A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 3 Part 5 pot

Problems 589
radiation but that it offers very little resistance to conduction.
Thus, the window temperature is almost uniform.)
10.28 A very effective low-temperature insulation is made by evacu-
ating the space between parallel metal sheets. Convection is
eliminated, conduction occurs only at spacers, and radiation
is responsible for what little heat transfer occurs. Calculate
q between 150 K and 100 K for three cases: (a) two sheets of
highly polished aluminum, (b) three sheets of highly polished
aluminum, and (c) three sheets of rolled sheet steel.
10.29 Three parallel black walls, 1 m wide, form an equilateral trian-
gle. One wall is held at 400 K, one is at 300 K, and the third is
insulated. Find Q W/m and the temperature of the third wall.
10.30 Two 1 cm diameter rods run parallel, with centers 4 cm apart.
One is at 1500 K and black. The other is unheated, and ε =
0.66. They are both encircled by a cylindrical black radiation
shield at 400 K. Evaluate Q W/m and the temperature of the
unheated rod.
10.31 A small-diameter heater is centered in a large cylindrical radi-
ation shield. Discuss the relative importance of the emittance
of the shield during specular and diffuse radiation.
10.32 Two 1 m wide commercial aluminum sheets are joined at a
120
◦
angle along one edge. The back (or 240
◦
angle) side is
insulated. The plates are both held at 120
◦
C. The 20
◦

C sur-
roundings are distant. What is the net radiant heat transfer
from the left-hand plate: to the right-hand side, and to the
surroundings?
10.33 Two parallel discs of 0.5 m diameter are separated by an infi-
nite parallel plate, midway between them, with a 0.2 m diame-
ter hole in it. The discs are centered on the hole. What is the
view factor between the two discs if they are 0.6 m apart?
10.34 An evacuated spherical cavity, 0.3 m in diameter in a zero-
gravity environment, is kept at 300
◦
C. Saturated steam at 1 atm
is then placed in the cavity. (a) What is the initial flux of radiant
heat transfer to the steam? (b) Determine how long it will take
for q
conduction
to become less than q
radiation
. (Correct for the
rising steam temperature if it is necessary to do so.)
590 Chapter 10: Radiative heat transfer
10.35 Verify cases (1), (2), and (3) in Table 10.2 using the string
method described in Problem 10.14.
10.36 Two long parallel heaters consist of 120
◦
segments of 10 cm di-
ameter parallel cylinders whose centers are 20 cm apart. The
segments are those nearest each other, symmetrically placed
on the plane connecting their centers. Find F
1–2

using the
string method described in Problem 10.14.)
10.37 Two long parallel strips of rolled sheet steel lie along sides of
an imaginary 1 m equilateral triangular cylinder. One piece is
1 m wide and kept at 20
◦
C. The other is
1
2
m wide, centered
in an adjacent leg, and kept at 400
◦
C. The surroundings are
distant and they are insulated. Find Q
net
. (You will need a
shape factor; it can be found using the method described in
Problem 10.14.)
10.38 Find the shape factor from the hot to the cold strip in Prob-
lem 10.37 using Table 10.2, not the string method. If your
instructor asks you to do so, complete Problem 10.37 when
you have F
1–2
.
10.39 Prove that, as the figure becomes very long, the view factor
for the second case in Table 10.3 reduces to that given for the
third case in Table 10.2.
10.40 Show that F
1–2
for the first case in Table 10.3 reduces to the

expected result when plates 1 and 2 are extended to infinity.
10.41 In Problem 2.26 you were asked to neglect radiation in showing
that q was equal to 8227 W/m
2
as the result of conduction
alone. Discuss the validity of the assumption quantitatively.
10.42 A 100
◦
C sphere with ε = 0.86 is centered within a second
sphere at 300
◦
C with ε = 0.47. The outer diameter is 0.3 m
and the inner diameter is 0.1 m. What is the radiant heat flux?
10.43 Verify F
1–2
for case 4 in Table 10.2.(Hint: This can be done
without integration.)
10.44 Consider the approximation made in eqn. (10.30) for a small
gray object in a large isothermal enclosure. How small must
A
1
/A
2
be in order to introduce less than 10% error in F
1–2
if
Problems 591
the small object has an emittance of ε
1
= 0.5 and the enclo-

sure is: a) commerical aluminum sheet; b) rolled sheet steel;
c) rough red brick; d) oxidized cast iron; or e) polished elec-
trolytic copper. Assume both the object and its environment
have temperatures of 40 to 90
◦
C.
10.45 Derive eqn. (10.42), starting with eqns. (10.36–10.38).
10.46 (a) Derive eqn. (10.31), which is for a single radiation shield
between two bodies. Include a sketch of the radiation net-
work. (b) Repeat the calculation in the case when two radia-
tion shields lie between body (1) and body (2), with the second
shield just outside the first.
10.47 Use eqn. (10.32) to find the net heat transfer from between two
specularly reflecting bodies that are separated by a specularly
reflecting radiation shield. Compare the result to eqn. (10.31).
Does specular reflection reduce the heat transfer?
10.48 Some values of the monochromatic absorption coefficient for
liquid water, as ρκ
λ
(cm
−1
), are listed below [10.4]. For each
wavelength, find the thickness of a layer of water for which
the transmittance is 10%. On this basis, discuss the colors one
might see underwater and water’s infrared emittance.
λ (µm) ρκ
λ
(cm
−1
) Color

