Chapter 1
For liquids of high volatility, A. Stefan (1879) devised a convenient means of measuring the
diffusivity DAB of their vapor A through a stagnant gas B. If the volatile substance A (e.g.
ethyl ether or ethanol) is placed in the lower part of a vertical capillary, then liquid A will
evaporate and, by the mechanism of diffusion, travel to the end of the capillary. Maintaining
the mouth of the capillary at a given composition automatically establishes the concentration
gradient in the capillary, and the falling rate of the meniscus in the capillary provides the rate
of transport. The capillary is placed in an envelope through which air is passed. At the
meniscus the gaseous phase composition is specified by the vapor pressure of liquid A, the
diffusing constituent. At the mouth of the capillary, the gaseous phase is essentially air. The
gradient in the capillary is thus obtained by circulating sufficient air to reduce the substance
A concentration at the mouth to a negligible quantity. The air rate should be low, constant,
and not turbulent. The falling rate of the meniscus can be observed remotely with a
cathetometer.
gas B
(air)
PA2
z=0
PA1
z = z1 at t = 0
z = zt at t
pure liquid A
Figure 9.3-1 Diffusion cell with moving liquid surface.
If the length of the diffusion path changes a small amount over a long period of time, a
pseudo-steady state assumption may be used. This condition can be quantified as follow:
Let td be the average time it takes for molecule A to travel the diffusion path z1. td can be
estimated by Einstein ‘s formula
z12
td =
2 DAB
Let ∆z be the change in diffusion path (or the drop in liquid level) during the time td. The
pseudo-steady state assumption may be used if
∆z/ z1 < 0.01
1-9
The molar flux of A is given by
NA,z = − cDAB
dy A
+ yA(NA,z + NB,z)
dz
We will assume that air is insoluble in liquid A, it is stagnant (or nondiffusing) and NB,z = 0.
Solving for NA,z give
NA,z(1 − yA) = − cDAB
z
NA,z ∫ dz = − cDAB
0
∫
dy A
cDAB
⇒ NA,zdz = −
dyA
dz
1 − yA
yA2
y A1
dy A
1 − y A2
⇒ NA,zz = cDABln
1 − yA
1 − y A1
Using the definition of log mean average,
y B 2 − y B1
(1 − y A2 ) − (1 − y A1 )
=
ln( y B 2 / y B1 )
ln[(1 − y A2 ) /(1 − y A1 )]
yB,lm =
we have
ln
1 − y A2
y − y A2
= A1
y B ,lm
1 − y A1
The molar flux of A is then written as
NA,z =
cDAB ( y A1 − y A2 )
zy B ,lm
The molar flux is related to the amount of A leaving the liquid by the material balance
NA,z =
ρ dz
cDAB ( y A1 − y A2 )
= A, L
zy B ,lm
M A dt
Separating the variables and integrating the equation from t = 0 to t gives
t
∫ dt
0
=
ρ A, L
MA
y B ,lm
cDAB ( y A1 − y A2 )
∫
zt
z1
zdz
We have
t=
ρ A, L
MA
y B ,lm
cDAB ( y A1 − y A2 )
 zt2 − z12 


2


1-10
Replacing c by P/RT yields
t=
ρ A, L
MA
y B ,lm RT
DAB P ( y A1 − y A2 )
y B ,lm RT
 zt2 − z12  ρ A, L

 =
M A DAB ( PA1 − PA2 )
 2 
 zt2 − z12 


 2 
Since
yB,lm =
P ( y A1 − y A2 )
P
y A1 − y A2
=
= B ,lm
ln[(1 − y A2 ) /(1 − y A1 )]
P
P ln[ P (1 − y A2 ) / P (1 − y A1 )]
yB ,lm
PB ,lm
=
PA1 − PA2
P ( PA1 − PA2 )
Therefore
t=
ρ A, L
MA
PB ,lm RT
DAB P ( PA1 − PA2 )
 zt2 − z12 


