Fundamentals of Transport Phenomena ChE 715

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 19
Ch 3 cont’d
•Singular Perturbation
Spring 2011

2. Regular versus Singular Perturbation
Regular perturbation method: solution obtained
by setting ε 0 is similar to that for ε > 0by setting ε = 0 is similar to that for ε --> 0.
How it works: Develop solution as a power series
expansion Θ = Θ0 + εΘ1 + ε2Θ2 +expansion, Θ = Θ0 + εΘ1 + ε Θ2 + …
Singular perturbation method: solution obtained
by setting ε = 0 has distinctly different
characteristics than that for ε --> 0.
How it works: Rescale the problem and develop
solution as a more complicated power series
expansion.p

3. Regular versus Singular Perturbation
Let us find the roots of εx2+x-1=0, where ε<<1
Like reg perturbation, let x= a0+εa1+ε2a2+…g p , 0 1 2
O(1) problem: a0-1=0 or x=1 Any problems? We lost the second root!
Need to rescale, define X=εa x
Obtain X2 + εa-1 X ε2a-1 =0Obtain X2 + εa 1 X- ε2a 1 =0
We now need to identify the most important terms (dominant balance) as ε→0
Choice 1: X2 and X Choice 2: X2 and
term involving ε
εa-1 =1, or a=1
Obtain X2 + X- ε =0
ε2a-1 =1, or a=1/2
Obtain X2 + ε-1/2 X- 1=0Obtain X + X ε 0 Obtain X + ε X 1 0
Which one to use?
Contradicts assumed
dominant balance as ε→0

4. Singular Perturbation Primer
let X= a0+εa1+ε2a2+…
X2 + X- ε =0
0 1 2
O(1) bl
(a0+εa1+ε2a2+…)2 + (a0+εa1+ε2a2+…)- ε =0
(1) 1 d (2) 0O(1) problem: a0
2+a0=0 a0
(1) =-1; and a0
(2) =0
O(ε) problem: 2a0 a1 + a1 -1=0 or a1
(1) =-1; and a1
(2) =1
Therefore X1 = -1-ε +O(ε2)
( ) p 0 1 1
f X1 ( )
X2 = ε +O(ε2)

5. Singular Perturbation Primer
KEY INFO for SINGULAR PERTURBATION PROBSKEY INFO for SINGULAR PERTURBATION PROBS
No uniformly valid approximation thru entire domain of prob
ε multiplies highest order term (derivative)
Setting ε=0 reduces order of eqn, so not possible to satisfy all BCs
Need new scaling and identifying dominant terms
Two regimes have to be consideredTwo regimes have to be considered
Solutions of various regions have to be asympotically matched

6. Example Problem – Singular Perturbation
CA(R) = Co
R k C
R
r
Cylindrical catalyst pellet
Sites uniformly distributed throughout
1st order rxn, Fast rxn
RA = -k1CA
2
1 1
[ ] 0A A AC C C
D r R
∂ ∂ ∂∂ ∂⎛ ⎞ ⎛ ⎞
+ + + =⎜ ⎟ ⎜ ⎟
1
vA2 2
[ ] 0ABD r R
r r r r zθ θ
+ + + =⎜ ⎟ ⎜ ⎟
∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
CA/C0
A
1
0A
AB
C
D r kC
r r r
∂∂ ⎛ ⎞
− =⎜ ⎟
∂ ∂⎝ ⎠
2
1k R
Da = >> 1 (fast reaction) 0
A
Da
D
= >> 1 (fast reaction)
0 1 r/R
1 -01
0
r

7. Example Problem – Singular Perturbation
CA(R) = Co
R k C
R
r
1
0
; ;AC r
r Da
C R
ε −
Θ = = =
RA = k1CA
0C R
dΘd dε Θ⎛ ⎞
(1) 1Θ =BC’s:
0
0
r
d
dr =
Θ
=;0
d d
r
r dr dr
ε Θ⎛ ⎞
− Θ =⎜ ⎟
⎝ ⎠
2
2
1
[ ] 0
d d
dr r dr
ε
Θ Θ
+ − Θ =
(trivial solution)ε = 0 Θ = 0
Define:
( )1 b
rξ ε≡ − (b < 0)

8. Example Problem – Singular Perturbation
CA(R) = Co
R k C
R
r
( )1 b
rξ ε≡ − 1 b
r ξε −
= −
b
d dξ−
RA = k1CA
b
dr dε ξ= −
bd d
ε
Θ Θ
= −
2 2
2bd d
ε
Θ Θ
dr d
ε
ξ
= −
2 2
b
dr d
ε
ξ
=
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 =
−

