1 Cambridge University Engineering Department Engineering Tripos Part IIA 3A6: Heat and Mass Transfer Mass Transfer By N. Swaminathan Lent 2009 Course Objectives : • understand the principles of mass transfer • understand the process of mass diffusion in gases • understand the relevant non-dimensional groups and their analogues in heat and mo- mentum transfers • apply these understanding to solve simple problems • study catalytic converter via elementary modelling and applying the principles of heat and mass transfer Further reading: Discussion on mass transfer in Refs. [1] and [2].
Transcript
1
Cambridge University Engineering Department
Engineering Tripos Part IIA
3A6: Heat and Mass Transfer
Mass TransferBy N. Swaminathan
Lent 2009
Course Objectives:
• understand the principles of mass transfer
• understand the process of mass diffusion in gases
• understand the relevant non-dimensional groups and their analogues in heat and mo-mentum transfers
• apply these understanding to solve simple problems
• study catalytic converter via elementary modelling and applying the principles ofheat and mass transfer
Further reading: Discussion on mass transfer in Refs. [1] and [2].
Mass transfer process deals with the transfer of mass from one point to another. Also, thisprocess makes sense only when there is more than one component or chemical substanceor species involved. For example consider an simple experiment, where a small crystal ofpotassium permanganate (KMnO4) is dropped into a beaker of water. As KMnO4 beginsto dissolve, it creates a concentration gradient, which assists the mass transfer process viadiffusion. This mass transfer process can be observed with the naked eyes via the (perhapsvery slow) growth of purple region, signifying the concentration of KMnO4, in the jar.This diffusion mass transfer is called ordinary diffusion to distinguish it from pressurediffusion–motion of a component because of pressure gradient, thermal diffusion–motion ofa component resulting from thermal gradient, and forced diffusion–motion of a componentbecause of unequal external forces on the components. It is also important to note that thereis no bulk motion involved in the above experiment. This type of mass transfer is analogousto heat conduction. This analogy naturally leads us to think of a situation, where the masstransfer is analogous to convection heat transfer, involving bulk fluid motions. For thesesituation, the concentration boundary layers will naturally play a role. Convective masstransfer problems play an important role in nature and in many engineering applications.For example, evaporation of water at the ocean surface, dissolving of a sugar cube in a cupof coffee with a stirrer, cooling towers of power plants, ablative cooling for reentry vehicles,combustion, etc. In this part of our discussion, we will study the principles governing theabove two types of, diffusion and convection, mass transfer. First, let us introduce somedefinitions to describe a mixture and to avoid ambiguity during the analysis. As we notedearlier, in mass transfer problems a multicomponent mixture, at least a binary mixture, isinvolved. For the purpose of generality, we will consider a multicomponent mixture tointroduce the definitions which are equally applicable for a simple case of binary mixture.
Let us consider a mixture contained in volume, V , with component masses, m1,m2, · · · ,mN .The mass of an arbitrary component is mi and its density is ρi = mi/V . Now, the totalmass of the mixture and its density are respectively
m =N∑
i=1
mi, and ρ =m
V=
N∑i=1
ρi. (1.1)
ρi is also known as mass concentration (kg/m3) of species i. The molar concentration(kmol/m3) is ci = ρi/Mi = ni/V , with ni as the number of moles of species i and Mi is itsmolecular weight (kg/kmol). (Can you recall what one mole of a species means?)
5
6 CHAPTER 1. INTRODUCTION
One can define mass fraction and mole fraction for species i as
Yi =ρi
ρ=
mi
m, and Xi =
ci
c=
ni
n. (1.2)
The mass fraction is realted to the mole fraction via
Yi =Xi Mi
M, M =
∑Xi Mi is the mixture molecular weight (1.3)
If each of the species in a multicomponent gas mixture behaves ideally, then
pi = ρiRiT = ρiTRMi
,
where pi is the partial pressure of component i and R = 8.314 kJ/kmol−K is the univer-sal gas constant. Now, the total pressure p of the mixture is given by Dalton’s law of partialpressure:
p =∑
i
pi.
If the volume containing the above mixture moves with a velocity u, the underscoresignifies that u is a vector, while each species is moving with its own velocity ui, then thebulk mass flux is
G ≡ ρu =N∑
i=1
ρiui, ⇒ u =1
ρ
N∑i=1
ρiui =1
ρ
N∑i=1
Gi. (1.4)
The velocity u is called mass average velocity. One can also define molar average velocityas
u∗ =1
c
N∑i=1
ciui =1
c
N∑i=1
J i. (1.5)
We should note that these average velocity are local–they vary in space and time. In ouranalysis later, we shall see that the mass flow rate of species i per unit area in a frame ofreference which moves with the flow will be required. This mass flux is called diffusiveflux and it is defined as
Gdif,i = ρi(ui − u)︸ ︷︷ ︸diffusion velocity
, =⇒∑
i
Gdif,i = 0. (1.6)
Here, it should be noted that the mass flux definition is incomplete until both units and theframe of reference are given. This is because the diffusion velocity is relative. For a binarymixture involving species A and B, the above relation yields Gdif,A = −Gdif,B.
The absolute flux of species i is defined as the sum of diffusive and convective flux.Thus, the absolute mass flux is
Gi = Gdif,i + ρiu = Gdif,i + YiG. (1.7)
If medium is stationary (G = 0) then the absolute flux is equal to the diffusive flux. Theabsolute molar flux is
J i = Jdif,i + XiJ. (1.8)
7
In the next chapter we shall establish the governing equations for diffusive and convec-tive mass transfer problems along with the required constitutive relations.
Problem 1.0:To check your understanding of the above definitions, work out the following example.
liquid B
vapor A
y, v
xA
t = 0
t > 0
At a specific time :xA = 1/3, v* = 10, vA – v* = 5, MA = 5MB
Ans.: vB = 15/2; vB-v* = -2.5, v = 90/7, vB-v = -75/14, vA-v = 15/7, sum of diffusive fluxes is zero.
8 CHAPTER 1. INTRODUCTION
Chapter 2
Conservation Laws and ConstitutiveRelations
The conservation laws include mass, momentum, energy and species conservation. Theconservation equations for mass, momentum and energy are familiar to you and have beenderived in chapter 2 of convection heat transfer part. In mass transfer problems, we dealwith multicomponent fluids and thus the conservation of each chemical species is need tobe considered. The species conservation equation is derived below. One should also beaware that the energy equation, Eq. (2.12) in the convection heat transfer part, will haveadditional terms when multicomponent mixtures are considered. Here, we shall restrictourselves to isothermal mass transfer cases.
