Chapter 1
1.4 Diffusion with Chemical Reaction
Example 1.4-1 -----------------------------------------------------------------------------A fluidized coal reactor operates at 1145 K and 1 atm. The process will be limited by the
diffusion of oxygen countercurrent to the carbon dioxide, CO2, formed at the particle surface.
Assume that the coal is pure solid carbon with a density of 1280 kg/m3 and that the particle is
spherical with an initial diameter of 1.5×10-4 m. Air (21% O2 and 79% N2) exists several
diameters away from the sphere. The diffusivity of oxygen in the gas mixture at 1145 K is
1.3×10-4 m2/s. If a quasi-steady state process is assumed, calculate the time necessary to
reduce the diameter of the carbon particle to 5.0×10-5 m.
(Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001,
pg. 496.)
Solution ---------------------------------------------------------------------------------------------The reaction at the carbon surface is
C(s) + O2(g) → CO2(g)
We have diffusion of oxygen (A) toward the surface and diffusion of carbon dioxide (B)
away from the surface. The molar flux of oxygen is given by
NA,r = − cDAmix
dy A
+ yA(NA,r + NB,r)
dr
In this equation, r is the radial distance from the center of the carbon particle. Since NA,r = −
NB,r, we have
NA,r = − cDAmix
dy A
dr
The system is not at steady state, the molar flux is not independent of r since the area of mass
transfer 4πr2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer
rate, 4πr2NA,r, is assumed to be independent of r at any instant of time.
WA = 4πr2NA,r = − 4πr2cDAmix
dy A
= constant
dr
R
r
yA,R
yA,inf
1-23
At the surface of the coal particle, the reaction rate is much faster than the diffusion rate to
the surface so that the oxygen concentration can be considered to be zero: yA,R = 0.
Separating the variables and integrating gives
WA ∫
∞
R
y A ,∞
dr
=
−
4π
cD
Amix
∫0 dy A
r2
∞
− WA
1
= − 4π cDAmixyA,∞ => WA = − 4πcDAmixyA,∞R
rR
Since one mole of carbon will disappear for each mole of oxygen consumed at the surface
WC = − WA = 4πcDAmixyA,∞R
Making a carbon balance gives
ρC d  4
ρC
dR
3
4πR2
= − 4πcDAmixyA,∞R
 πR  =
dt  3
MC
dt

MC
Separating the variables and integrating from t = 0 to t gives
t
∫ dt
0
t=
=−
ρC
M C cDAmix y A,∞
ρC
2 M C cDAmix y A,∞
∫
Rf
Ri
RdR
(Ri2 − Rf2)
The total gas concentration can be obtained from the ideal gas law
c=
P
1
=
= 0.0106 kmol/m3
RT
(0.08206)(1145)
Note: R = 0.08206 m3⋅atm/kmol⋅oK
The time necessary to reduce the diameter of the carbon particle from 1.5×10-4 m to 5.0×10-5
m is then
2
2
(1280)  ( 7.5 × 10−5 ) − ( 2.5 × 10−5 ) 

 = 0.92 s
t=
−4
2(12)(0.0106)(1.3 × 10 )(0.21)
1-24
Example 1.4-2 -----------------------------------------------------------------------------Pulverized coal pellets, which may be approximated as carbon spheres of radius R = 1 mm,
are burned in a pure oxygen atmosphere at 1450 K and 1 atm. Oxygen is transferred to the
particle surface by diffusion, where it is consumed in the reaction C(s) + O2(g) → CO2(g).
The reaction rate is first order and of the form Rɺ " = − k1”CO2|R where k1” = 0.1 m/s. This is
the reaction rate per unit surface area of the carbon pellets. Neglecting change in R,
determine the steady-state O2 molar consumption rate in kmol/s. At 1450 K, the binary
diffusion coefficient for O2 and CO2 is 1.71×10-4 m2/s.
(Ref. Fundamentals of Heat Transfer by Incropera and DeWitt.)
Solution ---------------------------------------------------------------------------------------------We have diffusion of oxygen (A) toward the surface and diffusion of carbon dioxide (B)
away from the surface. The molar flux of oxygen is given by
NA,r = − cDAB
dy A
+ yA(NA,r + NB,r)
dr
In this equation, r is the radial distance from the center of the carbon particle. Since NA,r = −
NB,r, we have
NA,r = − cDAB
dy A
dr
The system is not at steady state, the molar flux is not independent of r since the area of mass
transfer 4πr2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer
rate, 4πr2NA,r, is assumed to be independent of r at any instant of time.
WA = 4πr2NA,r = − 4πr2cDAB
dy A
= constant
dr
R
r
yA,R
yA,inf
The oxygen concentration at the surface of the coal particle, yA,R, will be determined from the
reaction at the surface. The mole fraction of oxygen at a location far from the pellet is 1.
Separating the variables and integrating gives
WA ∫
∞
R
y A ,∞
dr
= − 4π cDAB ∫ dy A
2
y A, R
r
∞
− WA
1
= − 4π cDAB(yA,∞ − yA,R) => WA = − 4πcDAB(1 − yA,R)R
rR
1-25
The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed
by the chemical reaction
WA = 4πR2 Rɺ " = − 4πR2k1”CO2|R = − 4πR2k1” c yA,R
− 4πcDAB(1 − yA,R)R = − 4πR2k1” c yA,R
DAB(1 − yA,R) = Rk1”yA,R => yA,R =
yA,R =
D AB
D AD + Rk1 "
1.71 × 10−4
= 0.631
1.71 × 10 −4 + 10 −3 × .1
The total gas concentration can be obtained from the ideal gas law. (Note: R = 0.08206
m3⋅atm/kmol⋅K)
c=
P
1
=
= 0.008405 kmol/m3
RT
(0.08206)(1450)
The steady-state O2 molar consumption rate is
WA = − 4πcDAB(1 − yA,R)R = − 4π(0.008405)( 1.71×10-4)(1 − 0.631)(10-3)
WA = − 6.66×10-9 kmol/s
Example 1.4-3 -----------------------------------------------------------------------------A biofilm consists of living cells immobilized in a gelatinous matrix. A toxic organic solute
(species A) diffuses into the biofilm and is degraded to harmless products by the cells within
the biofilm. We want to treat 0.1 m3 per hour of wastewater containing 0.1 mole/m3 of the
toxic substance phenol using a system consisting of biofilms on rotating disk as shown
below.
Waste water
feed stream
biofilm
CA0
CA(z)
Inert
solid
surface
CA0
Well-mixed contactor
biofilm
Treated
z=0
waste water
Determine the required surface area of the biofilm with 2 mm thickness to reduce the phenol
concentration in the outlet stream to 0.02 mole/m3. The rate of disappearance of phenol
(species A) within the biofilm is described by the following equation
1-26
rA = − k1cA where k1 = 0.019 s-1
The diffusivity of phenol in the biofilm at the process temperature of 25oC is 2.0×10-10 m2/s.
Phenol is equally soluble in both water and the biofilm.
(Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001,
pg. 496.)
Solution ---------------------------------------------------------------------------------------------The rate of phenol processed by the biofilms, WA, is determined from the material balance on
the process unit
WA = 0.1 m3/h(0.1 − 0.02) mol/m3 = 8.0×10-3 mol/h
WA is then equal to the rate of phenol diffused into the biofilms and can be calculated from

dc
WA = S·NA,z = S·  − DAB A
dz



z =0 
In this equation, S is the required surface area of the biofilm and NA,z is the molar flux of
phenol at the surface of the biofilm. The molar flux of A (phenol) is given by
NA,z = − cDAB
dy A
+ yA(NA,z + NB,z)
dz
Since the biofilm is stagnant (or nondiffusing), NB,z = 0. Solving for NA,z give
NA,z(1 − yA) = − cDAB
dy A
dz
The mole fraction of phenol in the biofilm, yA, is much less than one so that c can be
considered to be constant. Therefore
NA,z = − cDAB
dy A
dc
= − DAB A
dz
dz
Biofilm
Solid surface
NA,z
z
Making a mole balance around the control volume S·∆z gives
1-27
S·NA,z|z − S·NA,z|z+∆z + S·∆z·rA = 0
Dividing the equation by S·∆z and letting ∆z → 0 yields
dN A, z
= rA = − k1cA
dz
Substituting NA,z = − DAB
DAB
(E-1)
dc A
into equation (E-1) we obtain
dz
d 2cA
d 2cA
k
=
k
c
=>
= 1 cA
1
A
2
2
dz
dz
DAB
(E-2)
The solution to the homogeneous equation (E-2) has two forms

 k1 
k1 
z  + C2exp 
z
1) cA = C1exp  −
 DAB 
 DAB 
 k1 
 k1 
z  + B2cosh 
z
2) cA = B1sinh 
D
D
AB 
AB 


The first exponential form (1) is more convenient if the domain of z is infinite: 0 ≤ z ≤ ∞
while the second form using hyperbolic functions (2) is more convenient if the domain of z is
finite: 0 ≤ z ≤ δ. The constants of integration C1, C2, B1, and B2 are to be determined from
the two boundary conditions. We use the hyperbolic functions as the solution to Eq. (E-2).
 k1 
 k1 
z  + B2cosh 
z
cA = B1sinh 
 DAB 
 DAB 
At z = 0, cA = cAs = cA0 = B2
At z = δ,
 k1 
 k1 
k1
k1
dc A
δ  + B2
δ
= 0 = B1
cosh 
sinh 
DAB
D
D
D
dz
AB
AB
AB




Therefore
 k1 
 k1 
sinh 
δ
sinh 
δ
D
D
AB
AB

 =−c


B1 = − B2
A0
 k1 
 k1 
cosh 
δ
cosh 
δ
 DAB 
 DAB 
Equation (E-3) becomes
1-28
(E-3)
 k1 
sinh 
δ
DAB 
 k1 
 k1 

z  + cA0cosh 
z
cA = − cA0
sinh 
DAB 
DAB 
 k1 


cosh 
δ
D
AB


 k1 
 k1 
 k1 
 k1 
cosh 
δ  cosh 
z  − sinh 
δ  sinh 
z
DAB 
DAB 
DAB 
DAB 




cA = cA0
 k1 
δ
cosh 
 DAB 
Using the identity cosh(A – B) = cosh(A)cosh(B) – sinh(A)sinh(B) we have
 k1

cosh 
(δ − z ) 
 DAB

cA = cA0
 k1 
δ
cosh 
 DAB 
dcA
dz
 k1

k1
sinh 
(δ − z ) 
D AB
 D AB

 k1 
δ
cosh 
D
AB


= – cA0
z =0
= – cA0
 k1 
k1
tanh 
δ
DAB
 DAB 
z =0
The molar flux of phenol at the biofilm surface is given by
NA,z = − DAB
dcA
dz
The dimensionless parameter δ
=
z =0
c A0 D AB
δ
δ
 k1 
k1
tanh 
δ
DAB
D
AB


k1
represents the ratio of reaction rate to diffusion rate.
DAB
For this problem we have
δ
k1
DAB
1
s
= 0.002 m
2 = 19.49
−10 m
2 × 10
s
0.019
This value indicates that the rate of reaction is very rapid relative to the rate of diffusion. The
flux of phenol into the biofilm is then
1-29
(0.02)(2 × 10 −10 )
NA,z =
(19.49) tanh(19.49) = 3.9×10-8 mol/(m2·s)
0.002
The required surface area of the biofilm is finally
S=
WA
8.0 × 10−3
=
= 57.0 m2
−8
N A, z
(3.9 × 10 )(3600)
Example 1.4-4. ---------------------------------------------------------------------------------Consider a spherical organism of radius R within which respiration occurs at a uniform
volumetric rate of rA = − k1CA. That is, oxygen (species A) consumption is governed by a
first-order, homogeneous chemical reaction.
(a) If a molar concentration of CA(R) = CA,0 is maintained at the surface of the organism,
obtain an expression for the radial distribution of oxygen, CA(r), within the organism.
(b) Obtain an expression for the rate of oxygen consumption within the organism.
(c) Consider an organism of radius R = 0.10 mm and a diffusion coefficient for oxygen
transfer of DAB = 10-8 m2/s. If CA,0 = 5×10-5 kmol/m3 and k1 = 20 s-1, what is the molar
concentration of O2 at the center of the organism? What is the rate of oxygen
consumption by the organism?
Solution -----------------------------------------------------------------------------------------(a) If a molar concentration of CA(R) = CA,0 is maintained at the surface of the organism,
obtain an expression for the radial distribution of oxygen, CA(r), within the organism.
r r+dr
R
Figure E-1 Illustration of a spherical shell 4πr2dr
The one-dimensional molar flux of A is given by the equation
N A" = − DA dC A
dr
(E-1)
Applying a mole balance on the spherical shell shown in Figure E-1 yields for steady state
4πr2 N A" − 4πr2 N A"
r
r + dr
+ RA4πr2dr = 0
1-30
Dividing the equation by the control volume (4πr2dr) and taking the limit as dr → 0, we
obtain
−
1 d 2 "
(r N A ) + RA = 0
r 2 dr
(E-2)
For a first order reaction, RA = − k1CA and substituting the molar flux from equation (E-1)
into the above equation, we have
−
1 d  2 dC A 
 − r DA
 − k1CA = 0
r 2 dr 
dr 
DA
1 d  2 dC A 
r
 − 1kCA = 0
dr 
r 2 dr 
(E-3)
In this equation, DA and k1 are constants independent of r. We want to transform this
equation into the form
d2y
− α2y = 0
2
dr
(E-4)
k1
, we can transform equation (4.6-3) into the form of equation (E-4) by the
DA
following algebraic manipulations
Let α2 =
1 d  2 dC A 
2
r
 − α rCA = 0 ⇒
r dr 
dr 
2
 dC A
d 2C A 
 2r
 − α2 rCA = 0
+ r2
2 
dr
dr


