Vapor diffusivity measurement
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.
g as B
(ai r)
PA2
z=0
PA1
z = z1 at t = 0
z = zt a t t
p ure l i qu id A
Vapor diffusivity measurement
The capillary is placed in an
envelope through which air is
passed.
g as B
(ai r)
PA2
z=0
PA1
z = z1 at t = 0
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.
z = zt a t t
p ure l i qu id A
Vapor diffusivity measurement
The air rate should be low,
constant, and not turbulent. The
falling rate of the meniscus can be
observed remotely with a
cathetometer.
If the length of the diffusion path
changes a small amount over a
long period of time, a pseudosteady state assumption may be
used.
Pseudo-steady state:
-Steady state transport equation
- Unsteady state mass balance
g as B
(ai r)
PA2
z=0
PA1
z = z1 at t = 0
z = zt a t t
p ure l i qu id A
Conditions for quasi or pseudo-steady state assumption:
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
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
dy A
cD AB
NA,z(1  yA) =  cDAB
 NA,zdz = 
dyA
dz
1  yA
z
NA,z  dz =  cDAB
0

yA2
y A1
dy A
1  y A2
 NA,zz = cDABln
1  yA
1  y A1
NA,zz = cDABln
1  y A2
1  y A1
Using the definition of log mean average,
yB,lm =
(1  y A2 )  (1  y A1 )
y B 2  y B1
=
ln( y B 2 / y B1 ) ln[(1  y A2 ) /(1  y A1 )]
1  y A2
y A1  y A2
ln
=
1  y A1
yB,lm
The molar flux of A is then written as
NA,z =
cDAB ( y A1  y A2 )
zyB ,lm
The molar flux is related to the amount of A leaving the liquid by the
material balance
cDAB ( y A1  y A2 )  A,L dz
NA,z =
=
zyB ,lm
M A dt
g as B
(ai r)
PA2
z=0
PA1
z = z1 at t = 0
Separating the variables and integrating
the equation from t = 0 to t gives
t
 A,L
0
MA
 dt =
t=
 A,L
MA
yB ,lm
cDAB ( y A1  y A2 )
yB ,lm
cDAB ( y A1  y A2 )

zt
z1
zdz
 zt2  z12 


 2 
z = zt a t t
p ure l i qu id
t=
 A,L
MA
yB ,lm
cDAB ( y A1  y A2 )
 zt2  z12 


 2 
Replacing c by P/RT yields
t=
 A,L
MA
yB ,lm RT
DAB P( y A1  y A2 )
yB ,lm RT
 zt2  z12   A,L

 =
M A DAB ( PA1  PA2 )
 2 
 zt2  z12 


 2 
P  y A1  y A2 
PB ,lm
y A1  y A2
yB,lm =
=
=
ln[(1  y A2 ) /(1  y A1 )]
P ln[ P(1  y A2 ) / P(1  y A1 )]
P
yB ,lm
PB ,lm
=
PA1  PA2
P( PA1  PA2 )
t=
 A,L
MA
PB,lm RT
DAB P( PA1  PA2 )
 zt2  z12 


