NUMERICAL ANALYSIS FOR ENGINEERING
TERM PROJECT
TRANSIENT HEAT CONDUCTION IN A
PLANE WALL
Author: Rene J Hernandez
Due:
April 17, 2001
1
TABLE OF CONTENTS
Introduction…………………………………………………………………………….3
Problem Description and Formulation…………………………………………………3
Numerical Solution Approaches……………………………………………………….4
Results………………………………………………………………………………….5
Discussion and Error Analysis…………………………………………………………8
Conclusions……………………………………………………………………………11
References……………………………………………………………………………..12
Appendix A : Matlab Code
Appendix B : CASE 1 Solution table
Appendix C : CASE 2 Solution table
LIST OF SYMBOLS
Bi
Fo
h
k
m
q
T
t
u
w
x





Biot number
Fourier number
Convection coefficient
Thermal conductivity
Node index
Heat transferred/Heat generated
Temperature
Time
Generic variable for differential equation representation
Generic variable for solution of differential equation
Position
Thermal diffusivity
Eigenvalue
Mean time value
Truncation error
Mean position value
2
INTRODUCTION
Although exact analytical solutions have been obtained for transient heat conduction
problems, they often apply to simple geometry cases. Most practical problems, however,
require the development of extensive solutions that are simplified by using appropriate
assumptions making these analytical solutions approximate. Numerical mathematics can
approach these problems in a faster and more convenient manner by also providing an
approximated solution whose error depends on the assumptions made and the number of
iterations. In this project, a numerical solution will be developed to solve transient heat
conduction problems for the specific case of the plane wall. Using finite-difference
methods for prescribed boundary conditions, this numerical solution will generate the
temperature distribution across the width of the wall with respect to time.
The numerical solution will be applied to the following baseline cases:
- Wall with initial uniform temperature distribution with convective boundaries.
- Wall with initial distribution of the form T(x)=sin(x) with fixed temperature
boundaries.
Since the analytical solution for these cases is known, the results from the numerical and
analytical methods will be compared, showing the accuracy and validity of the numerical
approach.
PROBLEM DESCRIPTION AND FORMULATION
A wall of infinite length, uniform thermal conductivity (k) and diffusivity (), and
thickness (w) has an initial temperature distribution. For given boundary conditions, the
problem consists of determining the one-dimensional transient temperature distribution of
the wall as the internal temperature changes with time due to the influence of the
boundary conditions.
The x locations are designated by the m index of each nodal point.
FIGURE 1: NODAL NETWORK ON A PLANE WALL (FLAT PLATE)
3
NUMERICAL SOLUTION APPROACHES
Using Parabolic Partial Differential Equations will solve the mathematical problem
described above, specifically the explicit method (Burden and Faires, chapter 12.2) is
implemented. The approach used to approximate the solution involves finite differences.
The heat diffusion equation looks like:
u
 2u
( x, t )   2 2 ( x, t ) for 0<x<1, t>0
t
x
subject to the initial conditions:
u(0,t)=u(l,t)=0, t>0, and u(x,0)=f(x), 0<=x<=l
After selecting an integer m>0, defining h=l/m, and selecting a time-step size k, the grid
points are (xi,tj), where xi=ih, for I=0,1,..m, and tj = jk, for j=0,1,… The difference
method is obtained by using the Taylor series in t to form the difference quotient:
u( x i , t j  k )  u( x i , t j ) k  2 u
u
( xi , t j ) 

( x i ,  j ),
t
k
2 t 2
for some j  (tj, t j+1), and the Taylor series in x to form the difference quotient:
u( x i  h, t j )  2u( x i , t j )  u( x i  h, t j ) h 2  4 u
 2u
(
x
,
t
)


( i , t j ),
i
j
12 x 4
x 2
h2
where i  (x i-1, x i+1),
Therefore for the interior gridpoint:
u
 2u
( x i , t j )   2 2 ( x i , t j )  0,
t
x
so the difference method using the difference quotients is:
wi , j 1  w j
wi 1, j  2 wi , j  wi 1, j
2
 0,
k
h2
Solving this equation for w i,j+1

2 2 k 
k
wi , j 1  1  2  wij   2 2 ( wi 1, j  wi 1, j )
h 
h

The explicit or Forward Difference method is conditionally stable, and it converges to a
solution with rate of convergence O(k+h2), provided that:
k
1
2 2 
2
h
The next step is to develop computer code in MATLAB. The computer code will use the
forward-difference method to obtain the temperature distribution in a plane wall for the
following cases:
Initial Temperature Distribution:
- Temperature vs. position as a function f(x) specified by the user.
- User input of the temperature at the specific nodes.
Boundary conditions:
- Convective
- Fixed temperature
- Adiabatic
4
By expressing the temperature gradients as a function of the nodal temperatures and by
assuming a one- dimensional system, the value of derivative 2T/x2 at the node m can be
approximated as (Incropera and DeWitt, Chapter 5):
T
T
m 1 / 2 
m 1 / 2
2
 T
x
x

m
x
x 2
or written in a “numerical form” for MATLAB coding.
T  Tm1  2Tm
 2T
For a node at a given instant t
 m 1
2 m
x
x 2
Likewise, integer p is introduced to establish a time index:
t  pt
After defining the time index p, the finite-difference approximation to the time derivative
can be expressed as:
Tmp 1  Tmp
T

m
t
t
By combining the previous equations and adding a heat generation term, the explicit
finite-difference method yields the following nodal equations.
FINAL EQUATIONS IN MATLAB CODE FOR INDIVIDUAL NODES:
For an interior node:
2

q x  
  1  2 FoTmp
Tmp 1  Fo Tmp1  Tmp1 
k 

Stability Criterion:
Fo<=1/4
For an external node with convection
2
 p
q x  
p 1

  1  2 Fo  2 BiFo Tmp
Tm  2 Fo Tm 1,n  BiT  
2k 

Stability Criterion:
Fo(2+Bi)<=1/2
For complete numerical solution in MATLAB refer to Appendix A.