 2 
Solving for the diffusivity we have
ρ A, L PB ,lm RT
 zt2 − z12 


DAB =
tPM A ( PA1 − PA2 )  2 
In this equation
PB,lm = [(P - PA1) - (P - PA2)]/ln [(P - PA1)/(P - PA2)]
PA1 = vapor pressure of liquid A at temperature T.
PA2 = partial pressure of vapor A at the mouth of the capillary.
R = gas law constant.
t = time during which the meniscus fall from z1 to zt.
z1 = distance from the mouth of the capillary to the meniscus at t = 0.
zt = distance from the mouth of the capillary to the meniscus at t.
P = ambient atmospheric pressure
ρA,L = density of liquid A at T.
1-11
Example 1.2-2 -----------------------------------------------------------------------------A long glass capillary tube, of diameter 0.01 cm, is in contact with water at one end and dry
air at the other. Water vapor evaporates at the wet end within the capillary, and the vapor
diffuses through the capillary toward the dry end. How long is required for one gram of water
to evaporate through this system? The vapor pressure of water is 17.5 mmHg at 20oC, the
temperature at which the entire system is maintained. Take the diffusivity of water vapor in
air at 20oC to be 0.20 cm2/s, and assume that the dry air is at a pressure of 760 mmHg.
Assume that the distance from the wet interface within the capillary to the dry end is always
10 cm. (Ref. An Introduction to Mass and Heat transfer by Stanley Middleman.)
Solution ---------------------------------------------------------------------------------------------The molar flux of A (water vapor) is given by
NA,z = − cDAB
z=L
yAL
z=0
yA0
dy A
+ yA(NA,z + NB,z)
dz
Since air is insoluble in water, it is stagnant (or nondiffusing) and NB,z = 0.
Solving for NA,z give
dy
cDAB
NA,z(1 − yA) = − cDAB A ⇒ NA,zdz = −
dyA
dz
1 − yA
L
NA,z ∫ dz = − cDAB
0
∫
y AL
yA0
dy A
1 − y AL
⇒ NA,zL = cDABln
1 − yA
1 − y A0
The ambient air is dry so yAL = 0. Vapor pressure of water at 20oC is 17.5 mmHg, therefore
yA0 = 17.5/760 = 0.023. Note: R = 82.057 cm3⋅atm/mol⋅oK
NA,z =
cD AB
1
0.20
1
=
= 1.935×10-8 mol/cm2⋅s
ln
ln
L
1 − y A0
82.057 × 293.15 × 10 1 − 0.023
Time for 1 g of water to evaporate through this system is then
t=
1/18
4(1/18)
=
= 9.14×109 s
−8
N A, z ( Area )
(1.935 × 10 )(π )(0.01^ 2)
The average time it takes for molecule A to travel 10 cm can be estimated by
td =
102
L2
=
= 250 s.
2(.2)
2 DAB
1-12
Example 1.2-3 -----------------------------------------------------------------------------Water evaporates at the surface of a column confined to a long
narrow capillary tube, closed at the bottom. The system is at 1
atm and 25oC as shown. The inside diameter of the tube is 1
mm. Dry air is blow over the top of the tube. For the case that
Lo = 10 cm and Lw(t = 0) = 10 cm, how long will it take for the
column of water to vanish. Diffusivity of water vapor in air is
0.22 cm2/s. Vapor pressure of water at 25oC is 3.17 kPa. (Ref.
An Introduction to Mass and Heat transfer by Stanley Middleman.)
Lo
Pure water
Lw(t)
Solution ---------------------------------------------------------------------------------------------The molar flux of A (water vapor) is
