This document provides an overview of steady-state heat conduction in multiple dimensions. It begins with an introduction to two-dimensional heat flow and the Laplace equation. It then discusses analytical solutions using separation of variables for some simple geometries. Numerical techniques for solving the heat equation are also covered, including discretization of the domain and developing finite difference equations at nodes. Examples are provided to illustrate the concepts.
3. CUMT
HEAT TRANSFER LECTURE
3-1 Introduction
In Chapter 2 steady-state heat transfer was calculated
in systems in which the temperature gradient and area
could be expressed in terms of one space coordinate. We
now wish to analyze the more general case of two-dimensional
heat flow. For steady state with no heat
generation, the Laplace equation applies.
2 2
2 2 T T 0
x y
¶ + ¶ =
¶ ¶
(3-1)
The solution to this equation may be obtained by analytical,
numerical, or graphical techniques.
Video.edhole.com
4. HEAT TRANSFER LECTURE
The objective of any heat-transfer analysis is usually to
predict heat flow or the temperature that results from a
certain heat flow. The solution to Equation (3-1) will give
the temperature in a two-dimensional body as a function
of the two independent space coordinates x and y. Then
the heat flow in the x and y directions may be calculated
from the Fourier equations
CUMT
3-1 Introduction
Video.edhole.com
5. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Analytical solutions of temperature distribution can be obtained
for some simple geometry and boundary conditions. The
separation method is an important one to apply.
Consider a rectangular plate.
Three sides are maintained at
temperature T1, and the upper
side has some temperature
distribution impressed on it.
The distribution can be a constant
temperature or something more
complex, such as a sine-wave.
Video.edhole.com
6. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Consider a sine-wave distribution on the upper edge, the
boundary conditions are:
Video.edhole.com
7. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Substitute:
2 2
2 2
1 T 1 T
X x Y y
- ¶ = ¶
¶ ¶
We obtain two ordinary differential equations in terms of
this constant,
2
2 X X 0
x
¶ +l 2
=
¶
2
2 Y Y 0
y
¶ -l 2
=
¶
where λ2 is called the separation constant.
Video.edhole.com
8. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
We write down all possible solutions and then see which
one fits the problem under consideration.
For l = 0 :
X = C +
C x
1 2
Y = C +
C y
T = C + C x C +
C y
( ) ( )
2
3 4
1 2 3 4
This function cannot fit the sine-function boundary
condition, so that the l 2 = 0 solution may be excluded.
Video.edhole.com
9. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
l l
< = +
Y C y C y
T C e C e C y C y
CUMT
HEAT TRANSFER LECTURE
For X C e C e
= +
= + +
( ) ( )
2
5 6
7 8
5 6 7 8
0 :
cos sin
cos sin
x x
x x
l l
l
l l
l l
-
-
This function cannot fit the sine-function boundary condition,
so that the l 2 < 0 solution may be excluded.
Video.edhole.com
10. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
= +
= + +
y y
l > = l +
l
Y l l
-
l l
-
T C x C x C e C e
CUMT
HEAT TRANSFER LECTURE
For 0 : X C cos x C sin
x
y y
C e C e
( ) ( )
2
9 10
11 12
l l
cos sin
9 10 11 12
It is possible to satisfy the sine-function boundary condition;
so we shall attempt to satisfy the other condition.
Video.edhole.com
11. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Let
The equation becomes:
Apply the method of variable separation, let
Video.edhole.com
12. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
And the boundary conditions become:
Video.edhole.com
13. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Applying these conditions,we have:
0 = ( C9 cosl x +C10 sinl x) ( C11 +C12 )
( ) 9 11 12 0 = C C e-l y +C el y
( ) ( ) 9 10 11 12 0 = C coslW +C sinlW C e-l y +C el y
( ) ( ) 9 10 11 12 sin cos sin H H
T x C x C x C e C e
m
æçp ö¸= l + l -l + l è W
ø
Video.edhole.com
14. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
accordingly,
and from (c),
This requires that
C11 = -C12
9 C = 0
( ) 10 12 0 = C C sinlW el y - e-l y
sinlW = 0
Video.edhole.com
15. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
then
We get
T T C n x n y
q p p
= - =å
The final boundary condition may now be applied:
p x ¥ T C n p x n p
H
=å
sin sin sinh m n
W W W
which requires that Cn =0 for n >1.
n
W
l = p
1
1
sin sinh n
n
W W
¥
=
1
n
=
Video.edhole.com
16. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
The final solution is therefore
( )
( ) 1
y W x T T T
p p
p
sinh /
= +
sin
m sinh /
H W W
The temperature field for this problem is shown. Note that the heat-flow
lines are perpendicular to the isotherms.