RESULTS
To validate the numerical solution, the results will be compared against existing
analytical solutions. Two cases will be considered:
CASE 1: Uniform initial temperature; solve for convection at exterior surface
The analytical solution for a flat plate is obtained by solving the heat equation by the
method of separation of variables (Geiger and Poirier, Chapter 9). For the case of initial
uniform temperature and convection at the surface, the analytical solution has the form:
5

sin( n L
T  Tf
 2
exp( 2nt ) cos( n x )
Ti  Tf
n 1 n L  sin( n L) cos( n L)
where the discrete values (eigenvalues) of n are the positive roots of the equation:
1
cot( n L) 
(  n L)
h
 L
k 
The analytical solution will be applied to the following problem:
Example 5.3, p231, Incropera and DeWitt
Known: Wall subjected to sudden change in convective surface conditions.
Data:
Plane Wall
material: steel AISI 1010, k=63.9 W/mK, =18.8e-6 m2/s.
thickness : L =40mm
initial temperature : Ti=-20 C
Fluid
Temperature : Tf = 60 C
convection coefficient : h = 500 W/m2K
Find : Plot temperature at the adiabatic surface during first 8 minutes.
The first step is to find the eigenvalues for the first 4 terms of the infinite series, which
will yield a highly accurate result. The first four eigenvalues are obtained by finding the
roots of the following equation for the corresponding interval (n,(n+1)).
1
f ( x )  cot( n L) 
(  n L)
h
 L
k 
Bisection Method
For (1L) in the interval (0,)
Approximate solution = 0.53189087
Number of iterations = 16 Tolerance = 1.00000000e-005
For (2L) in the interval (,2)
Approximate solution = 3.23796082
Number of iterations = 15 Tolerance = 1.00000000e-005
For (3L) in the interval (2,3)
Approximate solution = 6.33257446
Number of iterations = 16 Tolerance = 1.00000000e-005
For (4L) in the interval (3,4 )
Approximate solution = 9.45785522
Number of iterations = 16 Tolerance = 1.00000000e-005
After the first four eigenvalues are found, the analytical solution can be solved for the
first four terms of the infinite series.
6
The following graph shows a comparison of the analytical (pink color) and the numerical
(blue color)MATLAB solution for this case.
TRANSIENT TEMPERATURE AT THE ADIABATIC SURFACE
50
TEMPERATURE (C)
40
30
20
10
0
NUMERICAL SOLUTION
-10
ANALITYCAL SOLUTION
-20
0
40
80
120
160
200
240
280
320
360
400
440
480
TIME (seconds)
FIGURE 2 : TEMPERATURE DURING FIRST 8 MINUTES
As it can be observed, the numerical MATLAB solution closely matches the analytical
results.
CASE 2: Initial temperature distribution T=sin(x); solve for fixed temperature
boundary conditions
The analytical solution for CASE 2 is found at Burden and Faires, Chapter 12, page 706,
example 1. This example will also be solved with the numerical MATLAB solution.
Example 1, page 706, Burden and Faires
The analytical solution has the form:
T(x)=exp(-2t)sin(x)
Known:
Initial conditions
T(x,0)=sin(x), in the interval 0  x  1
Boundary conditions
T(0,t)=T(1,t)=0, for 0  t
Find: Plot the temperature distribution inside the wall at t=0 sec, t=0.04 sec, t=0.12 sec,
t=0.2 sec, t=0.4 sec.
7
The following graph shows a comparison of the analytical (pink color) and the numerical
(blue color)MATLAB solution for this case.
TEMPERATURE DISTRIBUTION
Analytical
Numerical
1
t = 0 sec
TEMPERATURE ( deg)
0.8
t = 0.04 sec
sec
0.6
0.4
t = 0.12 sec
0.2
t = 0.2 sec
t = 0.4 sec
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
POSITION ( x )
FIGURE 3 : TEMPERATURE DISTRIBUTION AT DIFFERENT TIMES
As it can be observed, the numerical MATLAB solution closely matches the analytical
results.
DISCUSSION AND ERROR ANALYSIS
The graphic results show that fast and highly accurate result can be obtained with the
MATLAB numerical solution. For CASE 1, the analytical solution involved the
calculation of the eigenvalues, which required the use of the bisection method. After the
eigenvalues were calculated, the series solution was found for the first four terms. This
process required more time and calculations than a quick run with the MATLAB
numerical solution. For CASE 2, the analytical and the numerical solution were as easy
to compute; however, it should be noted that this is a very simple case and that the
solution was already available.
While different methods and equations had to be considered for each of the analytical
cases, the numerical solution for both methods was obtained by running the same
MATLAB code.
8
When using the explicit finite-difference method, the truncation error for the difference
equation is:
2
4
k  2u
2 h  u
 i, j 
( xi ,  j )  
( i , t j )
2 t 2
12 x 4
A truncation error of order (k+h2) is expected when the explicit finite-difference method
is used. If an error is made in representing the initial condition , the error will propagate.
Moreover, if the time step is not carefully selected the error will grow as the number of
time steps increases, making this method conditionally stable. The time step in the
MATLAB program is calculated so the stability condition is respected.
Since the analytical solution was available for both cases, the absolute error produced by
the MATLAB numerical computations can be obtained for each time step for any
position.
In order to illustrate the effect of the mesh size on the accuracy of the results, the
temperature error at 10 seconds was computed for CASE 1 and the results shown in figure
4. The graph shape reflects the expected order (k+h2), as the mesh size h is expressed in
the x-axis as the percent of the nodal distance over the total wall thickness. However it
should be noted that even for the case of 50 % (nodal distance is one half of the total
thickness), the accuracy of the result was within 0.5 degrees. It should also be noted that
the as the nodal distance was changed, so was the time step k since the stability criterion
needs to be satisfied to guarantee convergence.
TEMPERATURE ERROR FOR DIFFERENT MESH SIZES
TEMPERATURE ERROR (C)
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
30
35
40
45
50
PERCENT NODAL DISTANCE OF THE TOTAL WALL THICKNESS (%)
FIGURE 4: TEMPERATURE ERROR FOR CASE 1 AT 10 SECONDS
9
The following table shows a comparison of the numerical vs. the analytical method, and
the absolute value of the absolute error for CASE 1.
time
numerical analytical
error
0
-20 -19.8054
0.1946
1
-20 -19.9602
0.0398
2
-20 -19.9929
0.0071
3
-20 -19.9983
0.0017
4
-20 -19.9953
0.0047
5
-19.994 -19.9829
0.0111
6 -19.9745 -19.9565
0.018
7
-19.937 -19.9128
0.0242
8 -19.8794 -19.8503
0.0291
9 -19.8016 -19.7688
0.0328
…………………………………………………….