2.1 Species Conservation
We follow the control volume approach and consider the fluid to be a multicomponentmixture. The rate of species i mass flux entering and leaving the control volume in xdirection is shown in Fig. 2.1. The rate of generation of species i inside the control volume
x, u
y,v
z,w
dx
dzdy
dxx
uYuY ii
ii ∂∂+ ρρiiii uYu ρρ =
iω&
x, u
y,v
z,w
dx
dzdy
dxx
uYuY ii
ii ∂∂+ ρρiiii uYu ρρ =
iω&
Figure 2.1: Control volume for conservation of species i in a flow of multicomponentmixture.
by some process, say chemical reactions for example, is denoted by ωi. Mass balance forspecies i gives
Rate of change = net flux︸ ︷︷ ︸in − out
+production− consumption
=⇒ ∂ρi
∂t∆x∆y∆z = −∂ρ Yi ui
∂x∆x∆y∆z + · · · · · ·+ ωi∆x∆y∆z,
9
10 CHAPTER 2. CONSERVATION LAWS AND CONSTITUTIVE RELATIONS
⇒ ∂ρYi
∂t+
∂ρ Yi ui
∂x+
∂ρ Yi vi
∂y+
∂ρ Yi wi
∂z= ωi
⇒ ∂ρYi
∂t+∇ · YiG = ωi −∇ ·Gdif,i, (2.1)
as the governing equation for the species i. Note that we have used Eq. (1.7). By summingEq. (2.1) over all the species in the mixture, one can easily see that the overall continuityequation , Eq. (2.1) in the convection heat transfer part, is obtained:
∂ρ
∂t+∇ ·G = 0, since
∑i
ωi = 0. (2.2)
2.2 Constitutive RelationThe constitutive equations establish relations between the fluxes involved in transfer pro-cesses and the respective driving potentials. Fourier law of heat conduction, Newton’s re-lation for viscous stresses are examples for the constitutive relations. In mass transfer, theconcentration gradient is the driving potential for the mass fluxes. In a moving multicom-ponent mixture, the constitutive relations become more intricate when there is simultaneousmass and heat transfer. For example, the composition gradient can drive energy fluxes –known as Dufour effect and the temperature gradient can drive mass fluxes – known as soreteffect. These effects are of smaller magnitude compared to ordinary diffusion in generalengineering problems. Thus, we shall neglect these effects from further consideration.
The constitutive relation for a simple mass diffusion is given by Fick’s law1 on massbasis:
Gdif,i = −ρDij∇Yi, (2.3)
in subscript form
Gdif,ik = −ρDij∂Yi
∂xk
, with i, j = 1, · · · , N ; k = 1, 2, and 3.
The symbol Dij is the diffusion coefficient which depends not only on temperature andpressure but also on species pair (species i – species j). The Fick’s law on molar basis is
Jdif,,i = −cDij∇Xi. (2.4)
The above two equations, Eqs. (2.3) and (2.4), are also known as Fick’s first law ofdiffusion. For binary mixtures we noted Gdif,1 = −Gdif,2 and Jdif,1 = −Jdif,2, which yields
D12 = D21 = D. (2.5)
In the above discussion, only the concentration gradient is considered to be the drivingpotential for the mass flux. However, even in isothermal systems it is known that pressuregradient and external forces acting unequally on the various components of the mixturecan lead to mass fluxes. For example, in swirling flow or centrifuge there will be a large
1Adolf Fick - 1855
2.3. ESTIMATION OF DIFFUSION COEFFICIENT 11
radial pressure gradient acting on the fluid. In ionic separation of liquids the externallyimposed electric field plays important role on the mass fluxes of ions present in the liquid.In non-isothermal systems, the temperature gradient also contributes to the mass flux andthis process is called thermal diffusion. For example, the diffusion of hydrogen which is animportant species involved in combustion is drastically influenced by the thermal diffusion.In multicomponent mixture all these driving potentials can interact resulting in interactionof heat and mass transport processes. Even a brief discussion of these topics is beyond thelevel of this module and so we shall restrict our curiosity to consider only binary mixtureswith Fickian type diffusion. The multicomponent mixtures can also be approximated to abinary mixture by considering the diffusion of one component into a mixture of the rest ofthe components. This approximation is often made in the analysis of heat and mass transferof a multicomponent mixture. However, one should be aware that the mass conservationmay be violated by this approximation and in order to ensure mass conservation, speciesvalues are obtained by solving the transport equations for N − 1 species while the N th
species value is obtained via YN
= 1 − ∑i Yi
. However, we shall note that the heat fluxvector εi (see section 2.3 of the convective heat transfer notes) gets modified as
εi = −κf∂T
∂xi
+ Gdif,1T (cp,1 − cp,2) (2.6)
for a binary mixture of ideal gases. This change is important while considering simulta-neous heat and mass transfer problems as in combustion. The first part is because of heatconduction while the second part is due to the enthalpy flux created by the diffusive massflux.
2.3 Estimation of Diffusion Coefficient
The binary diffusivity, D, is an important parameter in mass transfer problems. This quan-tity depends on temperature, pressure and the species pair. For dilute mixtures2 the diffu-sivity is almost composition independent, increases with temperature and varies inverselywith pressure. Liquid and solid diffusivities are strongly concentration dependent and gen-erally increases with temperature. Diffusivity of gases in solids are typically of order 10−14
at normal temperatures while for solids in solids are typically of order 10−20. For liquid-liquid combinations, typical values are of order 10−9 and for gas–gas combinations thevalues are of order 10−5. These typical values are in m2/s. From the kinetic theory of gases(the derivation is involved but we shall take the final result), the diffusivity may be writtenas
DAB = 1.883× 10−2
√T 3 (MA + MB) /MAMB
pσ2ABΩD
, (in m2/s) (2.7)
where
T absolute temperature (K)p Pressure (Pa)Mi molecular weight of component i (kg/kmol)σAB = 0.5(σA + σB) σA&σBare Lennard− Jones parameters (A)ΩD collision integral which depends on k
BT/ε
AB; ε
AB=√
εAε
B
2one of the species is present in large amount
12 CHAPTER 2. CONSERVATION LAWS AND CONSTITUTIVE RELATIONS
The values of Lennard–Jones potential parameters, σ and ε, for various gases are given inbooks on kinetic theory of gases and liquids. These values for few gases, which may beof our interest here, are given in Table 2.1. σ is a characteristic diameter of the molecule,
also known as collision diameter, and ε is a characteristic energy of interaction betweenthe molecule which is the maximum energy of attraction between a pair of molecule. k
Bis
r
P.E
.
