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Problem 1
We start with the liquid phase PDE (equation 5). Taking the Laplace transform and using the
initial condition (6) leads to
c 0 D
s
dy
d c L L
2
L
2 − = (14)
Note that, even though cL (and, consequently, cL ) is also a function of x, since there are no
derivatives with respect to x in this equation, we are considering x to be a fixed parameter and
thus have converted the particle derivative into a total derivative. The solution of this ODE is
L L
L D
s Bcosh y D
s c Asinh y (15)
Laplace transformation of the boundary condition (7) yields
dy
dc
L
L = , y=0 (16)
Taking the derivative of equation (15) and applying this condition implies A=0, and equation
(15) simplifies to
L
L D
s c Bcosh y (17)
Now we take the Laplace transform of equation (8),
k (c c ) dy
dc D (^) g s
L L =^ − ,^ y=h^ (18)
and equation (9),
cs = KcLy= h (19)
Substituting this equation into equation (18) and then substituting equation (17) in the resulting
equation leads to, after manipulations,
g L L
L
L D
s sinh h D
s
k
D
D
s Kcosh h
c B (20)
Substituting this equation into equation (17), evaluating the resulting equation at y=h and then
using equation (19) leads to
g L L
L
L
L s
D
s sinh h D
s
k
D
D
s Kcosh h
c D
s Kcosh h
c (21)
which allows us to express the mass transfer term in equation (1) as
k (^) g a(c− cs)=F(s) c (22)
where
g L L
L
g
D
s tanh h D
s
Kk
D
F( s) k a 1 (23)
Next, we take the Laplace transform of equation (1) and use equation (22) to get
F c dx
d c D dx
d c sc v 2
2
This equation can be rearranged as follows
c 0 D
(F s )
dx
d c
D
v
dx
d c
2
2
The solution is
c = Geλ^ +x^ +Eeλ−^ x (26)
where the roots of the characteristic equation are
D
(F s) 4 D
v
2 D
v
2
+
λ± = ± (27)
Note that λ+>0 and λ-<0 since F>0 (equation 23). According to equation (4), c =0 as x→∞. This
means that G=0 in equation (26). Also, from equation (3), we can see that c = ciat x=0, which
implies that E = ci. Letting λ-= λ, equation (26) becomes
x c cie = λ (28)
where
λ
ds
dF 1
D
(F s) 4 D
v
D
ds
d
2
Substituting into equation (37) and simplifying the limit leads to
μ = + → ds
dF 1 lim v
x
s 0
' 1 (39)
Now, from equation (23), we find
L L
2
g L L
L
L
D
s tanh h D
s
ds
d
D
s tanh h D
s
Kk
D
K 1
aD
ds
dF (40)
so that
→ s→ (^0) L L
L
s 0 D
s tanh h D
s
ds
d lim K
aD
ds
dF lim (41)
The derivative is better handled by performing the change of variables
D L
s ξ =h ⇒ = ξdξ h
2 D
ds 2
L (42)
Substituting into equation (41) we get, after manipulations,
ξ
ξ
ξ ξ ξ = ξ ξ
s → 0 ξ→ 0 ξ→ 0 cosh^2
tanh 1 lim 2 K
ah tanh d
1 d lim 2 K
ah
ds
dF lim (43)
Applying L’Hopital’s rule to the first limit, we get
cosh
lim
d
d
d
dtanh
lim
tanh lim 0 0 0 2
ξ
ξ
ξ
ξ
ξ
ξ
ξ
ξ → ξ→ ξ →
Hence
K
ah
ds
dF lim s 0
→
From equation (39) we get
μ = + K
ah 1 v
x ' 1 (46)
Note that ah/K is the retardation factor.
To find the second moment, we use
2 ' 1
' μ 2 =μ 2 −μ (47)
where
s 0
2
2
0 0
'^2 2 ds
d c
m
m
m
=
μ = = (48)
Using equation (31) we get, after manipulations,
2
2
s 0
2
s 0
' 2 2 ds
d xlim ds
d x lim
λ +
λ μ = → →
The first limit can be obtained from the development after equation (38). The second limit
involves extensive algebraic manipulations. Details are omitted here. After substituting the final
result into equation (47), we obtain
μ = + L g
(^22)
3 2 Kk
3 D
h
Kv
2 ah
K
ah 1 v
2 D
x (50)
Note that dispersion in the gas (D), diffusion in the liquid (DL) and interfacial mass transfer (kg),
all contribute to the spread of the pulse.
(c) Due to its nonlinear nature, this equation cannot be solved for yˆ (^) i for a generic n. To set up
an iterative procedure, we will place the reaction term on the right-hand side of this equation:
i 1 i
i 1 i
n i
2
i yˆ 2 x
h 1 2
yˆ 2 x
h 1 2
h yˆ yˆ^ + −
α = − (11)
This equation will be complemented with the discretized form of the boundary conditions (4)
and (5):
yˆ^0 = yˆ 1 (12)
yˆ^ N + 1 = 1 (13)
Direct application of equation (11) starting with initial values (here we will use yˆ^ i = 0 ,
i=1,2…N as first estimates) is the Gauss-Seidel method. Including relaxation in the iterative
procedure leads to
α = −ω +ω− −
−
− − (k) i 1 i
(k 1 ) i 1 i
(k 1 )n i
2 (k 1 ) i
(k) i
yˆ 2 x
h 1 2
yˆ 2 x
h 1 2
h yˆ yˆ^ ( 1 )yˆ (14)
where k is the iteration numbers and the values inside the square brackets are evaluated at the
current calculation stage, which proceeds by incrementing i. This scheme (Gauss-Seidel with
relaxation, GSR) was programmed in Visual Basic, and iterations were stopped using the
criterion
(k 1 ) 5 i
(k) i yˆ^ yˆ 10 −
− − ω < , for all i=1,2…N (15)
(d) The results obtained after convergence are shown in Figure 1 for various step sizes. We can
see that h=0.01 seems to give a good solution on the scale of the plot. Figures 2 and 3 show the
number of iterations required as a function of ω for two different step sizes. Note how the
iterative process becomes less efficient as the step size is decreased.
h=0.
h=0.
h=0.
h=0.
y
x
Figure 1. Effect of step size on dimensionless concentration profiles.
0
100
200
300
400
500
600
700
0 0.5 1 1.5 2 2. ω
M
Figure 2. Total number of iterations (M) vs. relaxation parameter for h=0.05. No convergence
was obtained as ω→2.
1
2 1 2 hx
h
b = − + (21)
(from equation 12), and
N
2
n N (^) N 2 hx
h
d = αyˆ − + (22)
(from equation 13).
The system of equations is solved by Thomas' algorithm in each iteration, and then relaxation is
applied as follows
(c ) i
(k 1 ) i
(k ) i yˆ^ = ( 1 −ω)yˆ +ωyˆ
− (23)
This scheme yields the same solution, but at a much lower computational effort (see Figure 4).
Also, the scheme is much less sensitive, in terms of number of iterations, to the step size.
h=0.
h=0.
ω
M
Figure 4. Total number of iterations (M) vs. relaxation parameter for the tri-diagonal scheme.
No convergence was obtained as ω→0.75.