Video.edhole.com
17. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Another set of boundary conditions
y
x
x W
q
q
q
q p
= 0 at =
0
= 0 at =
0
= 0 at
=
= sin æ ö at = m
çè ø¸
T x y H
W
Video.edhole.com
18. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Using the first three boundary conditions, we obtain the
solution in the form of Equation:
T T C n x n y
1
¥ p p
=
1
sin sinh n
n
W W
- =å
Applying the fourth boundary condition gives
T T C n x n H
2 1
¥ p p
=
1
sin sinh n
n
W W
- =å
Video.edhole.com
19. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
This series is
then
( ) ( ) 1
T T T T n x
2 1 2 1
2 1 1sin
1
n
n
p
n W
p
¥ +
=
- +
- = - å
+ - +
2 1 1 1
( ) ( )
( ) 1
= -
2 1
sinh /
n
n C T T
p np H W n
The final solution is expressed as
( ) ( )
¥ +
T - T 2 å
- 1 + 1 sinh n y / W
=
sin
n x T - T n W n H W
( )
1
1
2 1 1
sinh /
n
n
p p
p p
=
Video.edhole.com
20. 3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
HEAT TRANSFER LECTURE
Transform the boundary condition:
y
x
x W
q
q
q
q p
= 0 at =
0
= 0 at =
0
= 0 at
=
= sin æ ö at = m
çè ø¸
T x y H
W
Video.edhole.com
21. CUMT
HEAT TRANSFER LECTURE
3-3 Graphical Analysis
neglect
Video.edhole.com
22. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Consider a general one dimensional heat conduct-ion
problem, from Fourier’s Law:
let
then
Videow.ehdheorlee.c:om S is called shape factor.
23. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Note that the inverse hyperbolic cosine can be calculated from
cosh-1 x = ln ( x ± x2 -1)
For a three-dimensional wall, as in a furnace, separate shape
factors are used to calculate the heat flow through the edge and
corner sections, with the dimensions shown in Figure 3-4. when all
the interior dimensions are greater than one fifth of the thickness,
S A
wall
= edge S = 0.54D corner S = 0.15L
L
where A = area of wall, L = wall thickness, D = length of edge
Video.edhole.com
24. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
25. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
26. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
27. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
28. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
29. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
30. CUMT
3-4 The Conduction Shape Factor
HEAT TRANSFER LECTURE
Video.edhole.com
35. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
The most fruitful approach to the heat conduction is one
based on finite-difference techniques, the basic principles
of which we shall outline in this section.
Video.edhole.com
36. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
1、Discretization of the solving
Video.edhole.com
37. 3-5 Numerical Method of Analysis
T T x T x T x T
= + D ¶ + D ¶ + D ¶ +
( ) ( ) ...
+ ¶ x ¶ x ¶
x
m n m n m n
T T x T x T x T
= - D ¶ + D ¶ - D ¶ +
( ) ( ) ...
CUMT
HEAT TRANSFER LECTURE
2、Discrete equation
Taylor series expansion
2 2 3 3
m n m n 2 6
1, , 2 3
, , ,
2 2 3 3
m n m n 2 6
- x x x
1, , 2 3
¶ ¶ ¶
m , n m , n m ,
n
Video.edhole.com
38. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
2
2、Discrete equation
+ = + D 2
¶ +
2 ( ) ... m n m n m n
1, 1, , 2
m ,
n
T T T x T
+ - ¶
x
T T T T o x
x x
¶ 2
- + = m + n m n m - n
+ D
¶ D
1, , 1, 2
2 2
,
2
( )
( )
m n
2
T - 2
T +
T
T m n m n m n
¶ + -
, 1 , , 1
2
,
2
2
( )
( )
o y
y
y
m n
+ D
D
=
¶
Differential equation for two-dimensional steady-state heat flow
·
2 2
2 2 T T q 0
x y k
¶ + ¶ + =
¶ ¶
Video.edhole.com
39. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
2、Discrete equation
Discrete equation at nodal point (m,n)
·
T - 2 T + T T - 2
T +
T m + 1, n m , n m - 1, n + m , n + 1 m , n m , n - 1
+ q
=
0 2 2
x y k
D D
no heat generation
T - 2 T + T T - 2
T +
T
m + 1, n m , n m - 1, n + m , n + 1 m , n m , n - 1
=
0 2 2
x y
D D
Δx= Δy
1, 1, , 1 , 1 , 4 0 m n m n m n m n m n T T T T T + - + - + + + - =
Video.edhole.com
40. 3-5 Numerical Method of Analysis
D + D + D + D =
CUMT
HEAT TRANSFER LECTURE
2、Discrete equation
Thermal balance
(1) Interior points
steady-state & no heat
generation
1, 1, , 1 , 1 0 m n m n m n m n q q q q - + + - + + + =
T T T q kA k y
-
1, ,
1,
d
d
m n m n
m n
x x
-
-
= - = D
D
T -
T
q k y m n m n
m n D
x
= D +
+
1, ,
1,
, 1 ,
, 1
m n m n
m n
T T
q k x
y
+
+
-
= D
D
, 1 ,
, 1
m n m n
m n
T T
q k x
y
-
-
-
= D
D
T - T T - T T - T T -
T
k y m - 1,n m,n k y m + 1,n m,n k x m,n + 1 m,n k x
m,n - 1 m,n 0 Video.edhole.com
x x y y
D D D D
41. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
Thermal balance
(1) Interior points
Δx= Δy
1, 1, , 1 , 1 , 4 0 m n m n m n m n m n T T T T T + - + - + + + - =
· ·
= ´ = ´D D
gen q q V q x y
2
·
T T T T T q x
+ - + - + + + - + D =
1, 1, , 1 , 1 , 4 ( ) 0 m n m n m n m n m n
k
steady-state with heat generation
Video.edhole.com
42. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
2、Discrete equation
Thermal balance
(2) boundary
points
1, ,
1,
m n m n
m n
T T
q k y
x
-
-
-
= D
D
x T T q k
, 1 ,
, 1 2
m n m n
m n
y
+
+
D - =
D
x T T q k
, 1 ,
, 1 2
m n m n
m n
y
-
-
D - =
D
, ( ) w m n q h y T T¥ = ´D ´ -
T T T T T T k y m - 1, n m , n k x m , n + 1 m , n k x m , n -
1 m ,
n
h y T T
, ( )
2 2
m n
Video.edhole.com
x y y
¥
- D - D - D + + = ´D ´ -
D D D
43. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
Thermal balance
(2) boundary
points
Δx= Δy
h ´D x + T = T + T + T + h ´D
x T
k - + - k ¥
( 2) 1 ( )
m n 2 m n m n m n
, 1, , 1 , 1
D y T - T T - T k m - 1, n m , n + k D x m , n -
1 m ,
n
= h ´ D x ´ T - T + h ´ D y ´ T -
T
( ) ( )
m , n m , n
x y
¥ ¥
D D
2 2 2 2
h ´D x + T = T + T + h ´D
x T
k - - k ¥
( 1) 1 ( )
m n 2 m n m n
, 1, , 1
Δx= Δy
Video.edhole.com
44. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
Thermal balance
(2) boundary
points
T - T T - T T - T k D y + k D y +
k
D x m - 1, n m , n m + 1, n m , n m , n -
1 m ,
n
x x y
D D D
- D D + D = ´ ´ - + ´ ´ -
T T k x , 1 ,
h x T T h y T T
Δx= Δy
2 2
( ) ( )
, ,
2 2
m n m n
m n m n
y
+
¥ ¥
D
h ´D x + T = T + T + T + T + h ´D
x T
k - + - + k ¥
( 3) 1 (2 2 )
m n 2 m n m n m n m n
, 1, 1, , 1 , 1
Video.edhole.com
45. CUMT
3-5 Numerical Method of Analysis
HEAT TRANSFER LECTURE
3、Algebraic equation
a T + a T + + a T =
C
a T + a T + + a T =
C
11 1 12 2 1 1
21 1 22 2 2 2
a T + a T + + a T =
C
1 1 2 2
......
......
............................................
......
n n
n n
n n nn n n
Video.edhole.com
46. 3-5 Numerical Method of Analysis
é ù
ê ú
= ê ú
ê ú
ê ú
ë û
CUMT
HEAT TRANSFER LECTURE
Matrix notation
[ ]
a a a
a a a
é ê 11 12 1
ù
ú
= ê 21 22 2
ú
ê ú
ê ú
ë 1 2
û
...
...
... ... ... ...
...
n
n
n n nn
A
a a a
[ ]
T
T
é ê 1
ù
ú
= ê 2
ú
ê ...
ú
ê ú
ë n
û
T
T
[ ]
C
C
1
2
...
n
C
C
[ A] [T ] =[C]
Iteration
Simple Iteration & Gauss-Seidel Iteration
Video.edhole.com
47. CUMT
HEAT TRANSFER LECTURE
Example 3-5
Consider the square shown in the figure. The left face is
maintained at 100 ℃ and the top face at 500℃, while the
other two faces are exposed to a environment at 100℃.
h=10W/m2·℃ and k=10W/m·℃. The block is 1 m square.
Compute the temperature of the various nodes as indicated
in the figure and heat flows at the boundaries.
Video.edhole.com
48. CUMT
Example 3-5
HEAT TRANSFER LECTURE
[Solution]
The equations for nodes 1,2,4,5 are given by
T T T
+ + + - =
+ + + - =
+ + + - =
+ + + - =
500 100 4 0
2 4 1
500 T T T 4 T
0
1 3 5 2
100 T T T 4 T
0
1 5 7 4
T T T T T
4 0
2 4 6 8 5
Video.edhole.com
49. CUMT
Example 3-5
HEAT TRANSFER LECTURE
[Solution]
Equations for nodes 3,6,7,8 are
2 1 T = 1 (500 + 2 T + T
) + 1 ´
100
3 3 2 2 6
3
2 1 T = 1 ( T + 2 T + T
) + 1 ´
100
3 6 2 3 5 9
3
2 1 T = 1 (100 + 2 T + T
) + 1 ´
100
3 7 2 4 8
3
2 1 T = 1 ( T + 2 T + T
) + 1 ´
100
3 8 2 7 5 9
3
The equation for node 9 is
11 T = 1 ( T +T ) + 1 ´
100
3 9 2 6 8
3
Video.edhole.com
50. CUMT
HEAT TRANSFER LECTURE
Example 3-5
2 1 T = 1 (500 + 2 T +T ) + 1 ´
100
3 3 2 2 6
3
2 1 T = 1 ( T + 2 T +T ) + 1 ´
100
3 6 2 3 5 9
3
2 1 T = 1 (100 + 2 T +T ) + 1 ´
100
3 7 2 4 8
3
2 1 T = 1 ( T + 2 T +T ) + 1 ´
100
3 8 2 7 5 9
3
The equation for node 9 is
11 T = 1 ( T +T ) + 1 ´
100
3 9 2 6 8
3
Video.edhole.com
51. HEAT TRANSFER LECTURE
We thus have nine equations and nine unknown nodal temperatures. So
the answer is
CUMT
For the 500℃ face, the heat flow into the face is
q = k ´ A ´D T = ´ D x ´ - T + D x ´ - T + D x ´ -
T
1 2 3 10 [ (500 ) (500 ) (500 )]
D y D y D y 2
D
y
W m
in
å
= =
... 4843.4 /
The heat flow out of the 100℃ face is
q k A T y T y T y T
= ´ ´D = ´ D ´ - + D ´ - + D ´ -
1 4 7
1
10 [ ( 100) ( 100) ( 100)]
D D D D
2
å
... 3019 /
x x x x
W m
= =
Example 3-5
Video.edhole.com
52. CUMT
HEAT TRANSFER LECTURE
Example 3-5
The heat flow out the right face is
å
q h A T T
= ´ ´ ( -
)
2
¥ = ´ D y ´ T - + D y ´ T - + D y ´ T
-
= =
10 [ ( 100) ( 100) ( 100)]
3 6 9
2
W m
... 1214.6 /
The heat flow out the bottom face is
å
q h A T T
= ´ ´ ( -
)
3
¥ = ´ D ´ - + D ´ - + D ´ -
= =
y T y T y T
W m
10 [ ( 100) ( 100) ( 100)]
7 8 9
2
... 600.7 /
The total heat flow out is
1 2 3 ... 4834.3 / out q = q + q + q = = W m 4843.4 / in <q = W m
Video.edhole.com
53. 3-6 Numerical Formulation in Terms of Resistance Elements
CUMT
HEAT TRANSFER LECTURE
Thermal balance — the net heat input to node i must be zero
T -
T
+å =
j i 0
i
j i j
q
R
qi — heat generation, radiation, etc.
