Mathematical models for a chemical reactor

1. Mathematical Models for a Chemical Reactor
Ignatios Vakalis
Department of Computer Science
California Polytechnic and State University
San Luis Obispo, CA 93406
(email: ivakalis@calpoly.edu)
(Dated: Sept. 18, 2010)
Abstract
The module presents the development and solution methodology of mathematical models
for the concentration of a chemical in a reactor. The steady state case is first examined by
providing the development and the solution (both analytical and numerical) process of the
underlying models. The modeling process is extended to include time dependency by
presenting the formation of the mathematical model and its numeric solution process.
Keywords: mass balance, steady state, initial/boundary- value problem, parabolic
PDEs, implicit methods, finite difference, finite element
Work supported by the National Science Foundation under CCLI DUE 061825

2. Module Overview:
The main focus of the module is on modeling the concentration of a chemical in a
reactor. The module consists of two major parts: i) steady state; ii) time dependency,
presenting the development of the relevant models as well as the interplay of their
analytical and numerical solutions.
The first part begins with the development of the mathematical model that describes the
concentration of the chemical under steady-steady assumptions. It also presents both the
analytic and numerical solution of the modeling differential equations.
The second part of the module presents a detailed development and solution methodology
of the underlying models that capture “time dependency”. The development of the partial
differential equation (PDE) is presented along with the underlying assumptions. Since an
analytical solution is not possible, the module gives a detailed presentation and
comparison of numerical solutions.
Math and Science Level:
The module is suitable for a variety of courses, such as: Mathematical Modeling, Intro
to Chemical Engineering, Computational Modeling, and Numerical Analysis.
The module should be presented to students at least at the junior level in a science,
computational science, or mathematics undergraduate curriculum.
Background the Prerequisites:
The minimal mathematical background consists of a Calculus II course. In addition, an
understanding of differential equations is highly desirable. An intro course in Chemistry
will provide the necessary science background. An undergraduate course in Numerical
Analysis is not necessary, but it could add valuable and insightful experience to the
student reader.
The module requires familiarity with (and access to) a Computer Algebra System (i.e,
Maple, Mathematica) as well as Matlab

3. Modeling Concentration of a Chemical in a Reactor
The law of conservation of mass, states that the mass of a closed system (in the sense of a
completely isolated system) will remain constant over time. Thus the mass of an isolated
system cannot be changed as a result of processes acting inside the system. In other
words, mass cannot be created/destroyed, although it may be rearranged in space, and
changed into different types of particles.
In its generality, the above can be expressed as:
(rate of accumulation of a chemical in a system) =
(rate of flow into the system)
- (rate of flow out of the system)
+ (rate of generation of a chemical in the system)
To simplify the geometry in our modeling process, let’s assume a cylindrical reactor with
a single entry and exit point. We also assume that the chemical is well-mixed vertically
and laterally and also that the chemical is subject to first order decay. Chemical
Engineering literature [Cussler E., 1977] provides us with the following general model:
Vcz
x
xc
xx
xc
DS
x
xc
DSx
t
xc
xcFxFc
t
c
V γ−⎥⎦
⎤
⎢⎣
⎡
Δ
∂
∂
∂
∂
+
∂
∂
+
∂
∂
−⎥⎦
⎤
⎢⎣
⎡
Δ
∂
∂
+−=
Δ
Δ )()()()(
)()( (1)
where, V = volume, F= flow in, c = concentration, S = reactor’s cross sectional area, D
= dispersion coefficient, γ = first order decay coefficient. (See conceptual question #1)
Note that the concentration c is a function of space and time, thus c = c(x, t)
Allowing and , the above equation can be transformed into:0→Δx 0→Δt
),(
),(),(),(
2
2
txc
x
txc
W
x
txc
D
t
txc
γ−
∂
∂
−
∂
∂
=
∂
∂
(2)
(See problem #1)
where W = F / S
The above Eq. (2) is characterized as a Parabolic Partial Differential Equation [Allen,
1998].

