chap1 4

Prévia do material em texto

1-23 
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 A
dy
dr
 + yA(NA,r + NB,r) 
 
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 A
dy
dr
 
 
The system is not at steady state, the molar flux is not independent of r since the area of mass 
transfer 4pir2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer 
rate, 4pir2NA,r, is assumed to be independent of r at any instant of time. 
 
 WA = 4pir2NA,r = − 4pir2cDAmix A
dy
dr
 = constant 
 
R
r
yA,R yA,inf
 
 
 1-24 
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 2R
dr
r
∞
∫ = − 4pi cDAmix
,
0
Ay
Ady
∞
∫ 
 
 − WA
1
Rr
∞
 = − 4pi cDAmixyA,∞ => WA = − 4picDAmixyA,∞R 
 
Since one mole of carbon will disappear for each mole of oxygen consumed at the surface 
 
 WC = − WA = 4picDAmixyA,∞R 
 
Making a carbon balance gives 
 
 
C
CM
ρ 34
3
d R
dt
pi  
 
 = 
C
CM
ρ 4πR2 dR
dt
 = − 4picDAmixyA,∞R 
 
Separating the variables and integrating from t = 0 to t gives 
 
 
0
t
dt∫ = − 
Amix ,
C
C AM cD y
ρ
∞
f
i
R
R
RdR∫ 
 
 t =
Amix ,2
C
C AM cD y
ρ
∞
(Ri2 − Rf2) 
 
The total gas concentration can be obtained from the ideal gas law 
 
 c = 
P
RT
 = 
1
(0.08206)(1145) = 0.0106 kmol/m
3
 
 
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 
 
 t =
( ) ( )2 25 5
4
(1280) 7.5 10 2.5 10
2(12)(0.0106)(1.3 10 )(0.21)
− −
−
 × − ×  
×
 = 0.92 s 
 
 
 
 1-25 
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 A
dy
dr
 + yA(NA,r + NB,r) 
 
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 A
dy
dr
 
 
The system is not at steady state, the molar flux is not independent of r since the area of mass 
transfer 4pir2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer 
rate, 4pir2NA,r, is assumed to be independent of r at any instant of time. 
 
 WA = 4pir2NA,r = − 4pir2cDAB A
dy
dr
 = constant 
 
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 2R
dr
r
∞
∫ = − 4pi cDAB
,
,
A
A R
y
Ay
dy∞∫ 
 
 − WA
1
Rr
∞
 = − 4pi cDAB(yA,∞ − yA,R) => WA = − 4picDAB(1 − yA,R)R 
 
 1-26 
The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed 
by the chemical reaction 
 
 WA = 4piR2 "Rɺ = − 4piR2k1”CO2|R = − 4piR2k1” c yA,R 
 
 − 4picDAB(1 − yA,R)R = − 4piR2k1” c yA,R 
 
 DAB(1 − yA,R) = Rk1”yA,R => yA,R = 
1 "
AB
AD
D
D Rk+
 
 
 yA,R = 
4
4 3
1.71 10
1.71 10 10 .1
−
− −
×
× + ×
 = 0.631 
 
The total gas concentration can be obtained from the ideal gas law. (Note: R = 0.08206 
m
3
⋅atm/kmol⋅K) 
 
 c = 
P
RT
 = 
1
(0.08206)(1450) = 0.008405 kmol/m
3
 
 
The steady-state O2 molar consumption rate is 
 
 WA = − 4picDAB(1 − yA,R)R = − 4pi(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
Treated
waste water
biofilm
Well-mixed contactor
biofilm
z=0
C (z)A
CA0
CA0
Inert
solid
surface
 
 
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 
 1-27 
 
 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 
 
 WA = S·NA,z = S·
0
A
AB
z
dcD
dz
=
 
− 
 
 
 
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 A
dy
dz
 + yA(NA,z + NB,z) 
 
Since the biofilm is stagnant (or nondiffusing),NB,z = 0. Solving for NA,z give 
 
 NA,z(1 − yA) = − cDAB Adydz 
 
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 A
dy
dz
 = − DAB A
dc
dz
 
 
z
Solid surface
Biofilm
NA,z
 
 
Making a mole balance around the control volume S·∆z gives 
 1-28 
 
 S·NA,z|z − S·NA,z|z+∆z + S·∆z·rA = 0 
 
Dividing the equation by S·∆z and letting ∆z → 0 yields 
 
 
,A zdN
dz
= rA = − k1cA (E-1) 
 