i — solving node
j — adjoining node
Video.edhole.com
54. 3-6 Numerical Formulation in Terms of Resistance Elements
CUMT
HEAT TRANSFER LECTURE
R = D
x
kA
T -
T
+å =
j i 0
q
i
R
j i j
T - T T - T T - T T -
T
m - 1,n m,n + m + 1,n m,n + m,n - 1 m,n + m,n +
1 m,n =
0
R R R R
m m n n
- + - +
1, 1, , 1 , 1 , 4 0 m n m n m n m n m n T T T T T + - + - + + + - =
1
m m n n R R R R
- + - + k = = = =
so
Video.edhole.com
55. CUMT
HEAT TRANSFER LECTURE
3-7 Gauss-Seidel Iteration
T -
T
+å =
j i 0
i
j i j
q
R
q T R
( / )
(1/ )
i j i j
j
i
i j
j
T
R
+
=
å
å
Steps
Assumed initial set of values for Ti ;
Calculated Ti according to the equation ;
—using the most recent values of the Ti
Repeated the process until converged.
Video.edhole.com
56. 3-7 Gauss-Seidel Iteration
+ e -
£ e =10-3~10-6
CUMT
HEAT TRANSFER LECTURE
Convergence Criterion
T T
i n i n
i(n 1) i(n) T T d + - £ ( 1) ( )
i n
( )
T
Biot number
Bi = h D
x
k
Video.edhole.com
57. Apply the Gauss-Seidel technique to obtain the nodal temperature
for the four nodes in the figure.
[Solution]
All the connection resistance between
the nodes are equal, that is
CUMT
HEAT TRANSFER LECTURE
Example 3-6
R = D x =
1
k D
y k
Therefore, we have
å å å
å å å Video.edhole.com
q + T R q +
k T k T
( / ) ( ) ( )
(1/ ) ( ) ( )
i j i j i j j j j
j j j
i
= = =
i j j j
j j j
T
R k k
58. CUMT
HEAT TRANSFER LECTURE
Example 3-6
Because each node has four resistance connected to it and k is assumed
constant, so
åk = 4 k 1
j
j
T = åT
i 4 j
j
Video.edhole.com
59. CUMT
HEAT TRANSFER LECTURE
3-8 Accuracy Consideration
Truncation Error — Influenced by difference scheme
Discrete Error — Influenced by truncation error & △x
Round-off Error — Influenced by △x
Video.edhole.com
60. CUMT
HEAT TRANSFER LECTURE
Summary
(1)Numerical Method
Solving Zone
Nodal equations
thermal balance method — Interior & boundary point
Algebraic equations
q T R
Gauss-Seidel iteration
( / )
(1/ )
i j i j
j
i
i j
j
T
R
+
=
å
å
Video.edhole.com
61. CUMT
HEAT TRANSFER LECTURE
Summary
(2)Resistance Forms
T -
T
+å =
j i 0
i
j i j
q
R
(3)Convergence
Convergence Criterion
i(n 1) i(n) T T d + - £
i(n 1) i(n) T T d + - £
Video.edhole.com