Fundamentals of Transport Phenomena ChE 715

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 15
Mass Transfer problems, cont’d. (ch 2)
S i 2011Spring 2011

2. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
We are interested in concentration profile flux etcWe are interested in concentration profile, flux, etc.
0A BdN dN
dy dy
= = No rxn. term?
Because no homogeneous rxn.
• Conservation Eqn.
g
constant throughout gap LAN =
• Flux Eqn =x ( )A A A B AN N+ +N JFlux Eqn. x ( )A A A B AN N+ +N J
Is this first term negligible since we say stagnant film?
Not in this case! In fact, NB= -m NANot in this case! In fact, NB m NA
Last lecture

3. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
=x ( )N N+ +N J=x ( )A A A B AN N+ +N J
NB= -m NA, from rxn. stoichiometry
dx
=x (1 ) A
A A A AB
dx
N m CD
dy
− −N
AB ACD dx
N =
[1 x (1 )]
AB A
A
A m dy
−
− −
N
AB AD dC
N Assuming constant C=
[1 (C / )(1 )]
AB A
A
A
dC
C m dy
−
− −
N
Last lecture

4. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
D dC
=
[1 (C / )(1 )]
AB A
A
A
D dC
C m dy
−
− −
N
Case I; Assume surface rxn to be very fast
Note, rxn order does not enter the picture
Case I; Assume surface rxn. to be very fast
BC: x=L, CA = 0
x=0, CA= CAo
.
[1 (C / )(1 )]
AB A
A
D dC
const A
C m dy
− = =
− −
Solve this differential equation, get two constants, use 2 BCs
Last lecture

5. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
D dC
=
[1 (C / )(1 )]
AB A
A
A
D dC
C m dy
−
− −
N
Case II; General—finite rxn rate at surface
Flux at surface related to
reaction rate
Case II; General—finite rxn rate at surface
BC: x=L,
x=0, CA= CAo
( ) ( )][ n
A ASAN L R Lk C= − =
.
[1 (C / )(1 )]
AB A
A
A
D dC
N const
C m dy
− = = =
− −
( )][ n
A Lk C How?

6. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
.
[1 (C / )(1 )]
AB A
A
A
D dC
N const
C m dy
− = = =
− −
( )][ n
A Lk C
( ) ( )][ n
A ASAN L R Lk C= − =
C
[C ( )] 1 (1 )nA A
A
AB
dC k
L m
dy D C
⎛ ⎞ ⎡ ⎤
= − −⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠
x=0, CA= CAo
1
; = ; Da= ; =
n
AoA A
Ao AB Ao y L
kC LC Cy
Let
C L D C
θ η φ
−
=
=
θ(0)=1[1 (1 ) ]n
Ao
d
Da x m
d
θ
φ θ
η
= − − −

7. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
1n
kCC C
dθ
Da = Damkohler #, ratio of rxn
velocity to diffusion velocity
1
; = ; Da= ; =
n
AoA A
Ao AB Ao y L
kC LC Cy
Let
C L D C
θ η φ
−
=
=
θ(0)=1[1 (1 ) ]n
Ao
d
Da x m
d
θ
φ θ
η
= − − −
Solve thru separation of variables, get
implicit expressionimplicit expression
1 (1 )1
ln[ ], m 1
(1 ) 1 (1 )
n Ao
Ao Ao
x m
Da
x m x m
φ
φ
− −
= ≠
− − −
l
1 , m=1n
Da φ φ= − Da 0, rxn very slow, concentration const.
Da ∞, diffusion very slow, concentration
is zero at catalytic surface

8. Example—Diffusion in Gas w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B gases in stagnant film
H did t s i l st p ? surfacey=L
constant throughout gap LAN =[1 (1 ) ]n
Ao
d
Da x m
d
θ
φ θ
η
= − − −
How did we get answer in last page?
1
; = ; Da= ; =
n
AoA A
Ao AB Ao y L
kC LC Cy
Let
C L D C
θ η φ
−
=
=1
1 0[1 (1 ) ]
n
Ao
d
Da d
x m
φ θ
φ η
θ
= −
− −∫ ∫
1 (1 )1
ln[ ] , m 1
(1 ) 1 (1 )
nAo
Ao Ao
x m
Da
x m x m
φ
φ
− −
= ≠
− − −
For m=1
1
1 0
n
d Da d
φ
θ φ η= −∫ ∫
1n
Da φ φ= −