4. Steady State:
Model:
To derive the mathematical equation that models the steady state of the concentration, we
need to assume that the concentration is independent of time, and thus only depends on
spatial variation.
Thus, we set: 0
),(
=
∂
∂
t
txc
which implies that: )(xcc =
Therefore, Eq. (2) is reduced to the following second order Ordinary Differential
Equation (ODE):
02
2
=−− c
dx
dx
W
dx
cd
D γ (3)
Solution Methodology:
In this section we will present two types of solutions. Since Eq. (3) is a 2nd
order ODE,
an analytic solution is feasible and we will pursue to derive it. The module also presents
a numerical solution. Since a numerical solution can only provide an approximation to
the exact solution, its accuracy can be tested against the exact solution obtained by the
analytical method.
Our first step is to explicitly define the conditions for the problem at hand. Let’s assume
that the cylindrically shaped segment of the reactor has length L , at t = 0 no chemical
exists, and that the chemical is injected into the reactor’s inflow with constant level .inc
Thus,
dx
dc
DSFcFcin
)0(
)0( −= where, F= flow in, c = concentration, S = reactor’s
cross sectional area, D = dispersion coefficient.
To obtain a solution of the 2nd
order ODE (Eq. 3), an extra condition is imposed that
reflects an additional assumption. Specifically, 0
),(
=
=
dt
tLxdc
. In other words, the
dispersion in the reactor does not affect the exact rate.

5. Gathering the above observations, and observing that W = F / S , we can now list the
following 2nd
order ODE boundary value problem as:
02
2
=−− c
dx
dc
W
dx
cd
D γ for 0 < x < L
dx
dc
W
D
ccin
)0(
)0( −= (4)
0
)(
=
dt
Ldc
Analytic Solution:
From your course in ODEs, you recall that the solution takes the form:
rx
exc =)(
Substituting in Eq. (4), we obtain the characteristic equation: .02
=−− γWrDr
The discriminant of the above is: which is always positive. Thus the solution
of Eq. (4) can be represented as:
γDW 42
+
xrxr
eKeKxc 21
21)( += (5)
where the constants and can be obtained using the boundary conditions.1K 2K
Using a Computer Algebra System (i.e, Maple), we can obtain lengthy expressions for
the coefficients and (See problem # 2)1K 2K
Note that the mass fluxes for the steady-state solution can be computed using Fick’s fist
law. Thus, using Eq. (5), we obtain:
)(
)( 21
2211
xrxr
erKerKD
dx
xdc
DJ +−=−= (6)

6. Numerical Solution:
Finite Difference Method:
This section presents an overview of the Finite Difference Method for numerically
solving the ODE expressed in Eq. (4).
Using centered finite differences, [Cheney, Kincaid 2003], for the first and second
derivatives, we can obtain from Eq. (4) the following expression:
( )
0
2
2 11
2
11
=−
Δ
−
−
Δ
+− −+−+
j
jjjjj
c
x
cc
W
x
ccc
D γ (7)
Our goal is to compute the discrete values for all points of the discrete space. To that
end we can re-arrange the terms of Eq. (7) in order to form a system of linear algebraic
equations. Rearranging the terms, we obtain for:
jc
nj ≤≤0
0
2
12
2
1
11 =⎟
⎠
⎞
⎜
⎝
⎛
Δ
+−−⎟
⎠
⎞
⎜
⎝
⎛
Δ
+
Δ
+⎟
⎠
⎞
⎜
⎝
⎛
Δ
+− +− jjj c
xW
D
c
xW
D
W
x
c
xW
D γ
(8)
The above represents a liner system of (n+1) equations with (n+1) unknowns.
In order to solve the above system, we will use the discrete version of the boundary
conditions to calculate the values for (generated when j=0) as well as
(generated when j = n).
1−c 1+jc
Using the boundary condition (Eq. (4)) and discretizing the derivative, we obtain:
x
cc
W
D
ccin
Δ
−
−= −
2
11
0 which can be solved to obtain:
011
22
c
D
xW
c
D
xW
cc in
Δ
−
Δ
+=− (9)
Substituting the above expression into the first equations of the linear system (for j=0)
we obtain:
inc c
D
xW
c
xW
D
c
xW
D
D
xW
W
x
⎟
⎠
⎞
⎜
⎝
⎛
+
Δ
=⎟
⎠
⎞
⎜
⎝
⎛
Δ
−⎟
⎠
⎞
⎜
⎝
⎛
Δ
+
Δ
+
Δ
+ 2
22
2 1
γ
(10)

