14-1
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14-2
14-1C Examples of different kinds of diffusion processes:
(a) Liquid–to–gas: A gallon of gasoline left in an open area will eventually evaporate and diffuse into air.
14-2C (a) Temperature difference is the driving force for heat transfer, (b) voltage difference is the driving force for electric
current flow, and (c) concentration difference is the driving force for mass transfer.
14-3C The concentration of a commodity is defined as the amount of that commodity per unit volume. The concentration
gradient dC/dx is defined as the change in the concentration C of a commodity per unit length in the direction of flow x. The
14-4C (a) Homogenous reactions in mass transfer represent the generation of a species within the medium. Such reactions
are analogous to internal heat generation in heat transfer. (b) Heterogeneous reactions in mass transfer represent the
generation of a species at the surface as a result of chemical reactions occurring at the surface. Such reactions are analogous
14-6C Bulk fluid flow refers to the transportation of a fluid on a macroscopic level from one location to another in a flow
section by a mover such as a fan or a pump. Mass flow requires the presence of two regions at different chemical
14-3
14-7C We would carry out the hardening process of steel by carbon at high temperature since mass diffusivity increases with
temperature, and thus the hardening process will be completed in a short time.
14-9C (a) T (b) F (c) F (d) T (e) F
dw
dy
14-4
14-15 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents are to be
determined.
Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N2 and O2 only.
0.764=== 6.28
0.28
)78.0( :N 2
22
N
NN2 M
M
yw
0.224=== 6.28
0.32
)20.0( :O 2
22
O
OO2 M
M
yw
0.012=== 6.28
0.18
)02.0( :OH OH
OHOH2
2
22 M
M
yw
14-16 The mole numbers of the constituents of a gas mixture at a specified pressure and temperature are given. The mass
fractions and the partial pressures of the constituents are to be determined.
Assumptions The gases behave as ideal gases.
58.3%)(or
2.31
0.28
)65.0( :N 2
22
N
NN2 0.583=== M
M
yw
20.5%)(or
2.31
0.32
)20.0( :O 2
22
O
OO2 0.205=== M
M
yw
21.2%)(or
2.31
44
)15.0( :CO 2
22
CO
COCO2 0.212===
m
M
M
yw
Noting that the total pressure of the mixture is 250 kPa and the pressure fractions in an ideal gas mixture are equal to the
mole fractions, the partial pressures of the individual gases become
kPa162.5)kPa250)(65.0(
22 NN === PyP
78% N2
20% O2
2% H2 O
(Mole fractions)
290 K
250 kPa
14-5
14–17E The masses of the constituents of a gas mixture are given. The mass fractions, mole fractions, and the molar mass of
the mixture are to be determined.
Assumptions None.
Properties The molar masses of N2, O2, and CO2 are 28, 32, and 44 lbm/lbmol, respectively (Table A-1E)
2
0.28=== 25
7
:O 2
2
O
O2 m
m
w
10
CO
m
lbmol 0.286=== lbm/lbmol 28
lbm 8
:N
2
2
2
N
N
N2 M
m
N
lbmol 0.219=== lbm/lbmol 32
lbm 7
:O
2
2
2
O
O
O2 M
m
N
lbmol 0.227=== lbm/lbmol 44
lbm 10
:CO
2
2
2
CO
CO
CO2M
m
N
lbmol 732.0227.0219.0286.0
222 COON =++=++== NNNNN im
7 lbm O2
8 lbm N2
10 lbm CO2
14-6
14-7
14–20 The partial pressure ratio of CO and air is given. The CO level is to be determined whether it is above 35 ppm.
Assumptions 1 The gases behave as ideal gases.
Analysis The mole fraction of a species in a mixture is
𝑦𝑖=𝑁𝑖
𝑁=𝐶𝑖
𝐶
14-8
14-9
14-22 The molar concentration of PM10 soot particulates in ambient air is given. The level of PM10 soot particulates is to
be determined whether it is above the limit set by the NAAQS.
Assumptions 1 Air at 1 atm pressure.
Properties The density of air at 20°C is 1.204 kg/m3 (Table A-15). The molar mass of air is 28.97 kg/kmol (Table A-1).
Analysis The molar concentration of air is
14–10
14–11
14–24 The binary diffusion coefficients of CO2 in air at various temperatures and pressures are to be determined.
Assumptions The mixture is sufficiently dilute so that the diffusion coefficient is independent of mixture composition.
Properties The binary diffusion coefficients of CO2 in air at 1 atm pressure are given in Table 14-1 to be 0.7410-5, 2.6310–
14-25 The binary diffusion coefficient of O2 in N2 at various temperature and pressures are to be determined.
Assumptions The mixture is sufficiently dilute so that the diffusion coefficient is independent of mixture composition.
Properties The binary diffusion coefficient of O2 in N2 at T1 = 273 K and P1 = 1 atm is given in Table 14-2 to be 1.810-5
m2/s.
Analysis Noting that the binary diffusion coefficient of gases is proportional to 3/2 power of temperature and inversely
14–12
14–13
14–28 Prob. 14-27 is reconsidered. The diffusion coefficient as a function of the temperature is to be plotted.
T
[K]
DAB
[m2/s]
300
3.272E-13
350
2.967E-12
400
1.551E-11
450
5.611E-11
500
1.570E-10
550
3.643E-10
600
7.348E-10
650
1.330E-09
700
2.213E-09
750
3.439E-09
800
5.058E-09
850
7.110E-09
900
9.622E-09
950
1.261E-08
1000
1.610E-08
1050
2.007E-08
1100
2.452E-08
1150
2.944E-08
1200
3.482E-08
300 400 500 600 700 800 900 1000 1100 1200
0.00x100
6.96x10-9
1.39x10-8
2.09x10-8
2.78x10-8
3.48x10-8
T [K]
DAB [m2/s]
14–14
Boundary Conditions
14–30C When prescribing a boundary condition for mass transfer at a solid-gas interface, we need to specify the side of the
14–31C Three boundary conditions for mass transfer (on mass basis) that correspond to specified temperature, specified heat
flux, and convection boundary conditions in heat transfer are expressed as follows:
1)
0
)0( ww =
(specified concentration – corresponds to specified temperature)
AJ
dw
0
=
x
x
14–32C An impermeable surface is a surface that does not allow any mass to pass through. Mathematically it is expressed (at
x = 0) as
A
dw
14–15
14–16
14–39E Water is sprayed into air, and the falling water droplets are collected in a container. The mass and mole fractions of
air dissolved in the water are to be determined.
Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since water is constantly sprayed into it. 3 Air
is weakly soluble in water and thus Henry’s law is applicable.
psia79.135073.03.14
vaporairdry =−=−= PPP
From Henry’s law, the mole fraction of air in the water is determined to be
5
)psia696.14/atm1(psia79.13
sidegasair,dry
P
55
airdry
sideliquidair,drysideliquidair,dry 29
29
1029.1)0( −− === 101.29
m
M
M
yw
Water
14–17
14–18
14–19
14–20