Week 9

Dept of Chemical and Biomolecular Engineering

CN2125 Heat and Mass Transfer

Dr. Praveen Linga, E5-02-23, 6601-1487(Radiation, Mass Transfer)

Tutorial/HW Week #9WWWR Chapters 26-27

ID Chapter 14• Tutorial #9• WWWR #26.17,

26.27, 27.6, 27.22 & ID #14.33

• To be discussed during the week of Mar 19 - 23, 2012.

• By either volunteer or class list.

• Homework #9(self practice)

• WWWR # 26.22, 27.16 & ID #14.39

Molecular Diffusion

• General differential equation

• One-dimensional mass transfer without reaction

0

AA

A Rt

cN

)( ,,, zBzAAA

ABzA NNydz

dycDN

Unimolecular Diffusion• Diffusivity of gas can be measured in an

Arnold diffusion cell

• Assuming– Steady state, no reaction, and diffusion in z-

direction only

• We get

• And since B is a stagnant gas,

0, dz

dN zA

0, dz

dN zB

• Thus, for constant molar flux of A, when NB,z = 0,

– with boundary conditions:

at z = z1, yA = yA1

at z = z2, yA = yA2

• Integrating and solving for NA,z

)1(

)1(ln

)(1

2

12,

A

AABzA y

y

zz

cDN

dz

dy

y

cDN A

A

ABzA

1,

• since the log-mean average of B is

• we get

• This is a steady-state diffusion of one gas through a second stagnant gas;

lmB

AAAB

p

pp

zzRT

PD

,12

)(

)(21

lmB

AAABzA y

yy

zz

cDN

,12,

)(21

)1/()1(ln)/ln(12

21

12

12

,AA

AA

BB

BBlmB yy

yy

yy

yyy

• For film theory, we assume laminar film of constant thickness ,

• then, z2 – z1 =

• and

• But we know

• So, the film coefficient is then

)(21, AA

czA pp

RT

kN

lmB

ABc p

PDk

,

)(21

,, AA

lmB

ABzA pp

RTp

PDN

• To determine concentration profile,

• if isothermal and isobaric,

• integrated twice, we get

01,

dz

dy

y

cD

dz

dN

dz

d A

A

ABzA

01

1

dz

dy

ydz

d A

A

21)1ln( czcyA

– with boundary conditions:

at z = z1, yA = yA1

at z = z2, yA = yA2

• So, the concentration profile is:

)/()( 121

1

2

11

1

1

1zzzz

A

A

A

A

y

y

y

y

Example 1

Pseudo-Steady-State Diffusion

• When there is a slow depletion of source or sink for mass transfer

• Consider the Arnold diffusion cell, when liquid is evaporated, the surface moves,

• at any instant, molar flux is

lmB

AAABzA zy

yycDN

,,

)(21

• Molar flux is also the amount of A leaving

• Under pseudo-steady-state conditions,

• which integrated from t=0 to t=t, z=zt0 to z=zt becomes

dt

dz

MN

A

LAzA

,,

lmB

AAAB

A

LA

zy

yycD

dt

dz

M ,

, )(21

2)(

/ 22,, 0

21

tt

AAAB

AlmBLA zz

yycD

Myt

Example 2

Equimolar Counterdiffusion• Flux of one gaseous component is equal to

but in the opposite direction of the second gaseous component

• Again, for steady-state, no reaction, in the z-direction,

• the molar flux is

0, zANdz

d

zBzAAA

ABzA NNydz

dcDN ,,,

• In equimolar counterdiffusion, NA,z = -NB,z

Integrated at z = z1, cA = cA1 and at z = z2, cA = cA2 to:

• Or in terms of partial pressure,

dz

dcDN A

ABzA ,

)()( 21

12, AA

ABzA cc

zz

DN

)()( 21

12, AA

ABzA pp

zzRT

DN

• The concentration profile is described by

• Integrated twice to

With boundary conditions at z = z1, cA = cA1and at z = z2, cA = cA2

becomes a linear concentration profile:

02

2

, dz

cdN

dz

d AzA

21 CzCcA

21

1

21

1

zz

zz

cc

cc

AA

AA

Systems with Reaction

• When there is diffusion of a species together with its disappearance/appearance through a chemical reaction