NA,z = − cDAB
dy A
+ yA(NA,z + NB,z)
dz
Since air is insoluble in water, it is stagnant (or nondiffusing) and NB,z = 0.
Solving for NA,z give
NA,z(1 − yA) = − cDAB
L
NA,z ∫ dz = − cDAB
0
∫
dy A
cDAB
⇒ NA,zdz = −
dyA
dz
1 − yA
y AL
yA0
dy A
1 − y AL
⇒ NA,zL = cDABln
1 − yA
1 − y A0
The dry air is blow over the top yAL = 0, yA0 = 3.17/101.3 = 0.0313
NA,z =
cD AB
1
ln
L
1 − y A0
The molar flux is related to the amount of A leaving the liquid by the material balance
Ac
ρ A,L dLw
M A dt
= − AcNA,z = − Ac
cD AB
1
ln
L
1 − y A0
In this equation, Ac is the cross sectional area of the tube and Lw is the length of the liquid
dL
dL
column. Since w = −
, we have
dt
dt
ρ A,L dL
M A dt
=
cD AB
1
ln
L
1 − y A0
Separating the variables and integrating the equation from t = 0 to t gives
1-13
t
K ∫ dt =
0
K=
t=
∫
20
10
cDAB M A
ρ A, L
LdL where
ln
1
1
(0.22)(18)
=
ln
= 5.149×10-6 cm2/s
1 − y A0 (82.06)(298) 1 − 0.0313
1
(202 − 102) = 2.91×107 s = 8093 hr
2K
The average time it takes for molecule A to travel 10 cm can be estimated by
102
L2
td =
=
= 227 s.
2(.22)
2 DAB
At the end of this time the diffusion path can be determine from
227 =
1
(L2 − 102) => L = (100 + 227×5.149×10-6)0.5 = 10.0001 cm
2K
The pseudo-steady state assumption is valid since the diffusion path changes very little
during the average time the water molecule travel from the water interface to the top of the
tube.
1.3 Solute Diffusion in Homogeneous Solution
Solute can be transported by convection and by diffusion. In a bulk homogeneous solution,
diffusion of a solute is given by Fick’s law
JS = − DS
dC
dx
(1.3-1)
where S is the surface area normal to the direction of diffusion and D is the bulk solute
diffusivity. For a variety of solutes in dilute aqueous solution at 37oC, the following
correlation can be used to estimate the solute diffusivity from its molecular weight
D = 1.013×10-4 (MW)-0.46 (cm2/sec)
(1.3-2)
The movement of a spherical solute of radius a in dilute solution can be described by an
equation of motion, called the Langevin equation1
du A
m
= − ζ u A + F(t)
dt
1
Bird, Stewart, Lightfoot, Transport Phenomena, Wiley, p. 531 (2002)
1-14
where u A is the instantaneous velocity of the sphere of mass m, ζ u A is the Stokes’s law drag
force, ζ u A = 6πµa u A , and F(t) is the random fluctuating force due to the collisions between
the solutes and the solvent molecules. Solving the Langevin stochastic differential equation,
Einstein in 1906 arrived at the following expression for the diffusivity of solutes in dilute
solution
D=
RT
κT
=
6πµ rA N A
6πµ rA
(1.3-3)
In this equation, R is the ideal gas constant (8.314 J/mol⋅oK), T is the absolute temperature in
o
K, NA is Avogadro’s number (6.023×1023 molecules/mol), κ is the Boltzmann constant
(1.38×10-16 erg/K), and µ is the solvent viscosity (g/cm⋅sec). The viscosity of water at 37oC is
0.76cP. Equation (1.3-3) may be used to estimate the radius, rA, of the solute if its diffusivity
is known. If the diffusivity of the solute is not known, its radius might be estimated from the
molecular weight by the expression
 3MW 