7. Note that at the end of the segment (x = L), the following boundary condition my hold:
dx
dc
W
D
cc n
nn −= This simplifies to: 0=
dx
dcn
, and its discrete form: 0
2
11
=
Δ
− −+
x
cc nn
implies that: 11 −+ = nn cc
Substituting the above expression into the last equation of the linear system ( for j = n)
we obtain:
0
22
1 =⎟
⎠
⎞
⎜
⎝
⎛ Δ
+
Δ
+⎟
⎠
⎞
⎜
⎝
⎛
Δ
− − nn c
W
x
xW
D
c
xW
D γ
(11)
Note that Eqs. (8, 10, 11) form a tridiagonal system, which could easily be solved either
by hand, or by using a Computer Algebra System. (See problems #3, #4, #5).
Upon developing the graph of c(x) as function of the spatial distance x, we observe (as
expected), that the concentration decreases as the chemical flows though the cylindrical
reactor.
Appendix I presents an outline of the Thomas Algorithm for solving efficiently a
tridiagonal system
Note:
The Boundary Value Problem (BVP) expressed by Eq. (4) can also be solved using the
Numerical Solvers within Matlab. Appendix II, provides a sample Matlab code for the
solution of a BVP. Problem #6 asks the user to solve the BVP using the parameter
values specified in Problem #3.
Finite Element Method
Another very popular method for numerically solving PDEs is the Finite Element
Method (FEM). [ Logan, 1997].
The Eq. (4) can be re-written as follows, using the substitutions:
d
W
p −= and
D
q
γ
−= :
02
2
=++ qc
dx
dx
p
dx
cd
(12)
and the boundary conditions (see Eq. (4)) can be re-written as:
0
)(
=
dx
Ldc
and ( inccp
dx
dc
−−= 0
)0(
) (13)

8. In project #1, the reader is asked to develop the complete solution methodology using the
finite element method.
For any specific set of the problem parameters, graphing the numerical values of the
concentration c(x) obtained by the Finite Difference and the Finite Element method
respectively, depicts extremely similar graphs.
Time Dependency:
In the previous sections we have examined the modeling process of the concentration of a
chemical in a reactor under steady-state. We have observed that the governing parabolic
PDE (Eq. (2)) reduces to an ODE (Eq. (3)).
In this part of module, we extend the modeling process by considering time-dependency.
As we will see, an analytic solution is no longer possible. Thus, we present two numerical
methods.
Model:
We know that the law of mass concentration states that:
(rate of accumulation of a chemical in a system) =
(rate of flow into the system)
- (rate of flow out of the system)
+ (rate of generation of a chemical in the system)
which can be modeled by:
),(
),(),(),(
2
2
txc
x
txc
W
x
txc
D
t
txc
γ−
∂
∂
−
∂
∂
=
∂
∂
(14)
where F= flow in, c = concentration, S = reactor’s cross sectional area, D = dispersion
coefficient, γ = first order decay coefficient, and W = F / S.
The above mathematical model (Eq. 14), was formulated under the following
assumptions:
a) The reactor tank is well mixed in all directions
b) Dispersion does not affect the exit rate
c) Chemical is subject to first order-decay
d) Prior to t=0 no concentration of the chemical is present in the tank
e) Starting at t=0 the chemical is injected at a constant rate.

9. Using the above assumptions, we can formulate boundary conditions for the governing
model given by Eq. (14) as follows:
Assumption (e) can be modeled as:
),0(),0( '
tc
W
D
ctc in += (15)
Assumption (b) can be modeled as:
for t > 0 (16)0),('
=tLc
Assumption (d) can be modeled as:
for 0 < x < L (17)0)0,( =xc
Upon some investigation in the theory and solution methods of Differential Equations,
it is evident that NO analytical solution exists for the PDE given by Eq. (14) in
conjunction with the initial and boundary conditions described by Eqs. (15 – 17). Thus,
the following sections examine two numerical techniques.
Solution Methodology:
Implicit Method:
In this part we present an implicit method for solving PDE depicted by Eq. (14).
(See Glossary for the difference between explicit and implicit methods)
We use a forward finite difference to approximate the time derivative, and central
differences for the space derivatives, namely:
t
cc
t
c
l
j
l
j
Δ
−
=
∂
∂
+1
which has an error of order )( tO Δ (17a)
x
cc
x
c
l
j
l
j
Δ
−
=
∂
∂ −
+
+
2
1
1
1
and
( )2
1
1
11
1
2
2
2
x
ccc
x
c
l
j
l
j
l
j
Δ
+−
=
∂
∂
+
−
++
+
(17b)