470
471
472
473
474
475
476
477
478
479
480
42.4687
42.5269
42.585
42.6428
42.7005
42.758
42.8152
42.8723
42.9292
42.986
43.0425
42.444
42.5023
42.5603
42.6182
42.6759
42.7334
42.7907
42.8478
42.9047
42.9615
43.018
0.0247
0.0246
0.0247
0.0246
0.0246
0.0246
0.0245
0.0245
0.0245
0.0245
0.0245
TABLE 1 : SAMPLE OF RESULTS FOR CASE 1
Analysis of the data shows that the greatest error is observed at the initial condition, this
error is due to the fact that only the first four terms of the infinite series solution were
considered for the analytical case. Using more terms in the series solution would have
increased the accuracy of the analytical solution for the first 4 seconds, as shown for the
cases of 4 and 6 term series in table 4. Since this effect is negligible after 4 seconds and
the problem is to be solved for 8 minutes, the accuracy of 4-term-series solution is
considered to be valid for this particular case. Should higher accuracy be required for the
analytical solution, more terms could be added to the series until the desired accuracy is
reached.
time
0
1
2
3
4
5
6
7
8
9
10
numerical
-20
-20
-20
-20
-20
-19.994
-19.9745
-19.937
-19.8794
-19.8016
-19.7043
4-term-series
6-term-series
-19.8054
-19.9182
-19.9602
-19.9981
-19.9929
-19.9999
-19.9983
-19.9994
-19.9953
-19.9955
-19.9829
-19.9829
-19.9565
-19.9565
-19.9128
-19.9128
-19.8503
-19.8503
-19.7688
-19.7688
-19.669
-19.669
TABLE 2 : EFFECT OF USING MORE TERMS IN ANALYTICAL SERIES SOLUTION FOR CASE 1
10
After the first 4 seconds the maximum error observed is e=0.0382, which is caused by the
truncation effect of the numerical method. The solution at time = 480 seconds shows a
relative error of 0.05%, which is a highly accurate result in the context of the problem.
As in CASE 1, the error due to the truncation effect is observed in CASE 2. Even though
the absolute results show that the numerical and the analytical solution closely match, the
relative error increases. This is because the base value that the error is being compared to
is approaching zero as time steps are incremented.
The following table shows a comparison of the numerical vs. the analytical method, and
the absolute value of the absolute error.
time
0
0.04
0.08
0.12
0.16
0.2
0.24
0.28
0.32
0.36
0.4
0.44
0.48
numerical
1
0.6665
0.4442
0.296
0.1973
0.1315
0.0876
0.0584
0.0389
0.0259
0.0173
0.0115
0.0077
analytical
1
0.673825
0.454041
0.305944
0.206153
0.138911
0.093602
0.063071
0.042499
0.028637
0.019296
0.013002
0.008761
error
0
0.007325
0.009841
0.009944
0.008853
0.007411
0.006002
0.004671
0.003599
0.002737
0.001996
0.001502
0.001061
% error
0
1.1
2.2
3.3
4.3
5.3
6.4
7.4
8.5
9.6
10.3
11.6
12.1
TABLE 3 : SAMPLE OF RESULTS FOR CASE 2
CONCLUSIONS
The MATLAB numerical method based on the explicit finite-difference method provided
a solution that closely matched the analytical result. For the two cases solved, the error
was found to be so small that the result was not impacted by the use of the numerical
solution. Mesh size or truncation errors were also not significant for the two cases
considered; mesh sizes of up to 50 % the thickness of the wall would have yielded
acceptable solutions for the cases studied. Although the analytical solution is considered
to be an exact result, arriving at those solutions required a higher use of time, hand
calculations, and each case had to be solved individually. On the contrary, by
programming the numerical solution in MATLAB, the same code can find the solution
for an infinite number of cases and the computer makes all the computations.
In conclusion, the MATLAB numerical solution can be used to find the transient
temperature distribution for the case of the plane wall, since accurate and reliable results
can be computed faster and easier than with the use of analytical methods.
11
REFERENCES
1. Burden, R. L. and Faires, J. D., Numerical Analysis, 7th edition, Brooks/Cole, Pacific
Grove, 2001, chapter 12.
2. Geiger, G. H. and Poirier, D. R., Transport Phenomena in Metallurgy, AddisonWesley Publishing Co. Massachusetts.
3. Incropera, Frank P. and DeWitt, David P., Introduction to Heat Transfer, 3rd edition,
John Wiley & Sons, New York, 1996, chapter 5.
12
APPENDIX A
MATLAB CODE
% Numerical Analysis for Engineering
%
Term Project
%
MATLAB CODE
%
% author : Rene J. Hernandez
% Date : 04/10/2001
%--------------------------------------------------------------% This program calculates the transient temperature distribution
% for the specific case of the plane wall. No input file is
% needed; all inputs are interactively entered by the user.
% The numerical explicit finite-difference method is used.
% -------------------------------------------------------------% The user defines:
% - Wall properties
% - Number of nodes
% - Wall internal uniform heat generation
% - Boundary conditions
% The initial temperature for the wall can be defined:
% - User defined function T(x)=f(x)
% - Specific initial temperature for each node
% The boundary conditions for each side of the wall can be:
% - Convection
% - Fixed Temperature
% - Adiabatic surface
% Solution in : MATLAB variable tmatrixc(x,t)
% Note : The program assumes constant properties during the
% transient calculations
%--------------------------------------------------------------clear
syms('x','s')
% general data input
fprintf(1,'\n___________________________________________\n');
fprintf(1,'NUMERICAL SOLUTION FOR TRANSIENT CONDUCTION \n');
fprintf(1,'
PLANE WALL \n\n');
fprintf(1,'WALL DATA INPUT\n');
fprintf(1,'Enter wall thickness in (m)\n');
thickness = input(' ');
fprintf(1,'Enter wall thermal conductivity (k) in (W/mK)\n');
k = input(' ');
fprintf(1,'Enter wall thermal diffusivity (alpha) in (m^2/s)\n');
alpha = input(' ');
fprintf(1,'\nNUMERICAL SOLUTION INPUT .\n');
fprintf(1,'Enter the number of nodal points \n');
13
NNODES = input(' ');
deltax=thickness/(NNODES-1);
%
% Initial Conditions (prior to transient event)
%
badinput1=1;
while (badinput1)==1
fprintf(1,'\nWALL INITIAL TEMPERATURE (prior to transient event).\n\n');
fprintf(1,' 1 - User defined function T(x)=f(x)\n');
fprintf(1,' 2 - Specific initial temperature for each node\n');
fprintf(1,' Enter option\n');
OPTION = input(' ');
n = 0;
switch OPTION
case {1} % Specific Function
fprintf(1,'Input the function F(x) in terms of x (degree C)\n');
fprintf(1,'For example: cos(x)\n ');
s = input(' ','s');
F = inline(s,'x');
for m=1:NNODES
inimatrix(m)= F(n);
n = n + deltax;
end
badinput1=0;
case {2} % Manual Input
for m=1:NNODES
fprintf('\nTemperature (degree C) At Position %6.3f (m)\n',n);
manin = input(' ');
inimatrix(m)= manin;
n = n + deltax;
end
badinput1=0;
end
end
inimatrix=inimatrix+273;
%
%
TRANSIENT SECTION
%
fprintf(1,'\nTRANSIENT EVENT CONDITIONS.\n\n');
fprintf(1,'Enter wall internal uniform volumetric heat generation rate W/m^3\n');
qtr = input(' ');
fprintf(1,'\nSURFACE BOUNDARY OPTION NUMBER.\n');
fprintf(1,'Convection
1.\n');
fprintf(1,'Fixed Temperature 2.\n');
fprintf(1,'Adiabatic
3.\n');
fprintf(1,'Boundary Conditions at surface 1. Enter option\n');
14
NODETYPEF = input(' ');
fprintf(1,'Boundary Conditions at surface 2. Enter option\n');
NODETYPEL = input(' ');
switch NODETYPEF % uses the req. eq. based on node type
case {1} % Convection at surface f
fprintf(1,'Convection at surface 1.\n');
fprintf(1,'Input Fluid Temperature in deg C\n');
tinff = input(' ');
fprintf(1,'Input the convection coefficient h in W/m^2K\n ');
hf = input(' ');
tinff=tinff + 273;
Bif=hf*deltax/k; % Biot number
Fof=0.5/(1+Bif); % Fourier number for stability
deltatf=(Fof*deltax^2)/alpha;
case {2} % Prescribed emperature at surface f
fprintf(1,'Fixed Temperature at surface 1.\n');
fprintf(1,'Input T-surface in deg C\n');
tsurff = input(' ');
tsurff=tsurff + 273;
Fof = 0.5; % Fourier number for stability
deltatf=(Fof*deltax^2)/alpha;
case {3} % Adiabatic surface f
Fof = 0.5; % Fourier number for stability
deltatf=(Fof*deltax^2)/alpha;
end
switch NODETYPEL % uses the req. eq. based on node type
case {1} % Convection at surface l
fprintf(1,'Convection at surface 2.\n');
fprintf(1,'Input Fluid Temperature in deg C\n');
tinfl = input(' ');
fprintf(1,'Input the convection coefficient h in W/m^2K\n ');
hl = input(' ');
tinfl=tinfl + 273;
Bil=hl*deltax/k; % Biot number
Fol=0.5/(1+Bil); % Fourier number for stability
deltatl=(Fol*deltax^2)/alpha;
case {2} % Prescribed emperature at surface l
fprintf(1,'Fixed Temperature at surface 2.\n');
fprintf(1,'Input T-surface in deg C\n');
tsurfl = input(' ');
tsurfl=tsurfl + 273;
Fol=0.5; % Fourier number for stability
deltatl=(Fol*deltax^2)/alpha;
case {3} % Adiabatic surface l
Fol=0.5; % Fourier number for stability
deltatl=(Fol*deltax^2)/alpha;
15
end
% calculating the minimum deltat and Fo number
if deltatf<deltatl
deltat=deltatf;
else
deltat=deltatl;
end
deltatint=(0.5*deltax^2)/alpha;
if deltatint<deltat
deltat=deltaint;
end
deltatnc=deltat;
% Calculating a userfriendly time interval
if (deltat<1000)&(deltat>=500)
deltat=500;
elseif (deltat<500)&(deltat>=250)
deltat=250;
elseif (deltat<250)&(deltat>=100)
deltat=100;
elseif (deltat<100)&(deltat>=50)
deltat=50;
elseif (deltat<50)&(deltat>=25)
deltat=25;
elseif (deltat<25)&(deltat>=10)
deltat=10;
elseif (deltat<10)&(deltat>=5)
deltat=5;
elseif (deltat<5)&(deltat>=2.5)
deltat=2.5;
elseif (deltat<2.5)&(deltat>=1)
deltat=1;
elseif (deltat<1)&(deltat>=0.5)
deltat=0.5;
elseif (deltat<0.5)&(deltat>=0.25)
deltat=0.25;
elseif (deltat<0.25)&(deltat>=0.1)
deltat=0.1;
elseif (deltat<0.1)&(deltat>=0.05)
deltat=0.05;
elseif (deltat<0.05)&(deltat>=0.025)
deltat=0.025;
elseif (deltat<0.025)&(deltat>=0.01)
deltat=0.01;
elseif (deltat<0.01)&(deltat>=0.005)
deltat=0.005;
elseif (deltat<0.005)&(deltat>=0.0025)
16
deltat=0.0025;
elseif (deltat<0.0025)&(deltat>=0.001)
deltat=0.001;
else
deltat=deltat;
end
fprintf(1,'\nTRANSIENT EVENT TOTAL TIME \n');
fprintf(1,'Enter time for transient event (sec)\n');
TOTIME = input (' ');
if TOTIME <= deltat
TOTIME = deltat;
fprintf('\n Transient time too small.\n');
end
NTIMEINTR = TOTIME / deltat;
NTIMEINT = floor(NTIMEINTR) + 1;
% initiallizing the temperature matrix
for p=1:NTIMEINT
for m=1:NNODES
tmatrix(m,p)=inimatrix(m);
end
end
% Output selection
fprintf(1,'\n OUTPUT SELECTION \n');
fprintf(1,'\n Plot specific node vs time - 1 ');
fprintf(1,'\n Display in tabular form - 2 ');
fprintf(1,'\nEnter option\n');
DISPSEL = input(' ');
if DISPSEL==1
fprintf(1,'\nEnter node number (ex: 1,2,...)\n');
nodeplot = input(' ');
else
DISPSEL=2;
end
% Calculating Fourier Nunber
Fo = (alpha*deltat)/(deltax^2);
% solving the temperature matrix in degree Kelvin
for p=2:NTIMEINT
switch NODETYPEF % uses the req. eq. based on node type
case {1}
tmatrix(1,p)= 2*Fo*(tmatrix(2,p-1)+Bif*tinff+(qtr*(deltax^2)/(2*k)))+(1-2*Fo2*Bif*Fo)*tmatrix(1,p-1);
case {2}
tmatrix(1,p)=tsurff+(qtr*(deltax^2)/k);
case {3}
tmatrix(1,p)=Fo*(2*tmatrix(2,p-1)+(qtr*(deltax^2)/k))+(1-2*Fo)*tmatrix(1,p-1);
end
17
for m=2:(NNODES-1)
tmatrix(m,p)=Fo*(tmatrix(m+1,p-1)+tmatrix(m-1,p-1)+(qtr*(deltax^2)/k))+(12*Fo)*tmatrix(m,p-1);
end
switch NODETYPEL % uses the req. eq. based on node type
case {1}
tmatrix(NNODES,p)=2*Fo*(tmatrix(NNODES-1,p1)+Bil*tinfl+(qtr*(deltax^2)/(2*k)))+(1-2*Fo-2*Bil*Fo)*tmatrix(NNODES,p-1);
case {2}
tmatrix(NNODES,p)=tsurfl+(qtr*(deltax^2)/k);
case {3}
tmatrix(NNODES,p)=Fo*(2*tmatrix(NNODES-1,p-1)+(qtr*(deltax^2)/k))+(12*Fo)*tmatrix(NNODES,p-1);
end
end
% Converting to deg C
for p=1:NTIMEINT
for m=1:NNODES
tmatrixc(m,p)=tmatrix(m,p) - 273;
end
end
% Output
switch DISPSEL
case {1}
time=0;
for p=1:NTIMEINT
ptarray(p)=tmatrixc(nodeplot,p);
tim(p)=time;
time = time + deltat;
end
plot(tim,ptarray)
xlabel('TIME (sec)'),ylabel('TEMPERATURE (C)')
title('Temperature at the selected node vs. time')
case {2}
if NTIMEINT >= 30
OUTYPE = 2;
else
OUTYPE = 1;
end
pos=0;
fprintf('Time(s)');
for m=1:NNODES
fprintf(' x=%6.3f',pos);
pos = pos + deltax;
18
end
switch OUTYPE % To avoid excees output
case {1} % For less than 30 time intervals
time=0;
fprintf('\n');
for p=1:NTIMEINT
fprintf('%7.3f',time);
for m=1:NNODES
fprintf(' %8.1f',tmatrixc(m,p));
end
fprintf('\n');
time = time + deltat;
end
case {2} % For more than 30 time intervals
time=0;
fprintf('\n');
for p=1:10
fprintf('%7.3f',time);
for m=1:NNODES
fprintf(' %8.1f',tmatrixc(m,p));
end
fprintf('\n');
time = time + deltat;
end
fprintf('\n .....
Iterations not shown');
fprintf('\n .....
stored in variable');
fprintf('\n .....
tmatrixc(x,t)\n');
initimou=(NTIMEINT-10);
time=initimou*deltat;
fprintf('\n');
for p=initimou:NTIMEINT
fprintf('%7.3f',time);
for m=1:NNODES
fprintf(' %8.1f',tmatrixc(m,p));
end
fprintf('\n');
time = time + deltat;
end
end
end
19
APPENDIX B
CASE 1 data
time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
numerical analitycal error
-20 -19.8054
0.1946
-20 -19.9602
0.0398
-20 -19.9929
0.0071
-20 -19.9983
0.0017
-20 -19.9953
0.0047
-19.994 -19.9829
0.0111
-19.9745 -19.9565
0.018
-19.937 -19.9128
0.0242
-19.8794 -19.8503
0.0291
-19.8016 -19.7688
0.0328
-19.7043
-19.669
0.0353
-19.5891 -19.5522
