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Numerical methods with
chemical engineering
applications
Solutions Manual
Kevin D. Dorfman and Prodromos Daoutidis
Contents
List of illustrations page iv
List of tables v
1 Introduction 1
2 Linear equations 43
3 Nonlinear equations 115
4 Initial value problems 208
5 Dynamical systems 276
6 Boundary value problems 346
7 Partial differential equations 402
8 Interpolation and integration 468
Illustrations
2.1 Process schematic for Problem 2.80 61
5.1 Phase plane for Problem 5.17 280
5.2 Schematic of a CSTR bioreactor for Problem 5.27 284
7.1 Boundary conditions for Problem 7.12 405
7.2 Geometry for Problem 7.23 411
7.3 Grid for Problem 7.23 412
Tables
5.1 Conditions for phase plane plots in Problem 5.29 289
8.1 Rate data for Problem 8.12 472
8.2 Concentration data of effluent in Problem 8.13 473
8.3 Concentration data for breakthrough curve in Problem 8.14 474
1Introduction
12 Introduction
Problems
(1.1) The operator is
(1.2) The operator is
dx3+x2
and the forcing function is
(1.4) operator: (···)2
(1.5) operator: ∂/∂t−∂2∂x2−6
(1.6) This is a separable equation that can be integrated to give
y=cexp 3x2
(1.7) This is a separable equation. The first step of integration gives
(1.8) This is solved using the integrating factor
d
dx ye−3x2/2=xe−3x2/2
Using integration by parts, the first integration gives
Zxe−3x2/2=rπ
(1.9) Use the integrating factor to get
d
dx e−xy=−2e−x
The integral is
y=2+cex
14 Introduction
(1.12) You should use substitution u=dy/dx to turn this into the first order equation
du
dx =2u
The solution is
(1.13) If we convert to dimensionless form, this problem becomes
y′′ −y′−2y=0
with y′(0) =0 and y(1) =1. The characteristic polynomial is
r2−r−2=(r−2)(r+1) =0
so the solution is
(1.14) First you need to solve the homogenous equation to get
y=c1e2x+c2e−2x
The particular solution is easy to find, y=−1/4. So the solution is
y=c1e2x+c2e−2x−1/4
y(0) =c1=0
For the other boundary condition:
y(1) =c2sin 1 +(1 −π2) sin π=c2sin 1 =1
(1.16) Write this in matrix form as
We need the eigenvalues of the matrix, obtained from
Evaluating the determinant gives
Rewrite as
λ2−5λ−2=0
The roots are
For the positive eigenvalue, the eigenvector equation is
2x
If we pick x=1, then
For the negative eigenvalue, the eigenvector equation is
2x
If we again pick x=1, then
At the initial condition, the equation for xis
4
4
(1.17) The system of equations is
(1.18) The matrix we need to consider is
"2−3
−4 1 #
The eigenvalues of the matrix are λ=5 and λ=-2. For the first eigenvalue, the
(1.19) The dimensionless time is scaled with 4W2/α, where α=k/ρCpand the dimen-
sionless length is scaled with 2Wover the domain x∈[0,W]. The dimensionless
(1.20) The basis functions are
f=c1cos λx+c2sin λx
and their derivatives are
(1.21) The first step is to convert the problem to have homogeneous boundary conditions. If
we define
y=c−1
then the differential equation is still the same:
f(0) =0=c1
For the right boundary condition, we need to differentiate:
f′=−λc1sin λx+λc2cos λx
Since c1=0, the boundary condition requires that
f′(1) =λc2cos λ=0
domain from x∈[0,1], we have
an
2=Z1
ǫ
(−1) sin " n+1
2!πx#dx
The integration gives
n=0
2π
2!πx#exp
2!2
Converting back to concentration, we have
X
2 cos hn+1
2πǫi
(1.22) From the separation of variables, we have
f=c1cos(λx)+c2sin(λx)
With the homogeneous boundary conditions, c1=0 and λ=nπ. The solution is then
T(x,t)=∞
X
n=0
cnsin(nπx) exp(−n2π2t)
0
n=0
0
1−cos(nπ)
nπ=cn
2
So the solution is
T(x,t)=∞
X
2(1 −cos(nπ)
T(x,t)=∞
k=0
(2k+1)πsin[(2k+1)πx] exp[−(2k+1)2π2t]
and you just count in integers of k. If you like to start you sums at k=1, you can
write
X
k=1
4
(2k−1)πsin[(2k−1)πx] exp[−(2k−1)2π2t]
(1.23) If we start from the separation of variables solution
f(x)=c1sin(λx)+c2cos(λx)
the no-flux boundary conditions require that c1=0 and λ=nπfor n=0,1, . . .. The
solution then has the form
T(x,t)=∞
X
dncos(nπx) exp −n2π2t
1−cos(nπ)
n2π2=dn 1
2!
This result is indeterminate for n=0. You need to evaluate this integral separately,
d0=Z1
(1 −x) cos(0)dx =1
m=1
(1.24) The eigenvalue problem here is
f′′ −λ2f=0
f(1) =c2cos(λ)=0
which means
λ=(2n+1)π
2,n=0,1, . . .
The general solution is
X
"1
λn
sin(λnx)#1
0
=cn
2
which simplifies to
22 Introduction
n=0
(2n+1)πcos "(2n+1)πx
2#exp
2!2
(1.25) The eigenvalue problem here is
f′′ −λ2f=0
which has solutions
f=c1sin(λx)+c2cos(λx)
From the left boundary condition, we know that c2=0. From the right boundary
n=1
where we have excluded the eigenvalue λ0=0. For the zero eigenvalue, we need to
go back to the eigenfunction equation
f′′ −λ2f=0
which reduces to
0
c(x,0) sin(λnx)dx =Z1
0
cnsin2(λnx)dx
The integrals are
"−1
cos(λnx)#1
=cn"x
For the zero eigenvalue, we need to multiply by x
xc(x,0) =c0x2+∞
X
n=1
cnxsin(λnx)
Integrating over the domain with c(x,0) =1 gives
Z1
xdx =c0Z1
x2dx
0
xsin(λnx)dx =sin λn
λ2
n−cos λn
λn
which we can rewrite as
Z1
xsin(λnx)dx =sin λn−λncos λn
0
(1.27) (a) 12.6 (b) 5.3 (c) 4.5378
(1.28) 0.103 ×102+0.00442 ×102=0.10742 ×102=10.7
(1.30) x=3
(1.31) The matrix A has the form
(1.32) The matrix is
(1.34) k=4
(1.36) The vector xis [1; 4; 9; 16]. Starting the counter at i=3 means that the first iteration
(1.37) The files for this problem are contained in the folder s11c1p1 matlab.
The Matlab script is:
1function s11c1p1
2%computes the solution to homework #1, spring 2011
