PROBLEM 14.1
KNOWN: Mixture of O2 and N2 with partial pressures in the ratio 0.21 to 0.79.
FIND: Mass fraction of each species in the mixture.
SCHEMATIC:
pO 2
p N2
=
0.21
0.79
MO = 32.00 kg / kmol
2
M N = 28.01 kg / kmol
2
ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: From the definition of the mass fraction,
ρi
ρi
=
ρ Σρ i
m=
i
Hence, with
pi
Mi pi
=
.
( ℜ / M i ) T ℜT
=
ρi
pi
=
R iT
mi =
M i p i / ℜT
ΣM i p i / ℜT
Hence
or, canceling terms and dividing numerator and denominator by the total pressure p,
mi =
Mi x i
.
ΣM i x i
With the mole fractions as
0.21
=
= 0.21
x O2 p=
O2 / p
0.21 + 0.79
=
x N 2 p=
N 2 / p 0.79,
find the mass fractions as
=
mO
2
32.00 × 0.21
= 0.233
32.00 × 0.21 + 28.01× 0.79
m N2 =
1 − mO2 =
0.767.
<
<
PROBLEM 14.2
KNOWN: Mole fraction (or mass fraction) and molecular weight of each species in a mixture of n
species. Equal mole fractions (or mass fractions) of O2, N2 and CO2 in a mixture.
FIND: (a) Equation for determining mass fraction of species i from knowledge of mole fraction and
molecular weight of each of n species. Equation for determining mole fraction of species i from
knowledge of mass fraction and molecular weight of each of n species. (b) For mixture containing
equal mole fractions of O2, N2, and CO2, find mass fraction of each species. For mixture containing
equal mass fractions of O2, N2, and CO2, find mole fraction of each species.
SCHEMATIC:
x=
= 1/ 3
O2 x=
N 2 x CO
2
or
m
=
m=
mCO
= 1/ 3
O
N
2
2
2
MCO = 44.01 kg/kmol
2
=
MO
32.00
=
kg/kmol, M N
28.01 kg/kmol
2
2
ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: (a) With
m=
i
ρi
=
ρ
ρi
=
∑ ρi
pi / R i T
=
∑ pi / R i T
p i M i / ℜT
∑ p i M i / ℜT
i
i
i
and dividing numerator and denominator by the total pressure p,
mi =
Mi x i
.
∑ Mi x i
(1)
<
(2)
<
i
Similarly,
=
xi
pi
=
∑ pi
ρi R i T
=
∑ ρi R i T
( ρ i / M i ) ℜT
∑ ( ρ i / M i ) ℜT
i
i
i
or, dividing numerator and denominator by the total density ρ
xi =
m i / Mi
.
∑ m i / Mi
i
(b) With equal mole fractions of each species, xi = 1/3, using Eq. (1),
MO x O + M N x N + MCO x CO = (32.00 + 28.01 + 44.01) / 3 = 34.7 kg/kmol
2
2
2
2
2
2
=
mO2 0.31,
=
m N 2 0.27,
=
mCO2 0.42.
<
With equal mass fractions of each species, mi = 1/3, using Eq. (2),
mO / MO + m N / M N + mCO / M
=
(1/ 32.00 + 1/ 28.01 + 1/ 44.01) / 3 =
2.99 ×10−2 kmol/kg
2
2
2
2
2 CO2
find
=
x O2 0.35,
=
x N 2 0.40,
=
x CO2 0.25.
<
PROBLEM 14.3
PROBLEM 14.4
KNOWN: Mass fraction of O2, temperature, and pressure for a mixture of O2 and H2.
FIND: Mass density, mole fraction, and partial pressure of each species.
SCHEMATIC:
p = 1 bar
mA = 0.75
T = 318 K
A → O2, MA = 32.00 kg/kmol
B → H2, MB = 2.016 kg/kmol
ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: Since mA = 0.75, mB = 1 – mA = 0.25. The mole fractions can be related to mass
fractions as follows:
x=
A
CA
CA
mA / MA
0.75 / 32.00
=
=
=
= 0.159
C
CA + CB (m A / M A + m B / M B ) (0.75 / 32.00 + 0.25 / 2.016)
<
And
<
xB =
1− xA =
1 − 0.159 =
0.841
From Eq. 14.11,
<
p A = x A p = 0.159 ×1 bar = 0.159 bar
p B = 0.841 bar.
From the ideal gas law, ρi =
a
<
pi
.
R iT
Hence,
rA
0.159 bar
= 0.192 kg / m3 <
(8.314 ×10−2 m3 ⋅ bar / kmol ⋅ K / 32.00 kg / kmol) × 318 K
rB
0.841 bar
= 0.0641 kg / m3 <
2
3
−
(8.314 ×10 m ⋅ bar / kmol ⋅ K / 2.016 kg / kmol) × 318 K
COMMENTS: The mole fraction of hydrogen is greater than its mass fraction because its molecular
weight is less than that of oxygen.
PROBLEM 14.5
KNOWN: He-Xe mixture containing 0.75 mole fraction of He at 300 K and 1 atm.
FIND: Mass fraction of He and mixture mass density, molar concentration, and molecular weight.
Mass of coolant in 20 liters.
SCHEMATIC:
He-Xe mixture
0.75 mole fraction
He
T = 300 K, p = 1 atm
ASSUMPTIONS: Ideal gas mixture.
PROPERTIES: M He = 4.003 kg / kmol , M Xe = 131.3 kg / kmol
ANALYSIS: The molar concentration of the mixture can be found directly from the ideal gas law, in
the form
=
C
p
1 atm
=
= 0.0406 kmol / m3
−
2
3
ℜT 8.206 ×10 m ⋅ atm / kmol ⋅ K × 300 K
<
The mass density of one component in a mixture can be related to the mole fraction by combining Eqs.
14.11 and 14.1 to yield
ρi =Mi x i C
For He this results in
kmol / m3 0.122 kg / m3
=
ρHe 4.003 kg / kmol × 0.75 × 0.0406=
<
Then the total mass density can be found by summing the species mass densities,
ρ=
∑ ρ=i C∑ Mi x=i 0.0406 [ 4.003 kg / kmol × 0.75 + 131.3 kg / kmol × 0.25]
i
i
ρ =1.46 kg / m3
<
Thus the helium mass fraction is
ρHe 0.122 kg / m3
m=
=
= 0.0838
He
ρ
1.46 kg / m3
<
Finally, the molecular weight of the mixture can be found from
Continued…
PROBLEM 14.5 (Cont.)
ρ
1.46 kg / m3
M= =
= 35.8 kg / kmol
C 0.0406 kmol / m3
<
Finally, the mass corresponding to a 20 liter cooling system capacity would be
M = ρV = 1.46 kg/m3 × 20 liters × 10-3 m3/liter = 0.0291 kg
<
COMMENTS: (1) As you may recall from thermodynamics, the molar concentration of an ideal gas
is a function only of pressure and temperature, independent of the species. (2) The mass fraction of
helium is much less than its mole fraction because its molecular weight is so much less than that of
xenon.
PROBLEM 14.6
KNOWN: Mass diffusion coefficients of two binary mixtures at a given temperature, 298 K.
FIND: Mass diffusion coefficients at a different temperature, T = 350 K.
ASSUMPTIONS: (a) Ideal gas behavior, (b) Mixtures at 1 atm total pressure.
-4
2
PROPERTIES: Table A-8, Ammonia-air binary mixture (298 K), DAB = 0.28 × 10 m /s;
-4 2
Hydrogen-air binary mixture (298 K), DAB = 0.41 × 10 m /s.
ANALYSIS: According to treatment of Section 14.1.4, assuming ideal gas behavior,
D AB ~ T3/ 2
where T is in kelvin units. It follows then, that for
NH3 − Air :
D AB ( 350 K ) =
0.28 ×10−4 m 2 / s ( 350 K / 298 K )
3/ 2
D AB ( 350 =
K ) 0.36 ×10−4 m 2 / s
H 2 − Air :
<
D AB ( 350 K ) =
0.41×10−4 m 2 / s ( 350 / 298 )
3/ 2
D AB ( 350 =
K ) 0.52 ×10−4 m 2 / s
COMMENTS: Since the H2 molecule is smaller than the NH3 molecule, it follows that
D H 2 − Air > D NH3− Air
as indeed the numerical data indicate.
<
PROBLEM 14.7
KNOWN: Pressure and temperature. Substance A and Substance B.
FIND: Plot of DAB versus MA for NH3, H2O, CO2, H2, O2, acetone, benzene and naphthalene in air.
ASSUMPTIONS: Ideal gas behavior.
PROPERTIES:
Substance A (T, p)
DAB* (m2/s)
MA (kg/kmol)
NH3 (298 K, 1 atm)
H2O (298 K, 1 atm)
CO2 (298 K, 1 atm)
H2 (298 K, 1 atm)
O2 (298 K, 1 atm)
Acetone (273 K, 1 atm)
Benzene (298 K, 1 atm)
Naphthalene (300 K, 1 atm)
0.28 × 10-4
0.26 × 10-4
0.16 × 10-4
0.41 × 10-4
0.21 × 10-4
0.11 × 10-4
0.88 × 10-5
0.62 × 10-5
17.03 **
18.02 **
44.01 **
2.016 **
32.00 **
58.08 ***
78.11****
128.16 *****
* Table A.8
** Table A.4
*** J.R. Howell and R.O. Buckius, Fundamentals of Engineering Thermodynamics, 2nd ed.,
McGraw-Hill, 1992.
**** M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics, 6th ed.,
John Wiley and Sons, Hoboken, 2008.
***** Example 6.2
ANALYSIS: The mass diffusivity values must be corrected to account for the temperature and
pressure dependence. From Table A.8, DAB ∝ p −1T 3/ 2 and the corrected mass diffusivity for NH3 is
DABc =
0.28 × 10−4 m 2 /s × (1atm /1.5atm ) × ( 320K / 298K )
3/ 2
=×
0.21 10−4 m 2 /s
Repeating the calculation for the other substances yields the following.
Substance A
DABc (m2/s)
NH3
H2O
CO2
H2
O2
Acetone
Benzene
Naphthalene
0.21 × 10-4
0.19 × 10-4
0.12 × 10-4
0.30 × 10-4
0.16 × 10-4
0.93 × 10-5
0.65 × 10-5
0.46 × 10-5
A plot of the corrected mass diffusivities versus molecular weight of Substance A follows.
Continued...
PROBLEM 14.7 (Cont.)
°
According to kinetic theory, the mass diffusivity decreases with increasing molecular weight. This
behavior is readily evident in the plot, and therefore the kinetic theory is consistent with measured
values.
COMMENTS: Small molecules can diffuse through the host medium more readily than large
molecules.
<
PROBLEM 14.8
KNOWN: The inner and outer surfaces of an iron cylinder of 100-mm length are exposed to a
carburizing gas (mixtures of CO and CO2). Observed experimental data on the variation of the carbon
composition (weight carbon, %) in the iron at 1000°C as a function of radius. Carbon flow rate under
steady-state conditions.
SCHEMATIC:
3
PROPERTIES: Iron (1000°C). ρ = 7730 kg/m .Experimental observations of carbon composition
r (mm)
Wt. C (%)
4.49
1.42
4.66
1.32
4.79
1.20
4.91
1.09
5.16
0.82
5.27
0.65
5.40
0.46
5.53
0.28
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial diffusion in a stationary
medium, and (3) Uniform total concentration.
ANALYSIS: (a) For the one-dimensional, radial (cylindrical) coordinate system, Fick’s law is
jA = − D AB A r
dr A
dr
(1)
where Ar = 2πrL. For steady-state conditions, jA is constant, and if DAB is constant, the product
r
dr A
= C1
dr
(2)
must be a constant. Using the differential relation dr/r = d (ln r), it follows that
(3)
so that on a ln(r) plot, ρA is a straight line. See the graph below for this behavior.
Continued ...
PROBLEM 14.8 (Cont.)
(b) To determine whether DC-Fe is a constant for the experimental diffusion process, the data are
represented on a ln(r) coordinate.
Wt. carbon distribution - experimental observations
1.6
1.4
Exp data
Wt Carbon (%)
1.2
1
0.8
0.6
0.4
0.2
0
1.45
1.5
1.55
1.6
1.65
1.7
1.75
ln (r, mm)
Since the plot is not linear, DC-Fe is not a constant. From the treatment of part (a), if DAB is not a
constant, then
must be constant. We conclude that DC-Fe will be lower at the radial position where the gradient is
higher. Hence, we expect DC-Fe to increase with increasing carbon content.
(c) From a plot of Wt - %C vs. r (not shown), the mass fraction gradient is determined at three
locations and Fick’s law is used to calculate the diffusion coefficient,
where the mass flow rate is
3
and ρ = 7730 kg/m , density of iron. The results of this analysis yield,
Wt - C (%)
r (mm)
∆ Wt-C/∆r (%/mm)
1.32
0.955
0.37
4.66
5.04
5.47
-0.679
-1.08
-1.385
DC-Fe × 10
11
6.51
3.79
2.72
2
(m /s)
PROBLEM 14.9
KNOWN: Air is enclosed at uniform pressure in a vertical, cylindrical container whose top and
bottom surfaces are maintained at different temperatures.
FIND: (a) Conditions in air when bottom surface is colder than top surface, (b) Conditions when
bottom surface is hotter than top surface.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform pressure, (2) Ideal gas behavior.
ANALYSIS: (a) If T1 > T2, the axial temperature gradient (dT/dx) will result in an axial density
gradient. However, since dρ/dx < 0 there will be no buoyancy driven, convective motion of the
mixture.
There will also be axial species density gradients, dρ O / dx and dρ N / dx. However, there is no
(
2
2
)
gradient associated with the mass fractions=
dmO / dx 0,=
dm N / dx 0 . Hence, from Fick’s
2
2
law, Eq. 14.12, there is no mass transfer by diffusion.
(b) If T1 < T2 , dρ / dx > 0 and there may be a buoyancy driven, convective motion of the mixture.
However,
=
dmO / dx 0=
and dm N / dx 0, and there is still no mass transfer. Hence, although
2
2
there is motion of each species with the convective motion of the mixture, there is no relative motion
between species.
COMMENTS: The commonly used special case of Fick’s law,
jA = −D AB
dρ A
dx
would be inappropriate for this problem since ρ is not uniform. If applied, this special case indicates
that mass transfer would occur, thereby providing an incorrect result.
PROBLEM 14.10
PROBLEM 14.10 (Cont.)
PROBLEM 14.11
KNOWN: Evaporation of liquid A into a column containing vapor A and B. Species B cannot be
absorbed in liquid A.
FIND: The relationship between the ratio of the molar-average velocity to the species velocity of
species A to the mole fraction of species A.
SCHEMATIC:
ASSUMPTIONS: (1) Steady, one-dimensional diffusion, (2) No chemical reactions.
ANALYSIS: From Section 14.2.2, we know that N"B,x = 0 . From Eq. 14.27,
N"A,x = CA v A,x and=
N"B,x C=
0 or v B,x = 0
B v B,x
(1)
From Eq. 14.29,
v*x = x A v A,x
(2)
Therefore
v*x
= xA
v A,x
<
Ratio of molar-average to species velocity, vx*/vA,x
The relationship is shown in the graph below.
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Mole Fraction of Species A, xA
0.8
1
Continued…
PROBLEM 14.11 (Cont.)
COMMENTS: (1) When the mole fraction of Species A is small and Species B is not in motion, the
molar-average velocity is dominated by Species B and is negligible compared to the non-zero species
velocity of A. In other words, the vapor in the column can be treated as a stationary medium since,
although Species A is in motion, there is very little species A present. (2) When the mole fraction of
Species A is large, there is very little Species B present, and the velocity of the mixture is dominated
by the velocity of Species A. Hence, the velocity ratio approaches unity as mixture becomes
dominated by Species A.
PROBLEM 14.12
KNOWN: Water in an open pan exposed to prescribed ambient conditions.
FIND: Evaporation rate considering (a) diffusion only and (b) convective effects.
SCHEMATIC:
Air, B, φ∞ = 0.25, T∞ = 27°C, p = 1 atm
L = 100 mm
Water, A, 27°C
D = 5 mm
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion, (3) Constant
properties, (4) Uniform T and p, (5) Ideal gas behavior.
-4
2
PROPERTIES: Table A-8, Water vapor-air (T = 300 K, 1 atm), DAB = 0.26 × 10 m /s; Table A-6,
3
Water vapor (T = 300 K, 1 atm), psat = 0.03531 bar, vg = 39.13 m /kg.
ANALYSIS: (a) The evaporation rate considering only diffusion follows from Eq. 14.32 simplified
for a stationary medium. That is,
N A,x =
N′′A,x ⋅ A =
−DABA
dCA
.
dx
Recognizing that φ ≡ pA/pA,sat = CA/CA,sat, the rate is expressed as
CA,∞ − CA,s D ABA
N A,x =
−D ABA
=
CA,sat (1 − φ∞ )
L
L
=
N A,x
0.26 × 10
−4
(
m / s ( π / 4 )( 0.005 m )
2
100 × 10
−3
)
(1 − 0.25 )
2
= 5.43 × 10
39.13 m / kg × 18.02 kg / kmol
3
m
−12
kmol / s
<
where CA,s = 1/ vg MA with M A = 18.02 kg / kmol.
(b)The evaporation rate considering convective effects using Eq. 14.40 is
N A,x
= N′′A,x ⋅ =
A
CD ABA 1 − x AL
ln
.
L
1 − x A,0
Using the ideal gas law, the total concentration of the mixture is
C = p / ℜT = 1.0133 bar / 8.314 ×10−2 m3 ⋅ bar / kmol ⋅ K × 300K = 0.0406 kmol / m3
(
)
Continued…
PROBLEM 14.12 (Cont.)
where p = 1 atm = 1.0133 bar. The mole fractions at x = 0 and x = L are
x=
A,0
p A,s 0.03531 bar
=
= 0.0348
p
1.0133 bar
x=
A,L φ∞ x=
A,0 0.00871.
Hence
3
=
N A,x
0.0406 kmol / m × 0.26 × 10
−4
100 × 10
m / s ( π / 4 )( 0.005 m )
−3
2
m
2
ln
1 − 0.00872
−12
= 5.54 × 10
kmol / s.
1 − 0.0348
<
COMMENTS: For this situation, the advective effect is very small but does tend to increase (by 2%)
the evaporation rate as expected.
PROBLEM 14.13
KNOWN: Spherical droplet of liquid A and radius ro evaporating into stagnant gas B.
FIND: Evaporation rate of species A in terms of pA,sat, partial pressure pA(r), the total pressure p and
other pertinent parameters.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial, species diffusion, (3)
Constant properties, including total concentration, (4) Droplet and mixer air at uniform pressure and
temperature, (5) Ideal gas behavior.
ANALYSIS: From Eq. 14.32 for a radial spherical coordinate system, the evaporation rate of liquid
A into a binary gas mixture A + B is
dCA CA
N A,r =
N A,r
−D ABA r
+
dr
C
2
where Ar = 4πr and NA,r = NA, a constant,
dC
 C 
−D AB ⋅ 4π r 2 ⋅ A .
N A 1 − A  =
C 
dr