0.0369
-19.4575 -19.4197
0.0378
-19.3112
-19.273
0.0382
-19.1518 -19.1138
0.038
-18.9809 -18.9433
0.0376
-18.7998 -18.7628
0.037
-18.6098 -18.5736
0.0362
-18.412 -18.3767
0.0353
-18.2074 -18.1731
0.0343
-17.9968 -17.9637
0.0331
-17.7812 -17.7492
0.032
-17.5612 -17.5303
0.0309
-17.3374 -17.3075
0.0299
-17.1103 -17.0815
0.0288
-16.8804 -16.8528
0.0276
-16.6482 -16.6216
0.0266
-16.4141 -16.3885
0.0256
-16.1783 -16.1536
0.0247
-15.9411 -15.9174
0.0237
-15.7029 -15.6801
0.0228
-15.4638 -15.4418
0.022
-15.224 -15.2029
0.0211
-14.9837 -14.9634
0.0203
-14.7431 -14.7235
0.0196
-14.5023 -14.4834
0.0189
-14.2614 -14.2432
0.0182
-14.0205
-14.003
0.0175
-13.7797 -13.7629
0.0168
-13.5392 -13.5229
0.0163
-13.2989 -13.2832
0.0157
-13.0589 -13.0437
0.0152
-12.8193 -12.8046
0.0147
-12.5801
-12.566
0.0141
-12.3414 -12.3278
0.0136
-12.1032
-12.09
0.0132
-11.8655 -11.8528
0.0127
-11.6284 -11.6162
0.0122
-11.3919 -11.3801
0.0118
time
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
numerical
-11.156
-10.9207
-10.6861
-10.4522
-10.2189
-9.9863
-9.7544
-9.5232
-9.2927
-9.0629
-8.8338
-8.6054
-8.3778
-8.1509
-7.9247
-7.6992
-7.4745
-7.2504
-7.0271
-6.8046
-6.5827
-6.3616
-6.1412
-5.9216
-5.7026
-5.4844
-5.2669
-5.0501
-4.834
-4.6186
-4.404
-4.19
-3.9768
-3.7643
-3.5524
-3.3413
-3.1309
-2.9212
-2.7121
-2.5038
-2.2961
-2.0892
-1.8829
-1.6773
-1.4724
-1.2682
-1.0646
-0.8618
-0.6596
analitycal error
-11.1446
0.0114
-10.9097
0.011
-10.6755
0.0106
-10.4419
0.0103
-10.209
0.0099
-9.9768
0.0095
-9.7452
0.0092
-9.5144
0.0088
-9.2842
0.0085
-9.0547
0.0082
-8.826
0.0078
-8.5979
0.0075
-8.3706
0.0072
-8.144
0.0069
-7.9181
0.0066
-7.6929
0.0063
-7.4685
0.006
-7.2447
0.0057
-7.0217
0.0054
-6.7994
0.0052
-6.5779
0.0048
-6.357
0.0046
-6.1369
0.0043
-5.9175
0.0041
-5.6988
0.0038
-5.4808
0.0036
-5.2636
0.0033
-5.0471
0.003
-4.8312
0.0028
-4.6161
0.0025
-4.4017
0.0023
-4.188
0.002
-3.975
0.0018
-3.7627
0.0016
-3.5512
0.0012
-3.3403
0.001
-3.1301
0.0008
-2.9206
0.0006
-2.7118
0.0003
-2.5037
0.0001
-2.2963
0.0002
-2.0895
0.0003
-1.8835
0.0006
-1.6781
0.0008
-1.4735
0.0011
-1.2695
0.0013
-1.0661
0.0015
-0.8635
0.0017
-0.6615
0.0019
20
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
-0.458
-0.2572
-0.057
0.1425
0.3414
0.5396
0.7372
0.934
1.1303
1.3259
1.5208
1.7151
1.9087
2.1017
2.2941
2.4858
2.6769
2.8673
3.0571
3.2463
3.4349
3.6228
3.8101
3.9968
4.1828
4.3683
4.5531
4.7373
4.9209
5.1039
5.2863
5.4681
5.6492
5.8298
6.0098
6.1892
6.3679
6.5461
6.7237
6.9007
7.0771
7.2529
7.4282
7.6028
7.7769
7.9504
8.1234
8.2957
8.4675
8.6387
-0.4602
-0.2595
-0.0596
0.1398
0.3384
0.5364
0.7337
0.9304
1.1265
1.3218
1.5166
1.7106
1.9041
2.0969
2.289
2.4806
2.6715
2.8617
3.0513
3.2403
3.4287
3.6164
3.8035
3.99
4.1759
4.3612
4.5458
4.7299
4.9133
5.0961
5.2783
5.4599
5.6409
5.8213
6.0011
6.1803
6.3589
6.5369
6.7144
6.8912
7.0674
7.2431
7.4182
7.5927
7.7666
7.94
8.1127
8.2849
8.4565
8.6276
0.0022
0.0023
0.0026
0.0027
0.003
0.0032
0.0035
0.0036
0.0038
0.0041
0.0042
0.0045
0.0046
0.0048
0.0051
0.0052
0.0054
0.0056
0.0058
0.006
0.0062
0.0064
0.0066
0.0068
0.0069
0.0071
0.0073
0.0074
0.0076
0.0078
0.008
0.0082
0.0083
0.0085
0.0087
0.0089
0.009
0.0092
0.0093
0.0095
0.0097
0.0098
0.01
0.0101
0.0103
0.0104
0.0107
0.0108
0.011
0.0111
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
8.8093
8.9794
9.1489
9.3179
9.4862
9.6541
9.8213
9.988
10.1542
10.3198
10.4848
10.6494
10.8133
10.9767
11.1396
11.3019
11.4637
11.625
11.7857
11.9459
12.1055
12.2646
12.4232
12.5813
12.7388
12.8959
13.0523
13.2083
13.3638
13.5187
13.6731
13.8271
13.9805
14.1334
14.2857
14.4376
14.589
14.7399
14.8902
15.0401
15.1895
15.3383
15.4867
15.6346
15.782
15.9289
16.0753
16.2213
16.3667
16.5117
8.7981
8.968
9.1374
9.3062
9.4744
9.6421
9.8092
9.9758
10.1418
10.3072
10.4721
10.6365
10.8003
10.9636
11.1263
11.2885
11.4502
11.6113
11.7719
11.9319
12.0915
12.2505
12.4089
12.5669
12.7243
12.8812
13.0375
13.1934
13.3487
13.5035
13.6578
13.8116
13.9649
14.1177
14.27
14.4217
14.573
14.7237
14.874
15.0238
15.173
15.3218
15.4701
15.6178
15.7651
15.9119
16.0582
16.2041
16.3494
16.4943
21
0.0112
0.0114
0.0115
0.0117
0.0118
0.012
0.0121
0.0122
0.0124
0.0126
0.0127
0.0129
0.013
0.0131
0.0133
0.0134
0.0135
0.0137
0.0138
0.014
0.014
0.0141
0.0143
0.0144
0.0145
0.0147
0.0148
0.0149
0.0151
0.0152
0.0153
0.0155
0.0156
0.0157
0.0157
0.0159
0.016
0.0162
0.0162
0.0163
0.0165
0.0165
0.0166
0.0168
0.0169
0.017
0.0171
0.0172
0.0173
0.0174
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
16.6562
16.8002
16.9437
17.0867
17.2293
17.3714
17.513
17.6542
17.7949
17.9351
18.0748
18.2141
18.3529
18.4913
18.6292
18.7667
18.9037
19.0402
19.1763
19.3119
19.4471
19.5818
19.7161
19.8499
19.9833
20.1163
20.2488
20.3808
20.5125
20.6437
20.7744
20.9047
21.0346
21.1641
21.2931
21.4217
21.5499
21.6776
21.8049
21.9318
22.0583
22.1844
22.31
22.4352
22.56
22.6844
22.8084
22.9319
23.0551
23.1778
23.3002
16.6386
16.7825
16.926
17.0689
17.2114
17.3534
17.4949
17.636
17.7766
17.9167
18.0564
18.1956
18.3343
18.4726
18.6104
18.7477
18.8846
19.0211
19.1571
19.2926
19.4277
19.5624
19.6966
19.8303
19.9636
20.0965
20.2289
20.3609
20.4924
20.6236
20.7542
20.8845
21.0143
21.1437
21.2726
21.4011
21.5292
21.6569
21.7842
21.911
22.0374
22.1634
22.2889
22.4141
22.5388
22.6631
22.787
22.9105
23.0336
23.1563
23.2786
0.0176
0.0177
0.0177
0.0178
0.0179
0.018
0.0181
0.0182
0.0183
0.0184
0.0184
0.0185
0.0186
0.0187
0.0188
0.019
0.0191
0.0191
0.0192
0.0193
0.0194
0.0194
0.0195
0.0196
0.0197
0.0198
0.0199
0.0199
0.0201
0.0201
0.0202
0.0202
0.0203
0.0204
0.0205
0.0206
0.0207
0.0207
0.0207
0.0208
0.0209
0.021
0.0211
0.0211
0.0212
0.0213
0.0214
0.0214
0.0215
0.0215
0.0216
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
23.5436
23.6647
23.7855
23.9058
24.0257
24.1452
24.2643
24.3831
24.5014
24.6193
24.7369
24.854
24.9708
25.0872
25.2032
25.3188
25.434
25.5488
25.6633
25.7774
25.8911
26.0044
26.1173
26.2299
26.3421
26.4539
26.5654
26.6764
26.7872
26.8975
27.0075
27.1171
27.2263
27.3352
27.4437
27.5519
27.6597
27.7672
27.8742
27.981
28.0873
28.1934
28.299
28.4044
28.5093
28.614
28.7182
28.8222
28.9257
29.029
29.1319
23.5219
23.643
23.7636
23.8839
24.0037
24.1232
24.2423
24.3609
24.4792
24.5971
24.7146
24.8317
24.9484
25.0647
25.1806
25.2962
25.4114
25.5262
25.6406
25.7546
25.8682
25.9815
26.0944
26.2069
26.3191
26.4308
26.5422
26.6533
26.7639
26.8742
26.9842
27.0937
27.203
27.3118
27.4203
27.5284
27.6362
27.7436
27.8506
27.9573
28.0636
28.1696
28.2753
28.3805
28.4855
28.5901
28.6943
28.7982
28.9017
29.0049
29.1078
22
0.0217
0.0217
0.0219
0.0219
0.022
0.022
0.022
0.0222
0.0222
0.0222
0.0223
0.0223
0.0224
0.0225
0.0226
0.0226
0.0226
0.0226
0.0227
0.0228
0.0229
0.0229
0.0229
0.023
0.023
0.0231
0.0232
0.0231
0.0233
0.0233
0.0233
0.0234
0.0233
0.0234
0.0234
0.0235
0.0235
0.0236
0.0236
0.0237
0.0237
0.0238
0.0237
0.0239
0.0238
0.0239
0.0239
0.024
0.024
0.0241
0.0241
249
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
23.4221
29.3366
29.4385
29.5401
29.6412
29.7421
29.8426
29.9428
30.0427
30.1422
30.2414
30.3403
30.4388
30.537
30.6349
30.7325
30.8297
30.9266
31.0232
31.1195
31.2154
31.3111
31.4064
31.5014
31.5961
31.6904
31.7845
31.8782
31.9717
32.0648
32.1576
32.2501
32.3423
32.4342
32.5257
32.617
32.708
32.7987
32.889
32.9791
33.0689
33.1584
33.2475
33.3364
33.425
33.5133
33.6013
33.689
33.7764
33.8635
33.9504
34.0369
23.4004
29.3125
29.4143
29.5159
29.617
29.7178
29.8183
29.9185
30.0183
30.1178
30.217
30.3158
30.4144
30.5125
30.6104
30.7079
30.8051
30.902
30.9986
31.0948
31.1908
31.2864
31.3817
31.4766
31.5713
31.6656
31.7597
31.8534
31.9468
32.0399
32.1327
32.2252
32.3173
32.4092
32.5008
32.592
32.683
32.7737
32.864
32.9541
33.0438
33.1333
33.2224
33.3113
33.3999
33.4882
33.5761
33.6638
33.7512
33.8383
33.9252
34.0117
0.0217
0.0241
0.0242
0.0242
0.0242
0.0243
0.0243
0.0243
0.0244
0.0244
0.0244
0.0245
0.0244
0.0245
0.0245
0.0246
0.0246
0.0246
0.0246
0.0247
0.0246
0.0247
0.0247
0.0248
0.0248
0.0248
0.0248
0.0248
0.0249
0.0249
0.0249
0.0249
0.025
0.025
0.0249
0.025
0.025
0.025
0.025
0.025
0.0251
0.0251
0.0251
0.0251
0.0251
0.0251
0.0252
0.0252
0.0252
0.0252
0.0252
0.0252
301
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
29.2344
34.2091
34.2948
34.3802
34.4653
34.5502
34.6347
34.719
34.803
34.8867
34.9701
35.0533
35.1362
35.2188
35.3011
35.3832
35.465
35.5465
35.6277
35.7087
35.7894
35.8698
35.95
36.0299
36.1095
36.1889
36.268
36.3469
36.4254
36.5038
36.5818
36.6596
36.7372