10. We substitute the above into Eq. (14), and to the initial/boundary conditions in order to
obtain a system of linear equations, which then must be solved. Specifically, Eq (14)
becomes:
( )
1
1
1
1
1
2
1
1
11
1
1
2
2 +
+
−
+
+
+
−
++
+
+
−
Δ
−
−
Δ
+−
=
Δ
− l
j
l
j
l
j
l
j
l
j
l
j
l
j
l
j
c
x
cc
W
x
ccc
D
t
cc
γ (18)
Rearranging the terms of the above, we obtain:
( ) ( ) ( )
l
j
l
j
l
j
l
j c
t
c
x
W
x
D
c
x
D
t
c
x
W
x
D
Δ
−=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
−
Δ
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
++
Δ
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
+
+
++
−
1
2
2
1
2
1
12
1
2
1
12
γ (19)
Observe that the initial condition: c(x,0)= 0 translates in the discrete space as:
for j = 0,1, 2, …, n and l=0.0=l
jc
A complication arises, when we try to explicitly write out the set of equations described
by Eq. (19). Specifically, for j = 0 we generate the term: 01
1 =+
−
l
c
This term can be evaluated using one of the boundary conditions. Specifically, when we
discretize Eq. (15) , we obtain:
⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
Δ
−
+=
+
−
+
+
W
D
x
cc
cc
ll
j
in
l
2
1
1
1
1
0
Solving the above and substituting in Eq. (19) as applied for j=0, we obtain the first eq.
of our system that describes the concentration of the nodal points (discrete space):
( ) ( )
l
in
ll
c
t
c
x
W
D
W
c
D
W
x
W
x
D
t
c
x
D
0
2
1
0
2
2
1
12
1
2
2
2
12
Δ
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ
+−=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
Δ
+
Δ
++
Δ
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
++
γ (20)
Now, the boundary condition for t > 0 expressed by Eq. (16), can be
discretized as follows:
0),('
=tLc
0
2
1
1
1
1
=
Δ
− +
−
+
+
x
cc l
n
l
n
which produces: (21)
1
1
1
1
+
−
+
+ = l
n
l
n cc

11. Substituting Eq. (21), into Eq. (19) as applied for j=n, we obtain the last eq. of our
system that describes the concentration of the nodal points (discrete space):
( ) ( )
l
n
l
n
l
n c
t
c
x
D
t
c
x
D
Δ
−=⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
++
Δ
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
++
−
1
2
12 1
2
1
12
γ (22)
We are almost at the end…..
Observe that Eqs. (19, 20, 22) form a system of liner equations with unknowns the
discrete values of the concentration at the nodal points. Furthermore, Eqs. (19,20,22)
form a tridiagonal system, which could easily be solved either by hand, or by using a
Computer Algebra System. (See problem #8).
Upon developing the graph of c(x) as function of the spatial distance x, we observe (as
expected), that the concentration decreases as the chemical flows though the cylindrical
reactor. (Appendix I presents an outline of the Thomas Algorithm for solving efficiently
a tridiagonal system)
Note:
The modeling problem at hand governed by Eqs. (14-17), can also be solved using the
Numerical Solvers within Matlab. Appendix IV provides a sample Matlab code for the
solution of a parabolic PDE. Problem #7 asks the user to modify the code presented in
Appendix IV, in order to solve Eq. (14) using an implicit method.
Crank - Nicholson Method
From Eqs. (17a, 17b), we observe that the time difference approximation is first –order
accurate, and the spatial difference approximation is second-order accurate.
Can we do better?
Indeed yes!. The Crank –Nicholson method provides an improved implicit scheme that
is second –order accurate in both space and time (i.e, in both independent variables of
the 2nd
order PDE).
For this scheme, the first time derivative can be approximated at point (midpoint)
by the expression:
2/1+l
t
t
cc
t
c
l
j
l
j
Δ
−
=
∂
∂
+1

12. Compare the above with Eq. (17a) that depicts a forward finite divided difference for
point .
1+l
t
To estimate the first and second derivatives in space at midpoint we average the
difference approximations at points and . Thus:
l
t 1+l
t
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
Δ
−
+
Δ
−
=
∂
∂
+
−
+
+−+
x
cc
x
cc
x
c
l
j
l
j
l
j
l
j
222
1
1
1
1
111
and
( ) ( ) ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
Δ
+−
+
Δ
+−
=
∂
∂
+
−
++
+−+
2
1
1
11
1
2
11
2
2
22
2
1
x
ccc
x
ccc
x
c
l
j
l
j
l
j
l
j
l
j
l
j
Substituting into Eq. (14) and then rearranging the terms, we can obtain:
( ) ( ) ( )
1
2
1
12
1
12
1
24242
++
+
+
− ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
+−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
−
Δ
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
l
j
l
j
l
j c
tx
D
c
x
W
x
D
c
x
W
x
D γ
=
=
( ) ( ) ( )
l
j
l
j
l
j c
tx
D
c
x
W
x
D
c
x
W
x
D
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
−−−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
−
Δ
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
− +−
1
24242
21212
γ
(23)
Again, observe that the initial condition: c(x,0)= 0 translates in the discrete space as:
for j = 0,1, 2, …, n and l=0.0=l
jc
A complication arises, when we try to explicitly write out the set of equations described
by Eq. (23). Specifically, for j = 0 we generate the term: 01
1 =+
−
l
c
This term can be evaluated using one of the boundary conditions. Specifically, when we
discretize Eq. (15) , we obtain:

13. ⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
Δ
−
+=
+
−
+
+
W
D
x
cc
cc
ll
j
in
l
2
1
1
1
1
0
Solving the above, and substituting in Eq. (23) as applied for j=0, we obtain the first
equation of our system that describes the concentration of the nodal points (discrete
space):
( ) ( )
1
12
1
0
2
2
2
1
2
++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
Δ
+
Δ
+
Δ
+− ll
c
x
D
c
D
W
x
W
x
D
t
γ
=
=
( ) ( ) in
ll
c
x
DW
D
W
c
x
D
c
D
W
x
W
x
D
t
⎟
⎠
⎞
⎜
⎝
⎛
Δ
+⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
Δ
−
Δ
−
Δ
−−−
2
2
2
1
2
12
1
0
2
2
γ
(24)
Now, the boundary condition for t > 0 expressed by Eq. (16), can be
discretized as follows:
0),('
=tLc
0
2
1
1
1
1
=
Δ
− +
−
+
+
x
cc l
n
l
n
which produces: (25)
1
1
1
1
+
−
+
+ = l
n
l
n cc
Substituting Eq. (25), into Eq. (23) as applied for j=n, we obtain the last equation of our
system that describes the concentration of the nodal points (discrete space):
( ) ( )
1
2
1
12
1
22
++
− ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
+−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
l
n
l
n c
tx
D
c
x
D γ
=
( ) ( )
l
n
l
n c
tx
D
c
x
D
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
+
Δ
−−−⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
Δ
− −
1
22
212
γ
(26)
We are almost at the end…..
Observe that Eqs. (23, 24, 26) form a system of liner equations with unknowns the
discrete values of the concentration at the nodal points. Furthermore, these equations
form a tridiagonal system, which could easily be solved either by hand, or by using a
Computer Algebra System. (See problem #9).
Upon developing the graph of c(x) as function of the spatial distance x, we observe (as
expected), that the concentration decreases as the chemical flows though the cylindrical
reactor.

14. Conceptual Questions:
1) List and explain the necessary assumptions for the development of the
concentration of a chemical for the steady state case.
2) Provide a detailed explanation of each term in the formation on Eq. (1). How
Fick's first law of flux diffusion is embedded in Eq. (1)? (see Glossary)
3) Investigate and provide reasons for selecting centered finite differences in solving
Eq. (4) using the finite difference method. Are there other alternatives for
discretizing the derivatives? How do the various methods rank in terms of
accuracy?
4) List and explain the necessary assumptions for the development of the
concentration of a chemical for the time-dependency case.
Problems:
1) Derive Eq. (2) from Eq. (1). Show all steps.
2) Use Maple to obtain the analytic form of the expression for the 1K and 2K in
the solution Eq. (5). (Hint: Use the boundary conditions – Eq. (4), and solve the
resulting algebraic system for 1K and 2K )
3) Solve the tridiagonal system described by Eqs. 8, 10,11 to obtain a numerical
solution of the values: 1,1,2,3.0,10 ===Δ== DWxL γ . Then obtain the
values of the concentration of the reactor at points x = 0, 2, 4, 6, 8, 10. Plot the
graph of c(x) as a function of the spatial distance x