From ideal gas behavior, C
=
= p / ℜT,
A p A / ℜT and C
p dp A
N A ( p − pA ) =
−D AB ⋅ 4p r 2 ⋅
ℜT dr
Separating variables, setting definite limits, and integrating
− NA
r dr
p A,r dp A
1
ℜT
=
∫
∫
p A,r p − p A
p 4p D AB ro r 2
o
find that
N A = 4p ro D AB
p − pA ( r )
p
1
ln
ℜT 1 − ro / r
p − p A,o
where=
p A,o p=
A ( ro ) p A,sat , the saturation pressure of liquid A at temperature T.
COMMENTS: Compare the method of solution and result with the content of Section 14.2.2,
Evaporation in a Column.
<
PROBLEM 14.14
2
KNOWN: Clean surface with pure steam has condensate rate of 0.020 kg/m ⋅s for the prescribed
conditions. With the presence of stagnant air in the steam, the condensate surface drops from 28°C to
24°C and the condensate rate is halved.
FIND: Partial pressure of air in the air-steam mixture as a function of distance from the condensate
film.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties including pressure in air-steam
mixture, (3) Ideal gas behavior.
PROPERTIES: Table A-6, Water vapor: psat (28°C = 301 K) = 0.03767 bar; psat (24°C = 297 K) =
-4 2
0.02983 bar; Table A-8, Water-air (298 K, 1 bar): DAB = 0.26 × 10 m /s.
ANALYSIS: The partial pressure distribution of the air as a function of distance y can be found from
the species (A) rate expression, Eq. 14.40,
′′A,y
N=
( CDAB / y ) ln (1 − x A,y ) / (1 − x A,0 ) .
With C =
p / ℜT, x B,y =
1 − x A,y and x B,0 =
1 − x A,0, recognizing that xB = pB/p, find