 2 
t=
 A,L
MA
PB,lm RT
DAB P( PA1  PA2 )
 zt2  z12 


 2 
Solving for the diffusivity we have
 A,L PB ,lm RT
 zt2  z12 


DAB =
tPM A ( PA1  PA2 )  2 
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
A,L = density of liquid A at T.
1 ft
z
3 ft
2 ft
•
A tank containing water has its top open
to air. The tank is cylindrical with a
diameter of 2 ft. The liquid level is
maintained at a level of 3 ft below the top
of the tank as shown in sketch. The
surrounding air is at 310 K, 1 bar, and 40
percent relative humidity.
A. The molar concentration of water vapor is
independent of z.
• B. The total molar concentration is independent
of z.
A. A and B are true B. Only A is true
C. Only B is true
D. A and B are false
1 ft
z
3 ft
2 ft
•
A tank containing water has its top open
to air. The tank is cylindrical with a
diameter of 2 ft. The liquid level is
maintained at a level of 3 ft below the top
of the tank as shown in sketch. The
surrounding air is at 310 K, 1 bar, and 40
percent relative humidity.
A. If the surrounding air is saturated with water
vapor then the molar flux of H2O is zero.
• B. If air is insoluble in water then the molar flux
of air is zero.
A. A and B are true B. Only A is true
C. Only B is true
D. A and B are false
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.
The molar flux of A (water vapor) is NA,z =  cDAB
Lo
dy A
+ yA(NA,z + NB,z)
dz
Since air is insoluble in water, it is stagnant (or nondiffusing) and NB,z = 0.
dy A
cD AB
NA,z(1  yA) =  cDAB
 NA,zdz = 
dyA
dz
1  yA
NA,z(1  yA) =  cDAB
L
dy A
cD AB
 NA,zdz = 
dyA
dz
1  yA
NA,z  dz =  cDAB
0

y AL
y A0
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
 A,L dLw
cD AB
1
Ac
=  AcNA,z =  Ac
ln
L
1  y A0
M A dt
In this equation, Ac is the cross sectional area of the tube and Lw is the
length of the liquid column.
 A,L dLw
cD
1
Ac
=  AcNA,z =  Ac AB ln
L
1  y A0
M A dt
Since
dL
dLw
=
, we have
dt
dt
 A,L dL
cD AB
1
=
ln
L
1  y A0
M A dt
Separating the variables and integrating the equation from t = 0 to t
gives
t
K  dt =
0
K=
cDAB M A
 A,L
t=
ln

20
10
LdL where
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 DAB
2(.22)
At the end of this time the diffusion path can be determine from
K

227
0
dt =

L
10
LdL
227 =
1
(L2 − 102)
2K
L = (100 + 227×5.149×10-6)0.5 = 10.0001 cm
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.
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
dC
JS =  D S
dx
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)
The movement of a spherical solute of radius a in dilute solution can be
described by an equation of motion, called the Langevin equation


du A
m
=  u A + F(t)
dt
is the instantaneous velocity of the sphere of mass m


 u A = 6a u A = the Stokes’s law drag force
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
T
=
6 rA N A
6 rA
RT
T
D=
=
6 rA N A
6 rA
In this equation, R is the ideal gas constant (8.314 J/molK), T is the
absolute temperature in K, NA is Avogadro’s number (6.0231023
molecules/mol),  is the Boltzmann constant (1.38×10-16 erg/K), and  is
the solvent viscosity (g/cmsec). The viscosity of water at 37oC is 0.76cP.
This equation 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 = 
 4N A 
1/ 3
This equation assume that the
solute is a solid sphere with density
  1 g/cm3.
MW 4
=  r A 3
3
NA
Wilke and Chang have proposed the following correlation to estimate
the diffusivity of nonelectrolytes in an infinitely dilute solution:
0
DAB
 7.4 108 ( B M B )1/ 2
=
VbA0.6
T
0
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,  [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 B is the association factor of
solvent B:
2.26 for water as solvent


1.9 for methanol as solvent

B = 
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
1.048
V
VbA = 0.285 C
Hayduk and Minhas 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)T1.52
9.58
=
 1.12
VbA
0
DAB
[cm2/s] is the diffusivity of A in very dilute solution
T [K] is temperature,  [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:
1.29
0.27
0.125
T
V

0
DAB
= 1.55×10-8 bB0.42 0.92 B0.105
VbA 
A
 [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:
 = Pc2 / 3 Tc1/ 3 (0.132c  0.287)(1  Tbr)11/9
 Tbr ln  Pc /1.013 
Tb
c = 0.9076 1 
 , where Tbr =
Tc
1  Tbr