0.30.0067
0.40.00058 violet
0.50.00025 green
0.60.0023 orange
0.80.0196
1.00.363
2.069.1
2.6–10.0 > 100.
10.49 The sun has a diameter of 1.391 × 10
6
km. The earth has a
diameter of 12,740 km and lies at a mean distance of 1.496 ×
10
8
km from the center of the sun. (a) If the earth is treated as a
flat disk normal to the radius from sun to earth, determine the
view factor F
sun–earth
. (b) Use this view factor and the measured
solar irradiation of 1367 W/m
2
to show that the effective black
body temperature of the sun is 5777 K.
592 Chapter 10: Radiative heat transfer
References
[10.1] E. M. Sparrow and R. D. Cess. Radiation Heat Transfer. Hemi-
sphere Publishing Corp./McGraw-Hill Book Company, Washing-
ton, D.C., 1978.
[10.2] M. F. Modest. Radiative Heat Transfer. McGraw-Hill, New York,
1993.

[10.3] D. K. Edwards. Radiation Heat Transfer Notes. Hemisphere Pub-
lishing Corp., Washington, D.C., 1981.
[10.4] R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. Tay-
lor and Francis-Hemisphere, Washington, D.C., 4th edition, 2001.
[10.5] J. R. Howell. A Catalog of Radiation Heat Transfer Configuration
Factors. University of Texas, Austin, 2nd edition, 2001. Available
online at />[10.6] A. K. Oppenheim. Radiation analysis by the network method.
Trans. ASME, 78:725–735, 1956.
[10.7] W J. Yang, H. Taniguchi, and K. Kudo. Radiative heat transfer by
the Monte Carlo method. In T.F. Irvine, Jr., J. P. Hartnett, Y. I. Cho,
and G. A. Greene, editors, Advances in Heat Transfer, volume 27.
Academic Press, Inc., San Diego, 1995.
[10.8] H. C. van de Hulst. Light Scattering by Small Particles. Dover
Publications Inc., New York, 1981.
[10.9] P. W. Atkins. Physical Chemistry. W. H. Freeman and Co., New
York, 3rd edition, 1986.
[10.10] G. Herzberg. Molecular Spectra and Molecular Structure. Kreiger
Publishing, Malabar, Florida, 1989. In three volumes.
[10.11] D. K. Edwards. Molecular gas band radiation. In T. F. Irvine, Jr.
and J. P. Hartnett, editors, Advances in Heat Transfer, volume 12,
pages 119–193. Academic Press, Inc., New York, 1976.
[10.12] H. C. Hottel and A. F. Sarofim. Radiative Transfer. McGraw-Hill
Book Company, New York, 1967.
References 593
[10.13] D. K. Edwards and R. Matavosian. Scaling rules for total absorp-
tivity and emissivity of gases. J. Heat Transfer, 106(4):684–689,
1984.
[10.14] M. Iqbal. An Introduction to Solar Radiation. Academic Press, Inc.,
New York, 1983.
[10.15] J. A. Duffie and W. A. Beckman. Solar Engineering of Thermal

Processes. John Wiley & Sons, Inc., New York, 2nd edition, 1991.
[10.16] H. G. Houghton. Physical Meteorology. MIT Press, Cambridge, MA,
1985.
[10.17] P. Berdahl and R. Fromberg. The thermal radiance of clear skies.
Solar Energy, 29:299–314, 1982.
[10.18] A. Skartveit, J. A. Olseth, G. Czeplak, and M. Rommel. On the
estimation of atmospheric radiation from surface meteorological
data. Solar Energy, 56:349–359, 1996.
[10.19] P. Berdahl and M. Martin. The emissivity of clear skies. Solar
Energy, 32:663–664, 1984.
[10.20] J. A. Fay and D. S. Gollub. Energy and Environment. Oxford Uni-
versity Press, New York, 2002.
[10.21] J. Hansen, R. Ruedy, M. Sato, M. Imhoff, W. Lawrence, D. Easter-
ling, T. Peterson, and T. Karl. A closer look at United States and
global surface temperature change. J. Geophysical Research, 106:
23947, 2001.
[10.22] R. T. Watson, editor. Climate Change 2001: Synthesis Report.
Third assessment report of the Intergovernmental Panel on Cli-
mate Change. Cambridge University Press, New York, 2002. Also
available at .
[10.23] P. A. Stott, S. F. B. Tett, G. S. Jones, M. R. Allen, J. F. B. Mitchell,
and G. J. Jenkins. External control of 20th century temperature
by natural and anthropogenic forcings. Science, 290:2133–2137,
2000.
[10.24] F. Kreith and J. F. Kreider. Principles of Solar Engineering. Hemi-
sphere Publishing Corp./McGraw-Hill Book Company, Washing-
ton, D.C., 1978.
594 Chapter 10: Radiative heat transfer
[10.25] U.S. Department of Commerce. Solar Heating and Cooling of Res-
idential Buildings, volume 1 and 2. Washington, D.C., October