15. 4) Compare the values of the concentration obtained in problem #3 with those that
can be derived from the analytical /exact solution described by Eq. (5). How
accurate is the numerical solution?
5) Solve again the tridiagonal system described by Eqs. 8, 10,11 to obtain a
numerical solution of the values: 4,1,2,3.0,10 ===Δ== DWxL γ . Plot the
graph of c(x) as a function of the spatial distance x. Compare the graph with
that obtained in problem #3. How does the increase of the value of D (depicting
the turbulence mixing) affects the shape of the graph? Why?
6) Solve the BVP described by Eq. (4) using Matlab. Use the parameter values
given in problem #3. (Hint: See Appendix II)
7) Modify the Matlab code given in Appendix IV to numerically solve the PDE with
the appropriate initial and boundary conditions given by Eqs. (14-17)
8) Solve the tridiagonal system described by Eqs. 19, 20, 22 to generate a
numerical solution of the values: 1,1,2,3.0,10 ===Δ== DWxL γ . Then
obtain the values of the concentration of the reactor at points: x = 0, 2, 4, 6, 8, 10.
Plot the graph of c(x) as a function of the spatial distance x.
9) Solve the tridiagonal system described by Eqs. 23, 24, 26 to generate a
numerical solution of the values: 1,1,2,3.0,10 ===Δ== DWxL γ . Then
obtain the values of the concentration of the reactor at points: x = 0, 2, 4, 6, 8, 10.
Plot the graph of c(x) as a function of the spatial distance x.

16. Projects:
1) Upon consulting a Numerical Analysis book, develop a complete formulation of
the Finite Element Method (FEM) in solving the Parabolic PDE in Eqs. 12. 13.
Graph the results in obtaining the values of c(x) for the same parameter values
given in problem #3. Compare the results.
2) Consider a first order chemical reaction in a tabular flow rector. On the
assumptions of steady state, laminar flow [Rogers,1992], and negligible axial
diffusion, the material balance equation is given by:
0
1
1 2
2
2
2
0 =−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡
−− kc
r
c
rr
c
D
z
c
R
r
ν
where, v0 = velocity of central stream line, R = radius tube, k = reaction constant,
D = diffusion constant, c = concentration, z = axial distance, r = radial distance
from center.
Using the following substitutions,
R
r
V
kR
D
c
c
C
kz
==== ,,, 2
00
β
ν
α
the original PDE can be transformed to:
( ) C
V
C
VV
CC
V −⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
+
∂
∂
=
∂
∂
−
1
1 2
2
2
β
α
a) Choosing a set of appropriate boundary conditions, what class of PDE is the
above equation? ( Hint: See Appendix III)
b) Set up the equation for numerical solution using the finite difference method.
c) Will the above step result in an explicit or implicit set of equations?
d) What are the stability conditions? Explain.

17. References
Allen M., et.al., Numerical Analysis for Applied Science, Wiley, New York, (1998).
Cheney W., Kincaid D., Numerical Mathematics and Computing, Brooks Cole (2003).
Coombes K., et. al., Differential Equations with Matlab, John Wiley (2000).
Cussler E., Diffusion: Mass Transfer in Fluid Systems, Cambridge University Press,
(1997)
Logan D., A first Course in the Finite Element Method, PWS Publishing Company
(1997)
Rogers, D.F., Laminar flow analysis, Cambridge U. Press (1992)

18. Appendix I :
Thomas Algorithm
A tridiagonal system can be represented as:
A* X = * =
nn
n
ab
c
ab
cab
ca
00
00
00
1
33
222
11
L
OOOM
O
MO
L
−
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
n
n
x
x
x
x
1
2
1
M
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
n
n
d
d
d
d
1
2
1
M
Applying an LU decomposition, the above can be written as:
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
nn lb
lb
lb
l
00
0
00
0
000
33
22
1
L
OOOM
O
MO
L
*
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
1000
0100
10
001
1
2
1
L
OOOO
O
MO
L
nμ
μ
μ
The solution can be found efficiently by a forward and backward substitution as:
,
1
1
1
l
b
w =
i
iii
i
l
wbd
w 1−−
= for i =1, 2,3, …,n
nn wx = and
i
iii
i
l
ww
x 1+−
=
μ
for i =1, 2,3, …,n-1

19. Appendix II :
Solving BVPs using Matlab
The following Matlab function twopoint implements the solution of a second order BVP
given by:
)()()(
)(
)(
)(
)( 2
2
xFxzxE
dx
xdz
xD
dx
xzd
xC =++ (a1)
where z = z(x)
The user must provide the values of the parameters C(x), D(x), E(x) and F(x) at each
nodal point of the descritized space. The boundary conditions must also be supplied.
function y = twopoint(x,C,D,E,F,flag1, flag2, p1,p2)
%
% x is a row vector of n+1 nodal points
% C, D,E, F are vectors describing C(x), D(x), E(x) and F(x)
% If y is specified at node 1 , flag1 must be set to 1.
% If y’ is specified at node 1 , flag1 must be set to 0.
% If y is specified at node n+1 , flag2 must be set to 1.
% If y’ is specified at node n+1 , flag2 must be set to 0
%
n = length(x) – 1;
h(2:n+1) = x(2:n+1) – x(1:n);
h(1) = h(2);
h(n+2) = h(n+1);
r(1:n+1) = h(2:n+2). / h(1:n+1);
s=1+r;
if flag1 ==1
y(1) =p1;
else
slope0=p1;
if flag2 ==1
y(n+1) =p2;
else
slopen=p2;
end
W= zeros(n+1, n+1);
If flag1 ==1
c0=3;
W(2,2)=E(2) – 2*C(2)/(h(2)^2*r(2));