0σ
ε
r
P.E
.
0σ
ε
Figure 2.2: Typical variation of intermolecular potential energy with intermolecular dis-tance r. σ and ε are the Lennard–Jones potential parameters.
the Boltzmann constant. Typical variation of this potential, attraction, energy, between twospherical molecule is shown in Fig. 2.2 which also defines σ and ε. The collision integralis a slowly varying function of nondimensional temperature k
BT/ε which are tabulated in
books on kinetic theory of gases and in books on transport phenomena (for example see[4]). This table is given in the Appendix for our convenience.
We should note that the above formula for diffusivity is applicable for dilute (low den-sity) mixtures of gases. For high density cases, other expressions are available (see [4]).The diffusion in liquids is complex and the values of diffusivities are obtained from ex-periments. In this module, we are primarily concerned with gases and thus we shall limitour discussion to ideal gases only. For those having inquisitive nature, [4] has good dis-cussion on diffusion in liquids and also contains numerous references. For more advanceddiscussion see [5].
Problem 2.0:Calculate the diffusivity of CO in air at 300 K and 1 bar.
Solution:SpeciesA(Air) : M
A= 28.97; σ
A= 3.617 A; ε
A/k
B= 97
2.3. ESTIMATION OF DIFFUSION COEFFICIENT 13
SpeciesB(CO) : MB
= 28.01; σB
= 3.59 A; εB/k
B= 110
σAB
= 0.5(3.617 + 3.59) = 3.604 A; εAB
/kB
=√
97× 110 = 103.3 Kfor k
BT/ε
AB= 2.904 ΩD = 0.9572 from the table in appendix
⇒ DAB
= 2.085× 10−5 m2/s from Eq. (2.7)
A few points to note from Eq. (2.7) are
1. the diffusivity increases with temperature,
2. it decreases as the pressure increases, and
3. also it depends on the molecular weights of the species involved.
For binary mixtures diffusing in predominantly one dimensional situation, the Fick’slaw becomes
Gdif,i = −ρDdYi
dx(in kg/m2s) and (2.8)
Jdif,i = −cDdXi
dx(in kmol/m2s). (2.9)
One can follow either molar or mass basis definitions given above and in the previouschapter. Chemical engineers usually use molar basis definitions for their own reasons. But,it may be appropriate to use mass basis definitions, since mass is conserved in any physicaland chemical processes. The moles do not have to be conserved all the time. For example,chemical reactions do not conserve moles but conserve mass:
C + O2︸ ︷︷ ︸2moles, 44g
→ CO2︸︷︷︸1mole, 44g
.
Thus, we shall prefer to use mass basis definitions in our analysis.
Note: We shall also be clear that in our discussion we refer to the medium as stationarywhen there is no bulk mass flux, ie. G = 0. This does not generally mean that the bulkmolar flux, J , is also zero. This can be shown as below for a binary mixture having com-ponents with molecular weights M1 and M2.
no bulk mass flux implies : G = G1 + G2 = 0 ⇒ G1 = −G2
14 CHAPTER 2. CONSERVATION LAWS AND CONSTITUTIVE RELATIONS
Chapter 3
Diffusion Mass Transfer
The diffusion mass transfer originates from molecular activity and occurs in gases, liquids,and solids. The molecular activity is strongly influenced by the mean free path length.Thus, diffusion occurs readily in gases than in liquids and readily in liquids than in solids.The order of magnitudes of the diffusion coefficients given in the previous chapter areclearly in support this observation.
The diffusion mass transfer occurs in stationary systems and it is usually unsteady inthe absence of any other physical processes. In special circumstances, diffusion can occurin steady state condition also. To visualise the unsteady nature of this mass transfer process,one can revisit the simple KMnO4 experiment discussed in the Introduction. The drivingpotential, the concentration of KMnO4, is changing with time and thus the mass diffusionprocess. If there is chemical reaction in the system then the diffusive flux can be balancedby the chemical reaction rate for a steady state to occur. Such situations occur in biologicalreactors. In mechanical or aerospace applications, for example GT combustor, fluid flowis involved invariably. The mass transfer in the presence of fluid flow is called convectivemass transfer. We consider the diffusion mass transfer in this chapter and the convectivemass transfer in the next chapter.
3.1 Governing EquationSince the diffusive mass transfer occurs in stationary systems, the mass average velocity, uis zero. Thus, the species coservation equation, Eq. (2.1), becomes
ρ∂Yi
∂t= ωi + ρD∇2Yi, (3.1)
• with the Fick’s law used
• take the dilute mixture to be binary and the chemical species of interest is in lowconcentration so that the bulk fluid properties are not unduly influenced by it
• approximate ρD as a constant.
From the species equation given above three important cases are:
2. unsteady diffusion – analogous to unsteady heat conduction (∂Yi
∂t= ρD∇2Yi) and
3. diffusion with chemical reactions - called diffusive–reactive cases (ωi+ρD∇2Yi = 0)
For simplicity sack, one dimensional situation are considered to understand the essentialphysics involved.
3.2 Steady DiffusionConsider a gas (note that this a single component) is held at slightly different pressureson either side of a membrane and the gas pressures are maintained at constant level bysome means. If the gas is soluble, with solubility S (kmol/m3-bar), in the membrane thenthere will be diffusion of gases inside the membrane as shown in Fig. 3.1 which is slightlydifferent from the case when the membrane is permeable or semi–permeable. Inside themembrane the mixture becomes binary and the gas is treated to be component 1 and themembrane material to be component 2. The mass or molar concentrations of the gas ateither ends of the membrane are related to the solubility, S, and the gas partial pressuresas noted in the figure. Mass diffusion of the gas inside the membrane attains a steady state
Gas at p0
Gas at pL
L
x
Y1,0
Y1,L Mol. Wt. (kgl/kmol)
ρ1,0 = S p0 M1 = c1,0M1 (kg/m3)
Solubility (kmol/m3-bar)
ρ1L = S pL M1 = c1,LM1 (kg/m3)
Gas at p0
Gas at pL
L
x
Y1,0
Y1,L
Gas at p0
Gas at pL
L
x
Y1,0
Y1,L Mol. Wt. (kgl/kmol)
ρ1,0 = S p0 M1 = c1,0M1 (kg/m3)
Solubility (kmol/m3-bar)
ρ1L = S pL M1 = c1,LM1 (kg/m3)
ρ1,0 = S p0 M1 = c1,0M1 (kg/m3)
Solubility (kmol/m3-bar)
ρ1L = S pL M1 = c1,LM1 (kg/m3)
Figure 3.1: Steady diffusion in a one dimensional situation.
since the pressure levels are kept constant. This steady state yields
d2Y1
dx2= 0
(see Eq. (3.1) with ω = 0).1 This gives
Y1(x) =(Y1,L − Y1,0)
Lx + Y1,0, for 0 ≤ x ≤ L (3.2)
as noted in Fig. 3.1. Now the rate of mass diffusion of the gas as
A Gdiff,1 = ρY1,0 − Y1,L
L/AD = ρY1,0 − Y1,L
Rm
, (kg/s) (3.3)
where A is the area available for the mass diffusion. This is similar to the heat transferby conduction. Rm (in SI units - s/m3) is the resistance offered by the geometry and the
1one can also obtain this equation by considering a strip of size dx inside the membrane. At steady state,mass of the gas entering this strip should be equal to the mass leaving. Thus, dG1/dx = 0. CombiningEq. 1.7 and Fick’s law, one gets d2Y1/dx2 = 0.