1977.
Part V
Mass Transfer
595

11. An introduction to mass transfer
The edge of a colossal jungle, so dark-green as to be almost black, fringed
with white surf, ran straight, like a ruled line, far, far away along a blue
sea whose glitter was blurred by a creeping mist. The sun was fierce, the
land seemed to glisten and drip with steam.
Heart of Darkness, Joseph Conrad, 1902
11.1 Introduction
We have, so far, dealt with heat transfer by convection, radiation, and
diffusion (which we have been calling conduction). We have dealt only
with situations in which heat passes through, or is carried by, a single
medium. Many heat transfer processes, however, occur in mixtures of
more than one substance. A wall exposed to a hot air stream may be
cooled evaporatively by bleeding water through its surface. Water vapor
may condense out of damp air onto cool surfaces. Heat will flow through
an air-water mixture in these situations, but water vapor will diffuse or
convect through air as well.
This sort of transport of one substance relative to another is called
mass transfer; it did not occur in the single-component processes of the
preceding chapters. In this chapter, we study mass transfer phenomena
with an eye toward predicting heat and mass transfer rates in situations
like those just mentioned.
During mass transfer processes, an individual chemical species trav-
els from regions where it has a high concentration to regions where it has
a low concentration. When liquid water is exposed to a dry air stream, its
vapor pressure may produce a comparatively high concentration of wa-

ter vapor in the air near the water surface. The concentration difference
between the water vapor near the surface and that in the air stream will
drive the diffusion of vapor into the air stream. We call this evaporation.
597
598 An introduction to mass transfer §11.1
Figure 11.1 Schematic diagram of a natural-draft cooling
tower at the Rancho Seco nuclear power plant. (From [11.1],
courtesy of W. C. Reynolds.)
In this and other respects, mass transfer is analogous to heat trans-
fer. Just as thermal energy diffuses from regions of high temperature
to regions of low temperature (following the temperature gradient), the
mass of one species diffuses from regions high concentration to regions
of low concentration (following its concentration gradient.) Just as the
diffusional (or conductive) heat flux is directly proportional to a temper-
ature gradient, so the diffusional mass flux of a species is often directly
proportional to its concentration gradient; this is called Fick’s law of dif-
fusion. Just as conservation of energy and Fourier’s law lead to equations
for the convection and diffusion of heat, conservation of mass and Fick’s
law lead to equations for the convection and diffusion of species in a
mixture.
The great similarity of the equations of heat convection and diffusion
to those of mass convection and diffusion extends to the use of con-
vective mass transfer coefficients, which, like heat transfer coefficients,
relate convective fluxes to concentration differences. In fact, with sim-
ple modifications, the heat transfer coefficients of previous chapters may
often be applied to mass transfer calculations.
§11.1 Introduction 599
Figure 11.2 A mechanical-draft cooling tower. The fans are
located within the cylindrical housings at the top. Air is drawn
in through the louvres on the side.

Mass transfer, by its very nature, is intimately concerned with mix-
tures of chemical species. We begin this chapter by learning how to quan-
tify the concentration of chemical species and by defining rates of move-
ment of species. We make frequent reference to an arbitrary “species i,”
the ith component of a mixture of N different species. These definitions
may remind you of your first course in chemistry. We also spend some
time, in Section 11.4, discussing how to calculate the transport properties
of mixtures, such as diffusion coefficients and viscosities.
Consider a typical technology that is dominated by mass transfer pro-
cesses. Figure 11.1 shows a huge cooling tower used to cool the water
leaving power plant condensers or other large heat exchangers. It is es-
sentially an empty shell, at the bottom of which are arrays of cement
boards or plastic louvres over which is sprayed the hot water to be cooled.
The hot water runs down this packing, and a small portion of it evapo-
rates into cool air that enters the tower from below. The remaining water,
having been cooled by the evaporation, falls to the bottom, where it is
collected and recirculated.
The temperature of the air rises as it absorbs the warm vapor and, in
600 An introduction to mass transfer §11.2
the natural-draft form of cooling tower shown, the upper portion of the
tower acts as an enormous chimney through which the warm, moist air
buoys, pulling in cool air at the base. In a mechanical-draft cooling tower,
fans are used to pull air through the packing. Mechanical-draft towers
are much shorter and can sometimes be seen on the roofs of buildings
(Fig. 11.2).
The working mass transfer process in a cooling tower is the evapora-
tion of water into air. The rate of evaporation depends on the tempera-
ture and humidity of the incoming air, the feed-water temperature, and
the air-flow characteristics of the tower and the packing. When the air
flow is buoyancy-driven, the flow rates are directly coupled. Thus, mass

