx
R o T dx
From equations (9.34) and (9.35) we conclude that the molar diffusion rates:
Let us consider isothermal evaporation of water from a surface and the subsequent diffusion through a
(9.35)
.
N d,B 
Isothermal evaporation of water into air
Let us assume that the system is in steady state and the process is isothermal with the total pressure remaining
Nd,A  Nd,B
constant.
(9.36)
13
14
This requires that there be a slight air movement over the top of the tank to remove the water vapour which diffuses
.
mw  
to that point. However, the air movement should not, in any case, create turbulence or alter the concentration
DAM w dp w p w M w
D dp A

A
R o T dx
R oT
p A dx
(9.45)
Making use of Dalton’s law of partial pressure, p  p A  p w
profiles in the air inside the tank.
water vapour upward. Since it is the same air moving
Let both the air and water vapour behave as ideal
upward in the tank, its mass flow rate given by the
gases.
equation (9.40), in terms of the bulk mass velocity v,
As the water evaporates, it will diffuse upward
through the air, and at steady state this upward
movement must be balanced by a downward diffusion
Using p A  p  p w
can be written as:
.
m d,A  A Av  
of air so that the concentration at any x position will
Air
pA M A
Av
R oT
(9.42)
x
2
remain constant.
m d,A
 DM A A dp A

R o T dx
.
mw  
DAM w
R oT
(9.40)
m d,w
x2
pw
.
m
pA
m d,w
w
dx  
x1
pw 2

p w1
.
And the mass diffusion of water vapour upward is
m d,w
M dp w
  DA w
R o T dx
1
mc,w
1
p
bottom of the tank will return back with a velocity just
mw 
Also
mw 
p
Water
The air molecules on reaching the water surface at the
or
m c,w
(9.41)
(9.47)
pw pA  pw
p
, equation (9.45) may be written as:


pA
pA
p  pw
 p  dp w


 p  p w  dx
(9.48)
Total mass evaporated from the deep tank will, therefore, be [ Take p  p w  t
m dA m dA
Where A – X-sectional area of the tank
.
and
dp A dp w

dx
dx
Relation given by (9.48) is called Stefan’s law.
The diffusion of air downward, by Fick’s law will be,
.
Since total pressure p in the tank is constant, therefore,
(9.46)
.
DpM w A dp w
R oT  p  pw 
.
or
m w  x 2  x1  
so that,  dp w  dt ]
pw 2
DpM w A
ln  p  p w  
p w1
R oT 
 p  pw 2 
DpM w A
ln 

R o T  x 2  x1   p  p w1 
(9.49)
p 
DpM w A
ln  A2 
R o T  x 2  x1   p A1 
(9.50)
Fig. 9.6 Diffusion of water-vapour into air
large enough to balance the diffusion of air downward.
9.4 Convective Mass Transfer
This bulk mass movement of the air will then carry
9.4.1 Concentration Boundary Layer
with it additional mass of
As explained earlier, the convective mass transfer is quite similar to the convective heat transfer. When there is heat
transfer to a fluid flowing over a stationary surface, there is formation of hydrodynamic (velocity) and thermal
Thus, the equations (9.40) and (9.42) should balance each other so that,
(temperature) boundary layers due to retarded fluid velocity and hence, conduction heat transfer. The thermal
p A M A Av DAM A dp A

.
R oT
R oT
dx
boundary layer is the region where temperature gradient exists and is mathematically represented by the energy
equation:
D dp A
This gives the velocity with which the air moves upwards, v 
p A dx
(9.43)
u
T
T
 2T
v
 2
x
y
y
(9.51)
Since the air moving up with velocity v carries with it water vapour, therefore, the bulk transport of water vapour
The hydrodynamic (velocity) boundary layer, in which viscous effect is felt, is represented by the momentum
will be
equation:
.
m c,w   w Av 
pw M w
p M
D dp A
Av  w w A
R oT
R oT
p A dx
(9.44)
u
.
Thus, the total mass of the water vapour moving upward,
.
.
m w  md,w  mc,w
Substituting the expression from equations (9.41) and (9.44) we get,
u
u
 2u
v
 2
x
y
y
(9.52)
The equations (9.51) and (9.52) are analogous to each other. The momentum and thermal boundary layers during
external flow on flat surface are shown in Fig. 9.7(a).
15
16
.
U
y
u
u
U
x
u
T
 u 
 
 y  y 0
't
 T 


 y  y 0
Tw
.
mw
R oT mw
pM

; C=
A  Cw1  Cw 2  M w A  p w1  p w 2 
R oT
(9.56)
.
Substituting the expression for m w given by equation (9.50) we get
Stationary solid surface
x
Stationary solid surface
Fig. 9.7(a)
hD 
T
y T
U
h DW 
Boundary layers with velocity (u) and Temperature (T) profiles during exteranal flow
In the same manner, there will be concentration boundary layer forming on a flat plate if there is diffusion mass
transfer because of no slip condition as in case of thermal boundary layer due to conduction in the retarded fluid
Dp
x 2  x 1 p w1  p w 2 
 p  p w2
ln 
 p  p w1

, m/s


(9.57)
As on flat surfaces, there will be boundary layer formation inside pipes. They are known as internal flows. The
velocity, temperature and concentration boundary layers for internal flows are shown in Fig. 9.7(d).
particles. The concentration boundary layer over a flat surface, with concentration gradient is shown in Fig. 9.7(b).
Velocity profiles
An element of sides dx, dy and dz is selected in the boundary layer. The bulk mass transfer at each face of this
element is shown in the Fig. 9.7(c).
U
m A ydy
y
C A
y
C A
x
Fig. 9.7(b)
u
 h D  C Aw  C A 
RA
dz
CA
CA
'm
 CA 


 y  y 0
m A x
dx
m A zdz
CAw
Species A
u
T
m A x dx
m A y
Fully developed region
Concentration boundary layer
T
T
T
T
Thermal entry length
Concentration profiles
Species A
C
C
T
T
Fully developed region
C
C
C
C
C
C
Concentration entry length
Fully developed region
Fig. 9.7 (d) Momentum, thermal and concentration boundary layer during internal flow
dz
dx
u
u
Temperature profiles
T
m A z
dy
u
 CA 
mA
  D AB 

