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UNIVERSITY OF ARIZONA
DEPARTMENT OF CHEMICAL AND ENVIRONMENTAL ENGINEERING
CHEE 502 - ADVANCED CHEMICAL ENGINEERING ANALYSIS
FALL 2008
Test 2
Due: 8 am, 13 November
Problem 1 (50%)
A solid spherical particle dissolves into a large volume of liquid. The dissolving compound
decomposes in the liquid following a first order chemical reaction. The diffusion equation
governs the concentration of the dissolving compound in the liquid phase, c,
kc r
c r r r
D
t
c 2 2
, r>R
where D is its diffusivity of the dissolving compound in the liquid, k is the first order reaction
rate constant and R is the radius of the particle (assumed constant).
Initial and boundary conditions for this problem are
c=0, t=
c=c 0 , r=R
c=0, r→∞
where c 0 is the solubility of the compound in the liquid.
Find c(r,t) using Laplace transform. You may leave the solution expressed in terms of an
unresolved integral.
Solution
The Laplace transform of the PDE and boundary conditions leads to
k c dr
dc r dr
d
r
D
s c
2 2
s
c c
0 = , r=R (2)
c = 0 , r→∞ (3)
Rearrange the ODE to get
c 0 D
(s k)
dr
dc r dr
d
r
2
Let
r
u c = (5)
Transformation of equation (4) with this change of variables leads to, after manipulations,
u 0 D
(s k)
dr
d u 2
2
with boundary conditions
s
Rc u
0 = , r=R (7)
u = 0 , r→∞ (8)
The solution of ODE (6) is
D
s k Bexp r D
s k u Aexp r (9)
Boundary condition (8) implies B=0, and from (7) we find
D
s k exp R s
Rc A
0 (10)
Problem 2 (50%)
Consider the chromatography model proposed in Chapter 4. The model has two partial
differential equations:
a k (c Kcˆ ) z
c D z
c v t
c v m s 2
2 − − ∂
ε
k (c Kcˆ ) t
cˆ^ m s
s = − ∂
With initial and boundary conditions
c=0, t=
cˆs = 0 , t=
c=ci, z=
z
c
∂
, z=L
For the case of a constant inlet concentration, ci, propose a numerical solution for this problem
by finite differences. Your solution must contain:
A description of the finite difference discretization of the solution domain.
The finite difference form of the differential equations and initial and boundary conditions.
A detailed explanation on how the calculations would proceed in your numerical method.
Do not perform any calculations.
Solution
Here we will apply the backwards difference method to solve the problem. The discretization
grid has a spacing h in time, ti=ih, i=0,1,2…, and a spacing g in z, such that N+1 intervals are
created
N 1
L
g
with coordinates for the discretization points given by
z (^) j = jg, j=0,1…N+1 (2)
We approximate the time derivatives as follows,
h
c c (t ,z ) t
c (^) i,j i 1 ,j i j
h
cˆ cˆ (t ,z ) t
cˆ^ si,j si 1 ,j i j
s −
Spatial derivatives of the fluid concentration are discretized using central differences:
2 g
c c (t ,z ) z
c (^) i,j 1 i,j 1 i j
2
i,j 1 i,j i,j 1 i j 2
2
g
c 2 c c (t ,z ) z
c + − + − ≈ ∂
Substituting these finite difference equations into the PDEs leads to
a k (c Kcˆ ) g
c 2 c c D 2 g
c c v h
c c v m i,j si,j 2
i, j i 1 ,j i,j 1 i,j 1 i,j 1 i,j i,j 1 − −
ε
− + − + − (7)
k (c Kcˆ ) h
cˆ cˆ
m i,j si, j
si (^) ,j si 1 ,j = −
− (8)
Solving equation (8) for cˆ^ si,jyields
m
si 1 ,j m i, j si (^) ,j 1 hKk
cˆ k hc cˆ^
− (9)