ℜT 
p B=
y
( y ) pB,0 ⋅ exp  N′′A,y
pD AB 

p B,0 =p − p A,0 =psat ( 28°C ) − psat ( 24°C ) =( 0.03767 − 0.02983) bar =0.00784 bar.
With N ′′A,y =
3.45 × 10−4 kmol / m2 ⋅ s,
− ( 0.020 / 2 ) kg / m2 ⋅ s / 29 kg / kmol =

8.314 × 10−2 m3 ⋅ bar / kmol ⋅ K × 299 K 
−
4
2
p ( y ) = 0.0784 bar × exp  3.45 × 10 kmol / m ⋅ s
y
B
−4 m2 / s


×
×
0.03767
bar
6.902
10


p B ( y=
) 784 kPa × exp ( −0.330y )
<
with pB in [kPa] and y in [mm], where T = 26°C = 299 K, the average temperature of the air-steam
-1 3/2
-4 2
3/2
-4 2
mixture, and DAB ≈ p T = 0.26 × 10 m /s (1/0.03767) (299/298) = 6.902 × 10 m /s.
Selected values for the pressure are shown below and the distribution is shown above:
y (mm)
pB(y) (kPa)
0
784
5
151
10
29.0
15
5.6
COMMENTS: To minimize inert gas effects, the usual practice is to pass vapor over the surfaces so
that the inerts are eventually collected near the outlet region of the condenser. Our estimate shows that
the effective region to be swept is approximately 10 mm thick.
PROBLEM 14.15
PROBLEM 14.16
PROBLEM 14.17
KNOWN: Three-dimensional diffusion of species A in a stationary medium with chemical reactions.
FIND: Derive molar form of diffusion equation. Compare with Eq.14.48b.
SCHEMATIC:
ASSUMPTIONS: (1) Uniform total molar concentration, (2) Stationary medium.
ANALYSIS: The derivation parallels that of Section 14.4.2, except that Eq. 14.43 is applied on a
molar basis. That is,


N A,x + N A,y + N A,z + N
N
A,g − N A,x + dx − N A,y + dy − N A,z + dz =
A,st .
With
∂N A,x
N A,x + dx =
N A,x +
dx,
∂x
N A,y + dy =
....
∂C
N A,x =
−D AB ( dydz ) A ,
N A,y =
....
∂x

N
A,g
∂CA
=

N
N
dxdydz
A ( dxdydz ) ,
A,st
∂t
It follows that
∂ 
∂CA  ∂ 
∂CA  ∂ 
∂CA  
∂CA
 +  DAB
 D AB
 +  DAB
 + NA = .
∂x 
∂x  ∂y 
∂y  ∂z 
∂z 
∂t
If DAB is constant, the foregoing result reduces to Eq. 14.48b.
COMMENTS: Equation 14.48b could also be derived from Eq. 14.48a by using ρA = CAMA.
<
<
PROBLEM 14.18
KNOWN: Gas (A) diffuses through a cylindrical tube wall (B) and experiences chemical reactions at
 .
a volumetric rate, N
A
FIND: Differential equation which governs molar concentration of gas in plastic.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial diffusion, (2) Uniform total molar concentration, (3)
Stationary medium.
ANALYSIS: Dividing the species conservation requirement, Eq. 14.43, by the molecular weight, MA,
and applying it to a differential control volume of unit length normal to the page,


N A,r + N
N
A,g − N A,r + dr =
A,st
where
N A,r =
−2π rD AB
( 2π r ⋅1) N′′A,r =
N A,r=
+ dr N A,r +
∂N A,r
∂r
∂CA
∂r
dr

 ( 2π r ⋅ dr ⋅1)
−N
N
A,g =
A
∂ CA ( 2π rdr ⋅1) 

N
.
A,st =
∂t
Hence
 ( 2π rdr ) + 2π D
−N
A
AB
∂  ∂CA 
∂C
2π rdr A
r
 dr =
∂r  ∂r 
∂t
or
D AB ∂  ∂CA  
∂CA
r
 − NA = .
r ∂r  ∂r 
∂t
<
COMMENTS: (1) The minus sign in the generation term is necessitated by the fact that the reactions
deplete the concentration of species A.
 ( r, t ) , the foregoing equation could be solved for CA (r,t).
(2) From knowledge of N
A
(3) Note the agreement between the above result and the one-dimensional form of Eq. 14.49 for
uniform C.
PROBLEM 14.19
KNOWN: One-dimensional, radial diffusion of species A in a stationary, spherical medium with
chemical reactions.
FIND: Derive appropriate form of diffusion equation.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial diffusion, (2) Uniform total molar concentration, (3)
Stationary medium.
ANALYSIS: Dividing the species conservation requirement, Eq. 14.43, by the molecular weight, MA,
and applying it to the differential control volume, it follows that


N A,r + N
N
A,g − N A,r + dr =
A,st
where
N A,r = −D AB 4π r 2
N A,r=
+ dr N A,r +

N
A,g
∂CA
∂r
∂N A,r
∂r
dr
(
)
∂ CA 4π r 2dr 


2
=

N
N
.
A 4π r dr ,
A,st
∂t
(
)
Hence
(
)
∂
∂C
2
2 ∂CA  dr =

N
4π r 2 A dr
A 4π r dr + 4π
 D ABr

∂r 
∂r 
∂t
or
1
r2
∂
∂CA
2 ∂CA  + N
 =
.
A
 D ABr

∂r 
∂r 
∂t
<
COMMENTS: Equation 14.50 reduces to the foregoing result if C is independent of r and variations
in φ and θ are negligible.
PROBLEM 14.20
KNOWN: Pressure and temperature of hydrogen stored in a spherical steel tank of prescribed
diameter and thickness.
FIND: (a) Initial rate of hydrogen mass loss from the tank, (b) Initial rate of pressure drop in the tank.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, (2) Uniform total
molar concentration, C, (3) No chemical reactions.
ANALYSIS: (a) From Table 14.1
=
N A,r
CA,o − CA,L
=
R m,dif
CA,o
(1/ 4π DAB ) (1/ ri − 1/ ro )
)
(
4π 0.3 ×10−12 m 2 / s 1.5 kmol / m3
N A,r
=
= 7.35 ×10−12 kmol / s
(1/ 0.05 m − 1/ 0.052 m )
or
n A,r = M A N A,r = 2.016 kg / kmol × 7.35 ×10−12 kmol / s = 14.8 ×10−12 kg / s.
<
(b) Applying a species balance to a control volume about the hydrogen,