3.2. STEADY DIFFUSION 17
x
Gas 1
M1
Gas 2
M2p1p2
dx
Figure 3.2: Steady diffusion between two reservoirs
medium to the mass diffusion. It is straight forward to work out this resistance in threeco-ordinate systems:
Cartesian : Rm ≡ L
AD (3.4)
Cylindrical : Rm ≡ ln(r2/r1)
2πLD (3.5)
Spherical : Rm ≡ 1
4πD(
1
r1
− 1
r2
), (3.6)
whereD is the diffusivity of the gas. We should note that if the gas is insoluble in the mem-brane or the membrane is impermeable then the diffusivity D is zero and the resistance Rm
becomes infinite. This simply means that there is no mass diffusion through the membrane.
3.2.1 Special Cases: Equi–mass and Equi–molar counter diffusion
Let us consider the case in Fig. 3.2 where two different gases at different pressures are in thereservoirs. The connecting tube is considered to be long with sufficiently large diameter.A thin diaphragm. initially separates the Gas–1 with molecular weight M1, on the left sidereservoir, at pressure p1 from the Gas–2 with molecular weight M2 at pressure p2 in theright reservoir. Let us take p1 to be slightly larger than p2. As soon as the diaphragm ispunctured, the Gas–1 will rush through and fill the tube. The pressure difference over thetube length will become zero leading to zero bulk mass flux across any cross section inthe tube. However, there will be diffusion of gases occurring inside the tube because ofconcentration gradient and this diffusion process will be steady. The conservation of massacross dx yields
dG1/dx = 0, and dG2/dx = 0.
The solution to this equation with appropriate boundary conditions is the same as in Eq. 3.3.Also, since there is no bulk mass flux one gets G1 = −G2 across the thin strip. Thissituation is called as equi–mass counter diffusion.
The molar flux across the thin strip is J1 = G1/M1 and J2 = G2/M2 = −J1M1/M2.If the molecular weights of the two gases are the same then one gets J2 = −J1 . Thiscondition is called equi–molar counter diffusion. It is clear that the equi–mass counterdiffusion does not imply that the bulk molar flux, J , is zero and vice–versa. The bulk molarflux across the thin strip can be obtained via J = J1 + J2. The absolute molar flux of the
18 CHAPTER 3. DIFFUSION MASS TRANSFER
Gas–1 is
J1 = Jdif,1 + X1(J1 + J2) ⇒ J1 = − cD(1−X1
[1− M1
M2
]) dX1
dx(3.7)
The balance of molar flux across the thin strip dx is dJ1/dx = 0 when there is no chemicalreaction inside the strip. This implies that the molar flux is constant, ie., J1 = A. From theabove balance equation, one can obtain the variation of X1 across the tube length, L, as
1−X1[1−M1/M2]
1−X1,0[1−M1/M2]=
(1−X1,L[1−M1/M2]
1−X1,0[1−M1/M2]
)(x/L)
(3.8)
where X1,0 and X1,L are the mole fractions of Gas–1 at x = 0 and x = L. The molar fluxis
J1 =cD
L[1−M1/M2]ln
1−X1,L[1−M1/M2]
1−X1,0[1−M1/M2]
. (3.9)
(Can you work out which direction this molar flux will go?) Similarly, one can also workout the molar flux J2.
In the case of equi–molar diffusion, it is straight forward to write
J1 =cDL
(X1,0 −X1,L) =D
RTL(p1,0 − p1,L) . (in kmol/m2s) (3.10)
3.2.2 Evaporation of a liquid Column: Stefan ProblemIn the steady diffusion problems, the evaporation of a liquid inside a column is little deceiv-ing. One needs to pay attention to the details of the physical processes happening near thesurface of the liquid. Let us consider the situation shown in Fig. 3.3, where the liquid levelin a long container is maintained at the same level by some means and the details of whichis unimportant to us. To maintain a steady state, the supply of liquid and a slight flow ofgases are required as shown in Fig 3.3. The liquid A evaporates and its vapor diffuses out
Slight flow of gas B + vapor A
Liquid A
B
AA
y
0
H
supply
YA,0
YA,H
Slight flow of gas B + vapor A
Liquid A
B
AA
y
0
H
supply
Slight flow of gas B + vapor A
Liquid A
B
AA
y
0
H
supply
YA,0
YA,H
Figure 3.3: Evaporation of a liquid into a gas mixture.
to y = H and then carried away by the slight flow. The gas B will also diffuse towards theliquid because of the difference in its concentration. We take the gas B to be insoluble (ornegligibly soluble) in liquid A. Thus, the diffusion of B will establish a stagnant gas filmwhen the steady state condition is reached. This means that the downward flux of B near
3.2. STEADY DIFFUSION 19
the surface (y = 0) should be balanced by the upward flux of vapor A. This balance createsa bulk motion for vapor A. Using Eq. (1.7), the absolute flux of A is
GA
=−ρD
(1− YA)
dYA
dy. (3.11)
But under steady state condition, dGA/dy = 0. This equation yields
1− YA
1− YA,0
=
(1− YA,L
1− YA,0
)y/L
, (3.12)
after some algebra, for the conditions shown in Fig. 3.3. From this solution, the quantitiesof engineering interest, the evaporation flux (in kg/m2-s), can be obtained as
GA(y = 0) = − ρD
1− YA,0
(dYA
dy
)
y=0
=ρDH
ln
(1− YA,H
1− YA,0
)=
ρDH
ln
(YB,H
YB,0
). (3.13)
The average mass fraction of vapor A in the column can be obtained from
Y A =1
H
∫ H
0
YA dy.
