Fundamentals of Transport Phenomena ChE 715

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 18
Ch 3 cont’d
• Transient prob-similarity method
R l P t b ti• Regular Perturbation
• Singular Perturbation
Spring 2011

2. Characteristic time scales
Two relevant approximations for time dependent problems
SECTION 3.4
• Quasi-steady state approximation (QSSA):
At “long” times, neglect time derivative(s).
• “Penetration” approximation:
At “short” times, assume that certain boundaries
are far away from the perturbationare far away from the perturbation.
Time for concentration (diffusion) or temperature (conduction)
propagation over distance L:propagation over distance L:
tD ~
L2
Di
; tC ~
L2
α
α ≡
k
ρC
⎛
⎝
⎜
⎞
⎠
⎟.
Di α ρCp⎝ ⎠

3. Characteristic time scales: membrane pore
C1(t); C2(t);
volume V
(well-mixed)
volume V
(well-mixed)
In the pore: C(x,t)
x = 0 x = Lx 0 x L
Initial conditions (I.C.’s): C1(0) = KC0 (the perturbation)
C (0) = 0C2(0) = 0
C(x,0) = 0

4. Characteristic time scales: membrane pore
C1(t); C2(t);1
volume V
(well-mixed)
2
volume V
(well-mixed)
In the pore: C(x,t)
x = 0 x = Lx = 0 x = L
Develop solutions for the QSSA and penetration regimes.p Q p g

5. Characteristic time scales: membrane pore
C1(t); C2(t);
volume V
(well-mixed)
volume V
(well-mixed)
In the pore: C(x,t)
x = 0 x = L
Generally speaking, what is the essence of the QSSA?

6. Membrane pore example: QSSA solution
2
2
0;
C C
D
t x
∂ ∂
∂ ∂
= ≈
1 2 1( , ) (t) + [ ( ) - ( )]
x
C x t KC K C t C t
L
=
C1(t) C2(t)
C A Bx= +
L
What BCs have we used?
C1(t) 2( )
1 20, ( ); , ( )x C KC t x L C KC t= = = =1
1 00
; (0)x x
dC
V AN C C
dt =
= − =
( )1 2
x x
C C
N J DK
L
−
= ≈ From solution of C(x,t)
0
( )/0
1( ) 1 ;
2 2
tC LV
C t e
KAD
τ
τ−
= + = How?
1
1 2[ ]
dC ADK
C C
dt LV
= − −
x = 0 x = L
2
2; (0) 0x x L
dC
V AN C
dt =
= − =
C
dt LV
1
1 0
2
[ / 2]
dC ADK
dt
C C LV
= −
−
2 1dC dC
V V
dt dt
= −
( )0
2 ( ) 1
2
C
C t e τ−
= −
Integrate, and use BC

7. Membrane pore example: QSSA solution (cont.)
When is the QSSA justified for this problem?
1. From scaling analysis, the time is well beyond the penetration regime:
t >> L2/D
2. We also need the following (membrane vol << external vol):
2
2
1
L ALK
D Vτ
= <<

8. Membrane pore example: penetration solution
C1(t);
l V
C2(t);
l Vvolume V
(well-mixed)
volume V
(well-mixed)
In the pore: C(x,t)
x = 0 x = L
Clearly, “short” times means that t << L2/D
In this regime, the length of the pore is semi-infinite!

9. Membrane pore example: penetration solution
Solution of PDE’s by the Similarity Method
∂C
= D
∂2
C
;
SECTION 3.5
∂t
D
∂x2
;
C(0,t) = C0
; C(∞,t) = C(x,0) = 0.
∂C
∂t
=
∂η
∂t
dC
dη
= −
x ′g
g2
dC
dη
= −
η ′g
g
dC
dη
;
Propose a substitution variable, η = x/g(t):
∂t ∂t dη g dη g dη
∂2
C
∂x2
=
1
g2
d2
C
dη2
;
d2
C
dη2
+
ηg ′g
D
dC
dη
= 0.
Essence of the similarity method: must be a constantg ′g