M
−M
−n A,r
A,st =
A,out =
d ( ρ A V ) pp
D3 dρ A
D3 dp A p D3M A dp A

=
M A,st =
=
=
dt
6 dt
6R A T dt
6ℜT
dt
Hence
(
)
6 0.08314 m3 ⋅ bar / kmol ⋅ K ( 300 K )
dp A
6ℜT
n A,r =
=
−
−
×14.8 ×10−12 kg / s
3
3
dt
pp
D M
(0.11m) 2.016 kg / kmol
A
dp A
=
−2.63 ×10−7 bar / s.
dt
<
2
COMMENTS: If the spherical shell is appoximated as a plane wall, Na,x = DAB(CA,o) πD /L = 8.54
-12
× 10 kmol/s. This result is 16% higher than that associated with the spherical shell calculation.
PROBLEM 14.21
KNOWN: Diameter and wall thickness of spherical rubber container. Pressure, temperature and type
of gas inside. Pressure and temperature of air outside.
FIND: Initial rate of pressure change if the gas is pure nitrogen or pure oxygen.
SCHEMATIC:
Rubber (B)
L = 1 mm
Air, pO2/pN2 = 0.21/0.79
p = 1 bar
T = 293 K
D = 200 mm
∙ A,out
M
N2 or O2, pA = 5 bar
ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, (2) Since L << D,
diffusion may be treated as through a plane wall, (3) Quasi-steady diffusion (pressure variation is
sufficiently slow to permit assuming steady-state conditions for diffusion through the rubber at any
instant, (4) Uniform total mass density, ρ, (5) Ideal gas behavior, (6) Air is 71% nitrogen and 29%
oxygen.
PROPERTIES: Table A-8, O2 in Rubber, (T ≅ 298 K): DAB = 0.21 × 10-9 m2/s; N2 in Rubber, (T ≅
298 K): DAB = 0.15 × 10-9 m2/s. Table A-10, O2 in Rubber, (T ≅ 298 K): S = 3.12 × 10-3 kmol/m3⋅bar;
N2 in Rubber, (T ≅ 298 K): S = 1.56 × 10-3 kmol/m3⋅bar.
ANALYSIS: The solution follows the procedure of Example 14.4. The rate of change of gas pressure
inside the container may be obtained by applying the species conservation requirement, Equation
14.43, to a control volume about the inner gas (A). It follows that
M A,st = − M A,out
or
dM A d ρ AV
′′ A
M A,st =
=
= − nA,x
dt
dt
Recognizing that ρA = MACA and applying the ideal gas law, CA = pA/ ℜ T, the species balance
becomes
(1)
For a stationary medium the absolute flux of species A through the rubber is equal to the diffusion
flux, and from Eq. 14.41 in a planar geometry,
(2)
Continued…
PROBLEM 14.21 (Cont.)
The species densities ρA,1 and ρA,2 pertain to conditions within the rubber at its inner and outer
surfaces, respectively, and may be evaluated from knowledge of the solubility and partial pressure of
gas outside the rubber, through Eq. 14.62. Hence with ρA = MACA,
where pA,1 and pA,2 are the partial pressures of species A inside the container and in the air outside the
container, respectively. Hence Eq. (2) becomes
Substituting into the species balance (Eq. (1)),
(3)
where pA,1 = 5 bar and pA,2 = 0.21 × 1 bar = 0.21 bar for O2 and pA,2 = 0.79 bar for N2. Evaluating Eq.
(3) for oxygen, the initial rate of pressure decrease is:
<
Similarly, for nitrogen,
<
COMMENTS: (1) The assumption that the spherical wall can be treated as a plane wall can be
assessed by evaluating the ratio of diffusion resistances for plane and spherical walls (see Table 14.1).
The result is Rm,dif,plane/Rm,dif,sphere = (L/πD2) / [(1/r1 – 1/r2)/4π] = 1.01. Therefore the assumption is
associated with a 1% error. (2) The molecular weight cancels out in Eq. (3). The difference between
the pressure decrease rates for the two gases is due to their different diffusion coefficients, solubilities,
and partial pressures in the surrounding air, all three of which contribute to making the pressure
decrease rate smaller for nitrogen than for oxygen.
PROBLEM 14.22
PROBLEM 14.23
KNOWN: Oxygen pressures on opposite sides of a rubber membrane.
FIND: Molar diffusive flux of O2 and molar concentrations of O2 outside the rubber on both sides.
SCHEMATIC:
pA,1=1.5bar
L=0.75mm
pA,2=0.5bar
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Stationary medium of uniform
total molar concentration, C = CA + CB, (3) Ideal gas behavior.
-9
2
PROPERTIES: Table A-8, Oxygen-rubber (298 K): DAB = 0.21 × 10 m /s; Table A-10, Oxygen-3
3
rubber (298 K): S = 3.12 × 10 kmol/m ⋅bar.
ANALYSIS: (a) For the assumed conditions
C ( 0 ) − CA ( L )
dC
−D AB A =
N′′A,x =
J∗A,x =
D AB A
.
dx
L
From Eq. 14.62,
3
−3
CA =
( 0 ) Sp=
A,1 4.68 × 10 kmol / m
CA (=
L ) Sp A,2
= 1.56 × 10−3 kmol / m3.
Hence
N ′′A,x
= 0.21 × 10−9 m2 / s
(4.68 ×10−3 − 1.56 ×10−3 ) kmol / m3
0.00075 m
N ′′A,x =×
0.874 10−9 kmol / s ⋅ m2 .
<
(b) From the ideal gas law
C=
A,1
p A,1
=
ℜT
(
1.5 bar
)
= 0.0605 kmol / m3
0.08314 m3 ⋅ bar / kmol ⋅ K 298 K
=
CA,2 C=
=
/ m3 (0.5 /1.5) 0.0202 kmol / m3.
A,1(p A,2 / p A,1 ) 0.0605 kmol
COMMENTS: Recognize that the molar concentrations outside the membrane differ from those
within the membrane; that is, CA,1 ≠ CA(0) and CA,2 ≠ CA(L).
<
<
PROBLEM 14.24
PROBLEM 14.25
KNOWN: Diameter and wall thickness of spherical fused silica container. Temperature and initial
pressure of helium stored inside.
FIND: Time required for the pressure in the container to drop by 1% and 10%. Compare result with
time estimated using initial rate of pressure decrease.
SCHEMATIC:
Fused silica (B)
L = 2 mm
D = 200 mm
x
ρA,1
ρA,2
∙ A,out
M
Helium, pA, i = 4 bar, T = 20°C
ASSUMPTIONS: (1) One-dimensional species diffusion in a stationary medium, (2) Since t << D,
diffusion may be treated as through a plane wall, (3) Quasi-steady diffusion (pressure variation is
sufficiently slow to permit assuming steady-state conditions for diffusion through the fused silica at
any instant, (4) Uniform total mass density, ρ, (5) Ideal gas behavior for helium, (6) Partial pressure of
helium in outside air is negligible.
PROPERTIES: Table A-8, He in Fused Silica, (T = 293 K): DAB = 0.4 × 10-13 m2/s. Table A-10, He in
Fused Silica, (T = 293 K): S = 0.45 × 10-3 kmol/m3⋅bar.
ANALYSIS: The analysis is identical to Example 14.4 up until the final equation for the rate of
change of helium pressure, namely
(1)
where pA is the partial pressure of helium inside the container. Rather than evaluating this expression
using the initial partial pressure of helium on the right hand side, Eq. (1) can be integrated to find
pA(t), thereby accounting for the fact that as pA decreases with time, so does the leakage rate.
Separating variables and integrating from the initial time to an arbitrary time,
where pA,i is the initial partial pressure of helium. Thus,
Continued…
PROBLEM 14.25 (Cont.)
Solving for time yields
The time for the pressure to drop by 1% is therefore,
The initial rate of pressure decrease from Example 14.4 is 2.63 × 10-11 bar/s. Using this value, the time
for a 1% drop in pressure, that is, a drop of 0.04 bar is:
COMMENTS: (1) The leakage rate is very low and the pressure remains nearly constant for many
years. Realistically, any opening created in the sphere in order to fill it would likely have a much
higher leakage rate. (2) The approximate solution is very accurate for a 1% change in pressure, since
the use of the initial leakage rate is a good assumption. The discrepancy increases for a 10% change in
pressure. (3) The quasi-steady approximation assumes that the time for diffusion through the wall to
reach steady-state is short compared to the time scale over which the interior pressure changes. It may
be shown that the time to reach steady state can be roughly estimated as the time when the Fourier
number is equal to 1. For mass transfer, this means Fom = DABt/L2 = 1, or t = L2/DAB = (0.002 m)2/0.4 ×
10-13 m2/s = 108 s. The time for diffusion in the wall to reach steady-state is very long, but it is still
short compared to the time over which significant pressure changes occur. Therefore, the quasi-steady
assumption is valid.
PROBLEM 14.26
PROBLEM 14.27
KNOWN: Thickness of polymer packaging material, temperature and humidity conditions in gas on
either side of the material.
FIND: (a) Solubility of the packaging material, (b) Total water vapor transfer rate for a material that
has 10% of the diffusivity of the material in Example 14.3, (c) Total water vapor transfer rate for a
material that has 10% the solubility of the material in Example 14.3, (d) Total water vapor transfer rate
after coating the exterior surface with a thin film to reduce its solubility by a factor of 9, leaving the
interior surface untreated.
SCHEMATIC:
Polymer
material
Exterior
pA(x = 0)
T = 295 K
φ1 = 0.9
pA(x = L)
Interior
CA,s(x = 0)
= C A,s1
CA,s(x = L)
= C A,s2
T = 295 K
φ2 = 0.1
Surface 2
Surface 1
x
L
ASSUMPTIONS: (1) Constant properties and steady-state conditions, (2) Stationary medium.
PROPERTIES: Table A.6, water (T = 295 K): psat = 0.02617 bars.
ANALYSIS:
(a) For the exterior Surface 1, p A (x = 0) = φ1p A,sat = 0.9 × 0.02617 bars = 0.02355 bars. For the
interior Surface 2, p A (x = L) = φ2 p A,sat = 0.1 × 0.02617 bars = 0.002617 bars. From Example 14.3,
-3
3
=
CA,s2 C=
A,s (x L) = 0.5 × 10 kmol/m so that
S=
0.5 × 10−3 kmol / m3
-3 kmol
p A (x = L)
0.002617 bar = 191 × 10
m3bar
CA,s2
=
<
(b) From Example 14.3, N A,x,p
= 0.32 × 10−15 kmol/s . If the diffusivity is reduced to 10% of its
original value,
N A,x =0.1N A,x,p =0.1 × 0.32 × 10−15 kmol / s =0.32 × 10−16 kmol / s
<
Continued…
PROBLEM 14.27 (Cont.)
(c) If the solubility is reduced to 10% of its original value at both surfaces, CA,s1
= 0.5 × 10−4 kmol / m3
and CA,s2
= 4.5 × 10−4 kmol / m3 . Hence,
N=
0.1N A,x,p
= 0.32 × 10−16 kmol/s
A,x
<
(d) If the solubility of exterior Surface 1 is reduced by a factor of 9, CA,s1 = CA,s1,p/9 = 4.5 × 10-3
kmol/m3/9 = 0.5 × 10-3 kmol/m3 = CA,s2. Hence,
CA,s1 – CA,s2 = 0 and NA,x = 0
<
COMMENT: (1) The same value of the solubility may be found in part (a) by considering
conditions at Surface 1. (2) By manipulating the solubilities of the surfaces independently,
one may eliminate concentration gradients in the material and, in turn, completely eliminate
water vapor transfer by diffusion. Materials that have properties designed to change through
their thickness in order to promote desired behavior are known as functionally-graded
materials.
PROBLEM 14.28
PROBLEM 14.29
KNOWN: Dimensions of N =100 closed-end palladium tubes. Hydrogen (H2) pressures and
temperature on either side of tube wall. Mass diffusivity of atomic hydrogen (H) through the
palladium, and Sievert’s constant.
FIND: Hourly production rate of pure hydrogen (H2).
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties and steady-state conditions, (2) Stationary medium.
PROPERTIES: Hydrogen (H) in palladium, given: DAB = 7 × 10-9 m2/s.
ANALYSIS: The concentration of atomic hydrogen (H) on the outer and inner surfaces of the tube
are
=
C H,o 1.4
kmol
× (0.85 × 15 bar)1/2 = 5.00 kmol/m3
3
1/2
m bar
and
=
C H,i 1.4
kmol
× (6 bar)1/2 = 3.43 kmol/m3
m3bar1/2
The one-dimensional species diffusion resistances for the wall and end of one tube are
=
R m,dif ,w
--3
6
3
ln(r2 / r1 ) ln (0.8 × 10 m + 75 × 10 m) / 0.8 × 10 m 
= 25.5×106 s/m3
=
-3
-9
2
2 πLD AB
2 π × 80 × 10 m × 7 × 10 m /s
and
R m,dif ,e =
t
D ABA c
=
75 × 10-6 m
= 5.33 × 109 s/m3
-9
2
-3
2
7 × 10 m /s × π × (0.8 × 10 m)
The molar transfer rate of atomic hydrogen (H) in one tube is therefore
Continued…
PROBLEM 14.29 (Cont.)
(5.00 − 3.43) kmol
(5.00 − 3.43) kmol
kmol
m3 +
m3 =
NH =
61.9 × 10−9
6 s
9 s
s
25.5 × 10 m3
5.33 × 10 m3
The molar transfer rate of molecular hydrogen (H2) is therefore NH2 = 0.5NH = 31.0 × 10-9 kmol/s.
The total production rate, NH2,t, in kg/h is
NH2,t = NH2 × MH2 × N × t
= 31.0 × 10-9 kmol/s × 2.016 kg/kmol × 3600 s/h × 100 tubes = 0.022 kg/h
<
Comments: (1) The concentrations of hydrogen (H2) in the gas streams are 0.25 kmol/m3 and 0.12
kmol/m3, respectively. (2) Palladium and other nanostructured materials, such as carbon nanotubes,
can store very high concentrations of hydrogen within their atomic matrix.
PROBLEM 14.30
KNOWN: Conditions of the exhaust gas passing over a catalytic surface for the removal of NO.
FIND: (a) Mole fraction of NO at the catalytic surface, (b) NO removal rate.
SCHEMATIC:
xA,L=0.10
A=250cm2
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species diffusion through the
film, (3) Effects of bulk motion on NO transfer in the film are negligible (stationary medium), (4) No
homogeneous reactions of NO within the film, (5) Constant properties, including the total molar
concentration, C, throughout the film.
ANALYSIS: Subject to the above assumptions, the transfer of species A (NO) is governed by
diffusion in a stationary medium, and the desired results are obtained from Eqs. 14.69 and 14.70.
Hence
x A,s
1
0.10
= =
= 0.067. <
x A,s
x A,L 1 + ( Lk1′′ / D AB )
1 + 0.001 m × 0.05 m / s 10−4 m2 / s
Also
N′′A,s = −
k1′′Cx A,L
1 + ( Lk1′′ / D AB )
where, from the equation of state for an ideal gas,
C
=
p
1.2 bar
=
= 0.0187 kmol / m3.
2
3
−
ℜT 8.314 ×10 m ⋅ bar / kmol ⋅ K × 773 K
Hence
0.05 m / s × 0.0187 kmol / m3 × 0.10
−
=
−6.22 ×10−5 kmol / s ⋅ m 2
N′′A,s =
−
4
2
1 + 0.001 m × 0.05 m / s 10 m / s
)
(
or
(
)
M A N′′A,S =
−1.87 ×10−3 kg / s ⋅ m 2 .
n ′′A,S =
30.01 kg / kmol −6.22 ×10−5 kmol / s ⋅ m 2 =
where the molecular weight of nitric oxide (NO) is 30.01 kg/kmol. The molar rate of NO removal for
the entire surface is then
−6.22 ×10−5 kmol / s ⋅ m 2 × 0.025 m 2 =
−1.56 ×10−6 kmol / s
N A,s =
N′′A,s A =
or
n A,S =
−4.67 ×10−5 kg / s.
<
COMMENTS: Because bulk motion is likely to contribute significantly to NO transfer within the
film, the above results should be viewed as a first approximation.
PROBLEM 14.31
KNOWN: Radius of coal pellets burning in oxygen atmosphere of prescribed pressure and
temperature.
FIND: Oxygen molar consumption rate.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional diffusion in r, (2) Steady-state conditions, (3) Constant
properties, (4) Ideal gas behavior, (5) Uniform C and T, (6) Stationary medium.
ANALYSIS: From Equation 14.57,
d  2 dCA 
r
=0
dr 
dr 
dCA / dr =
C1 / r 2
CA =
−C1 / r + C2 .
or
The boundary conditions at r → ∞ and r = ro are, respectively,
CA ( ∞ )= C → C2= C
dx
dC
 ′′ =
N
−CD AB A =
−D AB A
A N′′A ( ro ) =
dr r
dr r
o
o
Hence
−k1′′ ( −C1 / ro + C ) =
−D ABC1 / ro2
)
(
2
k1′′ ( C1 / ro ) + D AB C1 / ro=
k1′′C
C
=
1
or
k1′′C
(
( k1′′ / ro ) + DAB / ro2
)
.
The oxygen molar consumption rate is
dC
k1′′C
N′′A ( ro ) =
−D AB A =
−D AB
dr r
k1′′ro + D AB
o
where
C
=
p
=
ℜT
(
= 8.40 ×10−3 kmol / m3.
8.206 ×10−2 m3 ⋅ atm / kmol ⋅ K 1450 K
1 atm
)
Hence,
−1.71×10−4 m 2 / s
N′′A ( ro ) =
0.1 m / s × 8.40 ×10−3 kmol / m3
)
(
10−4 + 1.71×10−4 m 2 / s
=
−5.30 ×10−4 kmol / s ⋅ m 2
2=
′′A ( ro ) 4π ( 0.001 m ) × 5.30 ×10−4 kmol / s ⋅ m 2
N=
A ( ro ) 4π ro N
2
N A (=
ro ) 6.66 ×10−9 kmol / s.
<
COMMENTS: The O2 consumption rate would increase with increasing k1′′ and approach a limiting
finite value as k1′′ approaches infinity.
PROBLEM 14.32
KNOWN: Pore geometry in a catalytic reactor. Concentration of reacting species at pore opening
and order of catalytic reaction.
FIND: (a) Differential equation which determines concentration of reacting species, (b) Distribution
of reacting species concentration along the pore.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion in x direction, (3)
Stationary medium, (4) Uniform total molar concentration.
ANALYSIS: (a) Apply the species conservation requirement to the differential control volume,
N A,x − k1′′CA (π D ) dx − N A,x + dx =
0, where
(
)
N A,x=
+ dx N A,x + dN A,x / dx dx
and from Fick’s law
π D2
dx A  π D 2
dC

−
=
−
N A,x =
CD
D AB A .
AB


dx  4
4
dx

Hence
−
dN A
π D2
d 2CA
=
− k1′′CA (π D ) dx
= 0
dx − k1′′CA (π D ) dx
D AB
dx
4
dx 2
d 2CA
dx 2
−
4 k1′′
CA =
0.
DD AB
<
(b) A solution to the above equation is readily obtained by recognizing that it is of exactly the
same form as the energy equation for an extended surface of uniform cross section. Hence for
boundary conditions of the form
CA ( 0 ) = CA,0 ,
−D AB ( dCA / dx ) x = L = k1′′CA ( L )
the solution must be analogous to that obtained for a fin with a convection tip condition. With the
analogous quantities
1/ 2
1/ 2
CA ↔ θ ≡ T − T∞ ,
m ≡ ( 4 k1′′ / DD AB )
↔ ( 4h / Dk )
D AB ↔ k,
k1′′ ↔ h
the solution is, by analogy to Eq. 3.75
CA ( x ) =
cosh m ( L − x ) + ( k1′′ / mD AB ) sinh m ( L − x )
.
cosh mL + ( k1′′ / mDAB ) sinh mL
2
<
COMMENTS: The total pore reaction rate is – DAB(πD /4) (dCA/dx)x=0, which can be inferred by
applying the analogy to Eq. 3.76.
PROBLEM 14.33
PROBLEM 14.34
KNOWN: Mass flow rate of gas containing palladium (species A), which flows through a tube and
deposits into pores of tube wall. Inlet mass concentration of palladium. Mass transfer coefficient
between gas and tube surface. Deposition rate is proportional to mass concentration of palladium at
tube surface.
FIND: (a) Expression for variation of mean species density of palladium with x. Expression for local
deposition rate for tube of diameter D. (b) Ratio of deposition rates at x = L and x = 0.
SCHEMATIC:
D
Gas
 , ρA,m,i
m
ρA,m, hm
x=0
x=L
−n′′A,s =ρ
k1 A,s
ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Constant mass flow rate, (4)
Negligible leakage of gas through porous walls.
ANALYSIS: (a) Section 8.9 develops the variation of mean species density, ρA,m, for the case in
which the surface species concentration, ρA,s, is uniform. Here, however, the surface concentration
will vary as the mean species density decreases with x. Under steady-state conditions, the mass flux
of palladium reaching the surface must equal the mass flux of palladium depositing into the pores.
Referring to Equation 8.82, where n′′A,s is the mass flux from the surface,
n′′A,s = h m (ρ A,s − ρ A,m ) = − k1ρ A,s
Solving for the surface concentration yields ρA,s = hmρA,m/(hm + k1). Then substituting this into either
expression for n′′A,s yields
n′′A,s =
− U mρ A,m ,
1 1/ h + 1/ k
U−
m =
m
1
Comparing this result with Equation 8.82, we see that they are analogous if we replace h m with Um
and ρA,s with 0. Applying the same analogy to Equation 8.86, the distribution of the mean species
density is
ρ A,m (x)
 U ρP 
= exp  − m x 

ρ A,m,i
m


(1)
<
where P is the perimeter, P = πD. Note that we could have found this same result by expressing mass
species conservation for species A. Noting that the rate at which species A is carried downstream by
 ρ A,m / ρ , and assuming ρ to be constant, we have
the flow is m
Continued…
PROBLEM 14.34 (Cont.)
 dρ A,m
m
=
n′′A,s P =
− U m Pρ A,m
ρ dx
Integrating with respect to x and applying the inlet condition yields the same result as Equation (1).
The local deposition rate is
x
 U ρP 

− n′′A,s =
U mρ A,m =
U mρ A,m,i exp  − m x  =
U mρ A,m,i exp  − B  (2) <

m
L



.
where B = UmρPL/ m
(b) The ratio of deposition rates at x = L and x = 0 is
ratio of deposition rates
= exp ( − B)
<
COMMENT: From Eq. (2), the deposition rate decreases exponentially with distance x. Therefore,
as the tube length increases, the deposit thickness at the outlet end will become thinner, and the
variation in deposit thickness between the inlet and outlet will increase.
PROBLEM 14.35
PROBLEM 14.35 (Cont.)
PROBLEM 14.36
PROBLEM 14.36 (Cont.)
PROBLEM 14.37
KNOWN: Combustion at constant temperature and pressure of a hydrogen-oxygen mixture adjacent
to a metal wall according to the reaction 2H2 + O2 → 2H2O. Molar concentrations of hydrogen,
3
oxygen, and water vapor are 0.10, 0.10 and 0.20 kmol/m , respectively. Generation rate of water
-2
3
vapor is 0.96 × 10 kmol/m ⋅s.
FIND: (a) Expression for CH
2
as function of distance from wall, plot qualitatively, (b) CH
wall, (c) Sketch also curves for CO
2(
2
at the
x ) and CH 2O ( x ) , and (d) Molar flux of water at x = 10mm.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion, (3) Stationary
medium, (4) Constant properties including pressure and temperature.
-5
PROPERTIES: Species binary diffusion coefficient (given, for H2, O2 and H2O): DAB = 0.6 × 10
2
m /s.
ANALYSIS: (a) The species conservation equation, Eq. 14.48b, and its general solution are


d 2CA N
N
A x+C x+C .
(1,2)
+ A =
0
CA ( x ) =
−
1
2
2
D
2D
dx
AB
AB
The boundary condition at the wall must be dCA(0)/dx = 0, such that C1 = 0. For the species


hydrogen, evaluate C2 from knowledge of CH (10 mm ) = 0.10 kmol / m3 and N
H 2 = − N H 2O ,
2
according to the chemical reaction. Hence,
−0.96 ×10−2 kmol / m3 ⋅ s )
(
0.10 kmol / m3 =−
( 0.010 m )2 + 0 + C2
2 × 0.6 ×10−5 m 2 / s
C2 = 0.02 kmol / m3.
Hence, the hydrogen species concentration distribution is

N
H2 2
−
CH 2 ( x ) =
x + 0.02 =
800x 2 + 0.02
2D AB
<
which is parabolic with zero slope at the wall; see sketch above.
(b) The value of CH at the wall is,
2
CH 2 ( 0 ) =
0.02 kmol / m3.
( 0 + 0.02 ) kmol / m3 =
<
Continued …
PROBLEM 14.37 (Cont.)
(c) The concentration distribution for water vapor species will be of the same form,

N
H 2O 2
CH 2O ( x ) =
−
x + C1x + C2
2D AB
(3)
With C1 = 0 for the wall condition, find C2 from CH O (10 mm ) ,
2
( 0.96 × 10 kmol / m ) ( 0.010 m ) + C
0.20 kmol / m =
−
−2
3
3
3
2
2 × 0.6 × 10
−5
2
C2 =
0.28 kmol / m .
2
m /s
Hence, CH O at the wall is,
2
CH 2O ( 0 ) = 0 + 0 + C2 = 0.28 kmol / m3


and its distribution appears as above. Recognizing that N
O = −0.5N H O , by the same analysis, find
2
2
CO2 ( 0 ) = 0.06 kmol / m3
and its shape, also parabolic with zero slope at the wall is shown above.
(d) The molar flux of water vapor at x = 10 mm is given by Fick’s law
N′′H 2O,x = −D AB
dCH 2O
dx
and using the concentration distribution of Eq. (3), find

d  N
H 2O 2 

−D AB
+N
N′′H 2O,x =
x =
−
H 2O x

dx  2D AB

and evaluation at the location x = 10 mm, the species flux is
N′′H O,
2
x
)
(
(10 mm ) =+ 0.96 × 10−2 kmol / m3 ⋅ s × 0.010 m =9.60 × 10−5 kmol / m 2 ⋅ s.
<
COMMENTS: Note that the generation rate of water vapor is a positive quantity. Whereas for H2


and O2, species are consumed and hence N
H and N O are negative. According to the chemical
2
2
reaction one mole of H2 and 0.5 mole of O2 are consumed to generate one mole of H2O. Therefore,

 O and N


N
−N
−0.5 N
H2 =
H2
O2 =
H 2O .
PROBLEM 14.38
KNOWN: Molar concentrations of oxygen at inner and outer surfaces of lung tissue. Volumetric rate
of oxygen consumption within the tissue.
FIND: (a) Variation of oxygen molar concentration with position in the tissue, (b) Rate of oxygen
transfer to the blood per unit tissue surface area.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional species transfer by diffusion
through a plane wall, (3) Homogeneous, stationary medium with uniform total molar concentration
and constant diffusion coefficient.
ANALYSIS: (a) From Eq. 14.71 the appropriate form of the species diffusion equation is
d 2CA
D AB
− ko =
0.
2
dx
Integrating,
dCA=
/ dx
( k o / DAB ) x + C1
=
CA
ko
x 2 + C1x + C2 .
2D AB
With CA CA
=
=
and CA CA=
( 0 ) at x 0=
( L ) at x L,
=
C2 CA =
C1
( 0)
CA ( L ) − CA ( 0 )
L
−
koL
.
2DAB
Hence
C
=
A (x)
ko
x
x ( x − L ) + CA ( L ) − CA ( 0 )  + CA ( 0 ) .
2D AB
L
<
(b) The oxygen assimilation rate per unit area is
N′′A,x ( L ) = −D AB ( dCA / dx ) x = L
 k L
k L  D
N′′A,x ( L ) =
−D AB  o − o  − AB CA ( L ) − CA ( 0 ) 
L
 D AB 2D AB 
k L D
N′′A,x =
− o + AB CA ( 0 ) − CA ( L )  .
2
L
COMMENTS: The above model provides a highly approximate and simplified treatment of a
complicated problem. The lung tissue is actually heterogeneous and conditions are transient.
<
PROBLEM 14.39
KNOWN: Ground level flux of NO2 in a stagnant urban atmosphere.
FIND: (a) Vertical distribution of NO2 molar concentration, (b) Critical ground level flux of NO2,
N′′A,0,crit .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional diffusion in a stationary medium,
(3) Total molar concentration C is uniform, (4) Ideal gas behavior.
ANALYSIS: (a) For the prescribed conditions the molar concentration of NO2 is given by Eq. 14.73,
subject to the following boundary conditions.
N′′A,0
dCA 
=−
.

dx  x = 0
D AB
CA ( ∞ ) =0,
From the first condition, C1 = 0. From the second condition,
−mC2 =
− N′′A,0 / D AB .
Hence
CA ( x ) =
N′′A,0
mD AB
where m = (k1/DAB)
1/2
e− mx
<
.
(b) At ground level, CA ( 0 ) =
ℜT
p A=
( 0 ) C A ( 0 )=
N′′A,0
mD AB
ℜTN′′A,0
mDAB
-4 1/2
Hence, with m = (0.03/0.15 × 10 )
N′′A,0,crit
=
. Hence, from the ideal gas law,
.
-1
-1
m = 44.7 m .
mDABp A ( 0 )crit 44.7 m −1 × 0.15 ×10−4 m 2 / s × 2 ×10−6 bar
=
ℜT
8.314 ×10−2 m3 ⋅ bar / kmol ⋅ K × 300 K
N′′A,0,crit =
5.38 ×10−11 kmol / s ⋅ m 2 .
<
COMMENTS: Because the dispersion of pollutants in the atmosphere is governed strongly by
convection effects, the above model should be viewed as a first approximation which describes a worst
case condition.
PROBLEM 14.40
KNOWN: Ground level flux of NO2 in a stagnant urban atmosphere.
FIND: (a) Governing differential equation and boundary conditions for the molar concentration of
NO2, (b) Concentration of NO2 at ground level three hours after the beginning of emissions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional diffusion in a stationary medium, (2) Uniform total molar
concentration, (3) Constant properties.
ANALYSIS: (a) Applying the species conservation requirement, Eq. 14.43, on a molar basis to a unit
area of the control volume,
∂CA
N′′A,x − ( k1CA ) dx − N′′A,x + dx =
dx.
∂t
)
(
With N′′A,x + dx =N′′A,x + ∂N′′A,x / ∂x dx and N′′A,x =− D AB ( ∂CA / ∂x ) , it follows that
∂ 2CA
∂CA
− k1CA =
.
∂t
∂x 2
<
Initial Condition:
CA ( x, 0 ) = 0.
<
Boundary Conditions:
∂C 
=
−D AB A 
N′′A,0 ,
∂x  x = 0
D AB
=
CA ( ∞, t ) 0.
<
(b) The present problem is analogous to Case (2) of Fig. 5.7 for heat conduction in a semi-infinite
medium. Hence by analogy to Eq. 5.62, with k ↔ D AB and a ↔ D AB ,
1/ 2
 t 
CA ( x, t ) 2N′′A,0 
=

 p D AB 



x 2  N′′A,0 x
x

−
exp  −
erfc 
 4D AB t  D AB
 2 ( D t )1/ 2 


AB


At ground level (x = 0) and 3h,
1/ 2
 t 
CA ( 0,3h ) = 2N′′A,0 

 π D AB 
(
C A ( 0, 3h ) = 2 3 × 10
−11
kmol / s ⋅ m
2
)(10, 800 s / π × 0.15 × 10 m / s )
−4
2
1/ 2
= 9.08 × 10
−7
3
kmol / m .
<
COMMENTS: The concentration decays rapidly to zero with increasing x, and at x = 100 m it is, for
all practical purposes, equal to zero.
PROBLEM 14.41
PROBLEM 14.42
PROBLEM 14.43
KNOWN: Consider Problem 14.42. How long should the bubbles remain in the water to achieve an
average vapor concentration that is 50% of the maximum value?
FIND: (a) Time to reach 95% of the maximum average water vapor concentration, (b) Time to reach
50% of the maximum average water vapor concentration.
SCHEMATIC:
ro = 1 mm
Water
T = 25°C
Air
CA,s
ASSUMPTIONS: (1) One-dimensional radial diffusion of vapor in air, (2) Constant properties, (3)
Air is initially dry, (4) Stationary medium.
PROPERTIES: Table A.8, Water vapor-air (300 K): DAB = 0.26 × 10-4 m2/s.
ANALYSIS: (a) We may employ the one-term approximation to the infinite series solution (Eq.
5.55)
Q
3θ *
=
C1 exp −ζ 12 Fo
1 − 3o [sin(ζ 1 ) − ζ 1 cos(ζ 1 )] ; θ o* =
Qo
ζ1
(
)
By analogy, the preceding equations may be written as
CA − CA,s
3γ *
1 − 3o [sin(ζ 1 ) − ζ 1 cos(ζ 1 )] ; γ o* =
C1 exp −ζ 12 Fom
=
=
CA,max
C
C
−
ζ1
A,i
A,s
(
CA
)
Using values of C1 = 2.000 and ζ1 = 3.1415 from Table 5.1 for Bim→ ∞, it follows that
3γ o*
0.95 =
1−
=
[sin(3.1415) − 3.1415cos(3.1415)] ;
CA,max
3.14153
CA
γ o*
=
CA − CA,s
= 2.0exp −3.14152 Fom
CA,i − CA,s
(
)
Solving the two equations yields γ o* = 0.1645, Fom = 0.2531. Since Fom > 0.2, the approximate solution
is valid. Hence,
t=
Fom ro2 / DAB =
0.2531( 0.001m ) / 0.26 × 10−4 m 2 / s =
9.7 × 10−3 s = 9.7 ms
2
<
(b) The time associated with an average water vapor concentration of 50% is expected to be
significantly shorter than in part (a). Hence, Fom may be less than 0.2 and the one-term approximation
to the exact solution may not be valid. Therefore, we employ the approximate solution of Section 5.8
and apply the analogy between heat and mass transfer.
Continued…
PROBLEM 14.43 (Cont.)
From Table 5.2a for Fo < 0.2,
1
q′′r
αt
q* = s o
=
− 1 ; Fo = 2
k (Ts − Ti )
π Fo
ro
Substituting the expression for Fo into the first equation yields
=
qs′′
k (Ts − Ti )  ro −1/2 
− 1
 πα t
ro


We desire an expression for Q/Qo. Hence,
t
4π ro2
∫
qs′′dt
t
Q
3α  ro −1/2 
t =0
=
=
− 1 dt
t

Qo (4 / 3)π ro3r c(Ts − Ti ) ro2
πα


t =0
∫
or
Q 3α  2 r

= 2  o t1/ 2 − t 
Qo ro  πα

 2
= 3
 π

Fo − Fo 

Applying the analogy between heat and mass transfer,
CA
 2
= 0.50
= 3
CA,max
 π

Fom − Fom 

from which Fom = 0.0305. Hence,
t=
Fom ro2 / DAB =
0.0305 ( 0.001m ) / 0.26 × 10−4 m 2 / s =
1.17 × 10−3 s = 1.17 ms
2
<
COMMENTS: (1) Use of the approximate solution of Section 5.8 is not valid for part (a) since its use
yields Fom = 0.366, which does not satisfy the criterion Fom < 0.2. (2) Use of the one-term
approximation to the exact solution for part (b) yields a mass transfer Fourier number of Fom = 0.0198,
which does not satisfy the criterion Fom > 0.2.
PROBLEM 14.44
PROBLEM 14.45
KNOWN: Thick plate of pure iron at 1000°C subjected to a carburizing process with sudden
exposure to a carbon concentration CC,s at the surface.
FIND: (a) Consider the heat transfer analog to the carburization process; sketch the mass and heat
transfer systems; explain correspondence between variables; provide analytical solutions to the mass
and heat transfer situation; (b) Determine the carbon concentration ratio, CC (x, t)/CC,s, at a depth of 1
mm after 1 hour of carburization; and (c) From the analogy, show that the time dependence of the
1/ 2
mass flux of carbon into the plate can be expressed as n C
; also, obtain an
′′ = ρ C,s D C − Fe / π t
b
g
expression for the mass of carbon per unit area entering the iron plate over the time period t.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient diffusion, (2) Thick plate approximates a semiinfinite medium for the transient mass and heat transfer processes, and (3) Constant properties, (4)
Stationary medium.
ANALYSIS: (a) The analogy between the carburizing mass transfer process in the plate and the heat
transfer process is illustrated in the schematic above. The basis for the mass - heat transfer analogy
stems from the similarity of the conservation of species and energy equations (Eqs. 14.77 and 5.29,
respectively), the general solution to the equations, and their initial and boundary conditions. For both
processes, the plate is a semi-infinite medium with initial distributions, CC (x, t ≤ 0) = CC,i = 0 and T
(x, t ≤ 0) = Ti, suddenly subjected to a surface potential, CC (0, t > 0) = CC,s and T (0, t > 0) = Ts. The
heat transfer situation corresponds to Case 1, Figure 5.7, from which the following relations were
obtained.
Heat transfer
Distributions
Mass transfer
T ( x, t ) − Ts
 x 
= erf 

Ti − Ts
 2 αt 
Cc ( x, t ) - Cc,s
0 - Cc,s


x
= erf 

 2 DC-Fe t 


or


Cc ( x, t )
x
= erfc 



Cc,s
 2 DC-Fe t 
Fluxes
qs" (t ) =
k (Ts − Ti )
πα t
nC" ,s (t ) =
DC-Fe ( ρC ,s )
π DC-Fe t
Continued …
PROBLEM 14.45 (Cont.)
(b) Using the concentration distribution expression above, with L = 1 mm, t = 1 h and
-11 2
DC-Fe = 3 × 10 m /s, find the concentration ratio,
(c) From the species flux expression above, the mass flux of carbon can be written as
n"C,s =
ρC,s ( DC− Fe / π t )
<
1/ 2
The mass per unit area entering the plate over the time period follows from the integration of the rate
expression
=
M′′C ( t )
t
t
1/2
1/2
=
n"C,s dt ρC,s ( DC-Fe / π )
=
t -1/2 dt 2 ρC,s ( DC - Fe t/π )
∫0
∫0
<
PROBLEM 14.46
KNOWN: Radius of pharmaceutical product, density of the active ingredient, partition coefficient,
and binary diffusion coefficient of the active ingredient in the gastrointestinal tract.
FIND: (a) Dosage delivered over 5 hours from a D = 6 mm diameter tablet, (b) Dosage delivered
over 5 hours from N = 200 small, spherical tablets of the same mass.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial diffusion of active ingredient, (2) Constant properties,
(3) Stationary medium, (4) Constant sphere radius.
PROPERTIES: Given: DAB = 0.40 × 10-10 m2/s, K = 3 × 10-2.
ANALYSIS: (a) The approximate solution of Chapter 5 for external conduction from an isothermal
sphere and the heat-mass transfer analogy will be used. From Table 5.2a,
1
+ 1 or,
π Fo
q *=
( Fo)
qs′′ro
=
k (Ts − T∞ )
ro
πα t
+1
(1,2)
By analogy, Eq. 2 may be written as
′′ s ro
nA,
=
DAB ( rr
A,s − A,∞ )
ro
+1
π DABt
(3)
Rearranging,
′′ s
=
nA,
DAB / π ( rrrr
DAB ( A,s − A,∞ )
A,s − A,∞ )
+
ro
t
For ρA,∞ = 0, the dosage (for As = 4πro2) is
t
=
D
∫
′′ s dt
=
4π ro2 nA,
0
t
2
∫ 4π ro  DAB / π ( r A,s )t
o
−1/2
+
DAB ( r A,s ) 
 dt
ro

With ρA,s = KρA, the preceding expression yields
=
D 4π ro K r A  2 ro DAB / π t1/ 2 + DABt 


(4)
Continued…
PROBLEM 14.46 (Cont.)
Substituting the appropriate values into Eq. 4 results in
 2 × 3 × 10−3 m 0.4 × 10−10 m 2 / s × 18000s / π 

D= 4π × 3 × 10−3 m × 3 × 10−2 × 15kg/m3 × 
 +0.4 × 10−10 m 2 / s × 18000s

D = 60.9 × 10-9 kg = 60.9 × 10-6 g = 60.9 µg
<
(b) For the same initial mass and N = 200 tablets,
4 3
N4 3
ro,1 / N 1/ 3 =
3 × 10−3 m / 2001/ 3 =
0.513 mm
π ro,1 =
π ro, N or ro, N =
3
3
The dosage is D = ND1 where D1 is the dosage for one tablet. Hence,
 2 × 0.513 × 10−3 m 0.4 × 10−10 m 2 / s × 18000s / π 

D= ND1= 200 × 4π × 0.513 × 10−3 m × 3 × 10−2 × 15kg/m3 × 
 +0.4 × 10−10 m 2 / s × 18000s

D = 703 × 10-9 kg = 703 × 10-6 g = 703 µg
<
COMMENTS: (1) The dosage is controlled by the tablet size. In this example, the medication dosage
is increased by over an order of magnitude by replacing the single tablet with the encapsulated, smaller
diameter spherical tablets. (2) The initial mass of the medication is M = 4/3ρπro3 =
4/3×15kg/m3×π×(3×10-3m)3 = 1.696×10-6 kg = 1.696×10-3 g = 1.696 mg = 1696 µg. For the smaller
tablets, the mass of medication left after 5 hours is 1696 µg – 703 µg = 993 µg. Hence, the tablet
radius after 5 hours is r5h = 0.513mm×(993/1696)1/3 = 0.429 mm. The assumption of a constant radius
is marginally valid.
PROBLEM 14.47
PROBLEM 14.47 (Cont.)
PROBLEM 14.48
KNOWN: Carbon dioxide concentration at water surface and reaction rate constant.
FIND: (a) Differential equation which governs variation with position and time of CO2 concentration
in water, (b) Appropriate boundary conditions and solution for a deep body of water with negligible
chemical reactions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional diffusion in x, (2) Constant properties, including total density
ρ, (3) Water is stagnant, (4) Stationary medium.
ANALYSIS: (a) From Eq. 14.47b, it follows that, for the prescribed conditions,
∂ 2 ρA
∂ρ A
D AB
− k1ρ A =
.
2
∂t
∂x
<
The first term on the left-hand side represents the net transport of CO2 into a differential control
volume by diffusion. The second term represents the rate of CO2 consumption due to chemical
reactions. The term on the right-hand side represents the rate of increase of CO2 storage within the
control volume.
(b) For a deep body of water, appropriate boundary conditions are
ρ A ( 0, t ) = ρ A,0
ρ A ( ∞, t ) =
0
and, with negligible chemical reactions, the species diffusion equation reduces to
∂ 2 ρA
∂x 2
=
∂ρ A
.
D AB ∂t
1
With an initial condition, ρA(x,0) ≡ ρA,i = 0, the problem is analogous to that involving heat transfer in
a semi-infinite medium with constant surface temperature. By analogy to Eq. 5.60, the species
concentration is then


rr
x
A ( x, t ) − A,0

= erf 
 2 ( D t )1/ 2 
− r A,0


AB

.
 2 ( D t )1/ 2 
AB


rr
A ( x, t ) = A,0erfc 
x

<
PROBLEM 14.49
PROBLEM 14.49 (Cont.)
PROBLEM 14.49 (Cont.)
PROBLEM 14.50
PROBLEM 14.50 (Cont.)
PROBLEM 14.50 (Cont.)
PROBLEM 14.51
KNOWN: Dimensions of polymer sheet. Temperature and relative humidity of environment.
Increase in mass of sheet over 24 and 48 hour periods.
FIND: Solubility and mass diffusivity of water vapor in polymer, assuming mass diffusivity is greater
than 7 × 10-13 m2/s.
SCHEMATIC:
2L = 1 mm
Polymer sheet
2
A `= (100 mm)
∆M = 0.012 mg, t = 24 hr
∆M = 0.016 mg, t = 48 hr
T = 300 K
φi = 0,
φ = 0.95, t > 0
ASSUMPTIONS: (1) Constant properties, (2) One-dimensional mass diffusion, (3) Mass gain is
solely due to water vapor diffusing into the sheet.
PROPERTIES: Table A.6, saturated water (T = 300 K), pA,sat = 0.03531 bar, MA = 18.02 kg/kmol.
ANALYSIS: The process of diffusion of water vapor (A) in the polymer (B) sheet is governed by Eq.
14.77 with boundary and initial conditions given by Eqs. 14.78 through 14.80. These equations can be
cast in nondimensional form as in Eqs. 14.83 through 14.86, and the analogy with Eqs. 5.37 through
5.40 is apparent (for Bi →∞), with the analogous quantities defined in Table 14.2.
Since the mass gained by the polymer sheet is known, it is more convenient to work the problem in
mass terms. Making use of Eq. 14.1, we recognize that γ * defined in Eq. 14.81 can be written in the
alternative form,
=
γ*
CA − CA,s
ρ A − ρ A,s
=
CA,i − CA,s ρ A,i − ρ A,s
(1)
Here ρ A,i =
0 since the sheet is initially dry, and ρ=
A,s M AC=
A,s M ASp A,∞ , or
ρ A,s = M ASφp A,sat
(2)
where S is the solubility (see Eq. 14.62).
The mass gained by the polymer sheet is then analogous to the energy transfer, Q, in the heat transfer
problem. Specifically, for the mass loss, we can write a sequence of equations analogous to Eqs. 5.46
through 5.48,
Continued…
PROBLEM 14.51 (Cont.)
Mass loss = ∆M A = −[M A (t) − M A (0)] = − ∫ [ρ A (x, t) − ρ A,i ]dV
∆M A,o ≡ V(ρ A,i − ρ A,s )
and
[ρ A (x, t) − ρ A,i ] dV 1
∆M A
(1 − γ*)dV
= −∫
=
V V∫
∆M A,o
ρ A,i − ρ A,s
If Fom > 0.2, the solution can be approximated by the first term in the series, and the result for the
mass loss would be analogous to Eq. 5.49. To determine if the first term approximation can be used,
we estimate the mass transfer Fourier number with knowledge that the mass diffusivity is greater than
7 × 10-13 m2/s,
Fom = DABt/L2,
Fom > 7 × 10-13 m2/s × 24 h × 3600 s/h / (0.0005 m)2 = 0.24
Thus the one-term approximation is valid and by analogy to Eqs. 5.49 and 5.44, the nondimensional
mass loss is given by
or
∆M A
sin ζ1
= 1−
C1 exp( −ζ12 Fo m )
∆M A,o
ζ1
(3)
−DM A
sin ζ1
D
= 1−
C1 exp( −ζ12 AB t)
ρ A,sV
ζ1
L2
(4)
From Table 5.1 for Bi →∞, we find ζ1 =1.5707 =π / 2 and C1 = 1.2733 . The quantity −∆M A is
the mass gain at the two stated times. The unknowns to be determined are DAB and S, which appears
in ρA,s (see Eq. (2)). From Eq. (4) evaluated at the two times, we have two simultaneous equations
which can be solved for the unknowns DAB and ρA,s, namely
0.012 × 10-6 kg
2
sin pp
/2
D AB
=
1 -1.2733exp(
24 h × 3600 s/h)
5
3
p/2
4 (0.0005 m)2
ρ A,s × 10 kg / m
0.016 × 10−6 kg
2
sin pp
/2
D AB
=
1−
1.2733exp( −
48 h × 3600 s/h)
p/2
4 (0.0005 m)2
ρ A,s × 10−5 kg / m3
There are two solutions to these two equations,
DAB = 1.92 × 10-13 m2/s, ρA,s = 0.003848 kg/m3
or
DAB = 8.5 × 10-13 m2/s, ρA,s = 0.001976 kg/m3
Continued…
PROBLEM 14.51 (Cont.)
Since we expect DAB to be greater than 7 × 10-13 m2/s, we choose the second solution. Thus,
DAB = 8.5 × 10-13 m2/s
<
rA,s / M A φp A,sat =
S=
0.001976 kg / m3 / (18.02 kg/kmol × 0.95 × 0.03531 bar)
S=
3.3 × 10−3 kmol / m3 ⋅ bar
<
COMMENTS: The system of equations has two solutions, but one of them would yield a mass
diffusivity less than 7 × 10-13 m2/s, and is therefore rejected. That solution also has Fom < 0.2, so the
solution is not valid. It does raise the question of whether there is another solution for which Fom <
0.2. If the problem is solved correctly for Fom < 0.2, it can be determined that there is no other
solution.
PROBLEM 14.52
KNOWN: Diameters of glass optical fiber and acrylate polymer coating. Mass diffusivity of water
vapor in the acrylate.
FIND: Whether microcracking would occur within several hot and humid days.
SCHEMATIC:
Di = 125 µm
`
Do = 250 µm
Humid environment
ASSUMPTIONS: (1) One-dimensional mass diffusion. (2) Use of acrylate properties throughout
the cylinder is sufficient for estimating the diffusion process in order to answer the question.
PROPERTIES: Water vapor in acrylate polymer (given): DAB = 5.5 × 10-13 m2/s.
ANALYSIS: We arbitrarily begin by considering a two-day period. Then the mass transfer Fourier
number is,
Fo m =D ABt / ro2 =5.5 × 10−13 m2 / s × 48 h × 3600 s / h (125 × 10−6 m)2 =6.1
Since Fom > 0.2, we can use the one-term approximation, analogous to Eq. 5.52a. Referring to Table
14.2 for the analogies,
=
γ*
CA (r, t) − CA,s
= C1 exp( −ζ12 Fo m )J o ( ζ1r*)
CA,i − CA,s
(1)
where from Table 5.1, as Bi→∞,
=
ζ1 2.4050,
=
C1 1.6018 . At the outer surface of the glass, r* =
0.5 and from Table B.4, J 0 (2.4050 × 0.5) ≈ J 0 (1.2) ≈ 0.67 . Thus
γ*= 1.6018exp( −2.40502 × 6.1) × 0.67= 5.6 × 10−16
Referring to the definition of γ* in Eq. (1), we see that this very small value means that the
concentration has essentially already reached the surface concentration. Therefore, careful storage of
the optical fiber will not prevent microcracking, since within two days (probably much less), the water
vapor has penetrated through the acrylate polymer coating.
<
COMMENTS: (1) Equation 5.52 assumes uniform properties throughout the cylinder. Since the
glass is impermeable to moisture, the build-up of moisture in the coating would be even more rapid
than this equation predicts. (2) The time required for the concentration to be within 5% of the surface
concentration (γ* = 0.05) is around four hours. (3) Development of hermetic coatings for use in fiber
optic and other high technology applications is an ongoing area of research.
PROBLEM 14.53
PROBLEM 14.53 (Cont.)
PROBLEM 14.54
KNOWN: Distance from bag of hot popcorn to students. Time for students to smell popcorn is one
second. Trace amount of aromatic needed for students to smell popcorn.
FIND: Whether the aromatic could have reached the students by advection in one second. Estimate of
time needed for students to smell popcorn using Fick’s law.
SCHEMATIC:
1
CA/CA,s
B → air
?
t = 30 s
0.001
0
30
x (m)
ASSUMPTIONS: (1) Diffusion in the room can be treated as one-dimensional diffusion in a semiinfinite medium.
PROPERTIES: Table A-8, typical value for diffusion of gas in air, DAB = 0.4 × 10-4 m2/s.
ANALYSIS: To reach the back of the room in one second, the aromatic must travel at a speed of:
=
V L=
/ t 30 m /=
1 s 30 m/s
= 67 mph
It is not plausible that the room air carried the aromatic at a speed of 67 mph, therefore advection
<
cannot explain the students smelling the popcorn in 1 second.
For diffusion based on Fick’s law, we can use the mass transfer analogy to Eq. 5.60:
CA ( x, t ) − CA, s
CA,i − CA, s


x
= erf 

2 D t 
AB 

where CA is the concentration of the aromatic that gives the popcorn its smell. With CA,i = 0,
this can be rewritten as:


CA ( x , t )
x
1 − erf 
1 − erf(η )
=
=
2 D t 
CA, s
AB 

For CA/CA,s = 0.0001, we require erf(η ) = 0.9999. From Table B.2, η ≅ 2.8. Solving for time,
Continued…
PROBLEM 14.54 (Cont.)
2
2
1  x 
1
 30 m 
t= 
7.2 × 105 s =
8.3 days
 =

 =
−4
2
DAB  2 × 2.8  0.4 × 10 m /s  2 × 2.8 
<
Fick’s law cannot be used to explain the speed with which students in the back of the room were able
to smell the popcorn.
<
COMMENTS: The transport of the aromatics must be occurring by a process other than advection or
diffusion as described by Fick’s law. Fick’s law breaks down in this problem because of the small
number of molecules; the distance between molecules is not small compared to the size of the room, as
illustrated in Figure 2.6b (with the dots representing molecules of the aromatic). Therefore the
aromatic cannot be treated as a continuum as assumed in Fick’s law. The actual explanation is based
on the mean free path concept and molecular velocity as discussed in Chapter 2. The molecular
velocity of an ideal gas is given by kinetic theory as u = 3ÂT / M A . Considering the molecular
weight of carbon dioxide (44.01 kg/kmol) as an example, for T = 300 K, u = 412 m/s. This suggests
that molecules of the aromatic can travel rapidly to the back of the room.