(How would you approach the problem if the gas B is soluble in the liquid with solubilityS?). Also, by defining a number called mass transfer number as
1 + B =1− YA,H
1− YA,0
,
the evaporation flux in Eq. (3.13) can be re-written as
GA(y = 0) =ρDH
ln(1 + B). (3.14)
It is clear from the above equation when B is zero the evaporation rate (flux × area) iszero and as B increases so does the evaporation rate. This makes physical sense since theevaporation rate depends on (YA,0 − YA,H) which is involved in the definition of B. Thus,B can be interpreted as a driving potential for mass transfer.
If one considers the evaporation of a droplet in a stagnant environment then the evapo-ration rate, not the evaporation flux, remains constant. This is because of the change in thearea available for mass transfer during the evaporation process. Thus, rewriting Eq. (3.11)in terms of evaporation rate, following the above principles of analysis and consideringthe droplet mass conservation one can deduce an important relationship, called d2 law, indroplet evaporation/combustion. This relationship is
d2(t) = d2o −Kt, (3.15)
where d(t) is the diameter of the droplet at time t, do is the initial droplet diameter and Kis the evaporation constant given by
K =8ρDρl
ln(1 + B),
with ρl is the liquid density. (see problem 2 in Example sheet 4 as an application of theabove principles).
20 CHAPTER 3. DIFFUSION MASS TRANSFER
3.3 Unsteady DiffusionLet us reconsider our little experiment with potassium permanganate. We noted that theinterface to move very slowly with time. This is because of unsteady diffusion process.The unsteady diffusion occurs in many engineering situations and in nature, for examplethe spread of contaminants in the atmosphere or an oil spill in the ocean. However, weshould note that these spreading process are invariably influenced by the wind and theocean currents. Thus, one could say that the unsteady diffusion process is influenced by theconvection process in the above cases. Another good example for a pure unsteady diffusionis the separation of isotopes.
y
Y10 1
0
t
y
Y10 1
0
y
Y10 1
0
t
Figure 3.4: Evolution of a species mass fraction when its diffusion is unsteady.
The process of unsteady diffusion in one dimensional planar geometry is governed by,see Eq. 3.1,
ρ∂Yi
∂t=
∂
∂y
(ρD∂Yi
∂y
). (3.16)
This equation is also known as Fick’s second law of diffusion. In one treats ρD as a constantthen
∂Yi
∂t= D∂2Yi
∂y2,
which is similar to the unsteady heat conduction equation, which can be solved after spec-ifying appropriate initial and boundary conditions. These conditions for our potassiumpermanganate (denoted by subscript 1) experiment are
Y1(y, 0) = Y in1 = 0; Y1(0, t) = Y1,0 = 1; Y1(∞, t) → Y in
1 = 0.
We have taken the length of the apparatus (the glass jar) to be very long compared to thesize of the potassium permanganate source. Also, we have taken the concentration at y = 0does not vary with time for simplicity purpose. One can obtain a solution to the above setusing error function, which may be written as
Y1 − Y1,0
Y in1 − Y1,0
= erf
(y
2√Dt
), (3.17)
which is shown in Fig. 3.4.The above solution is the similar to that of unsteady heat conduction. You may also
recall that the analysis of unsteady heat conduction involved two parameters, called Fourier
3.4. DIFFUSION WITH CHEMICAL REACTION 21
and Biot numbers. The Fourier number for heat conduction is Fo = αt/R2, where α is thethermal diffusivity and R is the characteristic length. Similar to this, one can also defineFom = Dt/R2, as the Fourier number for diffusion mass transfer. This number is theratio of diffusion length scale to characteristic geometric length scale. The Biot number formass transfer is defined as hmR/D, where hm is the convective mass transfer coefficient.The Biot number plays a role when there is convective mass transfer (discussed in the nextchapter) at the boundaries of the domain in which diffusion mass transfer is occurring.When the diffusivity, D, is very large the spatial non-uniformity in mass concentration canbe neglected and the lumped capacitance method may be employed for the analysis.
3.4 Diffusion with Chemical ReactionThe diffusion mass transfer can be influenced by the presence of chemical reactions. Thechemical reactions can be homogeneous or heterogeneous. Homogeneous reactions areusually volumetric phenomena while the heterogeneous reactions are surface reactions.The reactions in gaseous or liquid combustion in the absence of solid catalysts are goodexamples for homogeneous reactions. The chemical reactions which occur on the surface ofa solid catalyst, as in the catalytic converter, are good examples for heterogeneous reactions.One should recall that the catalyst does not have to be a solid all the time, for example,additives used to improve the octane number for high performance racing cars are usuallyliquids. In this section, we shall see how the chemical reactions influence the diffusionmass transfer in steady state conditions. These cases occur mostly in bio-chemical andbiological reactors. It is important for us to understand these relatively simple cases beforewe embark on analysing mechanical and aerospace engineering cases.
3.4.1 Diffusion with Homogeneous ReactionThe homogeneous reactions are usually represented by
R1 + R2 → P,
where R1 and R2 are reactants and P is product. An example for this is the combustion ofcarbon monooxide with oxygen forming carbon dioxide. The molar rate of production ofP in the above reaction is typically written in the form
ω∗P
= kcnR1
cmR2
, (kmol/m3 − s)
[ωP
= MPω∗
P, (kg/m3 − s)]
where k is known as rate constant and n is the order of the reaction with respect to reactantR1. The units for k depends on the overall order (n + m) of the reaction. ci is the molarconcentration of species i. For simplicity purpose we shall take m = 0 and consider onlyzeroth and first order reactions. For these reactions, one gets
zeroth order reaction, n = 0 : ω∗P
= ko ko in kmol/m3 − sfirst order reaction, n = 1 : ω∗
P= k1cR1
k1 in 1/s.
For the above reaction, mass conservation yields ωP
= −ωR1
= −ωR2
. Note that thereaction rates are in mass units not in molar units.
22 CHAPTER 3. DIFFUSION MASS TRANSFER
y
0
H
Y1,0
Y1,H
Liquid or gas - 1
Liquid -2y
0
H
Y1,0
Y1,H
Liquid or gas - 1
Liquid -2
Figure 3.5: Steady state variation of mass fraction of species 1 in a dilute binary mixturein the presence of simultaneous diffusion and homogeneous, zeroth order reaction. Thedashed line is for the case of diffusion only.
Let us consider the situation shown in Fig. 3.5, where a gas or a liquid, 1, is slowlydiffusing into another liquid-2 and undergoes a zeroth order reaction. The mixture is diluteand there are some arrangements to keep the system in steady state. The mass fraction ofthe gas or liquid 1 is governed by
ρDd2Y1
dy2+
−koM1︷︸︸︷ω1 = 0; with Y1(0) = Y1,0 and
(dY1
dy
)
y=L
= 0. (3.18)
The above set of equations yield the variation of Y1 as
Y1(y) =
(koM1
ρD)[
y2
2− Ly
]+ Y1,0,
which is also shown in Fig. 3.5. If there is no chemical reaction then Y1 = Y1,0. If the reac-tion is first order then ω takes an appropriate form and one requires to solve a second orderODE to obtain Y1(y). (Recall your IA mathematics on ODE, see problem 3 in examplesheet 4).
3.4.2 Heterogeneous ReactionHeterogeneous reactions are common in catalysis, which usually involve a solid phasecatalyst. The chemical reactions in this type of situation are usually surface reactions. Also,in situations such as burning of solid particles such as carbon, sulphur, etc., in a stagnantenvironment the chemical reactions occur in a region very close to the solid particle, whichmay be treated as surface reactions for our analysis here. These cases are shown in Fig. 3.6.The analysis of this problem is simple as we will see that the influence of the heterogeneousreaction comes via a boundary condition. We start with the conservation of species R in thethin strip of size dy. This conservation equation has to be written in appropriate co-ordinatesystem of interest. Here, we take the planar case and let the mass fraction of the reactantR at y = H as YR,H , where H is sufficiently far away from the catalyst surface. Since themedium is stagnant and there is no homogeneous reaction the mass conservation of speciesR is
dGdif,R
dy= 0; ⇒ Y
R=
( AρD
)(y −H) + Y
R,H, (3.19)
whereA = ρD(dYR/dy), which is nothing but the diffusive flux of reactant R. One should
also note that this diffusive flux is constant. In steady state conditions, this diffusive flux
3.4. DIFFUSION WITH CHEMICAL REACTION 23
y R
P
dy
0catalyst
R → P on the surface by
s)(kg/m
s)(kmol/m
2
21
−Ω=Ω
−Ω−=−=Ω**
*,
'*
&&
&&
RR
PsRR
M
ck
r
dr
Rp
R
P
y R
P
dy
0catalyst
R → P on the surface by
s)(kg/m
s)(kmol/m
2
21
−Ω=Ω
−Ω−=−=Ω**
*,
'*
&&
&&
RR
PsRR
M
ck
r
dr
Rp
R
P
Figure 3.6: Diffusion in a stagnant medium with a heterogeneous catalyst (a) as planarsurface and (b) as spherical pellet with radius Rp. The burning of a solid particle in astagnant atmosphere can be viewed as in (b) with R as oxygen.
must be balanced by the reaction rate at the surface. We make use of this observation toobtain the diffusive flux.
ρD(
dYR
dy
)
y=0
= ΩR
= ρk′1YR,s
A = ρk′1YR,s= ρk′1
(−AH
ρD + YR,H
)
⇒ A =ρk′1YR,H
[1 + k′1H/D], (3.20)
where YR,s
is obtained from Eq. (3.19) with y = 0. After some simple algebra, one gets
YR,s
YR,H
=1
1 + Hk′1/D. (3.21)
It is clear from the above equation that the reactant mass fraction on the surface, thusthe reaction rate Ω
R, strongly depends on the competition between the diffusion and the
chemical reaction. When the chemical reactions are infinitely fast, ie., k′1 → ∞, YR,s
goesto zero. Under this circumstance, the operation of the catalyst is said to be diffusion limited,which implies the diffusion is the slow (or the rate controlling) process. This means thatas soon as the species R arrives at the catalyst surface via diffusion it is consumed at onceby the reaction. In the other limit when the reaction is very slow, ie., k′1 → 0, one getsY
R,s= Y
R,H. This limit is called kinetic limit.
Two points to note in the above analysis are
1. the specific rate constant k′1 is taken to be a constant while it can depend on temper-ature, and
2. the fluid medium is considered to be stagnant. When there is a flow over the catalyst,the concept of convective mass transfer, discussed in the next chapter, needs to beapplied. However, the above principles of analysis will still apply.
24 CHAPTER 3. DIFFUSION MASS TRANSFER
Chapter 4
Convective Mass Transfer
This mode of mass transfer involves bulk motion of fluid and thus boundary layers playimportant role. The velocity and thermal boundary layers played their role in the transfer ofheat and momentum. Now, it is natural to expect the existence of concentration boundarylayer when there is mass transfer from a surface to the fluid flow or vice versa. One canexpect the flow to be induced by external means or via natural convection phenomena. Weshall consider both of these cases here.
4.1 Forced Convective Mass transfer
4.1.1 External Flow
In the forced convection case, the flow, which can be either external or internal, is setup by using a fan or a blower. Let us consider a dilute binary mixture flowing with auniform velocity U∞ parallel to the surface at y = 0. The surface is coated with a certainsubstance, already present in the binary mixture at a level of ρ1,∞, that is soluble in themixture stream. (A good example for this situation is the case of flow over a fuel pool ora fuel droplet.) This case of binary mixture flow over a flat surface is sketched in Fig. 4.1.The mass concentration of species 1 at the wall is considered to be constant at ρ1,s and the
ρ1,δρ
Ux
y
L
ρ1,s
ρ1,δρ
Ux
y
L
ρ1,s
Figure 4.1: Mass transfer from a flat plate to a laminar flow. δρ is the concentration bound-ary layer thickness.
free stream value is also constant at ρ1,∞. The variation of mass concentration of species1 shown in the figure indicates the highest gradient near the wall. Also, because of no-slipcondition (fluid is at rest ) at the wall, the fluid layer which is very next to the wall is at rest.
25
26 CHAPTER 4. CONVECTIVE MASS TRANSFER
Thus, in this thin layer one can invoke Fick’s law if the mixture is dilute. Thus,
Gdif,1,s = −D(
∂ρ1
∂y
)
y=0
. (4.1)
If one applies the analogy of Newton’s law of cooling to mass transfer problems then thediffusive flux may be defined using a convective mass transfer coefficient, hm, and concen-trations at wall and free stream. Hence,
In SI system of units, the mass transfer coefficient hm has the units of m/s. By recalling thedefinition of convective heat transfer coefficient, h, one may see that the convective masstransfer and heat transfer coefficients have similar physical meaning.