9. Modified Prob. —Diffusion in Dilute Liq w/. Heterogeneous Rxn.
A mB at surface
y=0
Catalytic
CAo
y=L
A,B liquid in stagnant film
Rxn. at surface given by
surfacey=L
n
ASAR kC= −
constant throughout gap LAN =
0AdN
=
=x ( )A A A B AN N+ +N J
0
dy
=
Dilute xA is small; liquid convection
negligiblen g g
= A
A A AB
dC
D
dy
= −N J
2
2
0Ad C
dy
=
AoB C=
From BCs
AC Ay B= +
[ ]n
AB AoD A k AL C− = +

10. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
Would like
concentration profile
Inert surfacey=L
1
1
k
k
A B
−
⎯⎯→←⎯⎯
1 1A B VBVAR Rk C k C−+ = −= − v refers to reaction
within vol as againstwithin vol., as against
surface
1 10; ; and K= /VB A BVAR R KC C k k−= ==At equilibrium:
2
Ad C
Problem Formulation: at y=L0A BdC dC
dy dy
= =BCs: CA=CAO; CB=CBO at y=0
2
2
2
0
0
A
A VA
B
B VA
d C
D R
dy
d C
D R
d
+ =
− =
2 2
2 2
0A B
A B
d C d C
D D
dy dy
+ =
2B VA
dy

11. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
at y=L0A BdC dC
dy dy
= =BCs: CA=CAO; CB=CBO at y=0
2 2
2 2
0A B
A B
d C d C
D D
dy dy
+ =
dy dy
2
2
( ) 0A A B B
d
D C D C
dy
+ =
Integrating again:
A B
A B
dC dC
D D a
dy dy
+ = BCs: From y=L condition, a=0
Integrating again:
A A B BD C D C b+ =
A AO OD C D C b+ = BCs: From y=0 conditionA AO B BOD C D C b+ BCs From y 0 condition

12. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
at y=L0A BdC dC
dy dy
= =BCs: CA=CAO; CB=CBO at y=0A A B BD C D C b+ =
dy dy
A AO B BOD C D C b+ =
NOW, combine these two:
We know from before
( ) ( ) 0A A AO B B BOD C C D C C− + − =
( )A
A AO BOB
D
C C CC
D
= − + 2
0Ad C
D R+BD 2
0A
A VAD R
dy
+ =
2
12
( ) 0A
A AB
d C
D k KC C
dy
+ − =
dy
2
1 1 12
( ) ( ) 0A A A
A AO BO A
B B
d C D D
D k K C C k K k C
dy D D
+ + − + =

13. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
dC dC
CA=CAO; CB=CBO at y=0
2
1 1 12
( ) ( ) 0A A A
A AO BO A
B B
d C D D
D k K C C k K k C
dy D D
+ + − + =
at y=L0A BdC dC
dy dy
= =
Non-dimensionalize the eqn. and solve

14. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
dC dC
CA=CAO; CB=CBO at y=0
2
1 1 12
( ) ( ) 0A A A
A AO BO A
B B
d C D D
D k K C C k K k C
dy D D
+ + − + =
at y=L0A BdC dC
dy dy
= =
Non-dimensionalize the eqn. and solve
A By C KC
; ;A B
A B
AO AO
y
L
C KC
C C
η θ θ= = =Let
2
( ) ( ) 0A A AA AO d D D
k K C C k
D C
K k C
θ
θ+ + − + =2 1 1 12
( ) ( ) 0AO BO AO A
B B
k K C C k K k C
d D DL
θ
η
+ + − + =
2 2
1 1
2
1
2
22
1
( ) ( ) 0BOA
A
k KCd KL k KLk L k Lθ
θ+ + − + =2
( ) ( ) 0A
AB A ABOd CD D DD
θ
η

15. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
; ;A B
A B
AO AO
y
L
C KC
C C
η θ θ= = =
2 2
1 1
2
1
2
22
1
( ) ( ) 0BOA
A
k KCd
d C
KL k KL
D
k L k L
D DD
θ
θ+ + − + =
A h i l i ifi
Let
2
( ) ( ) A
AB A ABOd CD D DDη
2 2
11
; =; BOk L KCk LK
β α γ==
Any physical significance
of α and β?
Damkohler # for forward
& backward rxn;;
B A AOCD D
β γ
2
2
( ) ( ) 0A
A
d
d
θ
γα αβ β θ
η
+ + − + = at y=L 1) 0(0A AddC
y dd
θ
η
⇒ ==
& backward rxn.
dη y dd η
at y=0 (0) 1A AO AC C θ⇒ ==
AP AHA θθ θ+=
particular homogeneous ( )
( )
AP
β αγ
θ
β α
+
=
+

16. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
2
2
( ) ( ) 0A
A
d
d
θ
γα αβ β θ
η
+ + − + = (1) 0Ad
d
θ
η
=(0) 1Aθ =
; ;A B
A B
AO AO
y
L
C KC
C C
η θ θ= = =
2 2
11
; =; BOk L KCk LK
β α γ==
AP AHA θθ θ+=
( )
( )
AP
β αγ
θ
β
+
= ;;
B A AOCD D
β α γ
( )
AP
β α+
2
2
( )A
A
d
d
θ
θβ
η
α= +For :AHθ
Now apply BCs on entire solution
1 2exp( ) exp( )AH C Cθ α β η α β η= − + + +
Now, apply BCs on entire solution

17. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
(1) 0Ad
d
θ
η
=(0) 1Aθ =AP AHA θθ θ+=
( )β αγ
θ
+
2 2
11
; =;
B A
BO
AO
k L KCk L
CD D
K
β α γ==
( )
APθ
β α
=
+
1 2exp( ) exp( )AH C Cθ α β η α β η= − + + +
(1 )
[cosh( ) tanh( )sinh( )]A
α γ
θ α β η α β α β η
α
αγ
β β
β
α
−
= + + − + +
+
+
+
(1 )
[cosh( ) tanh( )sinh( )]B
β γ
θ α β η α β α β η
α
αγ
β β
β
α
−
= − + − + +
+
+
+

18. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
(1) 0Ad
d
θ
η
=(1 )
[cosh( ) tanh( )sinh( )]A
αγ β α γ
θ α β η α β α β η
α β α β
+ −
= + + − + +
+ +
2 2
11
; =; BOk L KCk LK
β α γ==
(0) 1Aθ =
(1 )
[cosh( ) tanh( )sinh( )]B
αγ β β γ
θ α β η α β α β η
α β α β
+ −
= − + − + +
+ +
;;
B A AOCD D
β α γ
For very fast rxn, i.e.,
; ; / fixedα β β α→ ∞ → ∞
tanh 1x →
/ (1 )
exp( )A
γ β α γ
θ α β η
⎡ ⎤+ −
= + − +⎢ ⎥tanh 1
cosh sinh exp( )
x
x x x
→
− → −
exp( )
1 / 1 /
Aθ α β η
β α β α
+ +⎢ ⎥+ +⎣ ⎦
/ ( / )(1 )
exp( )
1 / 1 /
B
γ β α β α γ
θ α β η
β α β α
⎡ ⎤+ −
= + − +⎢ ⎥+ +⎣ ⎦1 / 1 /β α β α+ +⎣ ⎦

19. Reversible Homogeneous rxn – liquid film
A B
y=0
Inert surface
CAo, CBo
y=L
A,B in dilute,
stagnant liquid
1
1
k
k
A B
−
⎯⎯→←⎯⎯
Inert surfacey=L
2 2
11
; =;
B A
BO
AO
k L KCk L
CD D
K
β α γ==
(1) 0Ad
d
θ
η
=
For very fast rxn, i.e.,
; ; / fixedα β β α→ ∞ → ∞
(0) 1Aθ =tanh 1
cosh sinh exp( )
x
x x x
→
− → −
/ (1 )
( )
γ β α γ
θ β
⎡ ⎤+ − / ( / )(1 )
( )
γ β α β α γ
θ β
⎡ ⎤+ −/ ( )
exp( )
1 / 1 /
A
γ β α γ
θ α β η
β α β α
⎡ ⎤
= + − +⎢ ⎥+ +⎣ ⎦
( )( )
exp( )
1 / 1 /
B
γ β β γ
θ α β η
β α β α
⎡ ⎤
= + − +⎢ ⎥+ +⎣ ⎦
For most of the film the exponentialFor most of the film, the exponential
terms are negligible:
Rxn layer thickness: 1/(α+β)1/2
θΑ ~ θΒ
Rxn layer thickness: 1/(α+β)

20. Diffusion into an open cone
Solute released into open cone of angle θο
Solute released at a const. rate (moles/time)
Solute source at origin θ
θο
m
r
g
Cone boundaries are impervious
Need to find concentration Ci:
S l t
St. steady conservation eqn. in spherical coordinates
2
2
vi2 2 2 2 2
1 1 1
[ sin ] 0
sin sin
i i i
i
C C C
D r R
r r r r r
θ
θ θ θ θ φ
∂ ∂ ∂∂ ∂⎛ ⎞ ⎛ ⎞
+ + + =⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
Solute source
/m moles time=
Possible BCs:
What terms can we eliminate?
21
[ ] 0iC
D
∂∂ ⎛ ⎞
⎜ ⎟
0 at =0iC
θ
θ
∂
=
∂
C∂
2
2
[ ] 0i
iD r
r r
⎛ ⎞
=⎜ ⎟
∂ ∂⎝ ⎠
2
0iC∂∂ ⎛ ⎞
⎜ ⎟
How?
symmetry
fl h0 at =i
O
C
θ θ
θ
∂
=
∂
2
0i
r
r r
⎛ ⎞
=⎜ ⎟
∂ ∂⎝ ⎠
no flux thru
wall

21. Diffusion into an open cone
θ
θο
2
0iC
r
r r
∂∂ ⎛ ⎞
=⎜ ⎟
∂ ∂⎝ ⎠
2 iC
r a
r
∂
=
∂
r
S l t
2
iC a
r r
∂
=
∂
i
a
C b
r
= − +
Any BC?
Solute source
/m moles time=( ) 0iC r = ∞ =
i
a
C = − How do we get a?i
r
g
Let us look at the
point source at r=0 0
lim( )i
r
m N S
→
= S=area of spherical shell of radius r
2
sindS r d dθ θ φ=
2
2
sin
o
S r d d
π
θ
θ θ φ= ∫ ∫
2
0
2 sin
O
S r d
θ
π θ θ= ∫0
0
φ∫ ∫ 0∫
2
2 [1 cos ]OS rπ θ= −

22. Diffusion into an open cone
θ
θο
i
a
C
r
= − How do we get a?
2 r
S l t
0
lim( )i
r
m N S
→
=
i
i i
dC
N D
dr
= −
2
2 [1 cos ]OS rπ θ= −
Solute source
/m moles time=
dr
0
2
2 [1 coslim( )]i
i O
r
dC
D r
dr
m π θ
→
− −=
From last page:0
2
2 [1 cos ]lim( )
r
i
i O
dC
D
dr
m rπ θ
→
= − − 2 iC
r a
r
∂
=
∂
2 [1 cos ]i Om aDπ θ−= −
m
a = −
2 [1 cos ]i O
a
Dπ θ−

23. Directional Solidification– dilute binary alloy
Well-mixed
melt
SolidStagnant
film U
Solid front moving
Deen Example Problem 2.8-5
Y=0Y δ
Ci=C∞
Ci=CsU
Solid front moving
Mass transfer thru stagnant layer
ISSUES: Y 0Y=-δISSUES:
•Meaning of stagnant layer
•Conservation equation in this case
•BC condition at melt-solid interface 2iDC
D C R∇
Conservation Eqn:
B cond t on at me t so d nterface 2i
i i Vi
DC
D C R
Dt
= ∇ +
2
i i
i
y y
C C
U D
∂ ∂
∂ ∂
=
BC
C∂
( ) ( )M M
C= +N v J
y y∂ ∂
(0) (0) (0)i
i i s
i
C
UC D UC
y
∂
= − =
∂
N
Note extra term appearing on
left because of convection
Note how convection appears here!
A A AC= +N v J
Example not covered in class;
look problem over in book