rA = 
 4πρN A 
1/ 3
(1.3-4)
This equation assume that the solute is a solid sphere with density ρ ≈ 1 g/cm3.
MW
4
= π r A 3ρ
NA
3
Wilke and Chang1have proposed the following correlation to estimate the diffusivity of
nonelectrolytes in an infinitely dilute solution:
0
DAB
µ 7.4 × 10−8 (Φ B M B )1/ 2
=
T
VbA0.6
(1.3-5)
0
In this equation, DAB
[cm2/s] is the diffusivity of A in very dilute solution in solvent B, MB is
molecular weight of solvent B, T [K] is temperature, µ [cP] is viscosity of solvent, VbA
[cm3/mol] is solute molar volume at its normal boiling point (for water as solute, VbA = 75.6
cm3/mol), and ΦB is the association factor of solvent B:
2.26 for water as solvent


1.9 for methanol as solvent

ΦB = 
1.5 for ethanol as solvent

1 for unassociated solvent,e.g., benzene, ether, heptane
If data are not available, the molar volume at normal boiling point might be estimated from
the following equation
1
Wilke, C. R., and P. Chang, AIChE, 1, 264 (1955)
1-15
VbA = 0.285 VC1.048
(1.3-6)
Hayduk and Minhas2 proposed correlations for diffusivity of solutes in infinite dilute solution
depending on the type of solute-solvent system. For solutes in aqueous solutions:
0
DAB
= 1.25×10-8( VbA−0.19 − 0.292)T1.52µε
ε=
(1.3-7)
9.58
− 1.12
VbA
0
In this equation, DAB
[cm2/s] is the diffusivity of A in very dilute solution in solvent B, T [K]
is temperature, µ [cP] is viscosity of solvent, and VbA [cm3/mol] is solute molar volume at its
normal boiling point. For non-aqueous or non-electrolyte solutions:
0.27
D
0
AB
= 1.55×10
-8 VbB
VbA0.42
T 1.29 σ B0.125
µ 0.92 σ A0.105
(1.3-8)
In this equation σ [dyn/cm] is surface tension at the normal boiling-point temperature. If
values of the surface tension are not known, they may be estimated by the corresponding
states method which is limited to nonpolar liquids:
σ = Pc2 / 3 Tc1/ 3 (0.132αc − 0.287)(1 − Tbr)11/9
(1.3-9)
 T ln ( Pc /1.013) 
Tb
αc = 0.9076 1 − br
 , where Tbr =
1 − Tbr
Tc


In this equation, Tc [K] and Pc [bar] are critical temperature and pressure, respectively. The
restrictions for correlation (1.3-8) are:
-
2
The solvent viscosity should be less than 20 cP.
A dimer value of VbA = 37.4 cm3/mol should be use for water as the solute.
For solvent other than water, methanol, or butanol, the organic acid solute should be
considered a dimer with twice the expected value of VbA.
For nonpolar solutes diffusing into monohydroxy alcohols, VbA should be replaced by
8µ VbA.
If data are not available, use VbA = 0.285Vc1.048
Hayduk, W., and B.S. Minhas, Can. J. Chem. Eng., 60, 295 (1982)
1-16
Example 1.3-13 -----------------------------------------------------------------------------Estimate the diffusivity of ethanol (C2H6O) in a dilute solution of water at 288 K using both
Wilke-Chang equation and Hayduk-Minhas correlation
Data: Critical volume of ethanol, Vc = 167.1 cm3/mol; viscosity of water at 288 K, µ = 1.153
cP.
Solution ---------------------------------------------------------------------------------------------(a) Wilke-Chang equation
0
DAB
µ 7.4 × 10−8 (Φ B M B )1/ 2
=
VbA0.6
T
ΦB = 2.26 for water as solvent
VbA = 0.285 VC1.048 = 0.285(167.1)1.048 = 60.9 cm3/mol
0
DAB
= (7.4×10-8)(288)(2.26×18)1/2/(1.153×60.90.6) = 1.002×10-5 cm2/s
The experimental value for the diffusivity of ethanol in a dilute solution of water at 288 K is
1.0×10-5 cm2/s. The error of the estimate is 0.2% which is not typical.
(b) Hayduk-Minhas correlation
ε=
9.58
9.58
− 1.12 =
− 1.12 = − 0.963
VbA
60.9
0
DAB
= 1.25×10-8( VbA−0.19 − 0.292)T1.52µε
0
DAB
= 1.25×10-8(60.9-0.19 − 0.292)(288)1.52(1.153)-0.963 = 0.991×10-5 cm2/s
The error of the estimate in this case is −0.9%.
Example 1.3-24 -----------------------------------------------------------------------------Estimate the diffusivity of acetic acid (C2H4O2) in a dilute solution of acetone at 313 K using
both Wilke-Chang equation and Hayduk-Minhas correlation using the following data:
Tb, K
Acetic acid 390.4
Acetone
329.2
3
4
T c, K
594.8
508.0
Pc, bar
57.9
47.0
Vc, cm3/mol
171.0
209.0
µ, cP
0.264
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 26
Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 27
1-17
M
60
58
Solution ---------------------------------------------------------------------------------------------(a) Wilke-Chang equation
0
DAB
µ 7.4 × 10−8 (Φ B M B )1/ 2
=
T
VbA0.6
ΦB = 1 for acetone as solvent
VbA = 0.285 VC1.048 = 0.285(171)1.048 = 62.4 cm3/mol
0
DAB
= (7.4×10-8)(313)(1×58)1/2/(0.264×62.40.6) = 5.595×10-5 cm2/s
The experimental value for diffusivity of acetic acid in in a dilute solution of acetone at 313
K is 4.04×10-5 cm2/s. The error of the estimate is 38.5%.
(b) Hayduk-Minhas correlation for non-aqueous solution
σ = Pc2 / 3 Tc1/ 3 (0.132αc − 0.287)(1 − Tbr)11/9
 T ln ( Pc /1.013) 
αc = 0.9076 1 − br