• Homogeneous reaction occurs throughout a phase uniformly

• Heterogeneous reaction occurs at the boundary or in a restricted region of a phase

• Diffusion with heterogeneous first order reaction with varying area:– With both diffusion and reaction, the process can

be diffusion controlled or reaction controlled.– Example: burning of coal particles

– steady state, one-dimensional, heterogeneous

– 3C (s) + 2.5 O2 (g) 2 CO2 (g) + CO (g)

– Along diffusion path, RO2 = 0, then the general mass transfer equation reduces from

– to

– For oxygen,

0sin

1)sin(

sin

1)(1 2

AAAArA R

N

r

N

rr

Nr

rt

c

0)( 2

dr

Nrd Ar

0)(

2

2

dr

Nrd rO

– From the stoichiometry of the reaction,

– We simplify Fick’s equation in terms of oxygen only,

– which reduces to

COrrO NN 5.22

rCOrO NN22

25.1

)025.1

1

5.2

1(

2222

2

22 rOrOrOO

OmixOrO NNNy

dr

dycDN

dr

dy

y

cDN O

O

mixOrO

2

2

2

2 2.01

– The boundary conditions are:

at r = R, yO2 = 0 and at r = , yO2 = 0.21,

– Integrating the equation to:

– The oxygen transferred across the cross-sectional area is then:

042.1

1ln

2.0

12

2

2 mixOrO

cD

RNr

)042.1ln(2.0

44 2

22

2 mixOrOO

cDRNrW

– Using a pseudo-steady-state approach to calculate carbon mass-transfer

output rate of carbon:

accumulation rate of carbon:

input rate of carbon = 0

Thus, the carbon balance is

)042.1ln(2.0

45.2

3

5.2

3

2

32

22

mixOOCOC

cDRWWW

24C C

C C

dV dRR

M dt M dt

dt

dRR

M

cDR

C

CmixO 24)042.1ln(2.0

45.2

30 2

– Rearranging and integrating from

t = 0 to t = , R = Ri to R = Rf, we get

– For heterogeneous reactions, the reaction rate is

)042.1ln(12

2

22

mixO

fiC

C

cD

RRM

AssRrA ckN

– If the reaction is only

C (s) + O2 (g) CO2 (g)

and if the reaction is not instantaneous, then

for a first-order reaction, at the surface,

then,

)(42222 sOOmixOO yyRcDW

ck

Ny

s

ROsO

2

2

22

2

212

OmixOs

mixOrO yRcD

Rk

DNr

– Combining diffusion with reaction process, we get

Rk

D

yRcDW

s

mixO

OmixOO

2

22

2

1

4

Example 3

• Diffusion with homogeneous first-order reaction:– Example: a layer of absorbing liquid, with

surface film of composition A and thickness ,

assume concentration of A is small in the film,

and the reaction of A is

dz

dcDN A

ABzA ,

AA ckR 1

– Assuming one-direction, steady-state, the mass transfer equation reduces from

to

with the general solution

0, AzA RNdz

d

012

2

1

A

AABA

AAB ck

dz

cdDck

dz

dcD

dz

d

zDkczDkcc ABABA /sinh/cosh 1211

– With the boundary conditions:

at z = 0, cA = cA0 and at z = , cA = 0,

– At the liquid surface, flux is calculated by differentiating the above and evaluating at z=0,

AB

ABAABAA

Dk

zDkczDkcc

/tanh

/sinh/cosh

1

11

0

0

AB

ABAz

A

Dk

Dkc

dz

dc

/tanh

/

1

10

0

– Thus,

– Comparing to absorption without reaction, the second term is called the Hatta number.

– As reaction rate increases, the bottom term approaches 1.0, thus

AB

ABAAB

zzA Dk

DkcDN

/tanh

/

1

10,

0

)0(010, AABzzA ckDN

– Comparing with

we see that kc is proportional to DAB to ½ power.

• This is the Penetration Theory model, where a molecule will disappear by reaction after absorption of a short distance.

)(21, AAczA cckN

2/1ABc Dk

Example 4

2- and 3-Dimensional Systems

• Most real systems are two- and three-dimensional

• Analytical solution to general differential equation with the boundary conditions

• Requires partial differential equations and complex variable theories.