20. W(2,3)=2*C(2)/(h(2)^2*r(2)*s(2) + D(2)/h(2)*s(2));
b(2) = F(2) – y(1)*(2*C(2)/(h(2)^2*s(2)) – D(2)/(h(2)* s(2)));
else
c0=2;
W(1,1) = E(1) – 2*C(1) / (h(1)^2*r(1));
W(1,2) = 2*C(1)*(1+ 1/r(1))/(h(1)^2*s(1));
b(1) = F(1) + slope0*(2*C(1)/h(1) – D(1));
end;
if flag2 ==1
c1=n-1;
W(n,n)=E(n) – 2*C(n)/(h(n)^2*r(n));
W(n,n-1)=2*C(n)/(h(n)^2*r(n)*s(n) - D(n)/h(n)*s(n));
b(n) = F(n)– y(n+1)*(2*C(2)/(h(n)^2*s(n)) + D(n)/(h(n)* s(n)));
else
c1=n;
W(n+1,n+1) = E(n+1) – 2*C(+1) / (h(n+1)^2*r(n+1));
W(n+1,n) = 2*Cn+(1)*(1+ 1/r(n+1))/(h(n+1)^2*s(n+1));
b(n+1) = F(n+1) + slope0*(2*C(n+1)/h(n+1) + D(n+1));
end
for i= c0:c1
W(i,i)= E(i) – 2*C(i) / (h(i)^2*r(i));
W(i, i-1)= 2*C(i)/(h(i)^2**s(i)) - D(i)/h(i)*s(i));
W(i,i+1)= 2*C(i)/(h(i)^2*r(i)*s(i)) + D(i)/h(i)*s(i));
b(i)= F(i);
end
z = W(flag1 +1:n+1- flag2, flag1 +1 : n+1 – flag2) / b(flag1+1:n+1 – flag2);
if flag1 == 1 & flag2==1, y = [y(1); z; y(n+1)]; end
if flag1 == 1 & flag2==0, y = [y(1); z]; end
if flag1 == 0 & flag2==1, y = [z; y(n+1)]; end
if flag1 == 0 & flag2==0, y = z; end
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

21. Appendix III :
Classification of 2nd
order PDEs
The general form of the 2nd
order PDEs with two independent variables x, t can be
written as:
0
),(
,
),(
,,,
),(
),(
),(
),(
),(
),( 2
22
2
2
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
+
∂
∂
+
∂∂
∂
+
∂
∂
t
txz
x
txz
ztxf
t
txz
txC
tx
txz
txB
x
txz
txA
For the cases examined in this module, the coefficients A, B, C are constants.
The PDEs are classified into three categories, based on the following criterion:
• If 0 then the PDE is42
<− ACB elliptic
• If 0 then the PDE is42
=− ACB parabolic
• If 0 then the PDE is42
>− ACB hyperbolic

22. Appendix IV :
Solving a general Parabolic PDE using Matlab
The following Matlab code provides a numerical solution to the general parabolic PDE:
22
2
),(),(
x
txu
K
t
txu
∂
∂
=
∂
∂
for 0 < x < L
function u =differ-equation(nx, hx, nt, ht, init, lowb, hib, K)
%
Clear all;
alpaha = K * ht/hx^2;
A = zeros( nx – 1, nx – 1);
u = zeros(nt+1, nx+1);
u(:, 1)= lowb * ones(nt+1, 1);
u(:, nx+1)= hib*ones(nt+1, 1);
u(1,:) = init;
A(1,1) = 1 + 2*alpha;
A(1,2) = - alpha;
for i = 2 : nx – 2
A(i, i) = 1 + 2*alpha;
A(i, i-1) = - alpha *
A( i,i+1) = - alpha;
end
A(nx - , nx – 2) = - alpha;
A(nx – 1, nx – 1) = 1 + 2* alpha;
b(1,1) = init(2) + init(1) * alpha;
for i=2: nx – 2
b(i,1)= init(i+1);
end
b(nx -1, 1) = init(nx) + init(nx+1) *alpha;
[L, U] = lu(A);
for j=2: nt+1
y = Lb;
x = UY;
u(j,2:nx) = x’;
b=x;
b(1,1)=b(1,1) + lowb*alpha;
b(nx-1,1)= b(nx-1,1) + hib*alpha;
end