1 − Tbr


For acetone (B):
Tbr =
Tb 329.2
=
= 0.648 ⇒ αc = 7.319 ⇒ σB = 20.0 dyn/cm
Tc
508
For acetic acid (A):
Tbr =
Tb 390.4
=
= 0.656 ⇒ αc = 7.91 ⇒ σA = 26.2 dyn/cm
Tc 594.8
For solvent other than water, methanol, or butanol, the organic acid solute should be
considered a dimer with twice the expected value of VbA. Therefore
VbA = 2×62.4 = 124.8 cm3/mol
For acetone:
VbB = 0.285 VC1.048 = 0.285(209)1.048 = 77.0 cm3/mol
0
DAB
= 1.55×10-8
VbB0.27 T 1.29 σ B0.125
VbA0.42 µ 0.92 σ A0.105
1-18
D
0
AB
-8 
770.27 
= 1.55×10 
0.42 
 124.8 
 3131.29 

0.92 
 0.264 
 200.125 
= 3.84×10-5 cm2/s

0.105 
 26.2

The error of the estimate in this case is −5.0%.
Example 1.3-3 -----------------------------------------------------------------------------A water droplet having a diameter of 0.10 mm is suspended in still air at 50oC, 1.0132×105
Pa (1 atm), and 20% relative humidity. The droplet temperature can be assumed to be at 50oC
and its vapor pressure at 50oC is 7.38 kPa.
1) Calculate the initial rate of evaporation of water if DAB of water vapor in air is 0.288
cm2/s.
2) Determine the time for the water droplet to evaporate completely.
Solution ---------------------------------------------------------------------------------------------T
yd,w d
Tg
Bulk gas phase
yg,w
Droplet surface
The molar flux of water vapor (A) into the air (B) is
NA,r = − cDAB
dy A
+ yA(NA,r + NB,r)
dr
In this equation, r is the radial distance from the center of the drop. For still air NB,r = 0, we
have
NA,r = −
cDAB dy A
1 − y A dr
The system is not at steady state, the molar flux is not independent of r since the area of mass
transfer 4πr2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer
rate, 4πr2NA,r, is assumed to be independent of r at any instant of time.
WA = 4πr2NA,r = − 4πr2
cDAB dy A
= constant
1 − y A dr
Separating the variables and integrating gives
WA ∫
∞
R
y g , w dy
dr
A
= − 4π cDAB ∫
2
y
d ,w 1 − y
r
A
1-19
1 − y g ,w
1 − y g ,w
WA
= 4π cDAB ln
⇒ WA = 4πRcDAB ln
1 − yd , w
1 − yd , w
R
This equation provides the moles of water evaporated from the droplet when the radius of the
drop is R.
1) Calculate the initial rate of evaporation of water if DAB of water vapor in air is 0.288
cm2/s.
The initial rate of evaporation occurs when the diameter of the drop is 0.10 mm.
3 atm
82.057 cm .
mol. K
Rg
R
.005 cm
P
1 atm
T
( 50
273 ) K
2
P
c
c = 3.773 10
R g. T
PA
D AB
.288
3
cm
Mw
s
cm
7.38
101.32
atm
PA
y dw
For 20% relative humidity
WA
mol
5
4 . π . R. c . D AB. ln
y dw = 0.073
P
y gw
1
y gw
1
y dw
.2 .
PA
y gw = 0.015
P
W A = 4.161 10
The initial mass evaporation rate is
8
s
Making a material balance around the water droplet gives
1 − y g ,w
d 4

3
 π R ρ A  = − 4MwπRcDAB ln
1 − yd , w
dt  3

1 − y g ,w
dR
= − 4MwπRcDAB ln
1 − yd , w
dt
Simplifying and rearranging we have
RdR = − Mwc
D AB
ρA
ln
1 − y g ,w
1 − yd , w
dt = − Kdt
1-20
mol
8
7
4.161 . 10 . 18 = 7.49 10
2) Determine the time for the water droplet to evaporate completely.
4ρAπR2
1
g
s
18
g
mol
D AB
1 − y g ,w
. Integrating the equation from the initial radius to
1 − yd , w
zero gives the time for complete evaporation of the droplet.
In this equation, K = Mwc
0
ρA
ln
t
∫ RdR = − K ∫ dt => t =
0
Ri
Ri2
2K
Substituting the numerical values we have
2
y dw
.0728
D AB.ln
y gw
1
y gw
1
y dw
.0146
D AB
.288
cm
s
2
= 0.0175
cm
s
Density of liquid water at 50oC is 0.988 g/cm3
K=
t=
D AB
ρA
ln
1 − y g ,w
1 − yd , w
(Mw)c = (0.0175)(18)(3.773×10-5)/(0.988) = 1.207×10-5 cm2/s
Ri2
= 0.0052/(2×1.207×10-5) = 1.04 s
2K
Example 1.3-4 -----------------------------------------------------------------------------1
Consider the anaerobic fermentation of glucose to ethanol by yeast. Glucose (C6H12O6) is
converted into yeast, ethanol (C2H5OH), the byproduct glycerol (C3H8O3), carbon dioxide,
and water. An empirical chemical formula for yeast can be taken as CH1.74N0.2O0.45. We can
describe the fermentation by the following reaction:
C6H12O6 + aNH3 → b CH1.74N0.2O0.45 + c C2H5OH + d C3H8O3 + e CO2 + f H2O
Determine the stoichiometric coefficients of the above reaction if 0.21 moles of glycerol
were formed for each mole of ethanol produced and 0.13 moles of water were formed for
each mole of glycerol.
Solution ---------------------------------------------------------------------------------------------We can determine the stoichiometric coefficients of the reaction by making the atomic
balance and by using the constraints given in the problem.
Carbon balance:
6 = b + 2c + 3d + e
Hydrogen balance:
12 + 3a = 1.74b + 6c + 8d + 2f
Oxygen balance:
6 = 0.45b + c + 3d + 2e + f
Nitrogen balance:
a = 0.2b
Given constraints:
0.21 moles of glycerol were formed for each mole of ethanol produced:
0.13 moles of water were formed for each mole of glycerol:
1
d = 0.21c
f = 0.13d
Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p. 22.
1-21
We have 6 equations to solve for 6 unknowns a, b, c, d, e, and f. We can reduce the number
of equations to 3 by substituting a = 0.2b, d = 0.21 c, and f = 0.13d to solve for three
unknowns b, c, and e. The three equations are:
6 = b + 2c + 3(0.21c) + e ⇒
b + 2.63c + e = 6
12 + 3(0.2b) = 1.74b + 6c + 8(0.21c) + 2(0.13×0.21c)
1.14b + 7.7346c = 12
6 = 0.45b + c + 3(0.21c) + 2e + 0.13×0.21c ⇒
0.45b + 1.6573c + 2e = 6
The above three equations can be solved using the following Matlab statements:
>> A=[1 2.63 1;1.14 7.7346 0;.45 1.6573 2]
A=
1.0000
1.1400
0.4500
2.6300
7.7346
1.6573
1.0000
0
2.0000
>> b=[6;12;6]
b=
6
12
6
>> x=A\b
x=
0.4029
1.4921
1.6729
Therefore b = 0.4029, c = 1.4921, e = 1.6729, a = 0.2b = 0.0806, d = 0.21c = 0.3133, and f =
0.13d = 0.0407. The balanced equation for the anaerobic fermentation of glucose by yeast
can then be written as:
C6H12O6 + aNH3 → b CH1.74N0.2O0.45 + c C2H5OH + d C3H8O3 + e CO2 + f H2O
C6H12O6 + 0.0806 NH3 → 0.4029 CH1.74N0.2O0.45 + 1.4921 C2H5OH
+ 0.3133 C3H8O3 + 1.6729 CO2 + 0.0407 H2O
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