10. Membrane pore example: penetration solution
let
g ′g
D
= a;
d2
C dC
0
dη2
+ aη
dη
= 0
let Y =
dC
d
; ′Y + aηY = 0.
dη
η
Apply the integrating factor method:
Y exp aη2
2( )= constant ≡ A;
dC
d
= Aexp −aη2
2( ); let a = 2.
dη
p η( )
00 (0)(0, ) C CC t C= ⇒ =
BC
( , ) ( ,0) 0 ( ) 0CC t C x∞ = ⇒ ∞= =
BCs:

11. Membrane pore example: penetration solution
Solution of PDE’s by the Similarity Method (cont.)
C(η)
C0
=1−
2
π
e−s2
ds0
η
∫
≡ erfc η =1− erf η.
Apply B.C.’s and solve:
η η
0 8
1
0
0.6
0.8
0.2
0.4
0
0 0.5 1 1.5 2
η

12. Membrane pore example: penetration solution
Solution of PDE’s by the Similarity Method (cont.)
Not done yet: need to determine g(t):
Initial and boundary conditions must be consistent:
g ′g = 2D;
C(x,0) = C(∞,t) = 0; requires that g(0) = 0.
η = x/g(t)
g ′g =
1
2
g2
( )′;
g = 4Dt.g
C(η)
C
= erfc
x
4DtC0 4Dt

13. Regular Perturbation
Perturbation: Regular, Singularg g
Define small parameter: ε
Mathematical Order: O(1)
1 ~ 1 O(1)
15 ~ 10 O(10)
O( )ε ~ ε O(ε)
ε2 O(ε2)

14. Regular Perturbation
Solution when ε = 0 has the same
h t th l ti hcharacter as the solution when
ε 0:
Solution:
2
0 1 2 ...
n
ε ε
ε
∞
Θ = Θ + Θ + Θ +
= Θ∑0
n
n
ε
=
= Θ∑

15. Example Problem – Heated Wire
Ts = T∞
From Example 2.8-1
2
; ;
/
T T r hR
r Bi
H R k R k
∞−
Θ = = =
k, Hv (Bih >> 1)
v /H R k R k
2
2
1
4 2
V Vr
T T R R
R
H H
k h∞
⎡ ⎤⎛ ⎞
− − +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
=
2
1 1T T r∞− −
Θ = = +
0
How?
2
/ 4 2v nH R k Bi
Θ = = +
In this case let
How?
( )( ) 1k T k a T T∞ ∞= + −⎡ ⎤⎣ ⎦
( )⎡ ⎤
In this case, let
( )( ) 1vH T H a T T∞ ∞= + −⎡ ⎤⎣ ⎦

16. Example Problem – Heated Wire
1
( ) ( ) 0v
d dT
rk T H T
r dr dr
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
Ts = T∞
r dr dr⎝ ⎠
( )T R T=
k, Hv
BC’s:
0
dT;( )T R T∞BC s:
0
0
r
r
dr =
=;
Define:
2
2
; ;
/
T T H Rr
r a
H R k R k
ε∞ ∞
∞ ∞ ∞
−
Θ = = =
ε <<1 (basis for pert method)ε <<1 (basis for pert. method)

17. Example Problem – Heated Wire
Ts = T∞
1
( ) ( ) 0v
d dT
rk T H T
r dr dr
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
k, Hv
( )( ) 1k T k a T T∞ ∞= + −⎡ ⎤⎣ ⎦
( )( ) 1vH T H a T T∞ ∞= + −⎡ ⎤⎣ ⎦
2
2
; ;
/
H R
a
T T r
r
H kR k R
ε∞
∞ ∞
∞
∞
−
Θ == =
1
(1 ) (1 ) 0
d d
r
r dr dr
ε ε
Θ⎛ ⎞
+ Θ + + Θ =⎜ ⎟
⎝ ⎠
∞ ∞ ∞
⎝ ⎠
(1) 0Θ =BC’s:
0
d
r
Θ
=( )
0
0
r
r
dr =

18. Example Problem – Heated Wire
1
(1 ) (1 ) 0
d d
r
r dr dr
ε ε
Θ⎛ ⎞
+ Θ + + Θ =⎜ ⎟
⎝ ⎠
Ts = T∞
(1) 0Θ =
k, Hv
BC’s:
0
0
d
r
dr
Θ
=;
0rdr =
2
0 1 ( )Oε εΘ = Θ + Θ +Solve within O(ε) :
⎡ ⎤⎛ ⎞
Then:
( )2 3 20 1
0 1
1
1 ( ) ( )
d dd
r O O
r dr dr dr
ε ε ε ε ε
⎡ Θ ⎤Θ⎛ ⎞
+ Θ + Θ + + +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
( )2 3
0 11 ( ) 0Oε ε ε+ + Θ + Θ + =

