This document provides details of an assignment to determine the temperature distribution in a block using numerical analysis and MATLAB. It describes the block geometry, assumptions made, and steps taken to solve the problem using a numerical technique called the Gauss-Seidel method. The block is divided into a grid and the heat conduction equation is applied to each node to iteratively calculate the temperature at each point. The process is repeated over several iterations until the temperatures converge to a steady solution.
1. COMPUTATIONAL HEAT TRANSFER AND FLUID FLOW ASSIGNMENT
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BAHIRDAR UNIVERSITY
BAHIRDAR INSTITUTE OF TECHNOLOGY
FACULITY OF MECHANICAL AND INDUSTRIAL ENGINEERING
DEP’T (OF): MECHANICAL ENGINEERING
STREAM: THERMAL ENGINEERING
COMPUTATIONAL HEAT TRANSFER AND FLUID FLOW INDIVIDUAL
ASSIGNMENT
BY: LIJALEM TSIHAYE
ID.NO: 0800985
SECTION: D
SUBMITTED TO: MULUKEN .T (PhD Candi...)
Submission Date: 6/3/2012 E.C.
2. COMPUTATIONAL HEAT TRANSFER AND FLUID FLOW ASSIGNMENT
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Q1) The simple conduction heat transfer is constrained as shown in the following figure. Thermal
conductivity (k) of the material is 10 W/m*C and the block is assumed to be infinitely long.
Determine the temperature distribution along the block.
A. Using Numerical calculation (show every step while solving )
B. Write a Matlab code and show the temperature distribution and compare the result with the
Numerical result.
Figure 1: Block
SOLUTION
ASSUMPTION:
The block is isentropic material (i.e. having the same k value and made of the same
material )
2-D
Steady state heat transfer
No heat generation ( Qgen=0)
A.USING NUMERICAL ANALYSIS
Hint: To find the temperature within the plate, we divide the plate area by a grid as shown in
Figure 2(4).
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Figure 2: Plate area divided into a grid
The length L along the x axis is divided into m equal segments, while the width W along the
y axis is divided into n equal segments, hence giving
m
L
x ……..Eqn (1)
n
W
y ………Eqn (2)
5,4,3,2,1;4,3,2,1,
4
1,1,,1,1
,
ji
TTTT
T
jijijiji
ji ………….Eqn (3)
myx 25.0
Re-writing Equations (1) and (2) we have
4
25.0
0.1
x
L
m
,
4
25.0
0.1
y
W
n
NOTE: Use Gauss-Seidel method to solve the problem
Gauss-Seidel method:
Is an iterative method used to solve a linear system of equations.
Is also known as Liebmann method or the method of successive
displacement.
tT
rT
x
y
),( ji ),1( ji
)1,( ji
),1( ji
)1,( ji
)0,0(
)0,(m
),0( n
bT
lT ),( ji
x
y
xy
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Figure 3: A square plate with the dimensions and boundary temperatures
The interior nodes are shown in Figure 4.
Figure 4: Square plate with nodes
All the nodes on the left and right boundary have ani value of zero and m , respectively. All of
the nodes on the top or bottom boundary have a j value of either zero or n , respectively.
From the boundary conditions
x
y
0,0T
0,1T 0,2T 0,3T 0,4T
1,0T
2,0T
3,0T
4,0T
1,1T 1,2T 1,3T 1,4T
2,1T 2,2T 2,3T
2,4T
3,1T 3,2T 3,3T
3,4T
4,1T 4,2T 4,3T 4,4T
C100
C100
C500
C100
m0.1
m0.1
x
y
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3,2,1,500
3,2,1,100
4,3,2,1,100
4,3,2,1,100
4,
0,
,4
,0
iT
iT
jT
jT
i
i
j
j
(***)
The corner nodal temperature of 0,44,44,0 ,, TTT and 0,0T are not needed. Now to get the temperature
at the interior nodes we have to write Equation (3) for all of the combinations of i and j ,
1,...,1;1,....,1 njmi .
Iteration #1
For iteration 1, start with all of the interior nodes having a temperature of C0 .