Unsteady-State Diffusion

• Transient diffusion, when concentration at a given point changes with time

• Partial differential equations, complex processes and solutions

• Solutions for simple geometries and boundary conditions

0

AA

A Rt

cN

• Fick’s second law of diffusion

• 1-dimensional, no bulk contribution, no reaction

• Solution has 2 standard forms, by Laplace transforms or by separation of variables

2

2

z

cD

t

c AAB

A

• Transient diffusion in semi-infinite mediumuniform initial concentration CAo

constant surface concentration CAs

– Initial condition, t = 0, CA(z,0) = CAo for all z

– First boundary condition:

at z = 0, cA(0,t) = CAs for t > 0

– Second boundary condition:

at z = , cA(,t) = CAo for all t

– Using Laplace transform, making the boundary conditions homogeneous

AoA cc

– Thus, the P.D.E. becomes:

– with(z,0) = 0(0,t) = cAs – cAo

(,t) = 0

– Laplace transformation yields

which becomes an O.D.E.

2

2

zD

t AB

2

2

0dz

dDAB

02

2

ABD

s

dz

d

– Transformed boundary conditions:•

•

– General analytical solution:

– With the boundary conditions, reduces to

– The inverse Laplace transform is then

s

ccz AoAs )0(

0)( z

zDszDs ABAB eBeA /1

/1

zDsAoAs ABes

cc /)(

tD

zcc

AB

AoAs2

erfc)(

– As dimensionless concentration change,• With respect to initial concentration

• With respect to surface concentration

– The error function

is generally defined by

tD

z

tD

z

cc

cc

ABABAoAs

AoA

2erf1

2erfc

erf2

erf

tD

z

cc

cc

ABAoAs

AAs

tD

z

AB2

deerf 0

22

– The error is approximated by• If 0.5

• If 1

– For the diffusive flux into semi-infinite medium, differentiating with chain rule to the error function

and finally,

3

2erf

3

211erf

e

tD

cc

dz

dc

AB

AAsz

A

0

0

AoAsAB

zzA cct

DN 0,

• Transient diffusion in a finite medium, with negligible surface resistance– Initial concentration cAo subjected to sudden

change which brings the surface concentration cAs

– For example, diffusion of molecules through a solid slab of uniform thickness

– As diffusion is slow, the concentration profile satisfy the P.D.E.

2

2

z

cD

t

c AAB

A

– Initial and boundary conditions of• cA = cAo at t = 0 for 0 z L

• cA = cAs at z = 0 for t > 0

• cA = cAs at z = L for t > 0

– Simplify by dimensionless concentration change

– Changing the P.D.E. to

Y = Yo at t = 0 for 0 z L

Y = 0 at z = 0 for t > 0

Y = 0 at z = L for t > 0

AsAo

AsA

cc

ccY

2

2

z

YD

t

YAB

– Assuming a product solution,

Y(z,t) = T(t) Z(z)– The partial derivatives will be

– Substitute into P.D.E.

divide by DAB, T, Z to

t

TZ

t

Y

2

2

2

2

z

ZT

z

Y

2

2

z

ZTD

t

TZ AB

2

211

z

Z

Zt

T

TDAB

– Separating the variables to equal -2, the general solutions are

– Thus, the product solution is:

– For n = 1, 2, 3…,

tDABeCtT2

1

zCzCzZ sincos 32

tDABezCzCY2

)sin()cos( '2

'1

L

n

– The complete solution is:

where L = sheet thickness and – If the sheet has uniform initial concentration,

for n = 1, 3, 5…– And the flux at z and t is

dzL

znYe

L

zn

Lcc

ccY

n

L

oXn

AsAo

AsA D

sinsin2

1 0

)2/( 2

21x

tDX AB

D

1

)2/( 2

sin14

n

Xn

AsAo

AsA DeL

zn

ncc

cc

DXn

nAoAs

ABzA e

L

zncc

L

DN

2)2/(

1, cos

4

Example 1

Example 2

• Concentration-Time charts

Figure F.1 Unsteady-state transport in a large flat slab

Figure F.2 Unsteady-state transport in a long cylinder

Example 3