A
 y  y  0
dy
U
u
U
Momentum entry length
U
Thermal boundary layer
Momentum boundary layer
We can obtain the concentration boundary layer equation for the mass transfer in the same way as the
x
energy and momentum equations were developed. Considering Fig. 9.7(c), bulk mass flow rate conservation on the
Species A
Bounadry layer with velocity (u) and Concentration (C) profiles during exteranal flow
z
Fig. 9.7(c) Mass diffusion rates on the faces of an
element dxdydz in the cartisian co-ordinate system
element in the boundary layer will lead to,
[Mass flow in – mass flow out + mass production due to chemical reaction= Rate of change of mass of the
The diffusion mass transfer of species A, taking place in the concentration boundary layer is shown by the
species].
concentration gradient, which then equals to the convective mass transfers, are related as follows:
Let mass flux in all directions be: n Ax  A u, n Ay  A v and nAz  A w
.
m A  DAB A
CA
y
 h D A  CAw  CA  ; where,
y 0
CA
y

y 0
 CAw  CA 
'm
from the slope in Fig. (9.7b)
(9.53)
.
CA n Ax n Ay n Az



 RA  0
t
x
y
z
CA
CB
 .n A  R A  0 for species A and
 .n B  R B  0 for species B
Or
t
t
h D  mass transfer coefficient, m2/s
CAw, CA∞ = concentrations through which diffusion occurs.
If one considers a steady state diffusion across a layer of thickness, x, x  x 2  x1  , then
.
mA 
DAB A  CAw  CA 
 h D A  CAw  CA 
x
Again we get h D 
D AB
. This is similar to
x
(9.54)
Mw
p w1  p w 2 
R oT
So that mass transfer coefficient for this situation could be written as:
(9.59)
For a mixture of species A and B, adding the two expressions in the equation (9.59), we get
 (CA  CB )
 .(n A  n B )  R A  R B  0
t
h  k x in convective heat transfer
In a mixture one mole of B is produced for each mole of A disappearing, therefore R A=-RB.
For the vaporization of water from a deep tank, we can write
C w1  C w 2 
(9.58)
CA



n Az .dxdy   n Az  dxdy   n Az  dxdy  dz   R A dxdydz 
dxdydz
z
t


This, then gives: h D  D AB
which is similar to the heat transfer coefficient h  k (see Fig. 9.7a)
'm
't
m A = diffusive mass flux of component A, kg/m2s;
A






n Ax .dydz   n Ax  dydz   n Ax  dydz  dx   n Ay  dxdz   n Ay  dxdz   n Ay  dxdz  dy 
x
y




And with n=nA+nB = ρAV+ ρBV=ρV , the continuity equation for the mixture will become:
(9.55)
C

 .(Cv)  0 or in terms of density:
 .(v)  0
t
t
(9.60)
17
18
Since in the concentration boundary layer, both diffusion and convective mass transfer take place, therefore the mass

 1 . The dimensionless number   Sc    Momentum diffusivity , where, Sc = Schmidt number which plays
D
D
D
Mass diffusivity
transfer equation of the boundary layer will be sum of the diffusion mass transfer (given by equation 9.11) and bulk
a role similar to that of the Prandtl number, Pr    Momentum diffusivity , in the convective heat transfer. The

Thermal diffusivity
mass transfer (given by equation 9.58). Thus, we have


C A
C
C A 
C A
C 
 
.dydz  D AB  dydz  A 
 dxdz   D AB  dxdz  A 
D AB  dydz 
 dx  D AB
   D AB


x

x

x

x

y

y







  



C

C

C

C

A

   D AB  dxdz  A  dy  D AB

 dxdy   D AB  dxdy  A  D AB  dxdy  A  dz


y 
z
z
z 
z 
  y 
Diffusion 


 n .dydz   n dydz   n dydz dx   n dxdz   n dxdz   n dxdz dy  


 Ax    Ay    Ay   y  Ay    
Ax
Ax 
 

x






 


C A

 




  n Az .dxdy   n Az  dxdy   z n Az  dxdy  dz   R A dxdydz  t dxdydz




Convective 
 
Which will simplify as:
C A 
C A 
C A 
 
 
 
D AB  dydz 
 dx 
D AB  dxdz 
 dy  D AB  dxdy 
 dz
x 
x 
y 
y 
z 
z 
temperature and concentration profiles will be similar when   D or

 1, and the ratio  is called Lewis
D
D
number, Le    Thermal diffusivity .
D
Mass diffusivity
The solution of momentum and energy equations of the laminar flow, as in the chapter on Convection, has lead to
the following relation between velocity and thermal boundary layer thicknesses:
 velocity
 temperature
 Pr n . Now if the mass
transfer equation, being similar, is also solved the relative thicknesses of velocity, temperature and concentration
(9.61)
 
CA

 

  n Ax  dydz  dx    n Ay  dxdz  dy    n Az  dxdy  dz   R A dxdydz 
dxdydz
t
 x
  y

  z
boundary layers may be expressed as:
 velocity
cocentration
 Sc n and
 temperature
cocentration
 Le n where n=
1
for most cases .
3
C A n Ax n Ay n Az
C A   
C A   
C A 
 




D AB

D AB
  D AB
  RA
t
x
y
z
x 
x  y 
y  z 
z 
In convective heat transfer, the heat transfer coefficient (h) is related as:
Or
CA   CA u    CA v    C A w   
CA   
CA   
CA 




D AB

D AB
  D AB
  RA
t
x
y
z
x 
x  y 
y  z 
z 
Where, x is the distance in the direction of flow in case of external boundary layer, and d is the pipe diameter in case
Or
Nu 
Or
 u v w   
CA
C
C
CA
CA   
CA   
CA 
u A v A w
 CA 


D AB

D AB
  D AB
  RA

t
x
y
z
x  y 
y  z 
z 
 x y z  x 
 u v w 
 
  0 due to continuity, therefore
 x y z 
Since 
(9.62)
Nu  C Re m . Pr n
C
C
 CA
C
C
C
For Species A: u A  v A  DAB
or in general as : u
v
D 2
x
y
y 2
x
y
y
(9.65)
The similarity in the boundary layer equations for heat, mass and momentum transfer suggest that the empirical
correlations for the mass transfer coefficient (hD) would be similar to those for the heat transfer coefficient which
can be related as:
hDx
 f  Re,Sc 
D
(9.66)
For vaporization of liquids into air inside circular columns, where the liquid wets the surface and the air is forced
through the column, Gilliland has proposed the following equation:
With boundary layer approximation, equation (9.62) for 2D steady state condition will reduce to be
2
of internal boundary layer. The dimensionless number form of equation for heat transfer coefficient is of the form:
Sh 
CA
C
C
C A
C A   
C A   
C A 
 
u A v A w

D AB

D AB
  D AB
  RA
t
x
y
z
x 
x  y 
y  z 
z 
(9.64)
hx hd
or
 f  Re, Pr 
k
k
 Ud 
h Dd
 0.023 

D
  
2
(9.63)
Equation (9.63) is similar to the energy and momentum equations (9.51) and (9.52).
0.83

 
D
0.44
or Sh d  0.023  Re d 
0.83
Sc 
0.44
(9.67)
Here, D=diffusion coefficient and U=free stream velocity of the fluid.
Equation (9.67) is valid for 2000<Red <35,000 and 0.6<Sc<2.5 and is applicable to flow in smooth pipes.
A similar equation for heat transfer, known as Dittus–Boelter equation is,
9.4.2 Analogy between convective mass, momentum and energy equations
It has been seen through the equations (9.51), (9.52) and (9.63) that the convective mass, momentum and energy
equations are similar to each other. The concentration and velocity profiles will have the same shape when   D or
Nu d  0.023Re 0.8 Pr 0.4
(9.68)
19
20
The dimensionless groups described above, for forced convection, are:
The Colburn analogy for mass transfer over smooth plates will then be,
h x
h d
hx
hd
Sh  D or D is Sherwood number , Nu 
or
is Nusselt number
D
D
k
k
Laminar flow over a flat plate:
Ux Ud
h
Nu
Re 
or
is Reynolds number, St 

is Heat transfer Stanton number


UCp Re Pr
and St m  h D  Sh
U
ReSc
is Mass transfer Stanton number .
g  T  T  x 3
Grm 
2
g      x
 2
or 
3
g  T  T  L3
g      L
3
or 
 2
2
is Mass transfer Grashof number, Sh  f  GrmSc 
where,
y 0
T
y
 h  Tw  T 
From this we can write equation for Nusselt number as: Nu d 
y 0

f
8
1
f
hd
Re d Pr 3 
8
k
(9.75)
Equation (9.75) is known as the Colburn analogy for pipe flow which may be extended to express the mass transfer
2
3

generalized equation that can be applied when both heat and mass transfer are taking place simultaneously.
St Pr
C
and o  U f we will get
2
2
(9. 69)
For Prandtl number other than unity, the drag coefficient will be, Cf  o  0.332 Rex
2 U 2
1
1
3

Cf
1
 0.0296 Re x 5
2

2
2
2
2
2
Cf f
C
f
and St mSc 3  f  or St Pr 3  St mSc 3 or h Pr 3  h D Sc 3

2 8
2 8
UCp
U
h
 Sc 
 C p  
hD
 Pr 
2
3
or
h
/D
 Cp 

hD
 / 
2
3

 C p  
D
Cf
2
2
3
. This is referred to as Reynolds analogy.
(9.77)
(9.78)
9.5 Evaporative Cooling
When air, at higher temperature, flows over the surface of water evaporation of water takes place (Fig. 9.8a).
During this process there will be loss of energy (latent heat of water) and drop in temperature of the water producing
Combination of these two equations, lead to the following equation which will be for laminar flow over a flat plate:
2
3
2
2
Cf
1
 0.332 Re x 2 (This is referred to as Colburn analogy)
2
2
In the above equations if Pr=Sc=1 then, St  St m =
and the heat transfer equation for laminar flow over a flat plate, Nu x  0.332 Re x 2 Pr 3 .
A similar equation for turbulent flow over a flat plate is, St x Pr
3
1
h d
f
Cf f
and Sh d  Re d Sc 3  D
(9.76)

8
D
2 8
All the above described heat and mass transfer equations: (9.70) to (9.76), when combined, lead to a common
Equation (9.69) is known as Reynolds analogy and is valid for Pr=1.

2
coefficient in terms of the friction factor as:
St mSc
C
h
h
Nu
U  Cp or
 f  St 
U 2  Cf 2 
UCp
2
Re Pr
3
(9. 74)
The equation (9.74) may be modified by this factor to give: St Pr
T
2
Cf f

2 8
8
and

q
qo
du  C p  dT , which gives: o U  C p  Tw  T  .
o
o 0
Tw
St x Pr
(9.73)
For fluids with Prandtl number other than unity, the dependence on Prandtl number is of the order of Pr2/3 .

