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The University of Texas at Austin
PGE 310: Formulation and Solution in Geosystems Engineering
Homework #7: System of Linear Eqns(Iterative Methods)  System of Nonlinear Eqns(Newton’s Method)
By HosseinRoodi
1.
System of Linear Equations: Iterative Methods
(By HAND)
The below sketch shows three reactors linked by pipes. The rate of transfer of chemicals through each pipe is equal
to a flow rate (
Q
, with units of cubic meters per second) multiplied by the concentration of the reactor from which
the flow originates (c, with units of milligrams per cubic meter). If the system is at steady state,
the transfer into
each reactor
will balance
the transfer out
.
Writing massbalance equations for the reactors one gets a system of 3 linear equations in matrix form:
12
1
2
3
120
20
450
80
100
0
40
80
130
250
cc
c
c
c
!
"
!"
#
!
" !
a)
Determine if the system is guaranteed to converge in iterative methods and if they are not, write them in a different
order to guarantee its convergence.
As can be seen, in all rows diagonal elements are greater than the summation of off diagonal elements. So,
this matrix is guaranteed to converge in iterative method.
40
80
130
:
3
80
100
:
2
20
120
:
1
#
$
$
$
Row
Row
Row
b)
Solve for the concentrations using an initial guess of
c1=c2=c3=0
and the first two iterations using the following
methods:
Step 1: In all following iterative methods the first step is writing the equations in an explicit form:
)
80
40
250
(
130
1
)
(
100
80
)
20
450
(
120
1
2
1
3
1
2
2
1
c
c
c
c
c
c
c
#
#
"
"
#
"
a.
Jacobi
Initial Guess:
0
;
0
;
0
3
2
1
"
"
"
c
c
c
450 mg/s
250 mg/s
Q33=130
Q13=40
Q12=80
Q23=80
Q21=20
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First Iteration:
9231
.
1
)
0
80
0
40
250
(
130
1
0
)
0
(
100
80
75
.
3
)
0
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
9231
.
1
130
250
0
75
.
3
120
450
3
2
1
"
"
"
"
"
c
c
c
Second Iteration:
0769
.
3
)
0
80
75
.
3
40
250
(
130
1
3
)
75
.
3
(
100
80
75
.
3
)
0
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
0769
.
3
13
40
3
75
.
3
120
450
3
2
1
"
"
"
"
"
c
c
c
b.
Gauss Seidel
First Iteration:
9231
.
4
)
3
80
75
.
3
40
250
(
130
1
3
)
75
.
3
(
100
80
75
.
3
)
0
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
9231
.
4
13
64
3
75
.
3
120
450
3
2
1
"
"
"
"
"
c
c
c
Second Iteration:
3231
.
5
)
4
.
3
80
25
.
4
40
250
(
130
1
4
.
3
)
25
.
4
(
100
80
25
.
4
)
3
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
3231
.
5
65
346
4
.
3
5
17
25
.
4
4
17
3
2
1
"
"
"
"
"
"
c
c
c
c.
Gauss Seidel with relaxation with
2
.
1
"
&
.
First Iteration:
9662
.
5
)
32
.
4
80
5
.
4
40
250
(
130
1
6
.
3
)
5
.
4
(
100
80
75
.
3
)
0
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
1594
.
7
0
)
2
.
1
1
(
9662
.
5
2
.
1
)
1
(
32
.
4
0
)
2
.
1
1
(
6
.
3
2
.
1
)
1
(
5
.
4
0
)
2
.
1
1
(
75
.
3
2
.
1
)
1
(
3
3
3
2
2
2
1
1
1
"
%
!
#
%
"
!
#
"
"
%
!
#
%
"
!
#
"
"
%
!
#
%
"
!
#
"
old
new
old
new
old
new
c
c
c
c
c
c
c
c
c
Second Iteration:
4021
.
5
)
4214
.
3
80
464
.
4
40
250
(
130
1
5712
.
3
)
464
.
4
(
100
80
47
.
4
)
32
.
4
20
450
(
120
1
3
2
1
"
%
#
%
#
"
"
"
"
%
#
"
c
c
c
!
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 Spring '06
 Klaus

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