36.8145
36.8915
36.9683
37.0448
37.121
37.1971
37.2728
37.3483
37.4236
37.4986
37.5733
37.6478
37.7221
37.7961
37.8699
37.9434
38.0167
38.0897
38.1625
29.2103
34.1839
34.2696
34.355
34.4401
34.5249
34.6094
34.6937
34.7777
34.8614
34.9448
35.028
35.1108
35.1934
35.2758
35.3578
35.4396
35.5211
35.6023
35.6833
35.764
35.8444
35.9246
36.0045
36.0841
36.1635
36.2426
36.3214
36.4
36.4783
36.5564
36.6342
36.7117
36.789
36.8661
36.9428
37.0193
37.0956
37.1716
37.2474
37.3229
37.3982
37.4732
37.5479
37.6224
37.6967
37.7707
37.8445
37.918
37.9913
38.0643
38.1371
23
0.0241
0.0252
0.0252
0.0252
0.0252
0.0253
0.0253
0.0253
0.0253
0.0253
0.0253
0.0253
0.0254
0.0254
0.0253
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0255
0.0254
0.0255
0.0254
0.0254
0.0255
0.0255
0.0254
0.0255
0.0255
0.0254
0.0255
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
0.0254
353
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
34.1232
38.3074
38.3795
38.4513
38.5229
38.5942
38.6653
38.7362
38.8069
38.8773
38.9475
39.0174
39.0871
39.1566
39.2258
39.2949
39.3636
39.4322
39.5005
39.5686
39.6365
39.7042
39.7716
39.8388
39.9058
39.9726
40.0391
40.1054
40.1715
40.2374
40.303
40.3685
40.4337
40.4987
40.5635
40.6281
40.6924
40.7566
40.8205
40.8842
40.9477
41.011
41.0741
41.137
41.1997
41.2621
41.3244
41.3864
41.4483
41.5099
41.5713
41.6326
34.0979
38.282
38.3541
38.4259
38.4975
38.5689
38.64
38.7109
38.7815
38.8519
38.9221
38.9921
39.0618
39.1313
39.2005
39.2696
39.3384
39.4069
39.4753
39.5434
39.6113
39.6789
39.7464
39.8136
39.8806
39.9473
40.0139
40.0802
40.1463
40.2122
40.2779
40.3433
40.4086
40.4736
40.5384
40.603
40.6673
40.7315
40.7954
40.8592
40.9227
40.986
41.0491
41.112
41.1747
41.2372
41.2994
41.3615
41.4233
41.485
41.5464
41.6077
0.0253
0.0254
0.0254
0.0254
0.0254
0.0253
0.0253
0.0253
0.0254
0.0254
0.0254
0.0253
0.0253
0.0253
0.0253
0.0253
0.0252
0.0253
0.0252
0.0252
0.0252
0.0253
0.0252
0.0252
0.0252
0.0253
0.0252
0.0252
0.0252
0.0252
0.0251
0.0252
0.0251
0.0251
0.0251
0.0251
0.0251
0.0251
0.0251
0.025
0.025
0.025
0.025
0.025
0.025
0.0249
0.025
0.0249
0.025
0.0249
0.0249
0.0249
405
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
38.2351
41.7544
41.815
41.8754
41.9357
41.9957
42.0555
42.1151
42.1745
42.2337
42.2928
42.3516
42.4102
42.4687
42.5269
42.585
42.6428
42.7005
42.758
42.8152
42.8723
42.9292
42.986
43.0425
38.2097
41.7295
41.7902
41.8506
41.9108
41.9709
42.0307
42.0903
42.1498
42.209
42.2681
42.3269
42.3856
42.444
42.5023
42.5603
42.6182
42.6759
42.7334
42.7907
42.8478
42.9047
42.9615
43.018
24
0.0254
0.0249
0.0248
0.0248
0.0249
0.0248
0.0248
0.0248
0.0247
0.0247
0.0247
0.0247
0.0246
0.0247
0.0246
0.0247
0.0246
0.0246
0.0246
0.0245
0.0245
0.0245
0.0245
0.0245
APPENDIX C
CASE 2 data
X
time
0
0.04
0.08
0.12
0.16
0.2
0.24
0.28
0.32
0.36
0.4
0.44
0.48
0
0.04
0.08
0.12
0.16
0.2
0.24
0.28
0.32
0.36
0.4
0.44
0.48
ANALYTICAL SOLUTION
X
X
X
0 0.166667 0.333333
0.5
0
0.5 0.866025
1
0 0.336913 0.58355 0.6738255
0 0.22702 0.393211 0.4540407
0 0.152972 0.264955 0.3059442
0 0.103076 0.178534
0.206153
0 0.069456 0.120301 0.1389111
0 0.046801 0.081062 0.0936019
0 0.031536 0.054621 0.0630713
0 0.02125 0.036805 0.0424991
0 0.014318
0.0248 0.0286369
0 0.009648 0.016711 0.0192963
0 0.006501 0.01126 0.0130023
0 0.004381 0.007588 0.0087613
NUMERICAL SOLUTION
0
0.5
0.866
1
0
0.3332
0.5772
0.6665
0
0.2221
0.3847
0.4442
0
0.148
0.2564
0.296
0
0.0987
0.1709
0.1973
0
0.0658
0.1139
0.1315
0
0.0438
0.0759
0.0876
0
0.0292
0.0506
0.0584
0
0.0195
0.0337
0.0389
0
0.013
0.0225
0.0259
0
0.0086
0.015
0.0173
0
0.0058
0.01
0.0115
0
0.0038
0.0067
0.0077
X
0.666667
0.866025
0.58355
0.393211
0.264955
0.178534
0.120301
0.081062
0.054621
0.036805
0.0248
0.016711
0.01126
0.007588
X
0.833333
0.5
0.336913
0.22702
0.152972
0.103076
0.069456
0.046801
0.031536
0.02125
0.014318
0.009648
0.006501
0.004381
X
1
-4.10207E-10
-2.76408E-10
-1.86251E-10
-1.255E-10
-8.45654E-11
-5.69823E-11
-3.83961E-11
-2.58723E-11
-1.74334E-11
-1.17471E-11
-7.91548E-12
-5.33365E-12
-3.59395E-12
0.866
0.5772
0.3847
0.2564
0.1709
0.1139
0.0759
0.0506
0.0337
0.0225
0.015
0.01
0.0067
0.5
0.3332
0.2221
0.148
0.0987
0.0658
0.0438
0.0292
0.0195
0.013
0.0086
0.0058
0.0038
0
0
0
0
0
0
0
0
0
0
0
0
0
25