Cf
 o
2 U 2
Separating the variables and integrating from the wall surface to the free stream value, we have
qo  h  Tw  T 
(9.72)
Here also the equation (9.74) is known as Reynolds analogy and is valid for Pr=1.
For identical velocity and thermal boundary layers,    and Pr=1
Substituting

h
f
Nu
U  Cp or
  St 
U 2  f 8
UCp 8
Re Pr
C
The ratio of heat flux to shear stress: q o   k T  u    k Cp T   p dT =  Cp dT for Pr=1
y 
o  y
Cp
u
Pr du
du
U
3
h
u
y
Conduction heat flux: q y 0  q o  k
2
U
qo  h  Tw  T  and o  U 2 Cf  U 2 f we have
Now, considering the friction and conduction equations at the wall surface
Friction (Shear stress):  y 0  o  

3
Again, using the equation for the heat flux to shear stress ratio: q o U  C p  Tw  T  and substituting
o
is Heat transfer Grashof number, where Nu  f  Gr Pr  and
2
2
In pipe flow, the equations are expressed in terms of friction factor, which is related to the drag coefficient as:
In the way, the dimensionless terms for free convection will be
Gr 
h D  Sc 
Cf
1
 0.332 Re x 2
2
2
2
h
C
1
Turbulent flow over a flat plate:
St mSc 3  D  Sc  3  f  0.0296 Re x 5
U
2
Similar equations for heat transfer in pipe flow can also be obtained.
St mSc
(9. 70)
cooling effect in the water. This loss of energy must come by convection from the air flowing over the water. An
example of the evaporative cooling is that of the dessert cooler, shown in Fig.( 9.8b), being used in dry summer for
cooling buildings. As, drops of water drip down on the straw pads, mounted to the body of the dessert cooler,
(9.71)
21
22
evaporation of water takes place when the fan sucks in air. This results in temperature fall of the water. The air in
Here, total pressure (p) has been taken in place of the partial pressure of air (pa). This will be the case when mass of
contact of this cooled water also cools down producing cooling effect in the building.
air is greater than the mass of water vapour; that is,  ma 
Another important application where heat and mass transfer take place simultaneously is at damped cover of the
p  pa  p v  pa because p v
Wet-Bulb thermometer, shown in Fig. (9.8c). When dry or unsaturated air comes in contact of the damped cover or
wick enclosing the thermometer bulb, evaporation of the water in the cover takes place until the surrounding air gets
saturated. Corresponding to this condition, temperature read by the wet-bulb thermometer is known as the wet-bulb
temperature. While, actual temperature of the air read on a simple thermometer is the dry bulb-temperature. During
this process, heat of evaporation comes from the air.
Thus, under steady state condition, when the air becomes saturated, the heat transfer due to convection from the air
equals the heat of evaporation of the water, that is,
.
hA  T  Tw   m w h fg
(9.79)
.
p v
q evap
q conv

Tw
p vs
s
 p vs p v  M w
pM a

and  
into the equation (9.83) we have,

T
T
R
R
 
o
o Tf
 w
Substituting:  C w  C   
2
 pM
  p
p M
 3
a
C p   T  Tw    vs  v  w h fg

D
 R o Tf
  Tw T  R o
Assuming, Tf  Tw  T we can write equation (9.85) as:
 C p

 h fg

 
D
2
3

T
p v
Fan

Tw
p vs
s
Unsaturated
Air
Saturated
Air
Tw
p vs
T
p v

Straw Pad
(a) Evaporation of water
(b) Dessert Cooler
Damped Cover or wick
soaked in water
s

(9.86)
s 
p vs M w
p M
and   v w .
pM a
pM a
 p vs M w p v M w 

   s    and
pM a 
 pM a
Thus, the right hand side expression of the equation (9.86) will become: 
hence the equation (9.86) will reduce to,
Cp   
 
h fg  D 
But, m w  h D AC w  C  
mv pvMw

ma
pM a
Therefore, the specific humidity for the saturated and unsaturated air will be,
(c) Wet Bulb thermometer
Fig. 9.8 Heat and Mass Transfer between an unsaturated air and water
2
3

s   
T  Tw 
(9.86)
(9.80)
hAT  Tw   h D AC w  C h fg
Now, equating (9.77) and (9.81), we have
 p vs M w p v M w 


pM a 
 pM a
T  Tw   
Again assuming p a  >> p v  , we can make pa  p , and hence the mass ratio becomes:
Water
so that,
(9.85)
m v Mass of water vap our in the humid air

ma
Mass of air
R oT
R T
m
pM
Using the Perfect gas equation we can also write, p v V  m v
and p a V  m a o , so that v  v w
Mw
Ma
ma pa M a
Water Tray
Saturated Air
T
Saturated
Air
Unsaturated Air
Unsaturated
Air
pa .
Now, specific humidity is defined as:  
Where h – heat transfer coefficient of the air and m w is the mass of water evaporated.
Water dripping
down
 mv  , in the humid air so that
h

 C p  
hD
D
(9.81)
2
3

C w  C
h fg
T  Tw


 3
C p   T  Tw   C w  C  h fg 
D


p
p
M
p
M
p M
vs
w
where, C w 
and C   v w , so that C w  C     vs  v  w
R o Tw
R o T
T  R o
 Tw
The properties of humid air is to be evaluated at the film temperature, T  Tw  T .
f
(9.82)
Example 9.1
(9.83)
Determine the diffusion coefficients for Carbon dioxide, Oxygen, Hydrogen, Nitrogen, water and Ammonia in air at
the atmospheric and temperature of 25oC.
2
or
2
pM a
The density of air using perfect gas equation can be written as:  
R o Tf
Solved Examples
(9.84)
Solution: Let us consider species A as the air, therefore Molecular Volume and Weight for air (species) are
respectively, Vair  VA  29.9 and M air  M A  28.9 .
Using equation (9.10):
Dair  0.04357 
T
3
2
p  VA 3  VB 3 


1
1
2
1
1

MA MB
23
24
Substituting the values of p= 1.0132 x 105 N/m2, T=(273+25) K and VB and M B for the given gases as species B,
Let us assume that depth of ethyl alcohol as compared to the tube height is negligible. Therefore, the height of the
2
we get the diffusion coefficient D in m /s (as listed in the table Ex. 9.1).
stagnant air in which alcohol is diffusing is taken as:  x 2  x1   30 mm  0.03m . Therefore, rate of diffusion of ethyl
alcohol in air will be,
m
Table Ex. 9.1 Values of Molecular Volume and Molecular Weight of various gases and their diffusion coefficient in
The time required to evaporate the 2mm deep ethyl alcohol from the tube will then be,
air.
S.
No.
1.
2.
3.
4.
5.
6.
Gas (Species
B)
Carbon dioxide
Ammonia
Nitrogen
Water
Oxygen
Hydrogen
Molecular
Volume
Molecular
Weight
VB
MB
34
25.8
15.6
18.8
7.4
14.3
44.0
17.0
28.01
18.015
32.0
2.0

2
1.2 x 105 1.0132  105  46    0.01
 1.0132  105  0.0

4
ln 
 4.86x 109 kg/s
5
5
8315  298 x 0.03
1.0132  10  0.08  10 
1
1

MA MB
0.23943
0.30565
0.26514
0.30018
0.25661
0.73110
V 3  V 3 
B 
 A

1
40.2385
36.7073
31.3873
33.2096
25.5270
30.5920
1
2
Diffusion
coefficient D
in m2/s
1.316 x 10-5
1.841 x 10-5
1.868 x 10-5
1.999 x 10-5
2.236 x 10-5
5.286 x 10-5
Example 9.2
Time of evaporation 
density x volume
.
 
790 x  d 2  height 790 x   0.012  x 0.03
4 
4

 383002.81s  106.38 h
.
4.86x 109
m

m
Example 9.4
A coal particle of 3mm diameter burns in the atmosphere of oxygen at 800 oC and 1.0132 x 105 N/m2. The carbon
dioxide formed during the reaction, completely blankets the coal particles. If the combustion rate is such that all the
oxygen reaching the particle surface is instantaneously consumed so that the concentration of oxygen there is
effectively zero. Also, let the layer of carbon dioxide around the coal particle is large enough and that the
concentration of carbon dioxide far away is zero. Estimate burning rate of the coal assuming steady state equimolar
diffusion of oxygen and carbon dioxide. Let diffusion coefficient of oxygen in carbon dioxide be 0.8 cm 2/s.
Estimate the diffusion rate of water from the bottom of a well, 1m in diameter and 3 m deep, into dry atmospheric
air at 27oC. Take the diffusion coefficient D = 2.0 x 10-5m2/s.
Solution:
Diffusion of oxygen in to carbon dioxide and vice versa is shown in the figure. The concentration of oxygen is
decreasing from CA2 at r=r2 to CA1 at r=r1 while that of carbon dioxide is increasing from CB2 at r=r2 to CB1 at r=r1.
Solution: Since water is diffusing into the dry air at the top of the well, the water vapour pressure there may be
Diffusion of oxygen: U sin g Fick's law : m d,A  D AB A
.
taken as zero. The partial pressure at the bottom of the well will be the saturation pressure of water corresponding to
Thus,
pA1  p  pw1  1.0132 10  0.0353110  0.97789 10 N / m
5
5
5
2
And p A2  p  p w 2  1.0132  105  0  1.0132  105
Equation (9.50) then gives,

2
.
2.0  10  1.0132  10  18   1
1.0132 Thus, m w  1.3577  10 7 kg / s
4

 ln
8315  300  3.0
0.97789
5
.
mw 
DpM w A
p
ln A 2
R o T  x 2  x1  p A1
5
Example 9.3
Find the Diffusion rate of ethyl alcohol (C2H6O) forming a 2mm layer at the bottom of a tube of 10 mm diameter
and 30 mm height. Air flows at the top of the tube where concentration of alcohol is zero. The total pressure and
temperature, which are 25oC and one atmosphere, remain constant while the partial pressure of alcohol at this
temperature is 0.08 bar. If density of ethyl alcohol is790 kg/m3, find the time required to evaporate the total ethyl
alcohol from the tube. Assume, diffusion coefficient: D = 1.2 x 10-5 m2/s. Molecular weight of ethyl alcohol is 46.
Solution: Considering the equation (9.49): m 
 p  p b2  ;
DpAM
ln 

R o T  x 2  x1   p  p b1 
r2
CA 2
r1
CA1


Or md,A  dr2  DAB  4   dCA or md,A   1  1   4DAB  CA2  CA1 
r
r r
.
27oC. The saturation pressure of water vapour at 300K is 0.03531 bar.
CA
C
 D AB  4r 2  A
r
r
.

2
1

p M
pw A M A
In terms of partial pressure: C A  A A 
R oT
R oT
.
 1 1
pM A
Writing CA in terms of partial pressure we have, md,A      4DAB  CA2  CA1   4DAB
 w A2  w A1 
R oT
 r2 r1 
Referring to the figure, the boundary condition at r=r2 = is w A  w A2  1
Thus,
.
 1 1
pM A
m d,A      4D AB
1  0  or
R oT
  r1 
.
.
m d,A  4r1D AB
pM A
m d,A
p
or N A 
 4r1D AB
R oT
MA
R oT
.
Here,  b1  0.08bar and b2  0
and at r=r1 is w A  w A1  0
Similar equation can be written for diffusion of carbon dioxide: m d,B   D BA A
CB
C
 D BA  4r 2  B
r
r
25
.
 1 1
pM B
Writing CB in terms of partial pressure we have, md,B      4DBA  CB2  CB1   4D BA
 w B2  w B1 
R oT
 r2 r1 
26
Molar diffusion of oxygen,
N A  D  4r1 
p
0.8 x 104 4 x 1.5 x 103 x 1.0132 x 105
kmol

 1.712x 108
R oT
8315 x  273  800 
s
Since, NA=-NB for equimolar diffusion, therefore DAB = DBA= D. Hence, the rate of burning of carbon dioxide will
Concentration profile
of Carbon dioxide
CB2
CB1
CA2
.
md,B  D  4r1 
C A1
Concentration
profile of oxygen
Sp
ec
ies
B
Species B
is Carbon
dioxide
be
Considering the chemical reaction: C+O2=CO2 , it is also seen that 1 kmol of O2 is consumed by 12 kg of carbon.
Species A
is Oxygen
ies A
Spec
pM B 0.8 x 104 4 x 1.5 x 103 x 1.0132 x 105 x 12
kg