23. Partial Solutions-Hints (For Instructor’s Use ONLY)
Problem #1
Recall that:
x
xcxxc
x
c
Δ
−Δ+
=
Δ
Δ )()(
,
t
tctxc
t
c
Δ
−Δ+
=
Δ
Δ )()(
Also,
t
tcttc
t
c
t Δ
−Δ+
=
∂
∂
→Δ
)()(
lim
0
,
x
xcxtc
x
c
x Δ
−Δ+
=
∂
∂
→Δ
)()(
lim
0
and
x
x
xc
x
xxc
x
c
x Δ
∂
∂
−
∂
Δ+∂
=
∂
∂
→Δ
)()(
lim
02
2
Project #1
Develop a complete formulation of the Finite Element Method (FEM) is solving the
Parabolic PDE in Eqs. 12. 13.
Solution:
Let’s consider a discetization . Assume a tank of length L. Then Nh = L.
Multiply Eq. (12), with the smooth function v(x) and integrate over the interval [0, l]:
0)( '
0
''
=++∫ dxqcpcc
L
ν (b1)
Integrating by parts, we produce:
0)0()0()()(
00
'
0
''''
=++−− ∫∫∫
LLL
dxqcdxpcdxccLLc ννννν (b2)

24. From analysis, we know that c(x) can be written in the form:
where,)()(
0
xaxc
N
j
jj∑=
= φ )()( xx jφν = and )( ii xcc = at node i
0 if ji ≠
with =)( ji xφ
1, if i =j
Substituting to Eq. (b2), we obtain:
0)0()0(
00
'
0
'
0
'
00
'
=++−− ∫∑∑ ∫∫∑∑ ====
dxajqdxapdxaa j
L
i
N
j
j
N
j
L
o
ijj
L
i
N
j
ji
N
j
jj φφφφφφφφ (b3)
Since, , we have that:)()(
0
xaxc j
N
j
jφ∑=
= )0()0()(
0
''
0 ∑=
==−−
N
j
jin cccp φ
Thus, Eq. (b3) can be solved to produce:
0)0()(
00
'
0
'
0
'
0
0 =++−− ∫∑∑ ∫∫∑ ===
dxaqdxapdxaccp j
L
i
N
j
jj
N
j
L
o
ijj
L
i
N
j
jiin φφφφφφφ (b4)
Note that: hxx kk /)( 1−−=ϕ if ),( 1 kk xxx −∈
= ifhxxk /)( 1 −+
),( 1+∈ kk xxx
Clearly, Eqs. (b4) form a system of linear equations that can be solved to obtain the
discrete values of the concentration on various nodal points.jc

25. Glossary
Deterministic Systems /Models:
A deterministic system is a system in which no randomness is involved in the
development of future states of the system that the model describes. Deterministic models
produce the same output for a given starting condition
Fick's first law:
It relates the diffusive flux to the concentration field, by stating that the flux goes from
regions of high concentration to regions of low concentration, with a magnitude that is
proportional to the concentration gradient (spatial derivative). In one spatial dimension it
states:
x
c
DJ
∂
∂
−= where c= concentration, D = diffusion coefficient
Finite Difference:
Finite-difference methods are numerical methods for approximating the solutions to
differential equations using finite difference equations to approximate derivatives.
Laminar Flow:
Laminar flow, sometimes known as streamline flow, occurs when a fluid flows in parallel
layers, with no disruption between the layers. It is the opposite of turbulent flow. In
nonscientific terms laminar flow is "smooth," while turbulent flow is "rough”.
Explicit methods
Explicit methods calculate the state of a system at a later time from the state of the
system at the current time. Mathematically, if Y(t) is the current system state and
Y(t + Δt) is the state at the later time (Δt is a small time step), then, for an explicit
method:
Implicit Methods
Mathematically, if Y(t) is the current system state and Y(t + Δt) is the state at the later
time (Δt is a small time step), then, an implicit method one solves an equation
to find Y(t + Δt).
It is clear that implicit methods require an extra computation (solving the above
equation), and they can be much harder to implement. Implicit methods are used because
many problems arising in real life are stiff, for which the use of an explicit method
requires impractically small time steps Δt to keep the error in the result bounded .