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