 2.055 x 107
R oT
8315 x  273  800 
s
Therefore, the rate of burning of carbon dioxide will be 1.712 x 10-8 x 12= 2.055 x 10-7 kg/s.
Example 9.5
Carbon
dioxide
Coal
r1
A circular tube of 100 mm diameter and 5m long has its wall coated all around by a thin film of water at a
temperature
of 25oC. If dry air at one atmospheric pressure and 25oC, enters the tube at a velocity of 5 m/s, find concentration of
water vapour in the air leaving the tube.
r2
Fig. (Ex. 9.4) 1 D Mass diffusion of oxygen(A)
into carbon dioxide (B)
Solution:
W ater film
Dry Air
Referring to the figure, the boundary condition
at r=r2 = is w B  w B2  0
Thus,
and at r=r1 is w B  w B1  1
.
.
pM
m d,B
p
 1 1
pM B
md,B      4DBA
 0  1 or md,B  4r1DBA R TB or N B  M  4r1DBA R T
R oT
o
B
o
  r1 
Alternative method:
pA M A pw A M A
w A c1
c
1   2 CA 


 2 and w A   1  c 2
r
  0 and CA 
r 2 r 
r 
R oT
R oT
r
r
r
Applying the boundary conditions:
c
r
U sin g : w A   1  c 2 at r=, w A  1 gives: c 2  1 and at r=r1 , w A  0 gives: c1  r1 we have w A  1  1
r
r
c
r
Now u sin g : w B   1  c 2 at r=, w B  0 gives: c 2  0 and at r=r1 , w B  1 gives: c1   r1 we have w B  1
r
r
Therefore, w A  1  w B ;
w A
r

r  r1
c1 r1 1
w B
 
and
r 2 r12 r1
r

r  r1
.
Fick's law for molar diffusion in terms of partial pressure : N A 
m d,A
p w A
  D AB A
MA
R o T r
.
and N B 
m d,B
p w B
  D BA A
MB
R o T r
  D AB  4r1 
r  r1
p
R oT
  D AB  4r1 
r  r1
p
R oT
Cw
Fig. (Ex. 9.5) Convective Mass
The properties of air at 300 K are:
5 N
6
2
p 1.01325 10 m 2


 1.177kg / m 3 , Cp = 1.0057 kJ/kg K,   15.69 10 m / s;
RT
287  300
6
4
  22.16 10 m / s D  0.20 10 m / s, and Sc  0.7845, k=0.02624 W/mK
From the steam table at 27oC (300 K) saturation pressure p = 0.0353 bar and latent heat of vaporization, hfg = 2438
2
2
kJ/kg.
Therefore, Cw 
Re 
c1 r1
1


r 2 r12
r1
C e
Cb
27 o C
and C  i
.
Considering
Cw
5 m/s,
p w M w 0.0353x105 x18

 0.0254 kg/m3
R w Tw
8315  300
Ud
0.1 5

 0.3187 105  2100  Turbulent in pipe flow
 15.69 106
Using Sh d  h D d  0.023  Red 0.83 Sc 0.44 or Sh d  0.023  31870 0.83  0.78450.44  113.02
D
ShxD 113.02x0.2x104
hD 

 0.0226 m/s
d
0.1
Mass diffusion of water as the air moves along the pipe:
.
m  h D  dL C w  Cb  where, bulk concentration of water vapour in the air,Cb 
Cinlet  Cout
2
27
Since the air entering the pipe is dry, therefore, Cinlet  0, then Cb 
Cout
2
28
Sh  0.0228 Re0.8 Sc0.337  24091.91 
.
C
C


m  h D  dL C w  C b   h D  dL C w  exit   0.0226xx0.1 5x(0.02548  exit )
2 
2

.


2
Also, m  d 2 U Cexit  Cinlet    0.1 x5 Cexit 
4
4
.
C

2
m   0.1 x5 Cexit   0.027654xx0.1 5x(0.02548  exit .
4
2
This gives: Concentration of water vapour in the air leaving the pipe , Cexit  0.01586 kg/m 3
.
and mass of water diffused in the air m  6.23x10 4 kg/s
h Dm L
D
Csalt  380kg / m3 and Csalt  zero
hm = 3.10776 x 10-5 m/s;
mw  h m A  Csalt  Csalt   3.10776 x 105 x1x380=11.81x 103 kg/s
Example 9.8
A wet-bulb thermometer reads (WBT) 20oC when dry atmospheric air blows over it. Find,
(a) Dry bulb temperature (DBT) of the air
(b) Relative humidity of the air if its dry bulb temperature is 34 oC
Example 9.6
Dry air at 30 0C flows at a velocity of 6m/s over a plate 1m long and 0.5 wide. If the plate is maintained at 30oC and
its entire is surface covered with a water film, determine the average convective mass transfer coefficient and also
the mass of water evaporated per second.
concentration at the damped cover surface Cw is that corresponding to saturation conditions of water at the wet-bulb
temperature read by the thermometer and the concentration of water vapour in the dry air is zero, therefore, C  = 0.
From the steam table at 20oC (Tw=293 K) saturation pressure p = 0.02358 bar and latent heat of vaporization, hfg =
Solution:
For water vapor at 30oC ,    w  C w 
o
Properties of air at 30 C
Re 
Solution: (a) Let the bulb of the thermometer reading WBT be enclosed by a wick or damped cover. The
1
 0.0304kg / m3 ;
32.9
  1.165kg / m ;
3
  16  10
6
2
m / s,
in dry air  w  C w  zero
D  2.0 10 m / s, Sc    0.8,
D
5
2
 6 1  3.75 105  5 105  laminar flow
16 106
Therefore, C w 
p w M w 0.02358x105 x18

 0.01742 kg/m3
R w Tw
8315  293.0
The properties of air should be taken at the film temperature, Tf   T  Tw  / 2 .
Since DBT ( T ) is unknown, the calculation may be initiated assuming,
For laminar flow, average mass transfer coefficient is obtained for the equation for average Sherwood number:
Sh = 0.664 Re0.5 Sc0.333
Puting the values we have, Sh  377.496  Sh 
2454 kJ/kg.
h Dm L
 h Dm  7.55  103 m / s
D
Mass of water evaporated per second,
m 2  1.205kg / m 3
287  293.0
Cp = 1.0069 kJ/kg K,   15.267  10 6 m 2 / s;   21.57  10 6 m 2 / s and Pr  0.709,
Let
D = 0.256x 10-4 m2/s so that, Sc =  D  0.596 and Le   D  0.84
The properties of air at 293K are:  
p

RT
Thus, from equation (9.82), we get  T  Tw  
C w  C
C p  Le 
2
3
h fg
.
m  A  h Dm C w  C w   1x0.5  7.55 10 3 x0.0304  0.11475 10 3 kg/s=0.11475 g/s
Substituting the respective values, as above we have
 T  Tw  
Example 9.7
Pure water at 200C flows at a velocity of 2m/s over a slab of salt. If the concentration of salt at the interface is
380kg/m3, determine the average mass transfer coefficient and rate of diffusion of salt into water. Take area of the
salt slab as 1x1 m2.
or T  Tw  39.57o C  T  59.57o C = 332.57 K
Solution: Assume D = 1.29 x 10-9 m2/s;
Properties of air at 313K are:  
water  1000kg / m3 ,   1.006 10 6 m 2 / s ; therefore Sc 
Re L 
1.006  10
1.2  10 9
UL
2 1

 1.988 106  Turbulent flow ;

1.006 106
Let us use the following equation for turbulent flow:
6
 838.33;
Tf  Tw  293K .
1.01325  10 5 N
 0.01742  0   2454 103
2
1.205 1006.9   0.84  3
;
Now, the calculation is repeated by taking the properties of air at Tf  T  Tw  400 C(313K) ;
2
1.01325  10 5
 1.128kg / m 3 , Cp = 1.0075 kJ/kgK,  = 17.2 x 10-6m2/s,  
287  313
24.42 x 10-6 m2/s
and D = 0.256 x 10-4 m2/s (same as above), we have Le   D  0.954
The steam being at 20oC (293 K), the saturation pressure, Cw and latent heat of vaporization remain unchanged.
Thus, keeping p = 0.02358 bar, hfg = 2454 kJ/kg and Cw =0.01742 kg/m3, as above
29
T  Tw 
 0.01742  0   2454 103  38.82o C
1.128 1007.5   0.954 
2
30

Nu  0.1(GrPr)0.333 109 < GrPr < 1013
Thus, T  38.82  20  58.82o C
3
The value of T  58.82o C calculated now, is close to the T  59.57 o C obtained earlier. Therefore, this may
accepted as the required value. However, the calculation may be repeated for more precise result.


Nu  0.54(GrPr)0.25 ; 104 < GrPr < 107
(b) Relative Humidity:  
or

Partial pressure of water vapour at a temperature T
Partial pressure of water vapour when the air gets saturated at the same temperature T
p v v RT Cv
C
Actual concentration of vapour in air (C )


  
p vs vs RT Cvs C,s Concentration of vapour when the air is saturated (C,s )
Knowing the wet bulb and dry bulb temperatures, Properties of air are taken at Tf  T  Tw  27 0 C(300K)
2
Nu  0.15(GrPr)
0.333

7
; 10 < GrPr < 10

Nu  0.27(GrPr)0.25 ; 105 < GrPr < 1011


Sh  0.54(Gr Sc)0.25 ; 104 < GrSc < 107

Lower Surface of Horizontal Plate is hot (T s > Ta)
1.



Sh  0.15(Gr Sc)0.333 ; 107 < GrSc < 1011
Fluid near the lower horizontal surface is light,
(s  a ) :

Nu  0.27(Gr Sc)0.25 ; 105 < GrSc < 1011

For a system where heat and mass transfer take place simultaneously,
a.
Properties of saturated water at 200C remain same, therefore
3
0
Cw = 0.01742 kg/m and hfg = 2454 kJ/kg are taken (as above) at 20 C
Putting the respective values in C p  Le 
2
3
b.
 T  Tw    C w  C  h fg we have
1.1614 x 1007 x (0.879)2/3(34 – 20) = (0.01742 - C) x 2454 x 103
or C = 0.011298 kg/m3
2.
Table 9.2 SOME COMMON CORRELATIONS FOR HEAT AND MASS TRANSFER
Convective Heat Transfer
Convective Mass transfer
Forced Convection flow over a flat plate/surface
0.333
Local Nu x  0.332Re0.5
; (Re < 5 x 105 )
x Pr
3.
d.

C p  
D
0.333
Local Nu x  0.0298 Re0.8
; (Re > 5 x 105 )
x Pr
0.333
Local Sh x  0.0298Re0.8
; (Re > 5 x 105 )
x Sc
0.333
Average Nu L  0.664Re0.5
; (Re < 5 x 105 )
L Pr
0.333
Average Sh L  0.664 Re0.5
,
L Sc
b.
5
(Re < 5 x 10 and Sc < 0.5)
0.333
Average Nu L  0.037 Re0.8
; (Re > 5 x 105 )
L Pr
c.
0.333
Average Sh L  0.037 Re0.8
; (Re > 5 x 105 )
L Sc
Fully developed flow in smooth circular pipes
Nu  3.66 (for constant wall temperature)
Sh  3.66 (constant surface concentration)
Nu  4.36 (for constant wall heat flux)
Sh  4.36 (for constant wall mass flux)
(for Re < 2300)
(for Re < 2000)
Nu  0.023Re0.8 Pr 0.4 ; (Re  10000 and 0.7 < Pr < 160)
Sh  0.023Re0.83 Sc0.44 ;


Sh  0.59(Gr Sc)0.25 105 < GrSc < 109

2
3

C w  C
h fg
T  Tw

C w  C
h fg
T  Tw
Thermal diffusivity
Mass diffusivity
Momentum diffusivity
Mass diffusivity
Heat Transfer
Mass Transfer
Mass diffusivity
Momentum diffusivity
Ans.
Ans.
Ans.
Lewis number (Le) is
Thermal diffusivity
Mass diffusivity
Mass diffusivity
b.
Momentum diffusivity
a.
(2000 <Re< 35000 and 0.6 < Sc < 2.5)
Natural Convection over Vertical Plate/surface
d.
4.
3
Diffusion Mass Transfer is governed by
a.
Fourier’s law
b.
Fick’s law
c.
Poisson’s law
d.
Newton’s law
Schmidt number (Sc) is
a.
0.333
Local Sh x  0.332 Re0.5
; (Re < 5 x 105 )
x Sc
1
D
C p  

From the steam table, the saturation concentration at 34 C (p = 0.053156 bar)
0.053156  10 5
 0.03748kg / m 3 . Hence, R.H. = 0.011298 / 0.03748 = 0.30144 (30.144
 8315 

  273  34 
 18 
1
h
 C p  Le  3 h fg
hD
2
h
 C p  Sc  3 h fg
hD
c.
o


Questions of multiple choice
  1.1614kg / m3 , Cp = 1.007 kJ/kgoC,   0.225 x 10-4 m2/s, D = 0.256 x 10-4 m2/s; α / D =Le= 0.879
Nu  0.59(GrPr)0.25 ; 105 < GrPr < 109

11
Properties of air are taken at Tf  300K
C  ,s    , s 

Sh  0.1(Gr Sc)0.333 109 < GrSc < 1013
Natural Convection over horizontal Plate/surface
Upper Surface of Horizontal Plate is hot (T s > Ta)
Fluid near the upper horizontal surface is light,
(s  a ) :
Ans.
31
32
Questions for discussion
Momentum diffusivity
Mass diffusivity
Mass Transfer
d.
Heat Transfer
c.
5.
1.
2.
3.
4.
5.
6.
7.
8.
9.
The concentration boundary layer is related as:
a.
cocentration  velocitySc1/3
b.
cocentration  velocitySc1/3
c.
cocentration  velocitySc2/3
d.
cocentration  velocitySc2/3
Ans.
What is mass transfer?
Differentiate between diffusion and convective mass transfer.
Write a note on diffusion in solids, liquids and in gases.
How is mass diffusion analogous to conduction heat transfer.
Write Fick’s Law of diffusion for low pressure gases
How is convective mass transfer analogous to convective heat transfer.
What is wet bulb temperature.
Describe the process of evaporation in a desert cooler.
Give the significance of Sherwood number, Schmidt number and Lewis number.
Unsolved Questions
6.
7.
Sherwood number (Sh) is a function of
a.
f  Re, Pr 
b.
f  Re,Sc 
c.
f  Re, Le 
d.
f  Pr,Sc 
1.
2.
b.
c.
d.
In an industrial plant, water has been spilled out of a pipe line on the floor, the layer which is 1mm. and at
25oC. If the atmospheric air and the water both are at 25oC, determine the time required to evaporate the
complete water if the evaporation takes place by molecular diffusion through an air film of 5mm thickness.
Assume that the absolute humidity of air is 2g per kg of dry air.
Mass Transfer Grashoff number is
a.
Estimate the diffusion rate of water from the bottom of a test tube 10mm in diameter and 15cm long into
dry atmospheric air at 25oC.
Ans.
g      L3
H int : Find p w2 using, absolute humidity 
Ans.
 2
g  T  T  L3
3.
 2
 p

2
 0.622  w 2  and then find m w, total .
1000
 p  pw 2 
Air at 25oC and 50% relative humidity (ф) flows over water surface measuring 10m x 5m at a velocity of
2m/s. Determine the water loss per day considering flow direction along the 10m side.
g      L3
H int :

g      L
Find h D using, Sh  0.0584 Re
and then find p vl from  

4.
1
2
Sc
1
3
where, Sh 
hDL
UL
, Re 
D

pv
; p vs from steam table at 25o C;
p vs
m w  h D (10x5)
Mw
 p vs  p v 
R oT
Benzene (M =78) is stored in an open tank of 5m diameter forms a 1mm deep layer at its bottom. There is
diffusion of benzene through a stagnant air film of 5 mm thickness. If the operating pressure and
8.
Mass Transfer Stanton number is
a.
b.
c.
d.
Sh
ReSc
Sh
Re Pr
Sh
Pr Sc
Nu
ReSc
temperature of the system are 1 atmosphere and 25oC and the vapour pressure of benzene in the tank is 0.15
bar, calculate the time taken for the entire benzene to evaporate. For benzene take the density as 880kg/m3
Ans.
and diffusivity as 8 x 10-6 m2/s.
5.
Air at 20oC (= 1.205 kg/m3,  = 15.06 x 10-6 m2/s, D = 4.166 x 10-5 m2/s) flows over a tray (length = 0.3m,
width = 0.4m) full of water with a velocity of 2.5 m/s. The total pressure of moving air is 1 atmosphere and
the partial pressure of water present in the air is 0.007 bar. If the temperature on the water surface is 15oC,
calculate the evaporation rate of water.
6.
Air is contained in a tyre tube of surface area 0.5 m 2 and wall thickness 10 mm. The pressure of air drops
from 2.2 bar to 2.18 bar in a period of 5 days. The
solubility of air in the rubber is 0.07 m3 of air per m3
of rubber at 1 bar. Determine diffusivity of air in the rubber at the operating temperature of 300 K, if the
volume of air in the tube is 0.03 m3.