Substituting NA,z = − DAB A
dc
dz
into equation (E-1) we obtain 
 
 DAB
2
2
Ad c
dz
 = k1cA => 
2
2
Ad c
dz
 = 
1
AB
k
D
cA (E-2) 
 
The solution to the homogeneous equation (E-2) has two forms 
 
 1) cA = C1exp 1
AB
k
z
D
 
− 
 
 + C2exp 1
AB
k
z
D
 
 
 
 
 
 2) cA = B1sinh 1
AB
k
z
D
 
 
 
 + B2cosh 1
AB
k
z
D
 
 
 
 
 
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). 
 
 cA = B1sinh 1
AB
k
z
D
 
 
 
 + B2cosh 1
AB
k
z
D
 
 
 
 (E-3) 
 
At z = 0, cA = cAs = cA0 = B2 
 
At z = δ, Adc
dz
 = 0 = B1 1
AB
k
D
cosh 1
AB
k
D
δ  
 
 + B2 1
AB
k
D
sinh 1
AB
k
D
δ  
 
 
 
Therefore 
 
 B1 = − B2
1
1
sinh
cosh
AB
AB
k
D
k
D
δ
δ
 
 
 
 
 
 
 = − cA0
1
1
sinh
cosh
AB
AB
k
D
k
D
δ
δ
 
 
 
 
 
 
 
 
Equation (E-3) becomes 
 1-29 
 
 cA = − cA0
1
1
sinh
cosh
AB
AB
k
D
k
D
δ
δ
 
 
 
 
 
 
 sinh 1
AB
k
z
D
 
 
 
 + cA0cosh 1
AB
k
z
D
 
 
 
 
 
 cA = cA0
1 1 1 1
1
cosh cosh sinh sinh
cosh
AB AB AB AB
AB
k k k k
z z
D D D D
k
D
δ δ
δ
       
−       
       
 
 
 
 
 
Using the identity cosh(A – B) = cosh(A)cosh(B) – sinh(A)sinh(B) we have 
 
 cA = cA0
1
1
cosh ( )
cosh
AB
AB
k
z
D
k
D
δ
δ
 
− 
 
 
 
 
 
 
 
0
A
z
dc
dz
=
 = – cA0
1 1
1
0
sinh ( )
cosh
AB AB
AB z
k k
z
D D
k
D
δ
δ
=
 
− 
 
 
 
 
 = – cA0
1 1tanh
AB AB
k k
D D
δ  
 
 
 
The molar flux of phenol at the biofilm surface is given by 
 
 NA,z = 
0
A
AB
z
dcD
dz
=
− = 
0A ABc D
δ δ 
1 1tanh
AB AB
k k
D D
δ  
 
 
 
 
The dimensionless parameter δ 1
AB
k
D
represents the ratio of reaction rate to diffusion rate. 
For this problem we have 
 
 δ
1
AB
k
D
 = 0.002 m 2
10
10.019
s
m2 10
s
−×
 = 19.49 
 
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 
 1-30 
 
 NA,z = 
10(0.02)(2 10 )
0.002
−× (19.49) tanh(19.49) = 3.9×10-8 mol/(m2·s) 
 
The required surface area of the biofilm is finally 
 
 S = 
,
A
A z
W
N
 = 
3
8
8.0 10
(3.9 10 )(3600)
−
−
×
×
 = 57.0 m2 
 
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 r+dr
 
Figure E-1 Illustration of a spherical shell 4pir2dr 
 
The one-dimensional molar flux of A is given by the equation 
 
 
"
AN = − DA dr
dCA
 (E-1) 
 
Applying a mole balance on the spherical shell shown in Figure E-1 yields for steady state 
 
 4pir2
r
AN
"
 − 4pir2
drrA
N
+
"
 + RA4pir2dr = 0 
 
 1-31 
Dividing the equation by the control volume (4pir2dr) and taking the limit as dr → 0, we 
obtain 
 
 − 2
1
r dr
d (r2 "AN ) + RA = 0 (E-2) 
 
For a first order reaction, RA = − k1CA and substituting the molar flux from equation (E-1) 
into the above equation, we have 
 
 − 2
1
r dr
d






−
dr
dC
r AAD2 − k1CA = 0 
 
 DA 2
1
r dr
d






dr
dC
r A
2
 − 1kCA = 0 (E-3) 
 
In this equation, DA and k1 are constants independent of r. We want to transform this 
equation into the form 
 
 2
2
dr
yd
 − α2y = 0 (E-4) 
 
Let α2
 
= 
1
A
k
D
, we can transform equation (4.6-3) into the form of equation (E-4) by the 
following algebraic manipulations 
 
r
1
dr
d






dr
dC
r A
2
 − α2
 
rCA = 0 ⇒ 
r
1






+ 2
2
22
dr
Cd
r
dr
dC
r AA
 − α2
 
rCA = 0 
 
 2
dr
dCA
 + 2
2
dr
Cd
r A − α2
 
rCA = 0 
 
Since 
dr
d





 )( ArCdr
d
 = 
dr
d






+
dr
dC
rC AA = dr
dCA
 + 
dr
dCA
 + 2
2
dr
Cd
r A , the above equation 
becomes 
 
 
dr
d





 )( ArCdr
d
− α2
 
rCA = 0 
 
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
A
k
D
 
 1-32 
 
The two constants of integration B1 and B2 can be obtained from the boundary conditions 
 
 At r = 0, CA = finite or dr
dCA
 = 0 
 
 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 = 0
sinh( )
ARC
Rα
 
 
Therefore the concentration profile for species A within the organism is 
 
 CA = CA0
r
R
)sinh(
)sinh(
R
r
α
α
 (E-5) 
 
At the center of the organism, the concentration is given by CA(r = 0) = CA0 )sinh( R
R
α
α
 
 
(b) Obtain an expression for the rate of oxygen consumption within the organism. 
 
 Rate of oxygen consumption within the organism. = 4piR2(−DA A
r R
dC
dr
=
) 
 
The oxygen concentration within the organism is given by equation (E-5) 
 
 
 CA = CA0
r
R
)sinh(
)sinh(
R
r
α
α
 (E-5) 
 
 
AdC
dr
 = 
0
sinh( )
AC R
Rα 


 +− )cosh()sinh(12 rrrr α
α
α 
 
 
A
r R
dC
dr
=
= 
0
sinh( )
AC R
Rα 



− )sinh(1)cosh( 2 RRRR αα
αA
r R
dC
dr
=
= 
0AC
R
[ ])1)coth()( −RR αα 
 
 1-33 
Let φ = αR = 
1/ 22
1
A
k R
D
 
 
 
= Thiele modulus for a first order reaction. Ignoring the minus sign, 
the rate of oxygen consumption within the organism is then 
 
 Rate of oxygen consumption = 4pi R2DA 0A
C
R
 (φ cothφ - 1) 
 
 Rate of oxygen consumption = 4pi 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 )sinh( R
R
α
α
 
 
 α
 
= 
1/ 2
1
A
k
D
 
 
 
= 
1/ 2
8
20
10−
 
 
 
 = 4.4721×104 m 
 
 αR = 
1/ 22
1
A
k R
D
 
 
 
= 
( ) 1/ 224
8
20 10
10
−
−
 ×
 
 
 
 = 4.4721 
 
 CA(r = 0) = CA0 )sinh( R
R
α
α
 = 5×10-5 4.4721
sinh(4.4721) = 5.11×10
-6
 kmol/m3 
 
Rate of oxygen consumption = 4pi RDACA0 (φ cothφ - 1) 
 
 
Rate = 4pi(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; 
 1-34 
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
R
)sinh(
)sinh(
R
r
α
α
 (1.4-1) 
 
In this equation CR is the concentration of species A at the surface of the catalyst particle and 
α is defined by the expression α2
 
= 
A
k
D
, where k is the first order rate constant and DA is the 
diffusivity of A in the particle. At the center of the bead, the concentration is given by 
 
 CA(r = 0) = CR )sinh( R
R
α
α
 
(1.4-2)