dC A
d 2C A
+ r
− α2 rCA = 0
2
dr
dr
d d
d

 ( rC A )  =
dr  dr
dr

becomes
Since
1
r
dC A 
dC A
dC A
d 2C A

=
+
+
r
, the above equation
C
+
r
 A

dr 
dr
dr
dr 2

d d

2
 ( rC A )  − α rCA = 0
dr  dr

Let y = rCA, the equation has the same form as equation (E-4) with the solution
y = B1sinh(αr) + B2cosh(αr)
or
rCA = B1sinh(αr) + B2cosh(αr), where α2 =
1-31
k1
DA
The two constants of integration B1 and B2 can be obtained from the boundary conditions
At r = 0, CA = finite or
dC A
=0
dr
At r = R, CA = CA0 (a known value)
Applying the boundary at r = 0 yields
0 = B2
Applying the boundary at r = R yields
RCR = B1sinh(αR) ⇒ B1 =
RC A0
sinh(α R )
Therefore the concentration profile for species A within the organism is
CA = CA0
R sinh(αr )
r sinh(αR )
(E-5)
At the center of the organism, the concentration is given by CA(r = 0) = CA0
αR
sinh(αR )
(b) Obtain an expression for the rate of oxygen consumption within the organism.
Rate of oxygen consumption within the organism. = 4πR2(−DA
dC A
dr
)
r=R
The oxygen concentration within the organism is given by equation (E-5)
CA = CA0
R sinh(αr )
r sinh(αR )
dC A
C A0 R
=
dr
sinh(α R )
dC A
dr
dC A
dr
α
 1

− r 2 sinh(αr ) + r cosh(αr )
1
α

 R cosh(αR ) − R 2 sinh(αR )
=
C A0 R
sinh(α R )
=
C A0
[(αR) coth(αR) − 1)]
R
r=R
r=R
(E-5)
1-32
1/ 2
 k R2 
Let φ = αR =  1  = Thiele modulus for a first order reaction. Ignoring the minus sign,
 DA 
the rate of oxygen consumption within the organism is then
Rate of oxygen consumption = 4π R2DA
C A0
(φ cothφ - 1)
R
Rate of oxygen consumption = 4π RDACA0 (φ cothφ - 1)
(c) Consider an organism of radius R = 0.10 mm and a diffusion coefficient for oxygen
transfer of DAB = 10-8 m2/s. If CA,0 = 5×10-5 kmol/m3 and k1 = 20 s-1, what is the molar
concentration of O2 at the center of the organism? What is the rate of oxygen consumption by
the organism?
At the center of the organism, the concentration is given by CA(r = 0) = CA0
1/ 2
 k 
α=  1 
 DA 
1/ 2
 20 
=  −8 
 10 
1/ 2
k R 
2
αR
sinh(αR )
αR =  1 
 DA 
= 4.4721×104 m
 20 × (10−4 )2 

=
10−8




CA(r = 0) = CA0
1/ 2
= 4.4721
αR
4.4721
= 5×10-5
= 5.11×10-6 kmol/m3
sinh(αR )
sinh(4.4721)
Rate of oxygen consumption = 4π RDACA0 (φ cothφ - 1)
Rate = 4π(10-4)(10-8)( 5×10-5) [4.4721 coth(4.4721) - 1] = 2.18×10-15 kmol/s
The following Matlab program plots the concentration of oxygen within the organism as a
function of position.
% Example 1.4-4
%
alfa=4.4721e4; % m
R=1e-4; % m
alfaR=4.4721;
CA0=5e-5; % kmol/m3
roR=(1:50)/50;
r=R*roR;
1-33
CA=CA0*sinh(alfa*r)./(roR*sinh(alfaR));
plot(roR,CA)
grid on
xlabel('r/R');ylabel('C_A(kmol/m^3)')
Figure E1.4-4 Oxygen concentration profile in a spherical organism.
We now consider the diffusion of species A into a spherical catalyst particle where
homogeneous first order chemical reaction occurs. The concentration profile for species A
within the spherical catalyst particle is then
CA = CR
R sinh(αr )
r sinh(αR )
(1.4-1)
In this equation CR is the concentration of species A at the surface of the catalyst particle and
k
α is defined by the expression α2 =
, where k is the first order rate constant and DA is the
DA
diffusivity of A in the particle. At the center of the bead, the concentration is given by
CA(r = 0) = CR
αR
sinh(αR )
(1.4-2)
1-34