transfer lies at the core of the complex design of a cooling tower.
11.2 Mixture compositions and species fluxes
The composition of mixtures
A mixture of various chemical species displays its own density, molecular
weight, and other overall thermodynamic properties. These properties
depend on the types and relative amounts of the component substances,
which may vary from point to point in the mixture. To determine the
local properties of a mixture, we must identify the local proportion of
each species composing the mixture.
One way to describe the amount of a particular species in a mixture is
by the mass of that species per unit volume, known as the partial density.
The mass of species i in a small volume of mixture, in kg, divided by that
volume, in m
3
, is the partial density, ρ
i
, for that species, in kg of i per
m
3
. The composition of the mixture may be describe by stating the partial
density of each of its components. The mass density of the mixture itself,
ρ, is the total mass of all species per unit volume; therefore,
ρ =

i
ρ
i
(11.1)
The relative amount of species i in the mixture may be described by
the mass of i per unit mass of the mixture, which is simply ρ

i
/ρ. This
ratio is called the mass fraction, m
i
:
m
i
≡
ρ
i
ρ
=
mass of species i
mass of mixture
(11.2)
§11.2 Mixture compositions and species fluxes 601
This definition leads to the following two results:

i
m
i
=

i
ρ
i
/ρ = 1 and 0 m
i
 1 (11.3)
The molar concentration of species i in kmol/m

3
, c
i
, expresses con-
centration in terms of moles rather than mass. If M
i
is the molecular
weight of species i in kg/kmol, then
c
i
≡
ρ
i
M
i
=
moles of i
volume
(11.4)
The molar concentration of the mixture, c, is the total number of moles
for all species per unit volume; thus,
c =

i
c
i
. (11.5)
The mole fraction of species i, x
i
, is the number of moles of i per mole

of mixture:
x
i
≡
c
i
c
=
moles of i
mole of mixture
(11.6)
Just as for the mass fraction, it follows for mole fraction that

i
x
i
=

i
c
i
/c = 1 and 0 x
i
 1 (11.7)
The molecular weight of the mixture is the number of kg of mixture
per kmol of mixture: M ≡ ρ/c. By using eqns. (11.1), (11.4), and (11.6)
and (11.5), (11.4), and (11.2), respectively, M may be written in terms of
either mole or mass fraction
M =


i
x
i
M
i
or
1
M
=

i
m
i
M
i
(11.8)
Mole fraction may be converted to mass fraction using the following re-
lations (derived in Problem 11.1):
m
i
=
x
i
M
i
M
=
x
i
M

i

k
x
k
M
k
and x
i
=
Mm
i
M
i
=
m
i
/M
i

k
m
k
/M
k
(11.9)
In some circumstances, such as kinetic theory calculations, one works
directly with the number of molecules of i per unit volume. This number
density, N
i

, is given by
N
i
= N
A
c
i
(11.10)
where N
A
is Avogadro’s number, 6.02214 × 10
26
molecules/kmol.
602 An introduction to mass transfer §11.2
Ideal gases
The relations we have developed so far involve densities and concentra-
tions that vary in as yet unknown ways with temperature or pressure. To
get a more useful, though more restrictive, set of results, we now com-
bine the preceding relations with the ideal gas law. For any individual
component, i, we may write the partial pressure, p
i
, exterted by i as:
p
i
= ρ
i
R
i
T (11.11)
In eqn. (11.11), R

i
is the ideal gas constant for species i:
R
i
≡
R
◦
M
i
(11.12)
where R
◦
is the universal gas constant, 8314.472 J/kmol·K. Equation (11.11)
may alternatively be written in terms of c
i
:
p
i
= ρ
i
R
i
T = (M
i
c
i
)