It is clear from the above equation that our task is now shifted to find the concentrationgradient at the wall, like in the convective heat transfer problem our interest to obtainthe heat transfer rate was shifted primarily to determine the temperature gradient at thewall. One can apply any of the techniques employed for the heat transfer problems tomass transfer problems also. This means that one needs to solve the concentration (speciesconservation ) boundary layer equation
u∂ρ1
∂x+ v
∂ρ1
∂y= D∂2ρ1
∂y2, (4.3)
subject to the boundary conditions noted in Fig. 4.1. These boundary conditions are
ρ1 = ρ1,s at y = 0; and ρ1 → ρ1,∞ as y →∞.
If one replaces ρ1 by temperature T and diffusivity D by the thermal diffusivity α thenEq. (4.3) and it boundary conditions become the temperature equation and its boundaryconditions. This observation allows us to make an important conclusion that the concentra-tion boundary layer behavior is analogous to the thermal boundary layer. Thus, solution tothermal field can be used to obtain concentration field. Mathematically this is true only ifD = α. This condition identifies a non-dimensional number Le ≡ α/D called Lewis num-ber. This number is the ratio of thermal diffusivity to mass diffusivity which plays impor-tant role in simultaneous heat and mass transfer problems such combustion. In isothermalmass transfer problems the Schmidt number, Sc ≡ ν/D, plays an important role. One canalso easily verify that Le Pr = Sc.
We shall recall our result on Reynolds-Colburn analogy between heat and momentumtransfer processes. In the above discussion, we established an analogy between mass andheat transfer. Thus, one can make use skin friction coefficient to deduce solutions to con-vective mass transfer problems. This is illustrated below. From Eq. 3.15 in convective heattransfer part, the dimensionless local heat flux is
Nu =qsx
(Ts − T∞)κf
= 0.332Pr1/3Re1/2x , (4.4)
for large Pr. The mass transfer equivalent of this expression is
Gdifsx
(ρs − ρ∞)D= 0.332Sc1/3Re1/2
x , (4.5)
4.1. FORCED CONVECTIVE MASS TRANSFER 27
by replacing ν/α by ν/D. The left hand side of Eq. (4.5) is called Sherwood number whichis nothing but dimensionless convective mass transfer coefficient:
Sh ≡ Gdifsx
(ρs − ρ∞)D =hmx
D .
This number plays a role in mass transfer cases which is equivalent to the role played byNu in heat transfer problems.
If one uses the Reynold-Colburn analogy then
StmSc2/3 =Cf
2, (4.6)
where Stm is local mass transfer Stanton number which is
Stm =Sh
Sc Re=
hm
U∞.
Equation (4.6) is powerful since the mass transfer rate can be obtained from the informationon the skin friction coefficient. Also, this expression can be used for laminar and turbulentflows as we noted in convective heat transfer discussion. One can also define an averageSherwood number like average Nusselt number. Even for other external flows, if we knowthe correlation for Nusselt number then a correlation for Sherwood number can be obtainedby simply replacing Nu by Sh and Pr by Sc.
4.1.2 Internal FlowIn the case of duct flow, the mass transfer coefficient, hm, is based on the difference betweenspecies concentration at the wall and the bulk mean concentration. Thus, the rate of massflux is
Gdif,i = hm(ρi,s − ρi,b), (in kg/m2 − s) (4.7)
where the bulk mean mass concentration, ρi,b, is defined in a way which is similar to thedefinition of bulk mean temperature. Thus,
UAcρi,b =
∫ρiui dAc ⇒ ρi,b =
1
UAc
∫ρiui dAc, (4.8)
where U is the bulk mean velocity (also known as the mixture average velocity) obtainedfrom the mass flow rate m = ρUAc and Ac is the cross sectional area for the flow.
If one replaces Nu by Sh because of the heat and mass transfer analogy discussedabove then for a fully developed laminar flow1
Sh =hmdh
D = constant, (4.9)
where dh is the hydraulic diameter of the tube. The values of the above constant for variouscases are given in Table 3.4 of the convective heat transfer lecture notes. When the internalflow is turbulent then the heat transfer correlation
Nu = 0.023Re4/5dh
Pr1/3
1one can obtain Eq. (4.9) via a formal approach by solving the species balance equation as we did withtemperature equation in heat transfer analysis.
28 CHAPTER 4. CONVECTIVE MASS TRANSFER
can be simply modified as
Sh = 0.023Re4/5dh
Sc1/3 (Sc ≥ 0.5, 2× 10−4 ≤ Redh≤ 106) (4.10)
for mass transfer coefficient calculation.
4.2 Natural Convective Mass TransferThe density of binary mixture depends on temperature, pressure and mixture composition.Thus, the mixture density can be written as
ρ = ρ(T, P, ρi)
⇒ ρ = ρ∞ +
(∂ρ
∂ρi
)
T,p
(ρi − ρi,∞) + · · · (4.11)
using Taylors series when the temperature and pressure is kept constant. By defining βc asthe composition expansion coefficient, one may write
ρ = ρ∞[1− βc(ρi − ρi,∞)], (4.12)
after neglecting higher order terms. We can follow the principles used in the natural convec-tion heat transfer analysis to show that the momentum and species i conservation equationsbecome
u∂v
∂x+ v
∂v
∂y= ν
∂2v
∂y2+ βcg(ρi − ρi,∞), (4.13)
and u∂ρi
∂x+ v
∂ρi
∂y= D∂2ρi
∂y2. (4.14)
An analogy to the natural convection will result from these equations if the following sub-stitution is effected:
ρi → T ; D → α.
This simply means that the results of natural convection heat transfer can be applied to nat-ural convection mass transfer with the Rayleigh number Ray replaced by the mass transferRayleigh number, Ram,y, ie.,
Ray =gβ(Ts − T∞)y3
να→ Ram,y =
gβc(ρi,s − ρi,∞)y3
νD .
4.3 With Chemical ReactionsThe occurrence of convective mass transfer with homogeneous chemical reactions is com-mon in engineering applications involving combustion. Invariably this kind of problemsinvolve simultaneous heat and mass transfer. For example, if we consider the evapora-tion of a fuel droplet in a hot oxidising environment, the fuel vapor coming from the droplet
4.3. WITH CHEMICAL REACTIONS 29
will burn involving homogeneous chemical reactions forming a flame front in the neigh-borhood of the droplet. The radiation from the flame front will heat up the droplet leadingto increased evaporation rate which results in the increased reaction rate. In this relativelysimple situation one can see the interplay between the heat and mass transfer process. Theprinciples involved in the analysis of this kind of problems is best explained via consideringan application. The application we consider here is the catalytic convertor.
4.3.1 Catalytic ConverterA catalytic convertor is a chemical reactor which converts the toxic gases emitted by in-ternal combustion gasoline engines to benign gases. The polluting gases coming out theengine include predominantly carbon monoxide, nitric oxide and unburnt hydrocarbons.The unburnt hydrocarbons are partially oxidised hydrocarbons and some of which are car-cinogens. The complete combustion of any hydrocarbon will result in only carbon dioxideand water. The above three pollutants are notorious among many to cause irreversible dam-age to our environment. For example, nitric oxide can react with ozone in the presence ofsunlight and produce nitrates which dissolves in water forming nitric acid. The catalyticconvertor oxidises carbon monoxide and unburnt hydrocarbons to cardon dioxide and waterwhile it reduces nitric oxide to nitrogen and oxygen. These chemical reactions are usuallyoccur at high temperatures (> 1000 K). But the presence of catalysts which are usually no-ble metals allows these reactions at a reduced temperature (400 to 500 K). These reactionsare heterogenous reactions, occur on the surface of the catalyst and their rates are stronglytemperature dependent.
Construction wise, the reactors are simple involving a large number of channels ina honeycomb structure as shown in Fig. 4.2. The channels are usually of square crosssection. A typical catalyst used in an automobile will have about 2000 to 2500 channelsthrough which the exhaust flows at a total flow rate of 0.1 kg/s. The channels are typicallyof 1 mm × 1 mm in size. A good estimate of the Reynolds number for the flow througha single channel is about 1000 and thus the flow is laminar. Here our interest is to set outthe principles involved in the analysis of heat and mass transfer in the single channel. It isstraight forward to extend this analysis to model multi-channel honeycomb reactor.
A cut section view of the catalyst is shown in the second part of Fig. 4.2 which alsoshows the typical length scales of various components involved. The third picture showsthe scanning electron microscopic image of two adjacent channels. The white border is theporous wash coat containing the noble metal catalysts and the center dark core is the flowpassage. The various processes happening inside this channel can be depicted as in the lastpart of Fig. 4.2, titled ”model for analysis”. This figure also shows a control volume whichwill be used to setup the balance equations for our analysis. Because of the laminar flowinside the channel, there will be a boundary layer through which heat and mass exchangeoccur between the bulk flow and the chemical process happening inside the porous washcoat. If we take the flow to be fully developed, the balance occurs between pressure andviscous forces as we have seen earlier. This balance is simply
−Ac dp = τs P dx ⇒ dp
dx= −τs(P/Ac), (4.15)
where Ac is the cross section area of the channel, p is the pressure, P is the wetted perime-ter. From this balance equation and recalling that τs is related to the skin friction coefficient,
30 CHAPTER 4. CONVECTIVE MASS TRANSFER
Model for analysis SEM - image
CV
Model for analysis SEM - imageModel for analysisModel for analysis SEM - image
CV
Figure 4.2: Modelling of a single channel catalytic reactor. The arrows are pointed in thedirection of simplification for improved understanding.
Cf = C/Re, it is easy to show
dp
dx=−CµmRT
8p
(P 2
A3c
), (4.16)
with µ as the dynamic viscosity of the gases flowing through the channel at a flow rate ofm and R is the characteristic gas constant.
If we consider the energy conservation inside the control volume then
ρUAccpT − ρUAccp(T + dT ) + dqsPdx = 0
⇒ ρUcpdT
dx+ h(T − Ts)
(P
Ac
)= 0. (4.17)
where h is the convective heat transfer coefficient. From the analysis of heat transfer inthe internal laminar flows, we know that h is constant (see Table 3.4 in the convective heattransfer notes) and thus, the above equation can be integrated appropriately to obtain thevariation of bulk fluid temperature T , only if Ts is known. But the surface temperature,Ts,of the channel is influenced by the heat released by the catalytic reaction happening insidethe porous wash coat and the radiation exchange between the gases and the walls. If oneconsiders the energy balance in another control volume inside the porous part then
κwδwd2Ts
dx2− qrad − h(Ts − Tb) + Qsη = 0. (4.18)
The wall thickness is of the channel is δw and the thermal conductivity of the porous wallmaterial is κw. The amount of heat released by the heterogeneous reactions per unit areaof the catalyst surface is Qs (w/m2 of catalyst) and η is the ratio of catalytic surface areato geometric surface area per unit volume of the reactor wall. Thus, to obtain the bulk
4.3. WITH CHEMICAL REACTIONS 31
temperature variation along the length of the channel one needs to solve both Eqs. (4.17)and (4.18) together.
By performing the balance of species i inside the control volume shown in Fig. 4.2, onecan write
−UAc dρi + PGdif,idx = 0 ⇒ Udρi
dx+ hm,i(ρi − ρi,s)
(P
Ac
)= 0, (4.19)
where ρi is the bulk mass concentration of species i. hm is the convective mass transfercoefficient which can be obtained from the Sherwood number which is constant for a fullydeveloped laminar internal flow as we noted in subsection 4.1.2. Thus, the above speciesbalance equation can be simply integrated to yield the variation of species i if the concen-tration at the wall, ρi,s, is known. If the operation of the catalyst is diffusion limited, ie thetemperature of the catalyst is high, then ρi,s = 0. This leads to a simple logarithmic vari-ation of ρi along the length of the channel. If the operation of the catalyst is not diffusionlimited then the concentration at the surface is related to the reaction rate as we noted insubsection 3.4.2 via the balance between the mass diffusion rate and the reaction rate. Thisbalance is
hm,i(ρi − ρi,s) = ηΩi. (4.20)
The reaction rate is typically given by
Ω∗i =
−k′1ci,sc
O2,s
G. kmol/m2 − s
The rate constant k′1 depends on the temperature via Arrhenius expression. To account forreaction inhibition due to surface adsorption of chemical species, G is introduced whichalso depends on temperature and species concentration on the surface.
Reference
1. Holman, J. P. Heat Transfer, McGraw Hill. 9th edition, 2002, ISBN 0-07-240655-0.
2. Incropera, F. and Dewitt, D. P. Fundamentals of Heat and Mass Transfer, John Wiley& Sons Inc., 5th edition, 2002, ISBN 0-471-38650-2.
3. Kays, W. M., Crawford, M. and Weigand, B. Convective Heat and Mass Transfer,McGraw-Hill, 4th edition, 2005, ISBN 007-123829-8.
4. Bird, R. B., Stewart, W. E. and Lightfoot, E. N. Transport Phenomena, John Wiley& Sons, New York, 1960.
5. Taylor, R. and Krishna, R. Multicomponent Mass Transfer, John Wiley & Sons, NewYork, 1993.
6. Hayes, R. E. and Kolaczkowski, S. T. Introduction to catalytic combustion, Gordonand Breach Science Publishers, 1997.