Tc [K] and Pc [bar] are critical temperature and pressure, respectively.
r
W ater
droplet
•
Consider a water droplet evaporating into
still air. The radius of the droplet is R. Let
A represents water vapor. Temperature of
the droplet is the same as temperature of
the ambient air.
A. The partial pressure of water vapor at the
water-air interface is usually assumed to equal
to the vapor pressure of water.
• B. If the air is dry, the partial pressure of water
vapor at at the water-air interface is zero.
A. A and B are true B. Only A is true
C. Only B is true
D. A and B are false
r
W ater
droplet
•
Consider a water droplet evaporating into
still air. The radius of the droplet is R. Let
A represents water vapor.
A. If the system is at steady state, the molar
flux of A is independent of r.
• B. For quasi steady state assumption, R is
independent of time.
A. A and B are true B. Only A is true
C. Only B is true
D. A and B are false
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.
T
yd,w d
Tg
Bulk gas phase
yg,w
Droplet surface
The molar flux of water vapor (A) into the air (B) is
dy A
NA,r =  cDAB
+ 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 = 
cD AB 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 4r2 is not a constant. Using quasi steady
state assumption, the mass (mole) transfer rate, 4r2NA,r, is assumed to
be independent of r at any instant of time.
WA = 4r2NA,r =  4r2
cD AB dy A
= constant
1  y A dr
cD AB dy A
WA = 4r NA,r =  4r
= constant
1  y A dr
2
2
Separating the variables and integrating gives
WA 

R
y g , w dy
dr
A
=

4
cD
AB
 yd , w 1  y A
r2
1  y g ,w
1  y g ,w
WA
= 4 cDAB ln
 WA = 4RcDAB ln
R
1  yd , w
1  yd , w
2) Determine the time for the water droplet to evaporate completely.
Making a material balance around the water droplet gives
1  y g ,w
d 4

3
  R  A  = − 4MwRcDAB ln
dt  3
1  yd , w

1  y g ,w
dR
4ρAR
= − 4MwRcDAB ln
dt
1  yd , w
2
Simplifying and rearranging we have
RdR = − Mwc
K = Mwc
DAB
A
DAB
A
ln
ln
1  y g ,w
1  yd , w
1  y g ,w
1  yd , w
dt = − Kdt
Integrating the equation from the initial radius to zero gives the time for
complete evaporation of the droplet.
0
Ri2
R RdR = − K 0 dt => t = 2 K
i
o
Air at 50 C, 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.
t
Substituting the numerical values we have
2
y dw
.0728
D ABln
y gw
1
y gw
1
y dw
.0146
D AB
2
 0.0175
cm
s
.288
cm
s
Density of liquid water at 50oC is 0.988 g/cm3
K=
DAB
A
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
t=
= 0.0052/(2×1.207×10-5) = 1.04 s
2K
•
z
O
2
.
L
C
O
,
C
O
2
In a hot combustion chamber, oxygen diffuses through air to
the carbon surface where it reacts to make CO and CO2.
The mole fraction of oxygen at z = 0 is 0.21. The
reaction may be assumed to be instantaneous. No
reaction occurs in the gas film. The two boundary
conditions require to solve for yO2(z) are
A. At z = 0, yO2 = 0.21 and at z = L, yO2 = 0
B. At z = 0, yO2 = 0.21 and at z = L, dyO2/dz = 0
C. At z = 0, yO2 = 0.21 and at z = L, yO2 = minimum
D. None of the above
zO
2
L
C
O
,
C
O
2
In a hot combustion chamber, oxygen diffuses through air to
the carbon surface where it reacts to make CO and CO2.
The mole fraction of oxygen at z = 0 is 0.21. The
reaction may be assumed to be instantaneous. No
reaction occurs in the gas film.
•
A. The mole fraction of carbon in the gas phase
at the interface is zero.
• B. The mole fraction of carbon in the solid phase
at the interface is equal to 1.
A. A and B are true B. Only A is true
C. Only B is true
D. A and B are false