R
◦

M
i

T
= c
i
R
◦
T (11.13)
Equations (11.5) and (11.13) can be used to relate c to p and T
c =

i
c
i
=

i
p
i
R
◦
T
=
p
R
◦
T
(11.14)
Multiplying the last two parts of eqn. (11.14)byR

◦
T yields Dalton’s law
of partial pressures,
1
p =

i
p
i
(11.15)
Finally, we combine eqns. (11.6), (11.13), and (11.15) to obtain a very
useful relationship between x
i
and p
i
:
x
i
=
c
i
c
=
p
i
cR
◦
T
=
p

i
p
(11.16)
in which the last two equalities are restricted to ideal gases.
1
Dalton’s law (1801) is an empirical principle (not a deduced result) in classical
thermodynamics. It can be deduced from molecular principles, however. We built the
appropriate molecular principles into our development when we assumed eqn. (11.11)
to be true. The reason that eqn. (11.11) is true is that ideal gas molecules occupy a
mixture without influencing one another.
§11.2 Mixture compositions and species fluxes 603
Example 11.1
The most important mixture that we deal with is air. It has the fol-
lowing composition:
Species Mass Fraction
N
2
0.7556
O
2
0.2315
Ar 0.01289
trace gases < 0.01
Determine x
O
2
, p
O
2
, c

O
2
, and ρ
O
2
for air at 1 atm.
Solution. To make these calcuations, we need the molecular weights,
which are given in Table 11.2 on page 616. We can start by checking
the value of M
air
, using the second of eqns. (11.8):
M
air
=

m
N
2
M
N
2
+
m
O
2
M
O
2
+
m

Ar
M
Ar

−1
=

0.7556
28.02 kg/kmol
+
0.2315
32.00 kg/kmol
+
0.01289
39.95 kg/kmol

−1
= 28.97 kg/kmol
We may calculate the mole fraction using the second of eqns. (11.9)
x
O
2
=
m
O
2
M
M
O
2

=
(0.2315)(28.97 kg/kmol)
32.00 kg/kmol
= 0.2095
The partial pressure of oxygen in air at 1 atm is [eqn. (11.16)]
p
O
2
= x
O
2
p = (0.2095)(101, 325 Pa) = 2.123 × 10
4
Pa
We may now obtain c
O
2
from eqn. (11.13):
c
O
2
=
p
O
2
R
◦
T
= (2.123 ×10
4

Pa)

(300 K)(8314.5J/kmol·K)
= 0.008510 kmol/m
3
Finally, eqn. (11.4) gives the partial density
ρ
O
2
= c
O
2
M
O
2
= (0.008510 kmol/m
3
)(32.00 kg/kmol)
= 0.2723 kg/m
3
604 An introduction to mass transfer §11.2
Velocities and fluxes
Each species in a mixture undergoing a mass transfer process will have
an species-average velocity,

v
i
, which can be different for each species in
the mixture, as suggested by Fig. 11.3. We may obtain the mass-average
velocity,

2

v, for the entire mixture from the species average velocities
using the formula
ρ

v =

i
ρ
i

v
i
. (11.17)
This equation is essentially a local calculation of the mixture’s net mo-
mentum per unit volume. We refer to ρ

v as the mixture’s mass flux,

n,
and we call its scalar magnitude
˙
m

; each has units of kg/m
2
·s. Likewise,
the mass flux of species i is


n
i
= ρ
i

v
i
(11.18)
and, from eqn. (11.17), we see that the mixture’s mass flux equals the
sum of all species’ mass fluxes

n =

i

n
i
= ρ

v (11.19)
Since each species diffusing through a mixture has some velocity rel-
ative to the mixture’s mass-average velocity, the diffusional mass flux,

j
i
,
of a species relative to the mixture’s mean flow may be identified:

j
i

= ρ
i


v
i
−

v

. (11.20)
The total mass flux of the ith species,

n
i
, includes both this diffusional
mass flux and bulk convection by the mean flow, as is easily shown:

n
i
= ρ
i

v
i
= ρ
i

v +ρ
i



v
i
−

v

= ρ
i

v +

j
i
= m
i

n

 
convection
+

j
i
  
diffusion
(11.21)
2

The mass average velocity,

v, given by eqn. (11.17) is identical to the fluid velocity,

u, used in previous chapters. This is apparent if one applies eqn. (11.17) to a “mix-
ture” composed of only one species. We use the symbol

v here because

v is the more
common notation in the mass transfer literature.
§11.2 Mixture compositions and species fluxes 605
Figure 11.3 Molecules of different
species in a mixture moving with
different average velocities. The velocity

v
i
is the average over all molecules of
species i.
Although the convective transport contribution is fully determined as
soon as we know the velocity field and partial densities, the causes of
diffusion need further discussion, which we defer to Section 11.3.
Combining eqns. (11.19) and (11.21), we find that

n =

i

n

i
=

i
ρ
i

v +

i

j
i
= ρ

v +

i

j
i
=

n +

i

j
i
Hence


i

j
i
= 0 (11.22)
Diffusional mass fluxes must sum to zero because they are each defined
relative to the mean mass flux.
Velocities may also be stated in molar terms. The mole flux of the
ith species,

N
i
,isc
i

v
i
, in kmol/m
2
·s. The mixture’s mole flux,

N,is
obtained by summing over all species

N =

i

N

i
=

i
c
i

v
i
= c

v
∗
(11.23)
where we define the mole-average velocity,

v
∗
, as shown. The last flux
we define is the diffusional mole flux,

J
i
∗
:

J
∗
i
= c

i


v
i
−

v
∗

(11.24)
606 An introduction to mass transfer §11.2
It may be shown, using these definitions, that

N
i
= x
i

N +

J
∗
i
(11.25)
Substitution of eqn. (11.25) into eqn. (11.23) gives

N =

i


N
i
=

N

i
x
i
+

i

J
∗
i
=

N +

i

J
∗
i
so that

i


J
∗
i
= 0. (11.26)
Thus, both the

J
∗
i
’s and the

j
i
’s sum to zero.
Example 11.2
At low temperatures, carbon oxidizes (burns) in air through the sur-
face reaction: C + O
2
→ CO
2
. Figure 11.4 shows the carbon-air in-
terface in a coordinate system that moves into the stationary carbon
at the same speed that the carbon burns away—as though the ob-
server were seated on the moving interface. Oxygen flows toward
the carbon surface and carbon dioxide flows away, with a net flow
of carbon through the interface. If the system is at steady state and,
if a separate analysis shows that carbon is consumed at the rate of
0.00241 kg/m
2
·s, find the mass and mole fluxes through an imagi-

nary surface, s, that stays close to the gas side of the interface. For
this case, concentrations at the s-surface turn out to be m
O
2
,s
= 0.20,
m
CO
2
,s
= 0.052, and ρ
s
= 0.29 kg/m
3
.
Solution. The mass balance for the reaction is
12.0kgC+32.0kgO
2
→ 44.0kgCO
2
Since carbon flows through a second imaginary surface, u, moving
through the stationary carbon just below the interface, the mass fluxes
are related by
n
C,u
=−
12
32
n
O

2
,s
=
12
44
n
CO
2
,s
The minus sign arises because the O
2
flow is opposite the C and CO
2
flows, as shown in Figure 11.4. In steady state, if we apply mass
§11.2 Mixture compositions and species fluxes 607
Figure 11.4 Low-temperature carbon
oxidation.
conservation to the control volume between the u and s surfaces, we
find that the total mass flux entering the u-surface equals that leaving
the s-surface
n
C,u
= n
CO
2
,s
+n
O
2
,s

= 0.00241 kg/m
2
·s
Hence,
n
O
2
,s
=−
32
12
(0.00241 kg/m
2
·s) =−0.00643 kg/m
2
·s
n
CO
2
,s
=
44
12
(0.00241 kg/m
2
·s) = 0.00884 kg/m
2
·s
To get the diffusional mass flux, we need species and mass average
speeds from eqns. (11.18) and (11.19):

v
O
2
,s
=
n
O
2
,s
ρ
O
2
,s
=
−0.00643 kg/m
2
·s
0.2 (0.29 kg/m
3
)
=−0.111 m/s
v
CO
2
,s
=
n
CO
2
,s

ρ
CO
2
,s
=
0.00884 kg/m
2
·s
0.052 (0.29 kg/m
3
)
= 0.586 m/s
v
s
=
1
ρ
s

i
n
i
=
(0.00884 − 0.00643) kg/m
2
·s
0.29 kg/m
3
= 0.00831 m/s
Thus, from eqn. (11.20),

j
i,s
= ρ
i,s

v
i,s
−v
s

=



−0.00691 kg/m
2
·s for O
2
0.00871 kg/m
2
·s for CO
2
608 An introduction to mass transfer §11.3
The diffusional mass fluxes, j
i,s
, are very nearly equal to the species
mass fluxes, n
i,s
. That is because the mass-average speed, v
s

, ismuch
less than the species speeds, v
i,s
, in this case. Thus, the convective
contribution to n
i,s
is much smaller than the diffusive contribution,
and mass transfer occurs primarily by diffusion. Note that j
O
2
,s
and
j
CO
2
,s
do not sum to zero because the other, nonreacting species in
air must diffuse against the small convective velocity, v
s
(see Sec-
tion 11.7).
One mole of carbon surface reacts with one mole of O
2
to form
one mole of CO
2
. Thus, the mole fluxes of each species have the same
magnitude at the interface:
N
CO

2
,s
=−N
O
2
,s
= N
C,u
=
n
C,u
M
C
= 0.000201 kmol/m
2
·s
The mole average velocity at the s-surface, v
∗
s
, is identically zero by
eqn. (11.23), since N
CO
2
,s
+N
O
2
,s
= 0. The diffusional mole fluxes are
J

∗
i,s
= c
i,s

v
i,s
− v
∗
s

=0

= N
i,s
=



−0.000201 kmol/m
2
·s for O
2
0.000201 kmol/m
2
·s for CO
2
These two diffusional mole fluxes sum to zero themselves because
there is no convective mole flux for other species to diffuse against
(i.e., for the other species J

∗
i,s
= 0).
The reader may calculate the velocity of the interface from n
c,u
.
That calculation would show the interface to be receding so slowly
that the velocities we calculate are almost equal to those that would
be seen by a stationary observer.
11.3 Diffusion fluxes and Fick’s law
When the composition of a mixture is nonuniform, the concentration
gradient in any species, i, of the mixture provides a driving potential for
the diffusion of that species. It flows from regions of high concentration
to regions of low concentration—similar to the diffusion of heat from
regions of high temperature to regions of low temperature. We have
already noted in Section 2.1 that mass diffusion obeys Fick’s law

j
i
=−ρD
im
∇m
i
(11.27)
§11.3 Diffusion fluxes and Fick’s law 609
which is analogous to Fourier’s law.
The constant of proportionality, ρD
im
, between the local diffusive
mass flux of species i and the local concentration gradient of i involves

a physical property called the diffusion coefficient, D
im
, for species i dif-
fusing in the mixture m. Like the thermal diffusivity, α, or the kinematic
viscosity (a momentum diffusivity), ν, the mass diffusivity D
im
has the
units of m
2
/s. These three diffusivities can form three dimensionless
groups, among which is the Prandtl number:
The Prandtl number, Pr ≡ ν/α
The Schmidt number,
3
Sc ≡ ν/D
im
(11.28)
The Lewis number,
4
Le ≡ α/D
im
= Sc/Pr
Each of these groups compares the relative strength of two different dif-
fusive processes. We make considerable use of the Schmidt number later
in this chapter.
When diffusion occurs in mixtures of only two species—so-called bi-
nary mixtures—D
im
reduces to the binary diffusion coefficient, D
12

.In
fact, the best-known kinetic models are for binary diffusion.
5
In binary
diffusion, species 1 has the same diffusivity through species 2 as does
species 2 through species 1 (see Problem 11.5); in other words,
D
12
=D
21
(11.29)
3
Ernst Schmidt (1892–1975) served successively as the professor of thermodynam-
ics at the Technical Universities of Danzig, Braunschweig, and Munich (Chapter 6, foot-
note 3). His many contributions to heat and mass transfer include the introduction of
aluminum foil as radiation shielding, the first measurements of velocity and temper-
ature fields in a natural convection boundary layer, and a once widely-used graphical
procedure for solving unsteady heat conduction problems. He was among the first to
develop the analogy between heat and mass transfer.
4
Warren K. Lewis (1882–1975) was a professor of chemical engineering at M.I.T. from
1910 to 1975 and headed the department throughout the 1920s. He defined the original
paradigm of chemical engineering, that of “unit operations”, and, through his textbook
with Walker and McAdams, Principles of Chemical Engineering, he laid the foundations
of the discipline. He was a prolific inventor in the area of industrial chemistry, holding
more than 80 patents. He also did important early work on simultaneous heat and
mass transfer in connection with evaporation problems.
5
Actually, Fick’s Law is strictly valid only for binary mixtures. It can, however, of-
ten be applied to multicomponent mixtures with an appropriate choice of D

im
(see
Section 11.4).
610 An introduction to mass transfer §11.3
A kinetic model of diffusion
Diffusion coefficients depend upon composition, temperature, and pres-
sure. Equations that predict D
12
and D
im
are given in Section 11.4. For
now, let us see how Fick’s law arises from the same sort of elementary
molecular kinetics that gave Fourier’s and Newton’s laws in Section 6.4.
We consider a two-component dilute gas (one with a low density) in
which the molecules A of one species are very similar to the molecules A

of a second species (as though some of the molecules of a pure gas had
merely been labeled without changing their properties.) The resulting
process is called self-diffusion.
If we have a one-dimensional concentration distribution, as shown in
Fig. 11.5, molecules of A diffuse down their concentration gradient in
the x-direction. This process is entirely analogous to the transport of
energy and momentum shown in Fig. 6.13. We take the temperature and
pressure of the mixture (and thus its number density) to be uniform and
the mass-average velocity to be zero.
Individual molecules move at a speed C, which varies randomly from
molecule to molecule and is called the thermal or peculiar speed. The
average speed of the molecules is
C. The average rate at which molecules
cross the plane x = x

0
in either direction is proportional to NC, where
N is the number density (molecules/m
3
). Prior to crossing the x
0
-plane,
the molecules travel a distance close to one mean free path, —call it a,
where a is a number on the order of one.
The molecular flux travelling rightward across x
0
, from its plane of
origin at x
0
−a, then has a fraction of molecules of A equal to the value
of N
A
/N at x
0
− a. The leftward flux, from x
0
+ a, has a fraction
equal to the value of N
A
/N at x
0
+a. Since the mass of a molecule of
A is M
A
/N

A
(where N
A
is Avogadro’s number), the net mass flux in the
x-direction is then
j
A



x
0
= η

NC


M
A
N
A


N
A
N





x
0
−a
−
N
A
N




x
0
+a

(11.30)
where η is a constant of proportionality. Since N
A
/N changes little in a
distance of two mean free paths (in most real situations), we can expand
the right side of eqn. (11.30) in a two-term Taylor series expansion about
§11.3 Diffusion fluxes and Fick’s law 611
Figure 11.5 One-dimensional diffusion.
x
0
and obtain Fick’s law:
j
A




x
0
= η

NC


M
A
N
A

−2a
d(N
A
/N)
dx




x
0

=−2ηa(
C)ρ
dm
A
dx





x
0
(11.31)
(for details, see Problem 11.6). Thus, we identify
D
AA

= (2ηa)C (11.32)
and Fick’s law takes the form
j
A
=−ρD
AA

dm
A
dx
(11.33)
The constant, ηa, in eqn. (11.32) can be fixed only with the help of a more
detailed kinetic theory calculation [11.2], the result of which is given in
Section 11.4.
The choice of j
i
and m
i
for the description of diffusion is really some-

what arbitrary. The molar diffusion flux, J
∗
i
, and the mole fraction, x
i
,
are often used instead, in which case Fick’s law reads

J
i
∗
=−cD
im
∇x
i
(11.34)
Obtaining eqn. (11.34) from eqn. (11.27) for a binary mixture is left as an
exercise (Problem 11.4).
612 An introduction to mass transfer §11.3
Typical values of the diffusion coefficient
Fick’s law works well in low density gases and in dilute liquid and solid
solutions, but for concetrated liquid and solid solutions the diffusion co-
efficient is found to vary with the concentration of the diffusing species.
In part, the concentration dependence of those diffusion coefficients re-
flects the inadequacy of the concentration gradient in representing the
driving force for diffusion in nondilute solutions. Gradients in the chem-
ical potential actually drive diffusion. In concentrated liquid or solid
solutions, chemical potential gradients are not always equivalent to con-
centration gradients [11.3, 11.4, 11.5].
Table 11.1 lists some experimental values of the diffusion coefficient

in binary gas mixtures and dilute liquid solutions. For gases, the diffu-
sion coefficient is typically on the order of 10
−5
m
2
/s near room tem-
perature. For liquids, the diffusion coefficient is much smaller, on the
order of 10
−9
m
2
/s near room temperature. For both liquids and gases,
the diffusion coefficient rises with increasing temperature. Typical dif-
fusion coefficients in solids (not listed) may range from about 10
−20
to
about 10
−9
m
2
/s, depending upon what substances are involved and the
temperature [11.6].
The diffusion of water vapor through air is of particular technical
importance, and it is therefore useful to have an empirical correlation
specifically for that mixture:
D
H
2
O,air
= 1.87 ×10

−10

T
2.072
p

for 282 K ≤ T ≤ 450 K (11.35)
where D
H
2
O,air
is in m
2
/s, T is in kelvin, and p is in atm [11.7]. The scatter
of the available data around this equation is about 10%.
Coupled diffusion phenomena
Mass diffusion can be driven by factors other than concentration gradi-
ents, although the latter are of primary importance. For example, tem-
perature gradients can induce mass diffusion in a process known as ther-
mal diffusion or the Soret effect. The diffusional mass flux resulting from
both temperature and concentration gradients in a binary gas mixture is
then [11.2]

j
i
=−ρD
12

∇m
1

+
M
1
M
2
M
2
k
T
∇ln(T )

(11.36)
§11.3 Diffusion fluxes and Fick’s law 613
Table 11.1 Typical diffusion coefficients for binary gas mix-
tures at 1 atm and dilute liquid solutions [11.4].
Gas mixture T (K) D
12
(m
2
/s)
air-carbon dioxide 276 1.42×10
−5
air-ethanol 313 1.45
air-helium 276 6.24
air-napthalene 303 0.86
air-water 313 2.88
argon-helium 295 8.3
628 32.1
1068 81.0
(dilute solute, 1)-(liquid solvent, 2) T (K) D

12
(m
2
/s)
ethanol-benzene 288 2.25×10
−9
benzene-ethanol 298 1.81
water-ethanol 298 1.24
carbon dioxide-water 298 2.00
ethanol-water 288 1.00
methane-water 275 0.85
333 3.55
pyridene-water 288 0.58
where k
T
is called the thermal diffusion ratio and is generally quite small.
Thermal diffusion is occasionally used in chemical separation processes.
Pressure gradients and body forces acting unequally on the different
species can also cause diffusion. Again, these effects are normally small.
A related phenomenon is the generation of a heat flux by a concentration
gradient (as distinct from heat convected by diffusing mass), called the
diffusion-thermo or Dufour effect.
In this chapter, we deal only with mass transfer produced by concen-
tration gradients.