19. Example Problem – Heated Wire
(1) 0Θ =
k H
Ts = T∞
BC:
k, Hv
then:
2
0 1(1) (1) ( ) 0Oε εΘ + Θ + =
( )2 3
1 ( ) 0OΘ Θ
( )2 3 20 1
0 1
1
1 ( ) ( )
d dd
r O O
r dr dr dr
ε ε ε ε ε
⎡ Θ ⎤Θ⎛ ⎞
+ Θ + Θ + + +⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
01
1 0
dd Θ⎛ ⎞
⎜ ⎟ 0dΘ
O(1) problem: 0 ?Θ =
;
( )2 3
0 11 ( ) 0Oε ε ε+ + Θ + Θ + =
0
1 0r
r dr dr
⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
0
0
0
r
d
r
dr =
Θ
=0 (1) 0Θ =
( )2
0
1
( ) 1
4
r rΘ = −
Solving :

20. Example Problem – Heated Wire
⎡ ⎤⎛ ⎞k H
Ts = T∞
O(ε) problem: 1 ?Θ =
0 1
0 0
1
0
d dd
r
r dr dr dr
⎡ Θ ⎤Θ⎛ ⎞
Θ + + Θ =⎜ ⎟⎢ ⎥
⎝ ⎠⎣ ⎦
k, Hv
1
0
0
r
d
r
dr =
Θ
=1(1) 0Θ =
;
1dΘ solving: 0r=
( )4
1
1
( ) 1
64
r rΘ = −
0 1
2
d
r
dr
Θ
= −
solving:
64
and the total
( ) ( )2 4 21
( ) 1 1 ( )O
ε
Θsolution is: ( ) ( )2 4 21
( ) 1 1 ( )
4 64
r r r O
ε
εΘ = − + − +

21. Example Problem 2 – Regular Perturbation
Diffusion and second order
reaction (slow) at steady state A B
y=0
Inert surfacey=L2
CAo
Liq film
L t
Inert surfacey=L2
1 AVAR k C= −
at x=L0AdC
dy
=BCs: CA=CAO at y=0
Let
2
2
0A
A VA
d C
D R
dy
+ =
2
d C
2
; x= ; Da= AoA
Ao AB
kC LC y
C L D
θ =
d2
Θ
Θ2
0
2
2
2
0A
A A
d C
D kC
dy
− = Assume Da <<1 and Da=є
dx2
−εΘ2
= 0;
Θ(0) =1; ′Θ (1) = 0.
2 3
1 2( ) ( ) ( ) ( ) ( )ox x x x Oε ε εΘ = Θ + Θ + Θ +

22. Example Problem 2 – Regular Perturbation
Example: Diffusion and second order reaction (slow) at
steady state.
d2
Θ
dx2
−εΘ2
= 0;
Θ(0) =1; ′Θ (1) = 0.
Θ2
= Θ0 +εΘ1 +ε2
Θ2 + O(ε3
)( )
2
= Θ 2
+ 2εΘ Θ +ε2
Θ 2
+ 2Θ Θ( )+O(ε3
)
O(1) problem:
2
= Θ0 + 2εΘ0Θ1 +ε2
Θ1 + 2Θ0Θ2( )+O(ε3
)
d2
Θ0
dx2
= 0;
Θ0 (0) =1; Θ0
′(1) = 0;0 ( ) ; 0 ( ) ;
Θ0 (x) =1

23. Example Problem 2 – Regular Perturbation
Example: Second order reaction (slow)
d2
Θ1
dx2
=Θ0
2
=1;
′
O(ε) problem:
Θ1(0) = 0; Θ1
′(1) = 0;
Θ1(x) =
x2
2
− x.
d2
Θ2
d 2
= 2Θ0Θ1 = x2
− 2x;
O(ε2) problem:
dx2 0 1 ;
Θ2 (0) = 0; Θ2
′(1) = 0;
Θ ( )
x4
x3
2
Θ2 (x) =
x
12
−
x
3
+
2
3
x.