9. 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 + =

10. 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

11. 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:

12. Example Problem – Singular Perturbation
CA(R) = Co
R k C
R
r
RA = k1CA
BC’s:
2
dd ΘΘΘ Aξ p −ξ
; 2
01
12
2
0
dd
d d
d ξ
ξ ξ
ΘΘ
− − Θ =
Θ
Θ1 = Aξ p
e ξ
;
Θ1
′ = A pξ p−1
− ξ p
( )e−ξ
1
12
0
d
e
d
ξ
ξ
−Θ
− Θ + =Θ1
′′ = A p p−1( )ξ p−2
− 2pξ p−1
+ ξ p
[ ]e−ξ
;
1( )ξ 2
2 ξ 1
ξ[ ] ξ 1
1
2
e ξ
ξ −
Θ =
A p p−1( )ξ p−2
− 2pξ p−1
+ ξ p
[ ]= Aξ p
−1;
A p p −1( )ξ p−2
− 2pξ p−1
[ ]= −1;
1; A 1 2p =1; A =1 2.

13. Example2 – Slab with non-homogenous rxn sites
Rectangular porous material
Catalyst (Cc) NOT uniformly
distributed throughout
x=0
CAo
CC(x)=CCO [1-(x/L)2]
Impermeable
u ug u
x=LRA = -k CC(x) CA
A B
Need flux NAX (x=0) for Da>>1
A B
Need flux NAX (x=0) for Da>>1
Solve for CA (x)
2
C∂2
2
2
[1 ( ) ] 0A
AB CO A
C x
D kC C
x L
∂
− − =
∂
dC
C
CC(x)
(0) ; ( ) 0A
A AO
dC
C C L
dx
= =
0
CA(x)
0 L x
0

14. Example2 – Slab with non-homogenous rxn sites
x=0
CAo
CC(x)=CCO [1-(x/L)2]
Dimensionless (not necessarily scaled) prob.
C x
2
2
2
[1 ( ) ] 0A
AB CO A
C x
D kC C
x L
∂
− − =
∂
Impermeable
x=L
A B
0
;A
A
C x
C L
ηΘ = =
2 2
2 2 2
A AOA
A
D Cd C d
D
d L d
θ
=
C
CC(x)
C ( )
2 2 2A
dx L dη
2 2
[1 ( ) ] (1 )CO A CO AO
x
kC C kC C
L
η θ− = −
0 L x
0
CA(x)22
2
2
(1 ) 0; CO
A
kC Ld
Da Da
d D
θ
η θ
η
− − = =
0 L
Θ∼1 and 1-η2~1; so d2θ/dη2 ~ Da
Si Δθ 1 d li i θ t k l i Δ
For Da>>1, the DFQ is not properly scaled
Rescale ηSince Δθ ∼1, decline in θ takes place over a region Δη
<<1; i.e, there is a concentration boundary layer
Rescale η

15. Example2 – Slab with non-homogenous rxn sites
x=0
CAo
CC(x)=CCO [1-(x/L)2]
A B
1
Let = 1 and X= a
Daε η ε−
a
Xη ε −
=
Impermeable
x=L
A B
ad d
d dX
θ θ
ε
η
=
2
2
2
(1 ) 0
d
Da
d
θ
η θ
η
− − =Substitute into →
2 2
2
2 2
ad d
d dX
θ θ
ε
η
=
2
2 1 2 2
2
(1 ) 0a ad
X
dX
θ
ε ε ε θ− −
− − =
Could we have
used the X2 term
instead 1 as the
dominant?
Assume
dominant
Get a=-1/2
2
2
(1 ) 0
d
X
θ
ε θ− − =
BCs:
( )θ2
(1 ) 0X
dX
ε θ = (0) 1θ =
( ) 0θ ∞ =

16. Example2 – Slab with non-homogenous rxn sites
x=0
CAo
CC(x)=CCO [1-(x/L)2]
A B
2
2
2
(1 ) 0
d
X
dX
θ
ε θ− − =
(0) 1θ =
( ) 0θ ∞ =
Impermeable
x=L
A B
2
0 1 ( )Oθ θ εθ ε= + +
2 2
d dθ θ 2 2 20 1
0 12 2
( ) (1 )( ( ) 0
d d
O X O
dX dX
θ θ
ε ε ε θ εθ ε+ + − − + + =
O(1) problem:
2
0
0 0
d θ
θ− = 0 0(0) 1; ( ) 0θ θ= ∞ =02
0
dX
θ 0 0(0) 1; ( ) 0θ θ ∞
0 ( ) X
X eθ −
=