4
0,12,11,01,2
1,1
TTTT
T
4
10001000
C 0000.50
4
1,13,12,02,2
2,1
TTTT
T
4
00.5001000
C 5000.37
4
2,14,13,03,2
3,1
TTTT
T
4
5000.375001000
C 3750.159
4
0,22,21,11,3
1,2
TTTT
T
4
10000000.500
C 5000.37
4
1,23,22,12,3
2,2
TTTT
T
C
7500.18
4
5000.3705000.370
4
2,24,23,13,3
3,2
TTTT
T
4
7500.185003750.1590
C 5313.169
4
0,32,31,21,4
1,3
TTTT
T
4
10005000.37100
C 3750.59
4
1,33,32,22,4
2,3
TTTT
T
4
3750.5907500.18100
C 5313.44
4
2,34,33,23,4
3,3
TTTT
T
4
5313.445005313.169100
C 5157.203
Iteration #2
For iteration 2, use the temperatures that obtained from iteration 1.
6. COMPUTATIONAL HEAT TRANSFER AND FLUID FLOW ASSIGNMENT
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Iteration #3
For iteration 3, use the temperatures that obtained from iteration 2.
4
0,12,11,01,2
1,1
TTTT
T
4
7188.862007188.61
C 1094.87
4
1,13,12,02,2
2,1
TTTT
T
C
9492.122
4
1094.870625.3146251.90
4
2,14,13,03,2
3,1
TTTT
T
4
9492.1226000508.252
C 7500.243
4
0,22,21,11,3
1,2
TTTT
T
4
6251.1901094.875625.76
C 5743.88
4
1,23,22,12,3
2,2
TTTT
T
C
2774.145
4
5743.880000.3755352.117
4
2,24,23,13,3
3,2
TTTT
T
4
2774.1457500.7433965.242
C 8560.282
4
0,32,31,21,4
1,3
TTTT
T
4
5352.1175743.88200
C 5274.101
4
1,33,32,22,4
2,3
TTTT
T
C
3003.147
4
5274.1013965.2422774.245
4
2,34,33,23,4
3,3
TTTT
T
4
3003.1478560.282600
C 5391.257
Iteration #4
For iteration 4, use the temperatures that obtained from iteration 3.
4
0,12,11,01,2
1,1
TTTT
T
4
1005000.371005000.37
C 7500.68
4
1,13,12,02,2
2,1
TTTT
T
C
7188.86
4
7500.683750.1597500.118
4
2,14,13,03,2
3,1
TTTT
T
4
7188.865001005313.169
C 0625.214
4
0,22,21,11,3
1,2
TTTT
T
4
7500.1187500.683750.59
C 7188.61
4
1,23,22,12,3
2,2
TTTT
T
C
6251.90
4
7188.617188.860626.214
4
2,24,23,13,3
3,2
TTTT
T
4
6251.900625.7145157.203
C 0508.252
4
0,32,31,21,4
1,3
TTTT
T
4
1005313.447188.61100
C 5625.76
4
1,33,32,22,4
2,3
TTTT
T
C
5352.117
4
5625.765157.2030625.190
4
2,34,33,23,4
3,3
TTTT
T
4
5352.1170508.252600
C 3965.242
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T=zeros(length(dx),length(dy));
T(:,1)=100;
T(:,end)=100;
T(end,:)=100;
T(1,:)=500;
for k=1:25;
for i=2:length(dx)-1;
for j=2:length(dy)-1;
T(i,j)=0.25*[T(i-1,j)+T(i+1,j)+T(i,j-1)+T(i,j+1)];
end
end
end
contourf(dx,dy,T,'ShowText','on')
colormap jet
T
T =
500.0000 500.0000 500.0000 500.0000 500.0000
100.0000 271.4286 310.7143 271.4286 100.0000
100.0000 175.0000 200.0000 175.0000 100.0000
100.0000 128.5714 139.2857 128.5714 100.0000
100.0000 100.0000 100.0000 100.0000 100.0000
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Figure 5: Temperature Distribution Of Simple Heat Conduction
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CONCLUSTION
The temperature value that obtained at the end in each interior nodes by using Numerical
Analysis and Matlab is the same. So, the temperature distribution value on each nodes is (i.e.
Answer (solution for temperature distribution)
128.57141,1 T 2857.1391,2 T
0000.1752,32,1 TT
0000.2002,2 T
4286.2713,1 T 7143.3103,2 T
4286.2713,3 T
THE END!!!
5714.1281,3 T