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 = − cDAmixAdy

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 = − cDAmixAdy

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 = − 4πr2cDAmixAdy

dr = constant

Rr

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

∞

∫ = − 4π cDAmix,

0

Ay

Ady∞

∫

− WA1

Rr

∞

= − 4π cDAmixyA,∞ => WA = − 4πcDAmixyA,∞R

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

C

CM

ρ 34

3

dR

dtπ

= C

CM

ρ4πR2 dR

dt = − 4πcDAmixyA,∞R

Separating the variables and integrating from t = 0 to t gives

0

tdt∫ = −

Amix ,

C

C AM cD y

ρ∞

f

i

R

RRdR∫

t =Amix ,2C

C AM cD y

ρ∞

(Ri2 − Rf

2)

The total gas concentration can be obtained from the ideal gas law

c = P

RT =

1

(0.08206)(1145) = 0.0106 kmol/m3

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 = − cDABAdy

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 = − cDABAdy

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 = − 4πr2cDABAdy

dr = constant

Rr

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

∞

∫ = − 4π cDAB,

,

A

A R

y

Aydy

∞

∫

− WA1

Rr

∞

= − 4π cDAB(yA,∞ − yA,R) => WA = − 4πcDAB(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 = 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 = 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 m3⋅atm/kmol⋅K)

c = P

RT =

1

(0.08206)(1450) = 0.008405 kmol/m3

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

Treatedwaste water

biofilm

Well-mixed contactorbiofilm

z=0

C (z)A

CA0

CA0

Inertsolidsurface

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

AAB

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 = − cDABAdy

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) = − cDABAdy

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 = − cDABAdy

dz = − DAB

Adc

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 = − DABAdc

dzinto equation (E-1) we obtain

DAB

2

2Ad c

dz = k1cA =>

2

2Ad c

dz = 1

AB

k

DcA (E-2)

The solution to the homogeneous equation (E-2) has two forms

1) cA = C1exp 1

AB

kz

D

−

+ C2exp 1

AB

kz

D

2) cA = B1sinh 1

AB

kz

D

+ B2cosh 1

AB

kz

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

kz

D

+ B2cosh 1

AB

kz

D

(E-3)

At z = 0, cA = cAs = cA0 = B2

At z = δ, Adc

dz = 0 = B1

1

AB

k

Dcosh 1

AB

k

Dδ

+ B21

AB

k

Dsinh 1

AB

k

Dδ

Therefore

B1 = − B2

1

1

sinh

cosh

AB

AB

kD

kD

δ

δ

= − cA0

1

1

sinh

cosh

AB

AB

kD

kD

δ

δ

Equation (E-3) becomes

1-29

cA = − cA0

1

1

sinh

cosh

AB

AB

kD

kD

δ

δ

sinh 1

AB

kz

D

+ cA0cosh 1

AB

kz

D

cA = cA0

1 1 1 1

1

cosh cosh sinh sinh

cosh

AB AB AB AB

AB

k k k kz z

D D D D

kD

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

kz

D

kD

δ

δ

−

0

A

z

dc

dz =

= – cA0

1 1

1

0

sinh ( )

cosh

AB AB

AB z

k kz

D D

kD

δ

δ=

−

= – cA01 1tanhAB AB

k k

D Dδ

The molar flux of phenol at the biofilm surface is given by

NA,z = 0

AAB

z

dcD

dz =

− = 0A ABc D

δδ 1 1tanh

AB AB

k k

D Dδ

The dimensionless parameter δ1

AB

k

Drepresents the ratio of reaction rate to diffusion rate.

For this problem we have

δ1

AB

k

D = 0.002 m 2

10

10.019

sm

2 10s

−× = 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 4πr2dr

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

4πr2

rAN " − 4πr2

drrAN+

" + RA4πr2dr = 0

1-31

Dividing the equation by the control volume (4πr2dr) 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

dCr A

AD2 − k1CA = 0

DA 2

1

r dr

d

dr

dCr A2 − 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

dCr A2 − α2

rCA = 0 ⇒ r

1

+

2

222

dr

Cdr

dr

dCr AA − α2

rCA = 0

2dr

dCA + 2

2

dr

Cdr A − α2

rCA = 0

Since dr

d

)( ArCdr

d =

dr

d

+dr

dCrC A

A = dr

dCA + dr

dCA + 2

2

dr

Cdr 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 = CA0r

R

)sinh(

)sinh(

R

r

αα

(E-5)

At the center of the organism, the concentration is given by CA(r = 0) = CA0)sinh( R

R

αα

(b) Obtain an expression for the rate of oxygen consumption within the organism.

Rate of oxygen consumption within the organism. = 4πR2(−DA A

r R

dC

dr =

)

The oxygen concentration within the organism is given by equation (E-5)

CA = CA0r

R

)sinh(

)sinh(

R

r

αα

(E-5)

AdC

dr = 0

sinh( )AC R

Rα

+− )cosh()sinh(12

rr

rr

ααα

A

r R

dC

dr =

= 0

sinh( )AC R

Rα

− )sinh(1

)cosh(2

RR

RR

ααα

A

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 = 4π R2DA0AC

R (φ cothφ - 1)

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)sinh( R

R

αα

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

αα

= 5×10-5 4.4721

sinh(4.4721) = 5.11×10-6 kmol/m3

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;

1-34

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 = CRr

R

)sinh(

)sinh(

R

r

αα

(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

αα

(1.4-2)