24. Example Problem 2 – Regular Perturbation
Example: Second order reaction (slow)
⎛ ⎞ ⎛ ⎞2 3 4
2 32
( ) 1 ( )
2 3 3 12
x x x
x x x Oε ε ε
⎛ ⎞ ⎛ ⎞
Θ = − − + − + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1
0.95
1
ε = 0.1
0.9
ε = 0.2( )xΘ
0.85 ε = 0.5
0.8
0 0.2 0.4 0.6 0.8 1
Position, x

25. Example Problem – Singular Perturbation
CA(R) = Co
R
2
1k R
RA = k1CA
R
r 1
A
k R
Da
D
= >> 1 (fast reaction)
1
CA/C0
0
0 1 r/R
1 - ñ01

26. Example Problem – Singular Perturbation
CA(R) = Co
R 1C r
RA = k1CA
R
r 1
0
; ;AC r
r Da
C R
ε −
Θ = = =
(1) 1Θ =BC’s: 0
d
r
Θ
=;0
d d
r
ε Θ⎛ ⎞
− Θ =⎜ ⎟
⎝ ⎠ 0
0
r
r
dr =
r dr dr
⎜ ⎟
⎝ ⎠
(trivial solution)ε = 0 Θ = 0
Define:
( )1 b
rξ ε≡ − (b < 0)

27. Example Problem – Singular Perturbation
CA(R) = Co
R
( )1 b
rξ ε≡ − 1 b
r ξε −
= −
RA = k1CA
R
r b
dr dε ξ−
= −
bd d
dr d
ε
ξ
Θ Θ
= −
2 2
2
2 2
bd d
dr d
ε
ξ
Θ Θ
=
dr dξ dr dξ
2
0
d dε
ε
Θ Θ
+ Θ =
Replacing in the
d d l 2
0
dr r dr
ε + − Θ =expanded original eq.
2 1b
d dε +
Θ Θ
We obtain:
2 1
2
0
1
b
b
d d
d d
ε
ε
ξ ξε ξ
+
−
Θ Θ
− − Θ =
−

28. Example Problem – Singular Perturbation
CA(R) = Co
R
First term (Diffusion in a
slab: O(1)
RA = k1CA
R
r
2 1/ 2
0
d dεΘ Θ
− − Θ =
Set: b = -1/2
2 1/ 2
0
1d dξ ξε ξ
Θ
−
1/2 3/2
0 1 2 ( )Oε ε εΘ = Θ + Θ + Θ + 21
1 ( ) ..... x 0
1
x O x for= + + + →
Replacing:
0 1 2 ( )
2 2
d dΘ Θ
1/2
1/2
1
1 ( ) ...
1
Oξε ε
ξε
= + + +
−
( )
1
f
x−
1/20 1
2
...
d d
d d
ε
ξ ξ
Θ Θ
+ +
1/ 2 0dΘ
ξ
1/ 2 0
...
d
ε
ξ
− −
( )1/2
0 1 ... 0ε− Θ + Θ + =

29. Example Problem – Singular Perturbation
CA(R) = Co
R
2
1/2 1/2 1/2 1/20 01 1
2
[1 ( )][ ]
d dd d
O
d d d d
ε ε ξε ε ε
ξ ξ ξ ξ
Θ ΘΘ Θ
+ − + + +
RA = k1CA
R
r 1/2
0 1 ( ) 0
d d d d
O
ξ ξ ξ ξ
ε ε− Θ − Θ + =
BC’s:
2
d Θ
O(1) problem:
(0) 1Θ =2
0
02
0
d
dξ
Θ
− Θ = Other BC:
dΘ0
(1) 0
d
dξ
Θ
= (?)
NO! because BL doesn’t
extend to cylindrical
center

30. Example Problem – Singular Perturbation
CA(R) = Co
R
2
0
02
0
d
dξ
Θ
− Θ =
RA = k1CA
R
r
BC’s:
dξ
(0) 1Θ =
New BC:
( ) 0Θ ∞ =
Solving:
0 e ξ−
Θ =
O(ε1/2) problem: 2
01
0
dd ΘΘ
+ + Θ BC’s:
1(0) 0Θ =01
12
0
d dξ ξ
+ + Θ = BC s:
1( ) 0Θ ∞ =
e ξ
ξ −
1
2
eξ
Θ =Solving: